A Dichotomy for Real Boolean Holant Problems
AA Dichotomy for Real Boolean Holant Problems
Shuai Shao ∗ [email protected] Jin-Yi Cai ∗ [email protected] Abstract
We prove a complexity dichotomy for Holant problems on the boolean domain with arbitrarysets of real-valued constraint functions. These constraint functions need not be symmetric nordo we assume any auxiliary functions as in previous results. It is proved that for every set F of real-valued constraint functions, Holant( F ) is either P-time computable or f , f and their associated families with extraordinary closure propertiesrelated to Bell states in quantum information theory play an important role in this proof. ∗ Department of Computer Sciences, University of Wisconsin-Madison. Supported by NSF CCF-1714275. a r X i v : . [ c s . CC ] M a y rologue The young knight Siegfried and his tutor Jeyoda set out for their life journey together. Their aim is to pacifythe real land of Holandica, to bring order and unity.In the past decade, brave knights have battled innumerable demons and creatures, and have conqueredthe more familiar portion of Holandica, called Holandica Symmetrica. Along the way they have also beenvictorious by channeling the power of various deities known as Unary Oracles. In the past few years thisbrotherhood of the intrepid have also gained great power from the beneficent god Orieneuler and enhancedtheir skills in a more protected world ruled by Count Seaspie.“But prepared we must be,” Jeyoda reminds Siegfried, “arduous, our journey will become.” As the realland of Holandica is teeming with unknowns, who knows what wild beasts and creatures they may encounter.Siegfried nods, but in his heart he is confident that his valor and power will be equal to the challenge.They have recently discovered a treasure sword. This is their second gift from the Cathedral Orthogonia,more splendid and more powerful than the first. In their initial encounters with the minion creatures in theirjourney, the second sword from Cathedral Orthogonia proved to be invincible.These initial victories laid the foundation for their journey, but also a cautious optimism sets in. Perhapswith their new powerful sword in hand, final victory will not be that far away.Just as they savor these initial victories, things start to change. As they enter the Kingdom of Degree-Sixeverything becomes strange. Subliminally they feel the presence of a cunning enemy hiding in the darkness.Gradually they are convinced that this enemy possesses a special power that eludes the ongoing campaign,and in particular their magic sword. After a series of difficult and protracted battles with many twists andturns, their nemesis, the Lord of Intransigence slowly reveals his face. The Lord of Intransigence has asuit of magic armor, called the Bell Spell, that hides and protects him so well that the sword of CathedralOrthogonia cannot touch him.Siegfried and Jeyoda know that in order to conquer the Lord of Intransigence, they need all the skills andwisdom they have. Although the Lord of Intransigence has a strong armor, he has a weakness. The armoris maintained by four little elfs called the Bell Binaries. The next battle is tough and long. Siegfried andJeyoda hit upon the idea of convincing the Bell Binaries to stage a mutiny. With his four little elfs turningagainst him, his armor loses its magic, and the Kingdom of Six-degree is conquered. In the aftermath ofthis victory, Siegfried and Jeyoda also collect some valuable treasures that will come in handy in their nextcampaign.After defeating the Lord of Intransigence, Siegfried and Jeyoda enter the Land of Degree-Eight. Now theyare very careful. After meticulous reconnaissance, they finally identify the most fearsome enemy, the Queenof the Night. Taking a page from their battle with the Lord of Intransigence they look for opportunities togain help from within the enemy camp. However, the Queen of the Night has the strongest protective coatcalled the Strong Bell Spell. This time there is no way to summon help from within the Queen’s own camp.In fact, her protective armor is so strong that any encounter with Siegfried and Jeyoda’s sword makes hermagically disappear in a puff of white smoke.But, everyone has a weakness. For the Queen, her vanishing act also brings the downfall. After plottingthe strategy for a long time, Siegfried and Jeyoda use a magical potion to create from nothing the helpersneeded to defeat the Queen.Buoyed by their victory, they summon their last strength to secure the Land of Degree-Eight and beyond.Finally they bring complete order to the entire real land of Holandica. At their celebratory banquet, theywant to share the laurels with Knight Ming and Knight Fu who provided invaluable assistance in theirjourney; but the two brave and generous knights have retreated to their Philosopher’s Temple and arenowhere to be found. i ontents Prologue i1 Introduction 12 Preliminaries 2 n (cid:54)
176 First Major Obstacle: 6-ary Signatures with Bell Property 19 (cid:98) f . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206.2 (cid:98) f . . . . . . . . . . . . . . . . . . . . 30 Holant b ( F ) (cid:98) f . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 698.2 Holant problems with limited appearance and a novel reduction . . . . . . . . . . . . 83 n (cid:62) ii Introduction
Counting problems arise in many different fields, e.g., statistical physics, economics and machinelearning. In order to study the complexity of counting problems, several natural frameworks havebeen proposed. Two well studied frameworks are counting constraint satisfaction problems ( { , } n → C for some n >
0. Let F be any fixedset of signatures. A signature grid Ω = ( G, π ) over F is a tuple, where G = ( V, E ) is a graphwithout isolated vertices, π labels each v ∈ V with a signature f v ∈ F of arity deg( v ), and labelsthe incident edges E ( v ) at v with input variables of f v . We consider all 0-1 edge assignments σ ,and each gives an evaluation (cid:81) v ∈ V f v ( σ | E ( v ) ), where σ | E ( v ) denotes the restriction of σ to E ( v ). Definition 1.1 (Holant problems) . The input to the problem
Holant( F ) is a signature grid Ω =(
G, π ) over F . The output is the partition function Holant(Ω) = (cid:88) σ : E ( G ) →{ , } (cid:89) v ∈ V ( G ) f v ( σ | E ( v ) ) . Bipartite Holant problems
Holant(
F | G ) are Holant problems over bipartite graphs H = ( U, V, E ) ,where each vertex in U or V is labeled by a signature in F or G respectively. When { f } is asingleton set, we write Holant( { f } ) as Holant( f ) and Holant( { f } ∪ F ) as Holant( f, F ) . Weighted symmetric , a full dichotomy is proved [18]. When asymmetric signaturesare allowed, some dichotomies are proved for special families of Holant problems by assumingthat certain auxiliary signatures are available, e.g., Holant ∗ , Holant + and Holant c [19, 1, 20, 2].Without assuming auxiliary signatures a Holant dichotomy is established for non-negative real-valued signatures [25], and for all real-valued signatures where a signature of odd arity is present[15]. In this paper, we prove a full complexity dichotomy for Holant problems with real values. Theorem 1.2.
Let F be a set of real-valued signatures. If F satisfies the tractability condition (T) in Theorem 2.22, then Holant( F ) is polynomial-time computable; otherwise, Holant( F ) is F . Since a dichotomy is proved when F contains a signatureof odd arity, we only need to consider signatures of even arity. For signatures of small arity 2 or4 (base cases) and large arity at least 10, we given an induction proof based on results of f and f (and their associatedfamilies) of arity 6 and 8, are discovered which have extraordinary closure properties; we call themBell properties [15]. These amazing signatures are wholly unexpected, and their existence presenteda formidable obstacle to the induction proof.All four binary Bell signatures (related to Bell states [5] in quantum information theory) arerealizable from f by gadget construction. We introduce Holant b problems where the four binaryBell signatures are available. This is specifically to handle the signature f . We prove a b ( f , F ). In this proof, we find other miraculous signatures with specialstructures such that all signatures realized from them by merging gadgets are affine signatures,while themselves are not affine signatures. In order to handle the signature f , we introduce Holantproblems with limited appearance, where some signatures are only allowed to appear a limitednumber of times in all instances. We turn the obstacle of the closure property of f in our favorto prove non-constructively a P-time reduction from Holant b ( f , F ) to Holant( f , F ). In fact, it isprovable that except = , the other three binary Bell signatures are not realizable from f by gadgetconstruction. However, we show that we can realize, in the sense of a non-constructive complexityreduction, the desired binary Bell signatures which appear an unlimited number of times. Thisutilizes the framework where these signatures occur only a limited number of times. Then, we givea b ( f , F ) similar to Holant b ( f , F ). Let f be a complex-valued signature. If f ( α ) = f ( α ) for all α where f ( α ) denotes the complexconjugation of f ( α ) and α denotes the bit-wise complement of α , we say f satisfies Arrow ReversalSymmetry ( ars ). We may also use f α to denote f ( α ). We use wt( α ) to denote the Hamming weightof α . The support S ( f ) of a signature f is { α ∈ Z n | f ( α ) (cid:54) = 0 } . We say f has support of size k if | S ( f ) | = k . If S ( f ) = ∅ , i.e., f is identically 0, we say f is a zero signature and denote it by f ≡ f is a nonzero signature. Let E n = { α ∈ Z n | wt( α ) is even } , and O n = { α ∈ Z n | wt( α )is odd } . A signature f of arity n has even or odd parity if S ( f ) ⊆ E n or S ( f ) ⊆ O n respectively.In both cases, we say that f has parity. Let H n = { α ∈ Z n | wt( α ) = n } . A signature f of arity2 n has half-weighted support if S ( f ) ⊆ H n . We call such a signature an Eularian orientaion (EO) signature. For α ∈ Z n and 1 (cid:54) i (cid:54) n , we use α i to denote the value of α on bit i .Counting constraint satisfaction problems ( n to denote the Equality signature of arity n , which takes value 1 on the all-0 and all-1inputs and 0 elsewhere. (We denote the n -bits all-0 and all-1 strings by (cid:126) n and (cid:126) n respectively.We may omit the superscript n when it is clear from the context.) Let EQ = { = , = , . . . , = n , . . . } denote the set of all Equality signatures.
Lemma 2.1 ([9]) . F ) ≡ T Holant(
EQ | F ) .
2e use (cid:54) = to denote the binary Disequality signature with truth table (0 , , , f of arity 2 n is called a Disequality signature of arity 2 n , denoted by (cid:54) = n , if f = 1 when ( x (cid:54) = x ) ∧ . . . ∧ ( x n − (cid:54) = x n ), and 0 otherwise.By permuting its variables the Disequality signature of arity 2 n also defines (2 n − n − · · · Disequality signatures. These signatures are equivalent for the com-plexity of Holant problems; once we have one we have them all. Let
DEQ = {(cid:54) = , (cid:54) = , . . . , (cid:54) = n , . . . } denote the set of all Disequality signatures.We use = − to denote the binary signature (1 , , , −
1) and (cid:54) = − to denote the binary signature(0 , , − , as = +2 and (cid:54) = as (cid:54) = +2 . Let B = { = +2 , = − , (cid:54) = +2 , (cid:54) = − } . We callthem Bell signatures which correspond to Bell states | Φ + (cid:105) = | (cid:105) + | (cid:105) , | Φ − (cid:105) = | (cid:105) − | (cid:105) , | Ψ + (cid:105) = | (cid:105) + | (cid:105) and | Ψ − (cid:105) = | (cid:105) − | (cid:105) in quantum information science [5].A signature f of arity n (cid:62) k × n − k matrix M S k ( f ) where S k is a setof k many variables among all n variables of f . The matrix M S k ( f ) lists all 2 n many entries of f with the assignments of variables in S k listed in lexicographic order (from (cid:126) k to (cid:126) k ) as row indexand the assignments of the other n − k many variables in lexicographic order as column index. Inparticular, f can be expressed as a 2 × n − matrix M i ( f ) which lists the 2 n entries of f with theassignments of variable x i as row index (from x i = 0 to x i = 1) and the assignments of the other n − M i ( f ) = (cid:20) f , ... f , ... . . . f , ... f , ... f , ... . . . f , ... (cid:21) = (cid:20) f i f i (cid:21) , where f ai denotes the row vector indexed by x i = a in M i ( f ). Similarly, f can also be expressed asa 4 × n − matrix with the assignments of two variables x i and x j as row index. Then, M ij ( f ) = f , ... f , ... . . . f , ... f , ... f , ... . . . f , ... f , ... f , ... . . . f , ... f , ... f , ... . . . f , ... = f ij f ij f ij f ij , where f abij denotes the row vector indexed by ( x i , x j ) = ( a, b ) in M ij ( f ). For = , it has the 2-by-2signature matrix M (= ) = I = [ ]. For (cid:54) = , M ( (cid:54) = ) = N = [ ] . To introduce the idea of holographic transformation, it is convenient to consider bipartite graphs.For a general graph, we can always transform it into a bipartite graph while preserving the Holantvalue, as follows. For each edge in the graph, we replace it by a path of length two. (This operationis called the of the graph and yields the edge-vertex incidence graph.) Each new vertexis assigned the binary
Equality signature = . Thus, we have Holant(= | F ) ≡ T Holant( F ).For an invertible 2-by-2 matrix T ∈ GL ( C ) and a signature f of arity n , written as a columnvector (covariant tensor) f ∈ C n , we denote by T f = T ⊗ n f the transformed signature. For asignature set F , define T F = { T f | f ∈ F } the set of transformed signatures. For signatureswritten as row vectors (contravariant tensors) we define f T − and F T − similarly. Whenever we Given a set of variables, without other specification, we always list them in the cardinal order i.e., from variableswith the smallest index to the largest index.
T f or T F , we view the signatures as column vectors; similarly for f T − or F T − as rowvectors. We can also represent T f as the matrix M S k ( T f ) with the assignments of variables in S k as row index and the assignments of the other n − k variables as column index. Then, we have M S k ( T f ) = T ⊗ k M S k ( f )( T T ) ⊗ n − k . Similarly, M S k ( f T − ) = ( T − T ) ⊗ k M S k ( f )( T − ) ⊗ n − k .Let T ∈ GL ( C ). The holographic transformation defined by T is the following operation:given a signature grid Ω = ( H, π ) of Holant(
F | G ), for the same bipartite graph H , we get a newsignature grid Ω (cid:48) = ( H, π (cid:48) ) of Holant( F T − | T G ) by replacing each signature in F or G with thecorresponding signature in F T − or T G . Theorem 2.2 ([29]) . For every T ∈ GL ( C ) , Holant(
F | G ) ≡ T Holant( F T − | T G ) . Therefore, a holographic transformation does not change the complexity of the Holant problemin the bipartite setting. Let O ( R ) ⊆ R × be the set of all 2-by-2 real orthogonal matrices. Wedenote O ( R ) by O . For all Q ∈ O , since (= ) Q − = (= ), Holant(= | F ) ≡ T Holant(= | Q F ) . A particular holographic transformation that will be commonly used in this paper is the trans-formation defined by Z − = √ (cid:2) − i i (cid:3) . Note that (= ) Z = ( (cid:54) = ). Thus, Holant(= | F ) ≡ T Holant( (cid:54) = | Z − F ) . We denote Z − F by (cid:98) F and Z − f by (cid:98) f . It is known that f and (cid:98) f have thefollowing relation. Lemma 2.3 ([14]) . A (complex-valued) signature f is a real-valued signature iff (cid:98) f satisfies ars . We say a real-valued binary signature f ( x , x ) is orthogonal if M ( f ) M T ( f ) = λI for somereal λ >
0. Since M ( f ) = M T ( f ), M ( f ) M T ( f ) = λI iff M ( f ) M T ( f ) = λI . The following factis easy to check. Lemma 2.4.
A binary signature f is orthogonal or a zero signature iff (cid:98) f has parity and ars .Proof. Consider M ( f ) and M ( (cid:98) f ) = M ( Z − f ) = Z − M ( f )( Z − ) T . Then, M ( f ) = (cid:2) a b − b a (cid:3) iff M ( (cid:98) f ) = (cid:2) a + b i a − b i (cid:3) , and M ( f ) = (cid:2) a bb − a (cid:3) iff M ( (cid:98) f ) = (cid:2) a − b i a + b i (cid:3) . Also, f ≡ (cid:98) f ≡ O denote the set of all binary orthogonal signatures and the binary zero signature. Then, (cid:98) O = Z − O is the set of all binary signatures with ars and parity (including the binary zerosignature). Note that B ⊆ O and (cid:98)
B ⊆ (cid:98) O . Here the transformed set (cid:98) B = (cid:110) (cid:99) = +2 , (cid:99) = − , (cid:99) (cid:54) = +2 , (cid:99) (cid:54) = − (cid:111) = {(cid:54) = , = , ( − i ) · = − , i · (cid:54) = − } . For every Q ∈ O , let (cid:98) Q = Z − QZ . Then, (cid:98) Q (cid:98) F = ( Z − QZ )( Z − F ) = Z − ( Q F ) = (cid:100) Q F . Thus,Holant( (cid:54) = | (cid:98) F ) ≡ T Holant(= | F ) ≡ T Holant(= | Q F ) ≡ T Holant( (cid:54) = | (cid:98) Q (cid:98) F ) . Let (cid:99) O = { (cid:98) Q = Z − QZ | Q ∈ O } . Then, (cid:99) O = { [ α
00 ¯ α ] , [ α ¯ α ] | α ∈ C , | α | = 1 } . Note that thenotation (cid:98) · on a matrix Q ∈ O is not the same as the notation (cid:98) · on a signature f ∈ O . Supposethat M ( f ) = Q ∈ O . Since Q = Z (cid:98) QZ − and Z − ( Z − ) T = N , M ( (cid:98) f ) = Z − Q ( Z − ) T = Z − ( Z (cid:98) QZ − )( Z − ) T = (cid:98) QN (cid:54) = (cid:98) Q. .3 Signature factorization Recall that by our definition, every (complex valued) signature has arity at least one. A nonzerosignature g divides f denoted by g | f , if there is a signature h such that f = g ⊗ h (with possiblya permutation of variables) or there is a constant λ such that f = λ · g . In the latter case, if λ (cid:54) = 0,then we also have f | g since g = λ · f . For nonzero signatures, if both g | f and f | g , then they arenonzero constant multiples of each other, and we say g is an associate of f , denoted by g ∼ f . Interms of this division relation, the notions of irreducible signatures and prime signatures have beendefined. They are proved equivalent and thus, the unique prime factorization (UPF) of signaturesis established [14].A nonzero signature f is irreducible if there are no signatures g and h such that f = g ⊗ h .A nonzero signature f is a prime signature if f | g ⊗ h implies that f | g or f | h . These notionsare equivalent. We say a signature f is reducible if f = g ⊗ h , for some signatures g and h . Allzero signatures of arity greater than 1 are reducible. A prime factorization of a signature f is f = g ⊗ . . . ⊗ g k up to a permutation of variables, where each g i is irreducible. Lemma 2.5 (Unique prime factorization [14]) . Every nonzero signature f has a prime factoriza-tion. If f has prime factorizations f = g ⊗ . . . ⊗ g k and f = h ⊗ . . . ⊗ h (cid:96) , both up to a permutationof variables, then k = (cid:96) and after reordering the factors we have g i ∼ h i for all i . Lemma 2.6 ([14]) . let f be a real-valued reducible signature, then there exists a factorization f = g ⊗ h such that g and h are both real-valued signatures.Equivalently, let (cid:98) f be a reducible signature satisfying ars , then there exists a factorization (cid:98) f = (cid:98) g ⊗ (cid:98) h such that (cid:98) g and (cid:98) h both satisfy ars . In the following, when we say that a real-valued reducible signature f has a factorization g ⊗ h ,we always assume that g and h are real-valued. Equivalently, when we say a signature (cid:98) f satisfying ars has a factorization (cid:98) g ⊗ (cid:98) h , we always assume that (cid:98) g and (cid:98) h satisfy ars .For a signature set F , we use F ⊗ k ( k (cid:62)
1) to denote the set { λ (cid:78) ki =1 f i | λ ∈ R \{ } , f i ∈ F } .Here, λ denotes a normalization scalar. In this paper, we only consider the normalization bynonzero real constants. Note that F ⊗ contains all signatures obtained from F by normalization.We use F ⊗ to denote (cid:83) ∞ k =1 F ⊗ k . If a vertex v in a signature grid is labeled by a reducible signature f = g ⊗ h , we can replacethe vertex v by two vertices v and v and label v with g and v with h , respectively. The incidentedges of v become incident edges of v and v respectively according to the partition of variables of f in the tensor product of g and h . This does not change the Holant value. On the other hand, Linand Wang proved that, from a real-valued reducible signature f = g ⊗ h (cid:54)≡ f by g and h while preserving the complexity of a Holant problem. Lemma 2.7 ([25]) . If a nonzero real-valued signature f has a real factorization g ⊗ h , then Holant( g, h, F ) ≡ T Holant( f, F ) and Holant( (cid:54) = | (cid:98) g, (cid:98) h, (cid:98) F ) ≡ T Holant( (cid:54) = | (cid:98) f , (cid:98) F ) for any signature set F ( (cid:98) F ) . We say g ( (cid:98) g ) and h ( (cid:98) h ) are realizable from f ( (cid:98) f ) by factorization. One basic tool used throughout the paper is gadget construction. An F -gate is similar to a signaturegrid ( G, π ) for Holant( F ) except that G = ( V, E, D ) is a graph with internal edges E and dangling5dges D . The dangling edges D define input variables for the F -gate. We denote the regular edgesin E by 1 , , . . . , m and the dangling edges in D by m + 1 , . . . , m + n . Then the F -gate defines afunction f f ( y , . . . , y n ) = (cid:88) σ : E →{ , } (cid:89) v ∈ V f v (ˆ σ | E ( v ) )where ( y , . . . , y n ) ∈ { , } n is an assignment on the dangling edges, ˆ σ is the extension of σ on E bythe assignment ( y , . . . , y m ), and f v is the signature assigned at each vertex v ∈ V . This function f is called the signature of the F -gate. There may be no internal edges in an F -gate at all. Inthis case, f is simply a tensor product of these signatures f v , i.e., f = (cid:78) v ∈ V f v (with possibly apermutation of variables). We say a signature f is realizable from a signature set F by gadgetconstruction if f is the signature of an F -gate. If f is realizable from a set F , then we can freelyadd f into F while preserving the complexity (Lemma 1.3 in [9]). Lemma 2.8 ([9]) . If f is realizable from a set F , then Holant( f, F ) ≡ T Holant( F ) . Note that, if we view Holant(= | F ) as the edge-vertex incidence graph form of Holant( F ), thenit is equivalent to label every edge by = ; similarly in the setting of Holant( (cid:54) = | (cid:98) F ), every edge islabeled by (cid:54) = . Lemma 2.9. If f is realizable from a real-valued signature set F (in the setting of Holant(= | F ) ),then f is also real-valued. Equivalently, if (cid:98) f is realizable from a signature set (cid:98) F satisfying ars (inthe setting of Holant( (cid:54) = | (cid:98) F ) ), then (cid:98) f also satisfies ars . A basic gadget construction is merging . In the setting of Holant(= | F ), given a signature f ∈ F of arity n , we can connect two variables x i and x j of f using = , and this operation gives asignature of arity n −
2. We use ∂ ij f or ∂ + ij f to denote this signature and ∂ ij f = f ij + f ij , where f abij denotes the signature obtained by setting ( x i , x j ) = ( a, b ) ∈ { , } . While in the setting ofHolant( (cid:54) = | (cid:98) F ), the above merging gadget is equivalent to connecting two variables x i and x j of (cid:98) f using (cid:54) = . We denote the resulting signature by (cid:98) ∂ ij (cid:98) f or (cid:98) ∂ + ij (cid:98) f , and we have (cid:100) ∂ ij f = (cid:98) ∂ ij (cid:98) f = (cid:98) f ij + (cid:98) f ij . If (cid:54) = is available (i.e., it either belongs to or can be realized from F ) in Holant(= | F ), we canalso connect two variables x i and x j of f using (cid:54) = . We denote the resulting signature by ∂ (cid:98) + ij f .The merging gadget (cid:98) ∂ + ij is the same as ∂ (cid:98) + ij , we use different notations to distinguish whether thisgadget is used in the setting of Holant(= | F ) or Holant( (cid:54) = | (cid:98) F ) . Also, if = − and (cid:54) = − are availablein Holant(= | F ), then we can construct ∂ − ij f and ∂ (cid:98) − ij f by connecting x i and x j using = − and (cid:54) = − respectively. We also call ∂ − ij and ∂ (cid:98) − ij merging gadgets. Without other specification, by default amerging gadget refers to ∂ ij in the setting of Holant(= | F ). Similarly by default a merging gadgetrefers to (cid:98) ∂ ij in the setting of Holant( (cid:54) = | (cid:98) F ) . The following lemma gives a relation between a signature (cid:98) f and signatures (cid:98) ∂ ij (cid:98) f . Lemma 2.10 ([15]) . Let (cid:98) f be a signature of arity n (cid:62) . If (cid:98) f ( α ) (cid:54) = 0 for some wt( α ) = k (cid:54) = 0 and k (cid:54) = n , then there is a pair of indices { i, j } such that (cid:98) ∂ ij (cid:98) f ( β ) (cid:54) = 0 for some wt( β ) = k − . Inparticular, if for all pairs of indices { i, j } , (cid:98) ∂ ij (cid:98) f ≡ , then (cid:98) f ( α ) = 0 for all α with wt( α ) (cid:54) = 0 and n . We use f abij to denote a function, and f abij to denote a vector that lists the truth table of f abij in a given order. (cid:98) f is an EO signature satisfying ars , the following relation between (cid:98) f and (cid:98) ∂ ij (cid:98) f canbe easily obtained following the proofs of Lemmas 4.3 and 4.5 in [14]. Let D = {(cid:54) = } . Then D ⊗ = { λ · ( (cid:54) = ) ⊗ k | λ ∈ R \{ } , k (cid:62) } is the set of tensor products of (cid:54) = up to nonzero real scalars. Lemma 2.11.
Let (cid:98) f be a n -ary EO signature satisfying ars . • When 2 n = 8, if for all pairs of indices { i, j } , (cid:98) ∂ ij (cid:98) f ∈ D ⊗ , and there exists some (cid:54) = ( x i , x j ) andtwo pairs of indices { u, v } and { s, t } where { u, v }∩{ s, t } (cid:54) = ∅ such that (cid:54) = ( x i , x j ) | (cid:98) ∂ uv (cid:98) f , (cid:98) ∂ st (cid:98) f ,then (cid:98) f ∈ D ⊗ and (cid:54) = ( x i , x j ) | (cid:98) f . • When 2 n (cid:62)
10, if for all pairs of indices { i, j } , (cid:98) ∂ ij (cid:98) f ∈ D ⊗ , then (cid:98) f ∈ D ⊗ .Another gadget construction that connects a nonzero binary signature b with a signature f iscalled extending . An extending gadget connects one variable of f with one variable of b using = in the setting of Holant(= | F ), and connects one variable of (cid:98) f with one variable of (cid:98) b using (cid:54) = inthe setting of Holant( (cid:54) = | (cid:98) F ). By extending an irreducible signature using = or (cid:54) = , we still get anirreducible signature. A particular extending gadget is to extend f with binary signatures in B ⊗ using = in the setting of Holant( F ) . We use { f } B = to denote the set of signatures realizable byextending some variables of f with binary signatures in B ⊗ using = (recall that B ⊗ allows allnonzero real normalization scalars). Equivalently, this gadget is to extend (cid:98) f with binary signaturesin (cid:98) B using (cid:54) = in the setting of Holant( (cid:54) = | (cid:98) F ) . We use { (cid:98) f } (cid:98) B(cid:54) = to denote the set of signaturesrealizable by extending some variables of (cid:98) f with binary signatures in (cid:98) B ⊗ using (cid:54) = . If (cid:98) g ∈ { (cid:98) f } (cid:98) B(cid:54) = ,then we can say that the extending gadget by (cid:98) B defines a relation between (cid:98) g and (cid:98) f . Clearly, byextending variables of (cid:98) f with (cid:54) = ∈ (cid:98) B (using (cid:54) = ), we still get (cid:98) f . Thus, (cid:98) f ∈ { (cid:98) f } (cid:98) B(cid:54) = . So this relationis reflexive. The following lemma shows that this relation is symmetric and transitive, thus it is anequivalence relation. Lemma 2.12. (cid:98) g ∈ { (cid:98) f } (cid:98) B(cid:54) = iff (cid:98) f ∈ { (cid:98) g } (cid:98) B(cid:54) = .
2. If (cid:98) h ∈ { (cid:98) g } (cid:98) B(cid:54) = and (cid:98) g ∈ { (cid:98) f } (cid:98) B(cid:54) = , then (cid:98) h ∈ { (cid:98) f } (cid:98) B(cid:54) = . Proof.
Note that for any (cid:98) b ∈ (cid:98) B ⊗ , if we connect any variable of (cid:98) b with another arbitrary variable ofa copy of the same (cid:98) b using (cid:54) = , then we get (cid:54) = after normalization. Also, by extending a variableof (cid:98) f with (cid:54) = (using (cid:54) = ), we still get (cid:98) f . Suppose that (cid:98) g ∈ { (cid:98) f } (cid:98) B(cid:54) = , and it is realized by extendingcertain variables x i of (cid:98) f with certain b i ∈ (cid:98) B . Then, by extending each of these variables x i of (cid:98) g withexactly the same b i ∈ (cid:98) B , we will get (cid:98) f after normalization. Thus, (cid:98) f ∈ { (cid:98) g } (cid:98) B(cid:54) = . The other directionis proved by exchanging (cid:98) f and (cid:98) g . Thus, (cid:98) g ∈ { (cid:98) f } (cid:98) B(cid:54) = iff (cid:98) f ∈ { (cid:98) g } (cid:98) B(cid:54) = . Also, note that for any (cid:98) b , (cid:98) b ∈ (cid:98) B ⊗ , by connecting an arbitrary variable of (cid:98) b with an arbitraryvariable of (cid:98) b using (cid:54) = , we still get a signature in (cid:98) B ⊗ . Suppose that (cid:98) h is realized by extendingsome variables x i of (cid:98) g with some b i ∈ (cid:98) B ⊗ . We may assume every variable x i of (cid:98) g has been soconnected as (cid:54) = ∈ (cid:98) B ⊗ . Similarly we can assume (cid:98) g is realized by extending every variable x i of (cid:98) f with some b i ∈ (cid:98) B ⊗ . Let b i be the signature realized by connecting b i and b i (using (cid:54) = ). Then, (cid:98) h can be realized by extending each variable x i of (cid:98) f with b i ∈ (cid:98) B ⊗ . Thus, (cid:98) h ∈ { (cid:98) f } (cid:98) B(cid:54) = . Remark:
As a corollary, if (cid:98) g ∈ { (cid:98) f } (cid:98) B(cid:54) = , then { (cid:98) g } (cid:98) B(cid:54) = = { (cid:98) f } (cid:98) B(cid:54) = . Lemma 2.13.
Let (cid:98) b ( x , x ) , (cid:98) b ( y , y ) ∈ (cid:98) O . If by connecting the variable x of (cid:98) b and the variable y of (cid:98) b using (cid:54) = , we get λ · (cid:54) = ( x , y ) for some λ ∈ R \{ } , then (cid:98) b ∼ (cid:98) b . Moreover, by connectingthe variable x of (cid:98) b and the variable y of (cid:98) b , we will get λ · (cid:54) = ( x , y ) . roof. We prove this lemma in the setting of Holant( F ) after the transformation Z back. Now, b = Z (cid:98) b ∈ O and b = Z (cid:98) b ∈ O .Consider matrices M ( b ) = M T ( b ) and M ( b ) = M T ( b ). Since b , b ∈ O , both M ( b ) and M ( b ) are real multiples of real orthogonal matrices, of which there are two types, either rotationsor reflections. For such matrices X, Y , to get X T Y = λI for some λ ∈ R \{ } , X and Y mustbe either both reflections, or both rotations of the same angle, up to nonzero real multiples. Firstsuppose M ( b ) = (cid:2) a bb − a (cid:3) , reflection. Then by connecting x of b and y of b using = we get λ · = ( x , y ), i.e., M T ( b ) M ( b ) = λI . This implies that b is the same reflection up to a nonzeroscalar, i.e., b ∼ b . Similarly, for a rotation M ( b ) = (cid:2) a b − b a (cid:3) , M T ( b ) M ( b ) = λI implies that b is also a rotation of the same angle as b up to a nonzero scalar, thus b ∼ b . In either case, byconnecting the variable x of b and the variable y of b , we will get M T ( b ) M ( b ) = M ( b ) M T ( b ) = λI . This means that we get the signature λ · = ( x , y ) . The statement of the lemma follows from thisafter a Z − transformation.A gadget construction often used in this paper is mating . Given a real-valued signature f ofarity n (cid:62)
2, we connect two copies of f in the following manner: Fix a set S of n − m variablesamong all n variables of f . For each x k ∈ S , connect x k of one copy of f with x k of the othercopy using = . The variables that are not in S are called dangling variables. In this paper, weonly consider the case that m = 1 or 2. For m = 1, there is one dangling variable x i . Then, themating construction realizes a signature of arity 2, denoted by m i f . It can be represented by matrixmultiplication. We have M ( m i f ) = M i ( f ) I ⊗ ( n − M T i ( f ) = (cid:20) f i f i (cid:21) (cid:104) f i T f i T (cid:105) = (cid:20) | f i | (cid:104) f i , f i (cid:105)(cid:104) f i , f i (cid:105) | f i | , (cid:21) (2.1)where (cid:104)· , ·(cid:105) denotes the inner product and | · | denotes the norm defined by this inner product. (Wewill use the same notation (cid:104)· , ·(cid:105) to denote the complex inner product (with conjugation) below.The notation is consistent.) Note that |(cid:104) f i , f i (cid:105)| (cid:54) | f i | | f i | by the Cauchy-Schwarz inequality.Similarly, in the setting of Holant( (cid:54) = | (cid:98) F ), the above mating operation is equivalent to connectingvariables in S using (cid:54) = . We denote the resulting signature by (cid:98) m i (cid:98) f , which is the same as (cid:100) m i f , andwe have M ( (cid:98) m i (cid:98) f ) = M i ( (cid:98) f ) N ⊗ n − M T i ( (cid:98) f ) = (cid:34)(cid:98) f i (cid:98) f i (cid:35) (cid:20) (cid:21) ⊗ ( n − (cid:104) (cid:98) f i T (cid:98) f i T (cid:105) . Note that (in general complex-valued) (cid:98) f satisfies the ars since f is real, we have N ⊗ ( n − (cid:98) f i T = ( (cid:98) f , ... , (cid:98) f , ... , . . . , (cid:98) f , ... ) T = ( (cid:98) f , ... , (cid:98) f , ... , . . . , (cid:98) f , ... ) = (cid:98) f i T . Thus, we have M ( (cid:98) m i (cid:98) f ) = (cid:34)(cid:98) f i (cid:98) f i (cid:35) (cid:20) (cid:21) ⊗ ( n − (cid:104) (cid:98) f i T (cid:98) f i T (cid:105) = (cid:34)(cid:98) f i (cid:98) f i (cid:35) (cid:104)(cid:98) f i T (cid:98) f i T (cid:105) = (cid:34) (cid:104) (cid:98) f i , (cid:98) f i (cid:105) | (cid:98) f i | | (cid:98) f i | (cid:104) (cid:98) f i , (cid:98) f i (cid:105) (cid:35) . (2.2)If there are two dangling variables x i and x j , we use m ij f and (cid:98) m ij (cid:98) f to denote the signatures realizedby mating f using = and mating (cid:98) f using (cid:54) = respectively.With respect to mating gadgets, the following first order orthogonality was introduced.8 efinition 2.14 (First order orthogonality [15]) . Let f be a complex-valued signature of arity n (cid:62) . It satisfies the first order orthogonality ( ) if there exists some µ (cid:54) = 0 such thatfor all indices i ∈ [ n ] , the entries of f satisfy the following equations | f i | = | f i | = µ, and (cid:104) f i , f i (cid:105) = 0 . Remark:
When f is a real-valued signature, the inner product is just the ordinary dot productwhich can be represented by mating using = . Thus, f satisfies iff there is some real µ (cid:54) = 0 such that for all indices i , M ( m i f ) = µI . On the other hand, when (cid:98) f is a signature with ars , by (2.2), the complex inner product can be represented by mating using (cid:54) = . Thus, (cid:98) f satisfies iff there is some real µ (cid:54) = 0 such that for all i , M ( (cid:98) m i (cid:98) f ) = µN . Moreover, f satisfies iff (cid:98) f satisfies it. Lemma 2.15 ([15]) . Let f be a real-valued signature of arity n . If for all indices i ∈ [ n ] , M ( m i f ) = µ i I for some real µ i (cid:54) = 0 , then f satisfies (i.e., all µ i have the same value). We give some known signature sets that define polynomial time computable (tractable) countingproblems.
Definition 2.16.
Let T denote the set of tensor products of unary and binary signatures. Definition 2.17.
A signature on a set of variables X is of product type if it can be expressed asa product of unary functions, binary equality functions ([1 , , , and binary disequality functions ([0 , , , each on one or two variables of X . We use P to denote the set of product-type functions. Note that the product in Definition 2.17 are ordinary products of functions (not tensor prod-ucts); in particular they may be applied on overlapping sets of variables.
Definition 2.18.
A signature f ( x , . . . , x n ) of arity n is affine if it has the form λ · χ AX =0 · i Q ( X ) , where λ ∈ C , X = ( x , x , . . . , x n , , A is a matrix over Z , Q ( x , x , . . . , x n ) ∈ Z [ x , x , . . . , x n ] is a multilinear polynomial with total degree d ( Q ) (cid:54) and the additional requirement that thecoefficients of all cross terms are even, i.e., Q has the form Q ( x , x , . . . , x n ) = a + n (cid:88) k =1 a k x k + (cid:88) ≤ i A signature f is local-affine if for each σ = s s . . . s n ∈ { , } n in the supportof f , ( T α s ⊗ T α s ⊗ · · · ⊗ T α sn ) f ∈ A . We use L to denote the set of local-affine signatures. Definition 2.21. We say a signature set F is C -transformable if there exists a T ∈ GL ( C ) suchthat (= )( T − ) ⊗ ∈ C and T F ⊆ C . This definition is important because if Holant( C ) is tractable, then Holant( F ) is tractable forany C -transformable set F . Then, the following tractable result is known [20, 2]. Theorem 2.22. Let F be a set of complex valued signatures. Then Holant( F ) is tractable if F ⊆ T , F is P -transformable, F is A -transformable, or F is L -transformable. (T) Lemma 2.23 ([15]) . Let F be a set of real-valued signatures. If F does not satisfy condition (T),then for every Q ∈ O , Q F also does not satisfy condition (T). Moreover, (cid:98) F (cid:54)⊆ P and (cid:98) F (cid:54)⊆ A . We give some known hardness results. We state these results for our setting. Theorem 2.24 ([20, 15]) . Let F be a set of real-valued signatures. If F does not satisfy condition (T). Then for every Q ∈ O and every k (cid:62) 2, ( Q F ) and Holant( (cid:54) = | = k , (cid:98) Q (cid:98) F ) are Theorem 2.25 ([15]) . Let F be a set of real-valued signatures containing a nonzero signature ofodd arity. If F does not satisfy condition (T), then Holant( F ) is polynomial interpolation . Lemma 2.26 ([9]) . Let f and g be nonzero binary signatures with M ( f ) = P − (cid:104) λ λ (cid:105) P and M ( g ) = P − [ ] P for some invertible matrix P . If λ (cid:54) = 0 and | λ λ | (cid:54) = 1 , then Holant( g, F ) (cid:54) T Holant( f, F ) for any signature set F . By Lemmas 2.7 and 2.26, we can always realize a nonzero unary signature from nonzero signa-tures not satisfying . We have the following emma 2.27. Let F be a set of real-valued signatures containing a nonzero signature that doesnot satisfy . If F does not satisfy condition (T), then Holant( F ) is Proof. Consider M i ( f ) for all indices i . Clearly, M ( m i f ) = M i ( f ) M T i ( f ) is a real symmetric positivesemi-definite matrix, which is diagonalizable with two non-negative real eigenvalues λ i (cid:62) µ i (cid:62) f is real valued and f (cid:54)≡ 0, and so M ( m i f ) (cid:54) = 0.Thus, λ i (cid:54) = 0. Then, | µ i λ i | = 1 iff λ i = µ i . In other words, M ( m i f ) = µ i I for some real µ i (cid:54) = 0.Since f does not satisfy , by Lemma 2.15, there is an index i such that M ( m i f ) (cid:54) = µ i I for any real µ i (cid:54) = 0. Thus, M ( m i f ) has two eigenvalues with different norms. By Lemma 2.26, wecan realize a nonzero binary signature g such that M ( g ) is degenerate. This implies that g can befactorized as a tensor product of two nonzero unary signatures. By Lemma 2.7, we can realize anonzero unary signature and hence by Theorem 2.25, Holant( F ) is eight-vertex models and Eulerian Orientation (EO) problems. Theorem 2.28 ([13]) . Let (cid:98) f be a signature with M ( (cid:98) f ) = (cid:20) c a d b b d a c (cid:21) . Then, Holant( (cid:54) = | (cid:98) f ) is • (cid:98) f has support , • (cid:98) f has support and the nonzero entries of M ( (cid:98) f ) do not have the same norm, or • (cid:98) f has support , all nonzero entries of M ( (cid:98) f ) are positive real numbers and are not all equal. Theorem 2.29 ([14]) . Let (cid:98) F be a set of EO signatures (i.e., with half-weighted support) satisfying ars . Then Holant( DEQ | (cid:98) F ) is (cid:98) F ⊆ P or (cid:98) F ⊆ A .The following reduction states that we can realize all EQ once we have = in Holant( F ). Lemma 2.30 ([9]) . ( F ) (cid:54) T Holant(= , F ) . The following reductions state that we can realize all DEQ once we have (cid:54) = in Holant( (cid:54) = | (cid:98) F ) . Lemma 2.31 ([14]) . Holant( DEQ | (cid:98) F ) (cid:54) T Holant( (cid:54) = | DEQ , (cid:98) F ) (cid:54) T Holant( (cid:54) = |(cid:54) = , (cid:98) F ) . We use the following Table 1 to summarize notations given in this section. In the left column, welist notations in Holant(= | F ) where F is a set of real-valued signatures, and in the right column,we list corresponding notations in Holant( (cid:54) = | (cid:98) F ) where (cid:98) F = Z − F is the set of complex-valuedsignatures with ars . Note that although EO also satisfies ars , we will only use it in Holant(= | F ).Similarly, we will only use DEQ and D in Holant( (cid:54) = | (cid:98) F ) although it is real-valued.Recall that F ⊗ denotes the set { λ (cid:78) ki =1 f i | λ ∈ R \{ } , k (cid:62) , f i ∈ F } for any signature set F .We remark that both O ⊗ and (cid:98) O ⊗ contain all zero signatures of even arity since the binary zerosignature is in O and (cid:98) O . However, B ⊗ , (cid:98) B ⊗ , and D ⊗ do not contain any zero signatures.In the following, without other specifications, we use f to denote a real-valued signature and F to denote a set of real-valued signatures. We use (cid:98) f to denote a signature satisfying ars and (cid:98) F todenote a set of such signatures. We use Q to denote a matrix in O , and (cid:98) Q to denote a matrix in (cid:99) O . Clearly, if F is real-valued, then Q F is also real-valued. Equivalently, if (cid:98) F satisfies ars , then (cid:98) Q (cid:98) F = (cid:100) Q F also satisfies ars . 11olant(= | F ) where F is real-valued Holant( (cid:54) = | (cid:98) F ) where (cid:98) F satisfies ars EQ = { = , = , . . . , = n , . . . } N / AN / A DEQ = {(cid:54) = , (cid:54) = , . . . , (cid:54) = n , . . . } , D = {(cid:54) = }O = { binary orthogonal and zero signatures } (cid:98) O = { binary signatures with ars and parity }B = { = , = − , (cid:54) = , (cid:54) = − } (cid:98) B = {(cid:54) = , = , ( − i ) · = − , i · (cid:54) = − } a holographic transformation Q F by Q ∈ O a holographic transformation (cid:98) Q (cid:98) F by (cid:98) Q ∈ (cid:99) O a merging gadget ∂ ij f = f ij + f ij a merging gadget (cid:98) ∂ ij (cid:98) f = (cid:98) f ij + (cid:98) f ij extending gadgets { f } B = with B extending gadgets { (cid:98) f } (cid:98) B(cid:54) = with (cid:98) B a mating gadget m ij f = M ij ( f ) I ⊗ n − M T ij ( f ) a mating gadget (cid:98) m ij (cid:98) f = M ij ( (cid:98) f ) N ⊗ n − M T ij ( (cid:98) f )Table 1: Comparisons of notations in Holant(= | F ) and Holant( (cid:54) = | (cid:98) F ) By Theorem 2.22, if F satisfies condition (T), then Holant( F ) is P-time computable. So, we onlyneed to prove that Holant( F ) is F does not satisfy condition (T). If F containsa nonzero signature of odd arity, then by Theorem 2.25, we are done. In the following withoutother specifications, when refer to a real-valued signature set F or a corresponding signature set (cid:98) F = Z − F satisfying ars , we always assume that they consist of signatures of even arity, and F does not satisfy condition (T).In Section 4, we generalize the notion of first order orthogonality ( ) to second orderorthogonality ( ). This property plays a key role in our proof. We show that all irreduciblesignatures in F satisfy , or else, we get for signatures with ars . These will be used throughout in the proof.In Section 5, we give the induction framework of the proof. Since F does not satisfy condition(T), F (cid:54)⊆ T . Also since O ⊗ ⊆ T , and by Lemma 5.1, we may assume that F contains a signature f of arity 2 n (cid:62) f / ∈ O ⊗ . We want to achieve a proof of n . When 2 n = 2, as a corollary of , we show that Holant( F ) is n = 4, by , we show that Holant( F ) is f / ∈ O ⊗ be a 6-ary signature in F . Weshow that Holant( F ) is f with the Bellproperty can be realized (Theorem 6.5). By gadget construction, all four Bell signatures B can berealized from f . Then we prove the b ( f , F ) = Holant( B , f , F ) (Theorem7.19 and Lemma 7.20). Combining these two results, we have Holant( F ) is f / ∈ O ⊗ be an 8-ary signature in F . We showthat Holant( F ) is f with the strongBell property can be realized (Theorem 8.5). One can prove that B cannot be realized from f bygadget construction. However, by introducing Holant problems with limited appearance and usingthe strong Bell property of f , we show Holant b ( f , F ) ≤ T Holant( f , F ) (Lemmas 8.10). Then, we12rove the b ( f , F ). Combining these results, we have Holant( F ) is n (cid:62) 10. Let f / ∈ O ⊗ be a 2 n -ary (2 n (cid:62) 10) signature in F . Then, Holant( F ) is n that is not in O ⊗ (Lemma 9.1). Then, by a sequence of reductionsof length independent of the problem instance size, we can eventually realize a signature of arityat most 8 that is not in O ⊗ . Finally, combining Lemmas 5.1, 5.2, 7.21, 8.12 and 9.1, we finish theproof of Theorem 1.2. In the actual proof, for convenience, many results are proved in the settingof Holant( (cid:54) = | (cid:98) F ) which is equivalent to Holant( F ) under the Z − transformation. In this section, we generalize the notion of first order orthogonality ( ) to second orderorthogonality ( ) (Definition 4.1). We show that for real-valued F that does not satisfycondition (T), every irreducible f ∈ F of arity at least 4 satisfies , or otherwise Holant( F )is forsignatures with ars . These will be used throughout in the following proof. Definition 4.1 (Second order orthogonality) . Let f be a complex-valued signature of arity n (cid:62) .It satisfies the second order orthogonality ( ) if there exists some λ (cid:54) = 0 such that for allpairs of indices { i, j } ⊆ [ n ] , the entries of f satisfy | f ij | = | f ij | = | f ij | = | f ij | = λ, and (cid:104) f abij , f cdij (cid:105) = 0 for all ( a, b ) (cid:54) = ( c, d ) . Remark: Similar to the remark of first order orthogonality (Definition 2.14), f satisfies iff there is some λ (cid:54) = 0 such that for all ( i, j ), M ( m ij f ) = λI = λI ⊗ , and (cid:98) f satisfies iff there is some λ (cid:54) = 0 such that for all ( i, j ), M ( (cid:98) m ij (cid:98) f ) = λN = λN ⊗ . Moreover, f satisfies iff (cid:98) f satisfies it. Clearly, implies .In the next, we will prove Lemma 4.4 based on dichotomies of (cid:54) = | (cid:98) F ), for convenience, we will consider the problem Holant( (cid:54) = | (cid:98) F ) whichis equivalent to Holant( F ). Recall that (cid:98) F = Z − F satisfies ars , and we assumed that F does notsatisfy condition (T). We first give the following lemma. Lemma 4.2. Holant( DEQ | (cid:98) F ) is Since F does not satisfy condition (T), by Lemma 2.23, (cid:98) F (cid:54)⊆ P and (cid:98) F (cid:54)⊆ A . If (cid:98) F is a setof EO signatures, then by Theorems 2.29, Holant( DEQ | (cid:98) F ) is (cid:98) F (cid:54)⊆ P and (cid:98) F (cid:54)⊆ A .Thus, we may assume that there is a signature (cid:98) f ∈ (cid:98) F whose support is not half-weighted. Supposethat (cid:98) f has arity 2 n . Since S ( (cid:98) f ) (cid:54)⊆ H n , by ars , there is an α ∈ Z n with wt( α ) = k < n suchthat (cid:98) f ( α ) (cid:54) = 0. We first show that we can realize a signature (cid:98) g of arity 2 n − k such that (cid:98) g ( (cid:126) (cid:54) = 0.If wt( α ) = k = 0, then we are done. Otherwise, we have n > k (cid:62) 1. Thus, 2 n (cid:62) α haslength at least 4. By Lemma 2.10, there is a pair of indices { i, j } such that (cid:98) ∂ ij (cid:98) f ( β ) (cid:54) = 0 for somewt( β ) = k − 1. Clearly, (cid:98) ∂ ij (cid:98) f has arity 2 n − 2. Since 0 (cid:54) k − < (2 n − / (cid:98) ∂ ij (cid:98) f is not an EOsignature. Now we can continue this process, and by a chain of merging gadgets using (cid:54) = , we canrealize a signature (cid:98) g of arity 2 m = 2 n − k such that (cid:98) g ( (cid:126) (cid:54) = 0. Denote by a = (cid:98) g ( (cid:126) m variables of (cid:98) g with 2 m variables of (cid:54) = m that always take the samevalue in S ( (cid:54) = m ) using (cid:54) = . We get a signature (cid:98) h of arity 2 m where (cid:98) h ( (cid:126) 0) = a , (cid:98) h ( (cid:126) 1) = ¯ a by ars ,and (cid:98) h ( γ ) = 0 elsewhere. Then, consider the holographic transformation by (cid:98) Q = (cid:104) m √ ¯ a m √ a (cid:105) ∈ (cid:98) O .It transforms (cid:98) h to (cid:54) = m , but does not change DEQ . Thus,Holant( DEQ | (cid:98) h, (cid:98) F ) ≡ T Holant( DEQ | = m , (cid:98) Q (cid:98) F ) . Since (cid:98) F does not satisfy condition (T), by Theorem 2.24, Holant( DEQ | = m , (cid:98) Q (cid:98) F ) is DEQ | (cid:98) F ) is (cid:98) m ij (cid:98) f realized by mating. Lemma 4.3. Let (cid:98) f ∈ (cid:98) F be a signature of arity n (cid:62) . Then, • Holant( (cid:54) = | (cid:98) F ) is • for all pairs of indices { i, j } , there exists a nonzero binary signature (cid:98) b ij ∈ (cid:98) O such that (cid:98) b ij ( x i , x j ) | (cid:98) f or M ( (cid:98) m ij (cid:98) f ) = λ ij N for some real λ ij (cid:54) = 0 .Proof. If (cid:98) f ≡ 0, then the lemma holds trivially since for all { i, j } and any (cid:98) b ij (cid:54) = 0, (cid:98) b ij ( x i , x j ) | (cid:98) f .Thus, we may assume that f (cid:54)≡ (cid:98) f does not satisfy , then f does not satisfy it. By Lemma 2.27, Holant( (cid:54) = | (cid:98) F ) ≡ T Holant(= | F ) is (cid:98) f satisfies . Then, for all indices i , we have M ( (cid:98) m i (cid:98) f ) = (cid:34) (cid:104) (cid:98) f i , (cid:98) f i (cid:105) | (cid:98) f i | | (cid:98) f i | (cid:104) (cid:98) f i , (cid:98) f i (cid:105) (cid:35) = µ (cid:20) (cid:21) . For any variable x i , we may take another variable x j ( j (cid:54) = i ) and partition the sum in the innerproduct (cid:104) (cid:98) f i , (cid:98) f i (cid:105) = 0 into two sums depending on whether x j = 0 or 1. Also, by ars we have (cid:104) (cid:98) f i , (cid:98) f i (cid:105) = (cid:104) (cid:98) f ij , (cid:98) f ij (cid:105) + (cid:104) (cid:98) f ij , (cid:98) f ij (cid:105) = (cid:104) (cid:98) f ij , (cid:98) f ij (cid:105) + (cid:104) (cid:98) f ij , (cid:98) f ij (cid:105) = 2 (cid:104) (cid:98) f ij , (cid:98) f ij (cid:105) = 0 . Thus, for all pairs of indices { i, j } , (cid:104) (cid:98) f ij , (cid:98) f ij (cid:105) = 0 and (cid:104) (cid:98) f ij , (cid:98) f ij (cid:105) = 0. (Note that by exchanging i and j we also have (cid:104) (cid:98) f ij , (cid:98) f ij (cid:105) = 0 and (cid:104) (cid:98) f ij , (cid:98) f ij (cid:105) = 0.) Also by ars , we have | (cid:98) f ij | = | (cid:98) f ij | = | (cid:98) f ij | and | (cid:98) f ij | = | (cid:98) f ij | = | (cid:98) f ij | . Now, consider (cid:98) m ij (cid:98) f for all pairs of indices { i, j } . M ( (cid:98) m ij (cid:98) f ) = (cid:98) f ij (cid:98) f ij (cid:98) f ij (cid:98) f ij (cid:104)(cid:98) f ij T (cid:98) f ij T (cid:98) f ij T (cid:98) f ij T (cid:105) = (cid:104) (cid:98) f ij , (cid:98) f ij (cid:105) | (cid:98) f ij | (cid:104) (cid:98) f ij , (cid:98) f ij (cid:105) | (cid:98) f ij | | (cid:98) f ij | (cid:104) (cid:98) f ij , (cid:98) f ij (cid:105) | (cid:98) f ij | (cid:104) (cid:98) f ij , (cid:98) f ij (cid:105) . Note that |(cid:104) (cid:98) f ij , (cid:98) f ij (cid:105)| (cid:54) | (cid:98) f ij | · | (cid:98) f ij | by Cauchy-Schwarz inequality. Clearly, (cid:98) m ij (cid:98) f has even parity, andthus it represents a signature of the eight-vertex model. If there exists a pair of indices { i, j } suchthat Holant( (cid:54) = | (cid:98) m ij (cid:98) f ) is (cid:54) = | (cid:98) m ij (cid:98) f ) (cid:54) T Holant( (cid:54) = | (cid:98) F ).Thus, we may assume all (cid:98) m ij (cid:98) f belong to the tractable family for eight-vertex models. Clearly,by observing its antidiagonal entries of the matrix M ( (cid:98) m ij (cid:98) f ), we have (cid:98) m ij (cid:98) f (cid:54)≡ (cid:98) f (cid:54)≡ 0. ByTheorem 2.28, there are three possible cases. 14 There exists a pair { i, j } such that (cid:98) m ij (cid:98) f has support of size 2. By Cauchy-Schwarz inequality, M ( (cid:98) m ij (cid:98) f ) is either of the form λ ij (cid:20) (cid:21) where λ ij = | (cid:98) f ij | = | (cid:98) f ij | (cid:54) = 0 or λ ij (cid:20) (cid:21) where λ ij = | (cid:98) f ij | = | (cid:98) f ij | (cid:54) = 0. In both cases, (cid:54) = is realizable since λ ij (cid:54) = 0. The formthat (cid:104) (cid:98) f ij , (cid:98) f ij (cid:105) (cid:54) = 0 while | (cid:98) f ij | = | (cid:98) f ij | = 0 cannot occur since |(cid:104) (cid:98) f ij , (cid:98) f ij (cid:105)| (cid:54) | (cid:98) f ij || (cid:98) f ij | . Also,the form that (cid:104) (cid:98) f ij , (cid:98) f ij (cid:105) (cid:54) = 0 while | (cid:98) f ij | = | (cid:98) f ij | = 0 cannot occur. Since (cid:54) = is available,by Lemma 2.31, Holant( DEQ | (cid:98) F ) (cid:54) T Holant(= | (cid:98) F ) . By Lemma 4.2, Holant(= | (cid:98) F ) is • There exists a pair { i, j } such that (cid:98) m ij (cid:98) f has support of size 8. We can rename the fourvariables of (cid:98) m ij (cid:98) f in a cyclic permutation. We use (cid:98) g to denote this signature. Then M ( (cid:98) g ) = M ( (cid:98) g ) = (cid:20) c d a b b a d c (cid:21) where a and b are positive real numbers and c and d are nonzero complexnumbers. Consider the signature (cid:98) m (cid:98) g realized by mating (cid:98) g . We denote it by (cid:98) h . Then, M ( (cid:98) h ) = M ( (cid:98) g ) N M T ( (cid:98) g ) = cd | c | + | d | ab a + b a + b ab | c | + | d | c ¯ d = c (cid:48) d (cid:48) a (cid:48) b (cid:48) b (cid:48) a (cid:48) d (cid:48) c (cid:48) , where a (cid:48) , b (cid:48) , and d (cid:48) are positive real numbers and c (cid:48) is a nonzero complex number. Supposethat the argument of c (cid:48) is θ , i.e., c (cid:48) = | c (cid:48) | e i θ . Consider the holographic transformation by (cid:98) Q = (cid:104) e − i θ/ e i θ/ (cid:105) ∈ (cid:99) O . Then,Holant( (cid:54) = | (cid:98) h, (cid:98) F ) ≡ T Holant( (cid:54) = | (cid:98) Q (cid:98) h, (cid:98) Q (cid:98) F ) . Note that M ( (cid:98) Q (cid:98) h ) = (cid:34) | c (cid:48) | d (cid:48) a (cid:48) b (cid:48) b (cid:48) a (cid:48) d (cid:48) | c (cid:48) | (cid:35) where all entries are positive real numbers. Notice that allweight 2 entries of (cid:98) h are unchanged in (cid:98) Q (cid:98) h . By Theorem 2.28, Holant( (cid:54) = | (cid:98) Q (cid:98) h ) is a (cid:48) = b (cid:48) = | c (cid:48) | = d (cid:48) . Thus, we may assume that M ( (cid:98) Q (cid:98) h ) = (cid:20) (cid:21) up to normalization.Notice that M ( Z ( (cid:98) Q (cid:98) h )) = Z ⊗ M ( (cid:98) Q (cid:98) h )( Z T ) ⊗ = (cid:20) (cid:21) , which is the arity-4 equality (= ).Consider the holographic transformation by Z which transfers (cid:54) = back to = . Rememberthat (cid:98) Q = Z − QZ . Then, Z ( (cid:98) Q (cid:98) F ) = Z ( Z − QZ )( Z − F ) = Q F . Since (cid:98) Q ∈ (cid:99) O , we have Q ∈ O . Thus, Holant( (cid:54) = | (cid:98) Q (cid:98) h, (cid:98) Q (cid:98) F ) ≡ T Holant(= | = , Q F ) . By Lemma 2.30, ( Q F ) (cid:54) T Holant(= | = , Q F ). Since F does not satisfy condition (T)and Q ∈ O , by Theorem 2.24, ( Q F ) is (cid:54) = | (cid:98) F ) is • For all { i, j } , (cid:98) m ij (cid:98) f has support of size 4. By Cauchy-Schwarz inequality, M ( (cid:98) m ij (cid:98) f ) is of theform (cid:20) b a a b (cid:21) or (cid:20) b a a ¯ b 00 0 0 0 (cid:21) where a − | b | = 0, or the form λ ij (cid:20) (cid:21) where λ ij = | (cid:98) f ij | = | (cid:98) f ij | (cid:54) = 0. If M ( m ij (cid:98) f ) = λ ij N , then we are done. Otherwise, M ( (cid:98) m ij (cid:98) f ) has rank one. Hence, M ij ( (cid:98) f ) also has rank one. Then, by observing the form of M ( (cid:98) m ij (cid:98) f ) especially the all zerorows, (cid:98) f can be factorized as (cid:98) b ij ( x i , x j ) ⊗ (cid:98) g where (cid:98) b ij ∈ (cid:98) O and (cid:98) g is a signature on the other n − Remark: We give a restatement of Lemma 4.3 in the setting of Holant( F ). Let f ∈ F be asignature of arity 2 n (cid:62) 4. Then, Holant( F ) is { i, j } , thereexists a nonzero binary signature b ij ∈ O such that b ij ( x i , x j ) | f or M ( m ij f ) = λ ij I for some real λ ij (cid:54) = 0.Now for an irreducible signature (cid:98) f of arity 2 n (cid:62) 4, we show that it satisfies or weget Lemma 4.4. Let (cid:98) f ∈ (cid:98) F be an irreducible signature of arity n (cid:62) . If (cid:98) f does not satisfy ,then Holant( (cid:54) = | (cid:98) F ) is Since (cid:98) f is irreducible, by Lemma 4.3, M ( (cid:98) m ij (cid:98) f ) = λ ij N for all { i, j } . Now, we show all λ ij have the same value. If we connect further the two respective pairs of variables of m ij f , which totallyconnects two copies of f , we get a value 4 λ ij . This value clearly does not depend on the particularindices { i, j } . We denote the value λ ij by λ . This value is nonzero because (cid:98) f is irreducible.We derive some consequences from the condition for signatures with ars . Supposethat (cid:98) f satisfies . First, by definition we have | (cid:98) f abij | = λ for any ( x i , x j ) = ( a, b ) ∈ { , } . Given a vector (cid:98) f abij , we can pick a third variable x k and partition (cid:98) f abij into two vectors (cid:98) f ab ijk and (cid:98) f ab ijk according to x k = 0 or 1. By setting ( a, b ) = (0 , | (cid:98) f ij | = | (cid:98) f ijk | + | (cid:98) f ijk | = λ. (4.1)Similarly, we consider the vector (cid:98) f ik and partition it according to x j = 0 or 1. We have | (cid:98) f ik | = | (cid:98) f ijk | + | (cid:98) f ijk | = λ. (4.2)Comparing equations (4.1) and (4.2), we have | (cid:98) f ijk | = | (cid:98) f ijk | . Moreover, by ars , we have | (cid:98) f ijk | = | (cid:98) f ijk | . Thus, we have | (cid:98) f ijk | = | (cid:98) f ijk | . Note that the vector (cid:98) f jk is partitioned into two vectors (cid:98) f ijk and (cid:98) f ijk according to x i = 0 or 1. That is | (cid:98) f jk | = | (cid:98) f ijk | + | (cid:98) f ijk | = λ. Thus, we have | (cid:98) f ijk | = | (cid:98) f ijk | = λ/ 2. Then, by equation (4.1), we have | (cid:98) f ijk | = λ/ 2, and againby ars , we also have | (cid:98) f ijk | = | (cid:98) f ijk | = λ/ 2. Note that indices i, j, k are picked arbitrarily, bysymmetry, we have | (cid:98) f abcijk | = λ/ x i , x j , x k ) = ( a, b, c ) ∈ { , } . Given a vector (cid:98) f abcijk , we can continue to pick a fourth variable x (cid:96) and partition (cid:98) f abcijk into twovectors (cid:98) f abc ijk(cid:96) and (cid:98) f abc ijk(cid:96) according to x (cid:96) = 0 or 1. By setting ( a, b, c ) = (0 , , | (cid:98) f ijk | = | (cid:98) f ijk(cid:96) | + | (cid:98) f ijk(cid:96) | = λ/ . (4.4)16imilarly, we consider the vector (cid:98) f ij(cid:96) and partition it according to x k = 0 or 1. We have | (cid:98) f ij(cid:96) | = | (cid:98) f ijk(cid:96) | + | (cid:98) f ijk(cid:96) | = λ/ . (4.5)Comparing equations (4.4) and (4.5), and also by ars , we have | (cid:98) f ijk(cid:96) | = | (cid:98) f ijk(cid:96) | = | (cid:98) f ijk(cid:96) | = | (cid:98) f ijk(cid:96) | (4.6)for all indices { i, j, k, (cid:96) } . Similarly, we can get | (cid:98) f ijk(cid:96) | = | (cid:98) f ijk(cid:96) | = | (cid:98) f ijk(cid:96) | = | (cid:98) f ijk(cid:96) | . (4.7)By the definition of second order orthogonality, we also have (cid:104) (cid:98) f abij , (cid:98) f cdij (cid:105) = 0 (4.8)for all variables x i , x j and ( a, b ) (cid:54) = ( c, d ).Equations (4.6), (4.7) and (4.8) will be used frequently in the analysis of signatures satisfying ars and . This is also a reason why we consider the problem in the setting underthe Z − transformation, Holant( (cid:54) = | (cid:98) F ), where we can express these consequences of elegantly, instead of Holant( F ) which is logically equivalent. By combining and ars ofthe signature (cid:98) f , we get these simply expressed, thus easily applicable, conditions in terms of normsand inner products. n (cid:54) In this section, we introduce the induction framework and handle the base cases (Lemmas 5.1 and5.2). Recall that (cid:98) O denotes the set of binary signatures with ars and parity (including the binaryzero signature), and (cid:98) O ⊗ denotes the set of tensor products of signatures in (cid:98) O . Since F does notsatisfy condition (T), (cid:98) F (cid:54)⊆ T . Also, since (cid:98) O ⊗ ⊆ T , (cid:98) F (cid:54)⊆ (cid:98) O ⊗ . Thus, there is a nonzero signature (cid:98) f ∈ (cid:98) F of arity 2 n such that (cid:98) f / ∈ (cid:98) O ⊗ . We want to achieve a proof of n . We first consider the base that 2 n = 2. Notice that a nonzero binary signature (cid:98) f satisfies iff its matrix form (as a 2-by-2 matrix) is orthogonal. Thus, (cid:98) f / ∈ (cid:98) O implies that it doesnot satisfy . Then, we have the following result. Lemma 5.1. Let F contain a binary signature f / ∈ O ⊗ . Then, Holant( F ) is (cid:98) F contain a binary signature (cid:98) f / ∈ (cid:98) O ⊗ . Then, Holant( (cid:54) = | (cid:98) F ) is We prove this lemma in the setting of Holant( F ). Since O ⊗ contains the binary zerosignature, f / ∈ O ⊗ implies that f (cid:54)≡ 0. If f is reducible, then it is a tensor product of two nonzerounary signatures. By Lemma 2.7, we can realize a nonzero unary signature by factorization, andwe are done by Theorem 2.25. Otherwise, f is irreducible. Since f / ∈ O ⊗ , f does not satisfy . By Lemma 2.27, Holant( F ) is (cid:98) f of arity 2 n (cid:62) (cid:98) O ⊗ , and realize a signature (cid:98) g of arity 2 k (cid:54) n − (cid:98) O ⊗ , or otherwise wecan directly show Holant( (cid:54) = | (cid:98) F ) is (cid:98) b / ∈ (cid:98) O . By Lemma 5.1, we are done.For the inductive step, we first consider the case that (cid:98) f is reducible. Suppose that (cid:98) f = (cid:98) f ⊗ (cid:98) f .If (cid:98) f or (cid:98) f have odd arity, then we can realize a signature of odd arity by factorization and we aredone. Otherwise, (cid:98) f and (cid:98) f have even arity. Since (cid:98) f / ∈ (cid:98) O ⊗ , we know (cid:98) f and (cid:98) f cannot both bein (cid:98) O ⊗ . Then, we can realize a signature of lower arity that is not in (cid:98) O ⊗ by factorization. We aredone. Thus, in the following we may assume that (cid:98) f is irreducible. Then, we may further assumethat (cid:98) f satisfies . Otherwise, we get (cid:54) = to realize signatures of arity 2 n − (cid:98) f . Consider (cid:98) ∂ ij (cid:98) f for all pairs of indices { i, j } . If thereexists a pair { i, j } such that (cid:98) ∂ ij (cid:98) f / ∈ (cid:98) O ⊗ , then we can realize (cid:98) g = (cid:98) ∂ ij (cid:98) f which has arity 2 n − 2, andwe are done. Thus, we may assume (cid:98) ∂ ij (cid:98) f ∈ (cid:98) O ⊗ for all { i, j } . We denote this property by (cid:98) f ∈ (cid:98)(cid:82) (cid:98) O ⊗ .We want to achieve our induction proof based on these two properties: and (cid:98) f ∈ (cid:98)(cid:82) (cid:98) O ⊗ .We consider the case that 2 n = 4. Lemma 5.2. Let (cid:98) F contain a -ary signature (cid:98) f / ∈ (cid:98) O ⊗ . Then, Holant( (cid:54) = | (cid:98) F ) is P-hard.Proof. Since (cid:98) f / ∈ (cid:98) O ⊗ , f (cid:54)≡ 0. First, we may assume that (cid:98) f is irreducible. Otherwise, we can realizea nonzero unary signature or a binary signature that is not in (cid:98) O . Then, by Theorem 2.25 andLemma 5.1, we have (cid:98) f is irreducible, we may further assume that (cid:98) f satisfies . Otherwise, by Lemma 4.4, we get (cid:98) ∂ ij (cid:98) f realized from (cid:98) f by merging using (cid:54) = . Under the assumptionthat (cid:98) f satisfies , we will show that there exits a pair { i, j } such that (cid:98) ∂ ij (cid:98) f / ∈ (cid:98) O . Thenby Lemma 5.1, we are done. For a contradiction, suppose that (cid:98) f ∈ (cid:98)(cid:82) (cid:98) O i.e., (cid:98) ∂ ij (cid:98) f ∈ (cid:98) O for all pairs { i, j } . Since (cid:98) f satisfies , by equations (4.6) and (4.7), we have | (cid:98) f ijk(cid:96) | = | (cid:98) f ijk(cid:96) | = | (cid:98) f ijk(cid:96) | and | (cid:98) f ijk(cid:96) | = | (cid:98) f ijk(cid:96) | respectively for any permutation ( i, j, k, (cid:96) ) of (1 , , , (cid:98) f on inputs of even weight { , , } have the same norm, and all entries of (cid:98) f on inputs of odd weight { , } have the same norm. We denote by ν and ν the norm squares of entries on inputs of evenweight and odd weight, respectively.Then, we consider the equation (cid:104) (cid:98) f , (cid:98) f (cid:105) = 0 from (4.8) by taking ( i, j ) = (1 , (cid:104) (cid:98) f , (cid:98) f (cid:105) = (cid:98) f (cid:98) f + f (cid:98) f + (cid:98) f (cid:98) f + (cid:98) f (cid:98) f = 0 . (Here for clarity, we omitted the subscript 1234 of (cid:98) f abcd .) By ars , we have (cid:98) f (cid:98) f = (cid:98) f (cid:98) f and (cid:98) f (cid:98) f = (cid:98) f (cid:98) f . Thus, we have (cid:98) f (cid:98) f + (cid:98) f (cid:98) f = 0 . (5.1)Note that by taking norm, | (cid:98) f (cid:98) f | = ν and | (cid:98) f (cid:98) f | = ν . Then, it follows that ν = ν .Thus, all entries of (cid:98) f have the same norm. We normalize the norm to be 1 since (cid:98) f (cid:54)≡ (cid:98) ∂ (cid:98) f . We have (cid:98) ∂ (cid:98) f = ( (cid:98) f + (cid:98) f , (cid:98) f + (cid:98) f , (cid:98) f + (cid:98) f , (cid:98) f + (cid:98) f ) , and by assumption (cid:98) ∂ (cid:98) f ∈ (cid:98) O . Thus, at least one of the two entries (cid:98) f + (cid:98) f and (cid:98) f + (cid:98) f is equal to zero. If (cid:98) f + (cid:98) f = 0, then we have (cid:98) f (cid:98) f = ( − (cid:98) f ) (cid:98) f = −| (cid:98) f | = − . (cid:98) f (cid:98) f = 1. Otherwise, (cid:98) f + (cid:98) f = 0. Then, we have (cid:98) f (cid:98) f = − (cid:98) f (cid:98) f = 1. Thus, among these two products (cid:98) f (cid:98) f and (cid:98) f (cid:98) f ,exactly one is equal to 1, while the other is − 1. Then, we have (cid:98) f (cid:98) f (cid:98) f (cid:98) f = − . Similarly, by considering (cid:98) ∂ (cid:98) f and (cid:98) ∂ (cid:98) f respectively, we have (cid:98) f (cid:98) f (cid:98) f (cid:98) f = − , and (cid:98) f (cid:98) f (cid:98) f (cid:98) f = − . Multiply these three products, we have | (cid:98) f | | (cid:98) f | | (cid:98) f | | (cid:98) f | | (cid:98) f | | (cid:98) f | = ( − = − . A contradiction! Remark: In this proof, we showed that there is no irreducible 4-ary signature (cid:98) f that satisfies both and (cid:98) f ∈ (cid:98)(cid:82) (cid:98) O ⊗ .If Lemma 5.2 were to hold for signatures of arity 2 n (cid:62) 6, i.e., there is no irreducible signature (cid:98) f of 2 n (cid:62) (cid:98) f satisfies both and (cid:98) f ∈ (cid:98)(cid:82) (cid:98) O ⊗ , then the induction proof holdsand we are done. We show that this is true for signatures of arity 2 n (cid:62) 10 in Section 9. However,there are extraordinary signatures of arity 6 and 8 with special closure properties (Bell properties)such that they satisfy both and (cid:98) f ∈ (cid:98)(cid:82) (cid:98) O ⊗ . We consider the following 6-ary signature (cid:98) f . We use χ S to denote the indicator function on set S .Let (cid:98) f = χ S · ( − x x + x x + x x + x x + x x + x x where S = S ( (cid:98) f ) = E = { α ∈ Z | wt( α ) ≡ } . One can check that (cid:98) f is irreducible, and (cid:98) f satisfies both and (cid:98) f ∈ (cid:98)(cid:82) (cid:98) O ⊗ . (cid:98) f has the following matrix form M , ( (cid:98) f ) = − − 10 1 − − − − − − − − − − . (6.1)We use Figure 1 to visualize this matrix. A block with orange color denotes an entry +1 and ablock with blue color denotes an entry − 1. Other blank blocks are zeros.19 Figure 1: A visualization of (cid:98) f (cid:98) f In this subsection, we show how this extraordinary signature (cid:98) f was discovered. We prove that if (cid:98) F contains a 6-ary signature (cid:98) f where (cid:98) f / ∈ (cid:98) O ⊗ , then Holant( (cid:54) = | (cid:98) F ) is (cid:98) f is realizablefrom (cid:98) f after a holographic transformation by some (cid:98) Q ∈ (cid:99) O (Theorem 6.5). The general strategyof this proof is to show that we can realize signatures with special properties from (cid:98) f step by step(Lemmas 6.1, 6.2, 6.3 and 6.4), and finally we can realize (cid:98) f , or else we can realize signatures thatlead to (cid:98) f emerges as essentially the unique (and true) obstacle to our proofof Lemma 6.1. Suppose that (cid:98) F contains a -ary signature (cid:98) f / ∈ (cid:98) O ⊗ . Then, Holant( (cid:54) = | (cid:98) F ) is (cid:98) f (cid:48) is realizable from (cid:98) f , where (cid:98) f (cid:48) ( α ) = 0 for all α with wt( α ) = 2 or . Moreover, (cid:98) f (cid:48) is realizable by extending variables of (cid:98) f with nonzero binary signatures in (cid:98) O that are realizable by factorization from (cid:98) ∂ (cid:98) f . Proof. Since (cid:98) f / ∈ (cid:98) O ⊗ , (cid:98) f (cid:54)≡ 0. Again, we may assume that (cid:98) f is irreducible. Otherwise, by factoriza-tion, we can realize a nonzero signature of odd arity, or a signature of arity 2 or 4 that is not in (cid:98) O ⊗ .Then by Theorem 2.25, or Lemmas 5.1 or 5.2, we get (cid:98) f is irreducible, we may further assume that (cid:98) f satisfies by Lemma 4.4. Also, we mayassume that (cid:98) f ∈ (cid:82) (cid:98) O ⊗ . Otherwise, there is a pair of indices { i, j } such that the 4-ary signature (cid:98) ∂ ij (cid:98) f / ∈ (cid:98) O ⊗ . Then by Lemma 5.2, Holant( (cid:54) = | (cid:98) F ) is { i, j } , (cid:98) ∂ ij (cid:98) f ≡ 0, then by Lemma 2.10, we have (cid:98) f ( α ) = 0 for all α withwt( α ) (cid:54) = 0 and 6. Since f (cid:54)≡ 0, clearly such a signature does not satisfy . Contradiction.Otherwise, there is a pair of indices { i, j } such that (cid:98) ∂ ij (cid:98) f (cid:54)≡ 0. By renaming variables, without loss ofgenerality, we assume that (cid:98) ∂ (cid:98) f (cid:54)≡ 0. Since (cid:98) ∂ (cid:98) f ∈ (cid:98) O ⊗ , in the UPF of (cid:98) ∂ (cid:98) f , by renaming variableswe assume that variables x and x appear in one nonzero binary signature (cid:98) b ( x , x ) ∈ (cid:98) O ⊗ , andvariables x and x appear in the other nonzero binary signature (cid:98) b ( x , x ) ∈ (cid:98) O ⊗ . Thus, we have (cid:98) ∂ (cid:98) f = (cid:98) b ( x , x ) ⊗ (cid:98) b ( x , x ) (cid:54)≡ . By Lemma 2.7, we know that these two binary signatures (cid:98) b and (cid:98) b are realizable by factoriza-tion. Note that for a nonzero binary signature (cid:98) b i ( x i +1 , x i +2 ) ∈ (cid:98) O ( i ∈ { , } ), if we connect thevariable x i +1 of two copies of (cid:98) b i ( x i +1 , x i +2 ) using (cid:54) = (mating two binary signatures), then we20et (cid:54) = up to a nonzero scalar. We consider the following gadget construction G on (cid:98) f . Recall thatin the setting of Holant( (cid:54) = | (cid:98) F ), variables are connected using (cid:54) = . For i ∈ { , } , by a slight abuseof variable names, we connect the variable x i +1 of (cid:98) f with the variable x i +1 of (cid:98) b i ( x i +1 , x i +2 ).We get a signature (cid:98) f (cid:48) of arity 6. Such a gadget construction does not change the irreducibilityof f . Thus, (cid:98) f (cid:48) is irreducible. Again, we may assume that (cid:98) f (cid:48) ∈ (cid:98)(cid:82) (cid:98) O ⊗ and (cid:98) f (cid:48) satisfies .Otherwise, we are done.Consider (cid:98) ∂ (cid:98) f (cid:48) . Since the above gadget construction G does not touch variables x and x of f , the operation of forming G commutes with the merging operation (cid:98) ∂ . Thus, (cid:98) ∂ (cid:98) f (cid:48) can berealized by performing the gadget construction G on (cid:98) ∂ (cid:98) f , which connects each binary signature (cid:98) b i ( i ∈ { , } ) of (cid:98) ∂ (cid:98) f with another copy of itself using (cid:54) = (in the mating fashion). Then, each (cid:98) b i in (cid:98) ∂ (cid:98) f is changed to (cid:54) = up to a nonzero real scalar. After normalization and renaming variables,we have (cid:98) ∂ (cid:98) f (cid:48) = ( (cid:54) = )( x , x ) ⊗ ( (cid:54) = )( x , x ) . Since (cid:98) ∂ (cid:98) f (cid:48) ∈ D ⊗ , for any { i, j } disjoint with { , } we have (cid:98) ∂ ( ij )(12) (cid:98) f (cid:48) ∈ D ⊗ , and hence (cid:98) ∂ ij (cid:98) f (cid:48) (cid:54)≡ { i, j } , (cid:98) ∂ ij (cid:98) f (cid:48) has even parity. We first consider the casethat { i, j } is disjoint with { , } . Connect variables x i and x j of (cid:98) ∂ (cid:98) f (cid:48) using (cid:54) = . Since (cid:98) ∂ (cid:98) f (cid:48) haseven parity, a merging gadget using (cid:54) = will change the parity from even to odd. Thus, (cid:98) ∂ ( ij )(12) (cid:98) f (cid:48) has odd parity. Consider (cid:98) ∂ ij (cid:98) f (cid:48) . Remember that (cid:98) ∂ ij (cid:98) f (cid:48) (cid:54)≡ (cid:98) ∂ ( ij )(12) (cid:98) f (cid:48) (cid:54)≡ 0. Since (cid:98) f (cid:48) ∈ (cid:98)(cid:82) (cid:98) O ⊗ ,We have (cid:98) ∂ ij (cid:98) f (cid:48) ∈ O ⊗ . Thus, (cid:98) ∂ ij (cid:98) f (cid:48) has (either odd or even) parity. For a contradiction, supposethat it has odd parity. Then, (cid:98) ∂ (12)( ij ) (cid:98) f (cid:48) has even parity since it is realized by merging using (cid:54) = . Asignature that has both even parity and odd parity is identically zero. Thus (cid:98) ∂ (12)( ij ) (cid:98) f (cid:48) is the zerosignature. However, since (cid:98) ∂ ( ij )(12) (cid:98) f (cid:48) ∈ D ⊗ , it is not the zero signature. Contradiction. Therefore, (cid:98) ∂ ij (cid:98) f (cid:48) has even parity for all { i, j } disjoint with { , } .Then, consider (cid:98) ∂ ij (cid:98) f (cid:48) for { i, j } ∩ { , } (cid:54) = ∅ . If { , } = { i, j } , then clearly, (cid:98) ∂ (cid:98) f (cid:48) has even parity.Otherwise, without loss of generality, we may assume that i = 1 and j (cid:54) = 2. Consider (cid:98) ∂ j (cid:98) f (cid:48) for3 (cid:54) j (cid:54) 6. If it is a zero signature, then it has even parity. Otherwise, (cid:98) ∂ j (cid:98) f (cid:48) (cid:54)≡ 0. Since (cid:98) ∂ j (cid:98) f (cid:48) ∈ (cid:98) O ⊗ ,we assume that it has the following UPF (cid:98) ∂ j (cid:98) f (cid:48) = (cid:98) b (cid:48) ( x , x u ) ⊗ (cid:98) b (cid:48) ( x v , x w ) . By connecting variables x u and x v of (cid:98) ∂ j (cid:98) f (cid:48) using (cid:54) = , we get (cid:98) ∂ ( uv )(1 j ) (cid:98) f (cid:48) . Since the merging gadgetconnects two nonzero binary signatures in (cid:98) O , the resulting signature is a nonzero binary signature.Thus, (cid:98) ∂ ( uv )(1 j ) (cid:98) f (cid:48) (cid:54)≡ 0. Notice that { u, v } is disjoint with { , } . As showed above, (cid:98) ∂ uv (cid:98) f (cid:48) has evenparity. Then, (cid:98) ∂ (1 j )( uv ) (cid:98) f (cid:48) has odd parity. For a contradiction, suppose that (cid:98) ∂ j (cid:98) f (cid:48) has odd parity.Then (cid:98) ∂ ( uv )(1 j ) (cid:98) f (cid:48) has even parity. But a nonzero signature (cid:98) ∂ ( uv )(1 j ) (cid:98) f (cid:48) cannot have both even parityand odd parity. Contradiction. Thus, (cid:98) ∂ j (cid:98) f (cid:48) has even parity.We have proved that (cid:98) ∂ ij (cid:98) f (cid:48) has even parity for all pairs of indices { i, j } . In other words, forall pairs of indices { i, j } and all β ∈ Z with wt( β ) = 1 or 3, we have ( (cid:98) ∂ ij (cid:98) f (cid:48) )( β ) = 0. Then, byLemma 2.10, (cid:98) f (cid:48) ( α ) = 0 for all α with wt( α ) = 2 or 4. Clearly, (cid:98) f (cid:48) is realized by extending (cid:98) f withnonzero binary signatures in (cid:98) O that are realized by factorization from (cid:98) ∂ (cid:98) f . Lemma 6.2. Suppose that (cid:98) F contains an irreducible -ary signature (cid:98) f (cid:48) where (cid:98) f (cid:48) ( α ) = 0 for all α with wt( α ) = 2 or . Then, Holant( (cid:54) = | (cid:98) F ) is S ( (cid:98) f (cid:48) ) = O = { α ∈ Z | wt( α ) is odd } and all nonzero entries of (cid:98) f (cid:48) have the same norm. roof. Since (cid:98) f (cid:48) is irreducible, again we may assume that (cid:98) f (cid:48) satisfies and (cid:98) f (cid:48) ∈ (cid:98)(cid:82) (cid:98) O ⊗ . Let { i, j, k, (cid:96) } be an arbitrarily chosen subset of indices from { , . . . , } , and { m, n } be the other twoindices. Then by equation (4.7), and the condition that (cid:98) f (cid:48) vanishes at weight 2 and 4, we have | (cid:98) f (cid:48) ijk(cid:96) | = | (cid:98) f (cid:48) ijk(cid:96)mn | + | (cid:98) f (cid:48) ijk(cid:96)mn | = | (cid:98) f (cid:48) ijk(cid:96)mn | + | (cid:98) f (cid:48) ijk(cid:96)mn | = | (cid:98) f (cid:48) ijk(cid:96) | . (6.2)Also, by considering indices { k, (cid:96), m, n } , we have | (cid:98) f (cid:48) k(cid:96)mn | = | (cid:98) f (cid:48) ijk(cid:96)mn | + | (cid:98) f (cid:48) ijk(cid:96)mn | = | (cid:98) f (cid:48) ijk(cid:96)mn | + | (cid:98) f (cid:48) ijk(cid:96)mn | = | (cid:98) f (cid:48) k(cid:96)mn | . (6.3)By ars , we have | (cid:98) f (cid:48) ijk(cid:96)mn | = | (cid:98) f (cid:48) ijk(cid:96)mn | , (6.4)and | (cid:98) f (cid:48) ijk(cid:96)mn | = | (cid:98) f (cid:48) ijk(cid:96)mn | . (6.5)By calculating (6.2) + (6.3) − (6.4) − (6.5), we have | (cid:98) f (cid:48) ijk(cid:96)mn | = | (cid:98) f (cid:48) ijk(cid:96)mn | . (6.6)By (6.2) − (6.6), we have | (cid:98) f (cid:48) ijk(cid:96)mn | = | (cid:98) f (cid:48) ijk(cid:96)mn | . (6.7)From (6.6), since the indices ( i, j, k, (cid:96), m, n ) can be an arbitrary permutation of (1 , , , , , α , β ∈ Z with wt( α ) = wt( β ) = 1, we have | (cid:98) f (cid:48) ( α ) | = | (cid:98) f (cid:48) ( β ) | . The same statement holds forwt( α ) = wt( β ) = 3, by (6.7).Let a = | (cid:98) f (cid:48) ( (cid:126) ) | ; by ars , a = | (cid:98) f (cid:48) ( (cid:126) ) | as well. It is the norm of entries of (cid:98) f (cid:48) on input of Hammingweight 0 and 6. We use b to denote the norm of entries of (cid:98) f (cid:48) on inputs of Hamming weight 1. By ars , b is also the norm of entries of (cid:98) f (cid:48) on inputs of Hamming weight 5. We use c to denote thenorm of entries of (cid:98) f (cid:48) on inputs of Hamming weight 3. Remember that by assumption, | (cid:98) f (cid:48) ( α ) | = 0if wt( α ) = 2 or 4.By equation (4.6), we have | (cid:98) f (cid:48) | = a + 2 b = | (cid:98) f (cid:48) | = 2 c . Clearly, we have 0 (cid:54) a, b (cid:54) c . If c = 0, then a = b = 0 which implies that (cid:98) f (cid:48) is a zero signature.This is a contradiction since (cid:98) f (cid:48) is irreducible. Therefore c (cid:54) = 0. We normalize c to 1. Then a + 2 b = 2 . We will show that b = 1 and a = 0. This will finish the proof of the lemma. For a contradiction,suppose that b < 1, then we also have a > (cid:98) f (cid:48) , (cid:98) f (cid:48) and (cid:98) ∂ (cid:98) f (cid:48) = (cid:98) f (cid:48) + (cid:98) f (cid:48) . Since (cid:98) f (cid:48) ( α ) = 0 for all α with wt( α ) =2 or 4, (cid:98) f (cid:48) ( β ) = 0 and (cid:98) f (cid:48) ( β ) = 0 for all β with wt( β ) = 1 or 3. Thus, (cid:98) f (cid:48) and (cid:98) f (cid:48) have evenparity. We also consider the complex inner product (cid:104) (cid:98) f (cid:48) , (cid:98) f (cid:48) (cid:105) . First we build the following table.In Table 2, we call these four rows by Row 1, 2, 3 and 4 respectively and these nine columns byColumn 0, 1, . . . and 8 respectively. We use T i,j to denote the cell in Row i and Column j . Table2 is built as follows. 22 f (cid:48) (cid:98) f (cid:48) (cid:98) f (cid:48) (cid:98) f (cid:48) (cid:98) f (cid:48) (cid:98) f (cid:48) (cid:98) f (cid:48) (cid:98) f (cid:48) (cid:98) f (cid:48) (cid:98) f (cid:48) (cid:98) f (cid:48) (cid:98) f (cid:48) (cid:98) f (cid:48) (cid:98) f (cid:48) (cid:98) f (cid:48) (cid:98) f (cid:48) (cid:98) f (cid:48) (cid:98) f (cid:48) (cid:98) ∂ (cid:98) f (cid:48) s s s s s s s s (cid:104) (cid:98) f (cid:48) , (cid:98) f (cid:48) (cid:105) p p p p p p p p Table 2: Entries of (cid:98) f (cid:48) , (cid:98) f (cid:48) , (cid:98) ∂ (cid:98) f (cid:48) and pairwise product terms in (cid:104) (cid:98) f (cid:48) , (cid:98) f (cid:48) (cid:105) on even-weighed inputs • In Row 1 and Row 2, we list the entries of signatures (cid:98) f (cid:48) and (cid:98) f (cid:48) that are on even-weightedinputs (excluding the first two bits that are pinned) respectively. Note that, those that didnot appear are 0 entries on odd-weighted inputs (excluding the first two bits that are pinned)of the signatures (cid:98) f (cid:48) and (cid:98) f (cid:48) , since (cid:98) f (cid:48) and (cid:98) f (cid:48) have even parity. • In Row 3, we list the corresponding entries of the signature (cid:98) ∂ (cid:98) f (cid:48) = (cid:98) f (cid:48) + (cid:98) f (cid:48) , i.e., T ,j = T ,j + T ,j for 1 (cid:54) j (cid:54) • In Row 4, we list the corresponding items in the complex inner product (cid:104) (cid:98) f (cid:48) , (cid:98) f (cid:48) (cid:105) , i.e., T ,j = T ,j · T ,j for 1 (cid:54) j (cid:54) (cid:54) j (cid:54) 8, we consider the entry in T ,j and the entry in T , − j . By ars , we have T ,j = T , − j because their corresponding inputs are complement of each other. Thus, T ,j = T ,j + T ,j = T , − j + T , − j = T , − j , and T ,j = T ,j · T ,j = T , − j · T , − j = T , − j . We use s , . . . , s to denote the values in T , , . . . , T , and p , . . . , p to denote the values in T , , . . . , T , . Correspondingly, the values in T , , . . . , T , are s , . . . , s and the values in T , , . . . , T , are p , . . . , p . We also use x j and y j (1 (cid:54) j (cid:54) 8) to denote the entries in T ,j and T ,j respectively.By , we have (cid:104) (cid:98) f (cid:48) , (cid:98) f (cid:48) (cid:105) = 2( p + p + p + p ) = 0. Also we have | p | = b and | p | = | p | = | p | = 1. Notice the fact that if x i + y i = 0, then x i · y i = x i · − x i = −| x i | = −| x i · y i | .Thus, if s = 0 then p = −| p | = − b and for any i = 2 , , 4, if s i = 0 then p i = − 1. Note that (cid:98) ∂ (cid:98) f (cid:48) ( β ) = (cid:98) f (cid:48) ( β ) + (cid:98) f (cid:48) ( β ) = 0 for all β with wt( β ) = 1 or 3. Among all 16 entries of (cid:98) ∂ (cid:98) f (cid:48) , s , . . . , s , s , . . . , s are those that are possibly nonzero. Since (cid:98) ∂ (cid:98) f (cid:48) ∈ (cid:98) O ⊗ , it has support of sizeeither 4 or 0. Thus, among s , s , s and s , either exactly two of them are zero or they are all zero.There are three possible cases. • s = s = s = s = 0. Then p + p + p + p = − b − (cid:54) − (cid:54) = 0. Contradiction. • s (cid:54) = 0 and two of s , s and s are zero. Without loss of generality, we may assume that s = s = 0. Then p = p = − 1. Since p + p + p + p = 0, we have p + p = − p − p = 2.Then, 2 = | p + p | (cid:54) | p | + | p | = b + 1 < 2. Contradiction. • s = 0 and one of s , s and s is zero. Without loss of generality, we may assume that s = 0.Then p = − b and p = − 1. Thus, p + p = − p − p = 1 + b < 2. Let θ = arccos b .23e know that 0 < θ < π . Recall that | p | = | p | = 1. Thus, p = e ± i θ and p = e ∓ i θ (and p = p ).Let P = {− , e i θ , e − i θ } . Thus, p , p , p ∈ P . Otherwise, we get a contradiction.Now, we consider signatures (cid:98) ∂ ij (cid:98) f (cid:48) for all pairs of indices { i, j } . By symmetry, the same conclu-sion holds. In other words, let { i, j } be an arbitrarily chosen pair of indices from { , . . . , } and { k, (cid:96), m, n } be the other four indices, and let β ∈ Z be an assignment on variables ( x k , x (cid:96) , x m , x n )with wt( β ) = 2. Then, we have (cid:98) f (cid:48) βijk(cid:96)mn · (cid:98) f (cid:48) βijk(cid:96)mn ∈ P. Since the indices ( i, j, k, (cid:96), m, n ) can be anarbitrary permutation of (1 , , , , , (cid:98) f (cid:48) ( α ) · (cid:98) f (cid:48) ( α (cid:48) ) ∈ P for any two assignments α and α (cid:48) on the six variables where wt( α ) = wt( α (cid:48) ) = 3 and wt( α ⊕ α (cid:48) ) = 2, because for any such twostrings α and α (cid:48) , there exist two bit positions on which α and α (cid:48) take values 01 and 10 respectively.We consider the following three inputs α = 100011, α = 010011 and α = 001011 of (cid:98) f (cid:48) . Wehave (cid:98) f (cid:48) ( α ) · (cid:98) f (cid:48) ( α ) = q ∈ P , (cid:98) f (cid:48) ( α ) · (cid:98) f (cid:48) ( α ) = q ∈ P and (cid:98) f (cid:48) ( α ) · (cid:98) f (cid:48) ( α ) = q ∈ P. Recall that | (cid:98) f (cid:48) ( α ) | = 1 since wt( α ) = 3. Then, q · q = (cid:98) f (cid:48) ( α ) · (cid:98) f (cid:48) ( α ) · (cid:98) f (cid:48) ( α ) · (cid:98) f (cid:48) ( α ) = | (cid:98) f (cid:48) ( α ) | · (cid:98) f (cid:48) ( α ) · (cid:98) f (cid:48) ( α ) = q ∈ P. However, since 0 < θ < π , it is easy to check that for any two (not necessarily distinct) elements in P , their product is not in P . Thus, we get a contradiction. This proves that b = c = 1 and a = 0.Therefore we have proved that, S ( (cid:98) f (cid:48) ) = O , and all its nonzero entries have the same normthat is normalized to 1. Lemma 6.3. Suppose that (cid:98) F contains an irreducible -ary signature (cid:98) f (cid:48) where S ( (cid:98) f (cid:48) ) = O and | (cid:98) f (cid:48) ( α ) | = 1 for all α ∈ S ( (cid:98) f (cid:48) ) . Then, Holant( (cid:54) = | (cid:98) F ) is (cid:98) Q = (cid:104) ρ ρ (cid:105) ∈ (cid:99) O where ρ = e i δ and (cid:54) δ < π/ , an irreducible 6-ary signature (cid:99) f (cid:48)(cid:48) and = are realizable from (cid:98) f (cid:48) where S ( (cid:98) f (cid:48)(cid:48) ) = O and there exists λ = 1 or i such that forall α ∈ S ( (cid:99) f (cid:48)(cid:48) ) , (cid:99) f (cid:48)(cid:48) ( α ) = ± λ , i.e., Holant( (cid:54) = | = , (cid:99) f (cid:48)(cid:48) , (cid:98) Q (cid:98) F ) (cid:54) T Holant( (cid:54) = | (cid:98) F ) where (cid:99) f (cid:48)(cid:48) = (cid:98) Q (cid:98) f (cid:48) .Moreover, the nonzero binary signature ( ρ , , , ρ ) ∈ (cid:98) O is realizable from (cid:98) ∂ ij (cid:98) f (cid:48) for some { i, j } .Proof. Again, we may assume that (cid:98) f (cid:48) satisfies and (cid:98) f (cid:48) ∈ (cid:98)(cid:82) (cid:98) O ⊗ . We first show that thereexists λ = 1 or i such that for all α ∈ S ( (cid:98) f (cid:48) ) with wt( α ) = 3, (cid:99) f (cid:48)(cid:48) ( α ) = ± λ , or else we get | p | = | p | = | p | = | p | = 1. Recall that for 1 (cid:54) i (cid:54) s i = 0 implies that p i = − 1. Since (cid:98) ∂ (cid:98) f (cid:48) ∈ (cid:98) O ⊗ , it has support of size 4 or 0. Thus, among s , s , s and s , either exactly two of them are zero or they are all zero. If they are all zero, then we have p + p + p + p = − (cid:54) = 0. This is a contradiction to our assumption that (cid:98) f (cid:48) satisfies .Thus, exactly two of s , s , s and s are zeros. Suppose that they are s i and s j . Recall that weuse x i and y i (1 (cid:54) i (cid:54) 8) to denote the entries in Row 1 and Row 2 of Table 2. Thus | x i | = | y i | = 1,for 1 (cid:54) i (cid:54) 8. Since s i = x i + y i = 0 and s j = x j + y j = 0, we have x i = − y i , and x j = − y j . Also,since s i = s j = 0, we have p i = p j = − 1. Let { (cid:96), k } = { , , , }\{ i, j } . Then, by , wehave p (cid:96) + p k = − p i − p j = 2. Since | p (cid:96) | = | p k | = 1, we have p (cid:96) = p k = 1. Note that p (cid:96) = x (cid:96) · y (cid:96) = 1and also 1 = | y (cid:96) | = y (cid:96) · y (cid:96) . Thus, we have x (cid:96) = y (cid:96) . Similarly, x k = y k . Thus, for all 1 (cid:54) i (cid:54) x i = ± y i . Consider (cid:98) ∂ ij (cid:98) f (cid:48) for all pairs of indices { i, j } . By symmetry, the same conclusion holds.Thus, (cid:98) f ( α ) = ± (cid:98) f ( α (cid:48) ) for any two inputs α and α (cid:48) on the six variables where wt( α ) = wt( α (cid:48) ) = 3and wt( α ⊕ α (cid:48) ) = 2. In particular, we have (cid:98) f (cid:48) = ε (cid:98) f (cid:48) = ε (cid:98) f (cid:48) = ε (cid:98) f (cid:48) , ε , ε , ε = ± ars , we have (cid:98) f (cid:48) = (cid:98) f (cid:48) . • If (cid:98) f (cid:48) = (cid:98) f (cid:48) = (cid:98) f (cid:48) , then (cid:98) f (cid:48) = ± . • If (cid:98) f (cid:48) = − (cid:98) f (cid:48) = (cid:98) f (cid:48) , then (cid:98) f (cid:48) = ± i . Thus, there exists λ = 1 or i such that (cid:98) f (cid:48) = ± λ and (cid:98) f (cid:48) = ± λ . Consider any α ∈ Z withwt( α ) = 3. If α ∈ { , } , then clearly, (cid:98) f (cid:48) ( α ) = ± λ . Otherwise, either wt( α ⊕ α ⊕ (cid:98) f (cid:48) ( α ) = ± λ . Thus, there exists λ = 1 or i such that for all α ∈ Z with wt( α ) = 3, (cid:98) f (cid:48) ( α ) = ± λ .Since (cid:98) f (cid:48) ( α ) (cid:54) = 0 for all α with wt( α ) = 1, by Lemma 2.10, there exists a pair of indices { i, j } such that ( (cid:98) ∂ ij (cid:98) f (cid:48) ) (cid:54) = 0. Since (cid:98) ∂ ij (cid:98) f (cid:48) ∈ O ⊗ , it is of the form ( a, , , ¯ a ) ⊗ ( b, , , ¯ b ), where ab (cid:54) = 0,since no other factorization form in O ⊗ has a nonzero value at 0000. By Lemma 2.7, we canrealize the signature (cid:98) g = ( a, , , ¯ a ). Here, we can normalize a to e i θ where 0 (cid:54) θ < π . Then, let ρ = e i θ/ . Clearly, 0 (cid:54) θ/ < π/ 2. Consider a holographic transformation by (cid:98) Q = (cid:104) ρ ρ (cid:105) . Note that( (cid:54) = )( (cid:98) Q − ) ⊗ = ( (cid:54) = ) and (cid:98) Q ⊗ (cid:98) g = (1 , , , . The holographic transformation by (cid:98) Q does not change (cid:54) = , but transfers (cid:98) g = ( a, , , ¯ a ) to (= ) = (1 , , , (cid:54) = | (cid:98) g, (cid:98) f (cid:48) , (cid:98) F ) ≡ T Holant( (cid:54) = | = , (cid:98) Q (cid:98) f (cid:48) , (cid:98) Q (cid:98) F ) . We denote (cid:98) Q (cid:98) f (cid:48) by (cid:99) f (cid:48)(cid:48) . Note that (cid:98) Q does not change those entries of (cid:98) f (cid:48) that are on half-weightedinputs. Thus, for all α with wt( α ) = 3, we have (cid:99) f (cid:48)(cid:48) ( α ) = ± λ for some λ = 1 or i . Also, (cid:98) Q does notchange the parity and irreducibility of (cid:98) f (cid:48) . Thus (cid:99) f (cid:48)(cid:48) has odd parity and (cid:99) f (cid:48)(cid:48) is irreducible. Again, wemay assume that (cid:99) f (cid:48)(cid:48) satisfies and (cid:99) f (cid:48)(cid:48) ∈ (cid:98)(cid:82) (cid:98) O ⊗ . Otherwise, we are done.In the problem Holant( (cid:54) = | = , (cid:99) f (cid:48)(cid:48) , (cid:98) Q (cid:98) F ), we can connect two (cid:54) = on the LHS using = on the RHS,and then we can realize = on the LHS. Thus, we can use = to merge variables of (cid:99) f (cid:48)(cid:48) . Therefore,we may further assume (cid:99) f (cid:48)(cid:48) ∈ (cid:82) (cid:98) O ⊗ , i.e., ∂ ij (cid:99) f (cid:48)(cid:48) ∈ (cid:98) O ⊗ for all pairs of indices { i, j } ; otherwise, thereexist two variables of (cid:99) f (cid:48)(cid:48) such that by merging these two variables using = , we can realize a 4-arysignature that is not in (cid:98) O ⊗ , and then by Lemma 5.2 we are done.Consider the signature ∂ (cid:99) f (cid:48)(cid:48) = (cid:99) f (cid:48)(cid:48) + (cid:99) f (cid:48)(cid:48) and the inner product (cid:104) (cid:98) f (cid:48)(cid:48) , (cid:98) f (cid:48)(cid:48) (cid:105) . Same as Table 2,we build the following Table 3. (cid:99) f (cid:48)(cid:48) (cid:99) f (cid:48)(cid:48) (cid:99) f (cid:48)(cid:48) (cid:99) f (cid:48)(cid:48) (cid:99) f (cid:48)(cid:48) (cid:99) f (cid:48)(cid:48) (cid:99) f (cid:48)(cid:48) (cid:99) f (cid:48)(cid:48) (cid:99) f (cid:48)(cid:48) (cid:99) f (cid:48)(cid:48) (cid:99) f (cid:48)(cid:48) (cid:99) f (cid:48)(cid:48) (cid:99) f (cid:48)(cid:48) (cid:99) f (cid:48)(cid:48) (cid:99) f (cid:48)(cid:48) (cid:99) f (cid:48)(cid:48) (cid:99) f (cid:48)(cid:48) (cid:99) f (cid:48)(cid:48) ∂ (cid:99) f (cid:48)(cid:48) t t t t t t t t (cid:104) (cid:98) f (cid:48)(cid:48) , (cid:98) f (cid:48)(cid:48) (cid:105) q q q q q q q q Table 3: Entries of (cid:99) f (cid:48)(cid:48) , (cid:99) f (cid:48)(cid:48) , ∂ (cid:99) f (cid:48)(cid:48) and pair-wise product terms in (cid:104) (cid:98) f (cid:48)(cid:48) , (cid:98) f (cid:48)(cid:48) (cid:105) onodd-weighed inputsSame as the proof of x i = ± y i for Table 2, we have (cid:99) f (cid:48)(cid:48) = ± (cid:99) f (cid:48)(cid:48) . Since (cid:99) f (cid:48)(cid:48) = ± λ , (cid:99) f (cid:48)(cid:48) = ± λ , (here ± can be either ± or ∓ ). Consider ∂ ij (cid:99) f (cid:48)(cid:48) for all pairs of indices { i, j } . By25ymmetry, the same conclusion holds. Thus, for every α ∈ Z with wt( α ) = 1, (cid:99) f (cid:48)(cid:48) ( α ) = ± λ. Therefore, using ars , there exists λ = 1 or i such that for all α ∈ S ( (cid:99) f (cid:48)(cid:48) ), (cid:99) f (cid:48)(cid:48) ( α ) = ± λ , and we havethe reduction Holant( (cid:54) = | = , (cid:99) f (cid:48)(cid:48) , (cid:98) Q (cid:98) F ) (cid:54) T Holant( (cid:54) = | (cid:98) F )for some (cid:98) Q ∈ (cid:99) O . Clearly, (cid:99) f (cid:48)(cid:48) = (cid:98) Q (cid:98) f (cid:48) where (cid:98) Q = (cid:104) ρ ρ (cid:105) ∈ (cid:99) O , and the nonzero binary signature( ρ , , , ρ ) ∈ (cid:98) O is realizable from (cid:98) ∂ ij (cid:98) f (cid:48) for some { i, j } .Finally, we go for the kill in the next lemma. Recall the signature (cid:98) f defined in (6.1). This Lordof Intransigence at arity 6 makes its appearance in Lemma 6.4. Lemma 6.4. Suppose that (cid:98) F contains an irreducible -ary signature (cid:99) f (cid:48)(cid:48) where S ( (cid:99) f (cid:48)(cid:48) ) = O , andthere exists λ = 1 or i such that for all α ∈ S ( (cid:99) f (cid:48)(cid:48) ) , (cid:99) f (cid:48)(cid:48) ( α ) = ± λ . Then, Holant( (cid:54) = | = , (cid:98) F ) is (cid:98) f is realizable from (cid:99) f (cid:48)(cid:48) and = , i.e., Holant( (cid:54) = | (cid:98) f , (cid:98) F ) (cid:54) T Holant( (cid:54) = | = , (cid:98) F ) . Moreover, (cid:98) f is realizable by extending variables of (cid:99) f (cid:48)(cid:48) with binary signatures in (cid:98) B , i.e., (cid:98) f ∈ { (cid:99) f (cid:48)(cid:48) } (cid:98) B(cid:54) = . Proof. Again, we may assume that (cid:99) f (cid:48)(cid:48) satisfies and (cid:99) f (cid:48)(cid:48) ∈ (cid:98)(cid:82) (cid:98) O ⊗ . Since = is available onthe RHS, given any signature (cid:98) f ∈ (cid:98) F , we can extend any variable x i of (cid:98) f with = ∈ (cid:98) B using (cid:54) = .This gives a signature (cid:98) g where (cid:98) g i = (cid:98) f i and (cid:98) g i = (cid:98) f i . We call this extending gadget constructionthe flipping operation on variable x i . Clearly, it does not change the reducibility or irreducibilityof (cid:98) f . But it changes the parity of (cid:98) f if (cid:98) f has parity. Once a signature (cid:98) f is realizable, we can modifyit by flipping some of its variables.We first show that we can realize a signature (cid:99) f ∗ from (cid:99) f (cid:48)(cid:48) having support S ( (cid:99) f ∗ ) = E = { α ∈ Z | wt( α ) ≡ } , and (cid:99) f ∗ ( α ) = ± α ∈ S ( (cid:99) f ∗ ) . Remember that = is available. If weconnect = with an arbitrary variable of (cid:99) f (cid:48)(cid:48) using (cid:54) = , we will change the parity of (cid:99) f (cid:48)(cid:48) from odd toeven. If (cid:99) f (cid:48)(cid:48) ( α ) = ± α ∈ S ( (cid:99) f (cid:48)(cid:48) ), then (cid:99) f ∗ can be realized by flipping an arbitrary variable of (cid:99) f (cid:48)(cid:48) . Otherwise, (cid:99) f (cid:48)(cid:48) ( α ) = ± i for all α ∈ S ( (cid:99) f (cid:48)(cid:48) ). Consider (cid:98) ∂ (cid:99) f (cid:48)(cid:48) . Look at Table 3. We use x i and y i (1 (cid:54) i (cid:54) 8) to denote entries in Row 1 and 2. As we have showed, x i = ± y i . Thus, t i = ± i or 0for 1 (cid:54) i (cid:54) 4. Remember that if t i = 0 (i.e., x i = − y i ), then q i = x i · y i = − x i · x i = −| x i | = − t i = 0 for all 1 (cid:54) i (cid:54) 4, then (cid:104) (cid:98) f (cid:48)(cid:48) , (cid:98) f (cid:48)(cid:48) (cid:105) = 2( q + q + q + q ) = − (cid:54) = 0 . This contradicts with our assumption that (cid:99) f (cid:48)(cid:48) satisfies . Thus, t i (1 (cid:54) i (cid:54) 4) are not allzeros. Then ( (cid:98) ∂ (cid:99) f (cid:48)(cid:48) ) (cid:54)≡ 0. Thus, S ( (cid:98) ∂ (cid:99) f (cid:48)(cid:48) ) (cid:54) = ∅ and ( (cid:98) ∂ (cid:99) f (cid:48)(cid:48) )( α ) = ± i for all α ∈ S ( (cid:98) ∂ (cid:99) f (cid:48)(cid:48) ).Since (cid:98) ∂ (cid:99) f (cid:48)(cid:48) ∈ (cid:98) O ⊗ and it has even parity, (cid:98) ∂ (cid:99) f (cid:48)(cid:48) is of the form 2 · ( a, , , ¯ a ) ⊗ ( b, , , ¯ b ) or2 · (0 , a, ¯ a, ⊗ (0 , b, ¯ b, a and b are normalized to 1. In both cases, we have ab, ¯ ab, a ¯ b, ¯ a ¯ b ∈ { i , − i } . Thus, ab · ¯ ab = ( a ¯ a ) b = b = ± 1. Then, b = ± ± i . If b = ± 1, then a = a ¯ b · b = ± i . Similarly, if b = ± i , then a = a ¯ b · b = ± 1. Thus, among a and b , exactly one is ± i . Thus, by factorization we can realize the binary signature (cid:98) g = ( i , , , − i ) or (0 , i , − i , 0) up to ascalar − 1. Connecting an arbitrary variable of (cid:98) f with a variable of (cid:98) g , we can get a signature whichhas parity and all its nonzero entries have value ± 1. If the resulting signature has even parity, thenwe get the desired (cid:99) f ∗ . If it has odd parity, then we can flip one of its variables to change the parity.Thus, we can realize a signature (cid:99) f ∗ by extending variables of (cid:99) f (cid:48)(cid:48) with binary signatures in (cid:98) B ⊗ suchthat S ( (cid:99) f ∗ ) = E , and (cid:99) f ∗ ( α ) = ± α ∈ S ( (cid:99) f ∗ ) . (cid:99) f ∗ . In Table 4, we list 16 entries of (cid:99) f ∗ with x x x =000 , , , 110 as the row index and x x x = 000 , , , 110 as the column index. We alsoview these 16 entries in Table 4 as a 4-by-4 matrix denoted by M r ( (cid:99) f ∗ ), and we call it the representa-tive matrix of (cid:99) f ∗ . Note that for any α ∈ S ( (cid:99) f ∗ ) such that the entry (cid:99) f ∗ ( α ) does not appear in M r ( (cid:99) f ∗ ), (cid:99) f ∗ ( α ) appears in M r ( (cid:99) f ∗ ). Since (cid:99) f ∗ ( α ) = ± ∈ R , (cid:99) f ∗ ( α ) = (cid:99) f ∗ ( α ). By ars , (cid:99) f ∗ ( α ) = (cid:99) f ∗ ( α ) = (cid:99) f ∗ ( α ).Thus, the 16 entries of the matrix M r ( (cid:99) f ∗ ) listed in Table 4 gives a complete account for all the 32nonzero entries of (cid:99) f ∗ . x x x x x x 000 (Col 1) 011 (Col 2) 101 (Col 3) 110 (Col 4)000 (Row 1) (cid:99) f ∗ (cid:99) f ∗ (cid:99) f ∗ (cid:99) f ∗ 011 (Row 2) (cid:99) f ∗ (cid:99) f ∗ (cid:99) f ∗ (cid:99) f ∗ 101 (Row 3) (cid:99) f ∗ (cid:99) f ∗ (cid:99) f ∗ (cid:99) f ∗ 110 (Row 4) (cid:99) f ∗ (cid:99) f ∗ (cid:99) f ∗ (cid:99) f ∗ Table 4: Representative entries of (cid:99) f ∗ We use ( m ij ) i,j =1 to denote the 16 entries of M r ( (cid:99) f ∗ ). We claim that any two rows of M r ( (cid:99) f ∗ ) areorthogonal; this follows from the fact that (cid:99) f ∗ satisfies and ars . For example, considerthe first two rows of M r ( (cid:99) f ∗ ). By , the inner product (cid:104) (cid:98) f ∗ , (cid:98) f ∗ (cid:105) for the real-valued (cid:99) f ∗ is (cid:88) ( x ,x ,x ,x ) ∈ Z (cid:99) f ∗ x x x x (cid:99) f ∗ x x x x = 0 , where the sum has 8 nonzero product terms. The first 4 terms given by x = 0 are the pairwiseproducts m j m j , for 1 (cid:54) j (cid:54) 4. The second 4 terms are, by ars , the pairwise products m j m j in the reversal order of 1 (cid:54) j (cid:54) 4, where we exchange row 1 with row 2 on the account of flippingthe summation index x from 0 to 1, and simultaneously flipping both x and x . This shows that (cid:80) j =1 m j m j = 0. Similarly any two columns of M r ( (cid:99) f ∗ ) are orthogonal.Also, we consider the inner product (cid:104) (cid:98) f ∗ , (cid:98) f ∗ (cid:105) = 0. It is computed using the following 16entries in M r ( (cid:99) f ∗ ), listed in Table 5. (cid:99) f ∗ (cid:99) f ∗ (cid:99) f ∗ (cid:99) f ∗ (cid:99) f ∗ (cid:99) f ∗ (cid:99) f ∗ (cid:99) f ∗ = m = m = m = m = m = m = m = m (cid:99) f ∗ (cid:99) f ∗ (cid:99) f ∗ (cid:99) f ∗ (cid:99) f ∗ (cid:99) f ∗ (cid:99) f ∗ (cid:99) f ∗ = m = m = m = m = m = m = m = m Table 5: Pair-wise product terms in (cid:104) (cid:98) f ∗ , (cid:98) f ∗ (cid:105) on even-weighed inputs27et M r ( (cid:99) f ∗ ) [1 , be the 2-by-2 submatrix of M r ( (cid:99) f ∗ ) by picking the first two rows and the firsttwo columns, and M r ( (cid:99) f ∗ ) [3 , be the 2-by-2 submatrix of M r ( (cid:99) f ∗ ) by picking the last two rows andthe last two columns. Indeed, (cid:104) (cid:98) f ∗ , (cid:98) f ∗ (cid:105) = 2(perm( M r ( (cid:99) f ∗ ) [1 , ) + perm( M r ( (cid:99) f ∗ ) [3 , ))= 2( m m + m m + m m + m m ) = 0 . Then, we show that by renaming or flipping variables of (cid:99) f ∗ , we may modify (cid:99) f ∗ to realize asignature whose representative matrix is obtained by performing row permutation, column permu-tation, or matrix transpose on M r ( (cid:99) f ∗ ). First, if we exchange the names of variables ( x , x , x )with variables ( x , x , x ), then the representative matrix M r ( (cid:99) f ∗ ) will be transposed. Next, con-sider the group G of permutations on the rows { , , , } effected by any sequence of operationsof renaming and flipping variables in { x , x , x } . By renaming variables in { x , x , x } , we canswitch any two rows among Row 2, 3 and 4. Thus S on { , , } is contained in G . Also, ifwe flip both variables x and x of (cid:99) f ∗ , then for the realized signature, its representative matrixcan be obtained by switching both the pair Row 1 and Row 2, and the pair Row 3 and Row 4of M r ( (cid:99) f ∗ ). Thus, the permutation (12)(34) ∈ G . It follows that G = S . Thus, by renaming orflipping variables of (cid:99) f ∗ , we can permute any two rows or any two columns of M r ( (cid:99) f ∗ ), or transpose M r ( (cid:99) f ∗ ). For the resulting signature, we may assume that its representative matrix A also satisfyperm( A [1 , ) + perm( A [3 , ) = 0, and any two rows of A are orthogonal and any two columns of A are orthogonal. Otherwise, we get M r ( (cid:99) f ∗ ) by permuting any two rows or any two columns, or taking transpose. We showthat it will give M r ( (cid:98) f ), after a normalization by ± 1. In other words, (cid:98) f is realizable from (cid:99) f ∗ byrenaming or flipping variables, up to a normalization by ± i and Row j , of M r ( (cid:99) f ∗ ). Recall that every entry of M r ( (cid:99) f ∗ ) is ± i and Row j differ in Column k if m ik (cid:54) = m jk , which implies that m ik = − m jk ;otherwise, they are equal m ik = m jk . In the former case, m ik · m jk = − 1, and in the latter case m ik · m jk = 1. Since Row i and Row j are orthogonal, they differ in exactly two columns and areequal in the other two columns. Similarly, for any two columns of M r ( (cid:99) f ∗ ), they differ in exactlytwo rows and are equal in the other two rows. Depending on the number of − M r ( (cid:99) f ∗ ), we consider the following two cases. • Every row and column of M r ( (cid:99) f ∗ ) has an odd number of − − − − M r ( (cid:99) f ∗ ) by multiplying the matrix with − 1. This does not changethe parity of the number of − − − , , , . Consider the number of − – If they all have exactly one − − M r ( (cid:99) f ∗ ) has the following form M r ( (cid:99) f ∗ ) = − − − − . M r ( (cid:99) f ∗ ) [1 , ) + perm( M r ( (cid:99) f ∗ ) [3 , ) = 2 + 2 = 4 (cid:54) = 0 . Contradiction. – Otherwise, among Rows 2, 3 and 4, there is one that has three − − − , , − , − − m = 1 and m = − 1, we have m = m , both+1 or − 1. Similarly, m = m . Also, since Column 3 and Column 4 differ in exactlytwo rows, and m = m and m = m , we have m = − m and m = − m .Thus, M r ( (cid:99) f ∗ ) [3 , = ± (cid:2) − − (cid:3) . In both cases, we have perm( M r ( (cid:99) f ∗ ) [1 , ) = − 2. No-tice that M r ( (cid:99) f ∗ ) [1 , = (cid:2) − − (cid:3) . Thus, perm( M r ( (cid:99) f ∗ ) [1 , ) + perm( M r ( (cid:99) f ∗ ) [3 , ) = − (cid:54) = 0.Contradiction. • There is a row or a column of M r ( (cid:99) f ∗ ) such that it has an even number of − M r ( (cid:99) f ∗ ), we may assume that it is a row, say Row i . For any other Row j , itdiffers with Row i in exactly two columns. Thus, Row j also has an even number of − M r ( (cid:99) f ∗ ) have exactly two − ± 1) multiple of the other, thus not orthogonal;this is a contradiction. Thus, there exists a row in which the number of − M r ( (cid:99) f ∗ )with − 1, we may assume that all entries of Row 1 are +1. Thus, Row 1 is (1 , , , − − , − , , − m and m , exactly one is − 1. By permuting Column 1 and Column 2 (which doesnot change Row 1 and Row 2), we may assume that m = − 1. Also, among m and m ,exactly one is − 1. By permuting Column 3 and Column 4 (still this will not change Row 1and Row 2), we may assume that m = − 1. Thus, Row 3 is ( − , , − , − − , , , − 1) or (1 , − , − , . If Row 4 is (1 , − , − , M r ( (cid:99) f ∗ ) = − − − − − − . Thus, perm( M r ( (cid:99) f ∗ ) [12] ) + perm( M r ( (cid:99) f ∗ ) [34] ) = − (cid:54) = 0 . Contradiction.Thus, Row 4 is ( − , , , − M r ( (cid:99) f ∗ ) = − − − − − − . This gives the desired M r ( (cid:98) f ).Therefore, (cid:98) f is realizable from (cid:99) f ∗ . 29ince (cid:98) f is realized from (cid:99) f ∗ by flipping (and permuting) variables, i.e., extending some variablesof (cid:99) f ∗ with = (using (cid:54) = ), we have (cid:98) f ∈ { (cid:99) f ∗ } (cid:98) B(cid:54) = . Since (cid:99) f ∗ is realized from (cid:99) f (cid:48)(cid:48) by extending somevariables of (cid:99) f (cid:48)(cid:48) with signatures in (cid:98) B , we have (cid:99) f ∗ ∈ { (cid:99) f (cid:48)(cid:48) } (cid:98) B(cid:54) = . By Lemma 2.12, we have (cid:98) f ∈ { (cid:99) f (cid:48)(cid:48) } (cid:98) B(cid:54) = . Theorem 6.5. Suppose that (cid:98) F contains a -ary signature (cid:98) f / ∈ (cid:98) O ⊗ . Then, • Holant( (cid:54) = | (cid:98) F ) is • there exists some (cid:98) Q ∈ (cid:99) O such that Holant( (cid:54) = | (cid:98) f , (cid:98) Q (cid:98) F ) (cid:54) T Holant( (cid:54) = | (cid:98) F ) . Proof. By Lemmas 6.1, 6.2 and 6.3, Holant( (cid:54) = | (cid:98) F ) is (cid:54) = | = , (cid:99) f (cid:48)(cid:48) , (cid:98) Q (cid:98) F ) (cid:54) T Holant( (cid:54) = | (cid:98) F ) for some (cid:98) Q where Q ∈ (cid:99) O , and some irreducible 6-ary signature (cid:99) f (cid:48)(cid:48) where S ( (cid:98) f (cid:48)(cid:48) ) = E and there exists λ = 1 or i such that for all α ∈ S ( (cid:99) f (cid:48)(cid:48) ), (cid:99) f (cid:48)(cid:48) ( α ) = ± λ . Remember that (cid:98) Q (cid:98) F = (cid:100) Q F where Q = Z (cid:98) QZ − ∈ O . Clearly, Q F is a set of real-valued signatures of even arity.Since F does not satisfy condition (T), by Lemma 2.23, Q F also does not satisfy it. Then, byLemma 6.4, Holant( (cid:54) = | = , (cid:99) f (cid:48)(cid:48) , (cid:98) Q (cid:98) F ) is (cid:54) = | (cid:98) f , (cid:98) Q (cid:98) F ) (cid:54) T Holant( (cid:54) = | = , (cid:99) f (cid:48)(cid:48) , (cid:98) Q (cid:98) F ) . Thus, Holant( (cid:54) = | (cid:98) F ) is (cid:54) = | (cid:98) f , (cid:98) Q (cid:98) F ) (cid:54) T Holant( (cid:54) = | (cid:98) F ) . Remark: Theorem 6.5 can be more succinctly stated as simply that a reductionHolant( (cid:54) = | (cid:98) f , (cid:98) Q (cid:98) F ) (cid:54) T Holant( (cid:54) = | (cid:98) F )exists, because when Holant( (cid:54) = | (cid:98) F ) is (cid:54) = | (cid:98) f , (cid:98) Q (cid:98) F ) is (cid:98) Q ∈ (cid:99) O and all (cid:98) F where F = Z (cid:98) F is a real-valued signature set that does not satisfy condition (T). If so, then we are done.Recall that for all (cid:98) Q ∈ (cid:99) O , (cid:98) Q (cid:98) F = (cid:100) Q F for some Q ∈ O . Moreover, for all Q ∈ O , and allreal-valued F that does not satisfy condition (T), Q F is also a real-valued signature set that doesnot satisfy condition (T). Thus, it suffices for us to show that Holant( (cid:54) = | (cid:98) f , (cid:98) F ) is F that does not satisfy condition (T). (cid:98) f In this subsection, we give three conditions (Lemmas 6.6, 6.8 and 6.9) which can quite straightfor-wardly lead to the (cid:54) = | (cid:98) f , (cid:98) F ). We will extract two properties from (cid:98) f , thenon- (cid:98) B hardness (Definition 6.7) and the realizability of (cid:98) B (Lemma 6.11). Later, we will prove the (cid:54) = | (cid:98) f , (cid:98) F ) based on these two properties. Lemma 6.6. Holant( (cid:54) = | (cid:98) f , (cid:98) F ) is (cid:98) F contains a nonzero binary signature (cid:98) b / ∈ (cid:98) B ⊗ .Proof. If (cid:98) b / ∈ (cid:98) O ⊗ , then by Lemma 5.1, we are done. Otherwise, (cid:98) b ∈ (cid:98) O ⊗ . Thus, (cid:98) b = ( a, , , ¯ a ) or (cid:98) b = (0 , a, ¯ a, (cid:98) b (cid:54)≡ a (cid:54) = 0. We normalize the norm of a to 1. Since (cid:98) b / ∈ (cid:98) B ⊗ , a (cid:54) = ± ± i .We first consider the case that (cid:98) b ( y , y ) = (0 , a, ¯ a, x and x of (cid:98) f withvariables y and y of (cid:98) b using (cid:54) = , we get a 4-ary signature (cid:98) g . We list the truth table of (cid:98) g indexedby the assignments of variables ( x , x , x , x ) from 0000 to 1111. (cid:98) g = (0 , a + ¯ a, − a + ¯ a, , a − ¯ a, , , − a − ¯ a, − a − ¯ a, , , − a + ¯ a, , a − ¯ a, a + ¯ a, . a has norm 1, and a (cid:54) = ± ± i , | a ± ¯ a | (cid:54) = 0. Thus, | S ( (cid:98) g ) | = 8. Clearly, every 4-arysignature that is in (cid:98) O ⊗ has support of size 0 or 4. Thus, (cid:98) g / ∈ (cid:98) O ⊗ . By Lemma 5.2, Holant( (cid:54) = | (cid:98) f , (cid:98) F )is (cid:98) b ( y , y ) = ( a, , , ¯ a ) similarly. By connecting variables x and x of (cid:98) f with variables y and y of (cid:98) b using (cid:54) = , we also get a 4-ary signature that is not in (cid:98) O ⊗ . Thelemma is proved. Definition 6.7. We say a signature set (cid:98) F is non- (cid:98) B hard, if for any nonzero binary signature (cid:98) b / ∈ (cid:98) B ⊗ , the problem Holant( (cid:54) = | (cid:98) b, (cid:98) F ) is F is non- B hard, if for any nonzero binary signature b / ∈ B ⊗ , the problem Holant( b, F ) is Clearly, Lemma 6.6 says that { (cid:98) f } ∪ (cid:98) F is non- (cid:98) B hard for any (cid:98) F (where F = Z (cid:98) F is a real-valuedsignature set that does not satisfy condition (T)). Before we give the other two (cid:98) B hardness. We will extract twoproperties from (cid:98) f to prove the (cid:54) = | (cid:98) f , (cid:98) F ). These are the non- (cid:98) B hardnessand the realizability of (cid:98) B . From Lemma 6.11 we get the redutcion Holant( (cid:54) = | (cid:98) f , (cid:98) B ∪ (cid:98) F ) (cid:54) Holant( (cid:54) = | (cid:98) f , (cid:98) F ). We will show that for any non- (cid:98) B hard set (cid:98) F where F does not satisfy condition(T), Holant( (cid:54) = | (cid:98) B ∪ (cid:98) F ) is (cid:54) = | (cid:98) f , (cid:98) F ) is F does not satisfy condition (T). This slightly more general Theorem 7.19 will alsobe used when dealing with signatures of arity 8. Now, let us continue to give two more B (Lemma 6.8 and 6.9). Lemma 6.8. Suppose that (cid:98) F is non- (cid:98) B hard. Then Holant( (cid:54) = | (cid:98) F ) is (cid:98) F contains anonzero 4-ary signature (cid:98) f / ∈ (cid:98) B ⊗ .Proof. If (cid:98) f / ∈ (cid:98) O ⊗ , then by Lemma 5.2, we are done. Otherwise, (cid:98) f = (cid:98) b ⊗ (cid:98) b , where the binarysignatures (cid:98) b , (cid:98) b ∈ (cid:98) O ⊗ . Since (cid:98) f / ∈ (cid:98) B ⊗ , (cid:98) b and (cid:98) b are not both in (cid:98) B ⊗ . Then, we can realize a binarysignature that is not in (cid:98) B ⊗ by factorization. Since (cid:98) F is non- (cid:98) B hard, we are done.Let H = √ (cid:2) − (cid:3) . Then (cid:98) H = Z − HZ = (cid:20) (1+ i ) √ (1 − i ) √ (cid:21) = (cid:104) e i π/ e − i π/ (cid:105) . Let (cid:99) f H = (cid:98) H (cid:98) f . Let (cid:99) F = { (cid:98) f } (cid:98) B (cid:54) = be the set of signature realizable by extending variables of (cid:98) f with binary signaturesin (cid:98) B using (cid:54) = , and (cid:100) F H = { (cid:99) f H } (cid:98) B (cid:54) = be the set of signature realizable by extending variables of (cid:99) f H with binary signatures in (cid:98) B using (cid:54) = . One can check that (cid:100) F H = (cid:98) H (cid:99) F (cid:54) = (cid:99) F . Lemma 6.9. Suppose that (cid:98) F is non- (cid:98) B hard. Then, Holant( (cid:54) = | (cid:98) F ) is (cid:98) F contains anonzero 6-ary signature (cid:98) f / ∈ (cid:98) B ⊗ ∪ (cid:99) F ∪ (cid:100) F H .Proof. If (cid:98) f is reducible, since (cid:98) f / ∈ (cid:98) B ⊗ , then by factorization, we can realize a nonzero signature ofodd arity or a nonzero signature of arity 2 or 4 that is not in (cid:98) B ⊗ . If we have a nonzero signatureof odd arity, then we are done by Theorem 2.25. If we have a nonzero signature of 2, then weare done because (cid:98) F is non- (cid:98) B hard. If we have a nonzero signature of 4, then we are done byLemma 6.8. Now we assume that (cid:98) f is irreducible. In particular, being irreducible, (cid:98) f (cid:54)∈ (cid:98) O ⊗ . For acontradiction, suppose that Holant( (cid:54) = | (cid:98) F ) is not (cid:98) f is realizablefrom (cid:98) f . Remember that we realize (cid:98) f from (cid:98) f by realizing (cid:98) f (cid:48) , (cid:99) f (cid:48)(cid:48) and (cid:99) f ∗ (Lemmas 6.1, 6.3 and 6.4).We will trace back this process and show that they are all in (cid:99) F ∪ (cid:100) F H , which contradicts with thecondition that (cid:98) f / ∈ (cid:99) F ∪ (cid:100) F H . This lemma and the following Theorem 7.19 are stated and proved in the setting of Holant( F ). 31. First, by Lemma 6.4, (cid:98) f ∈ { (cid:99) f (cid:48)(cid:48) } (cid:98) B(cid:54) = . Then, by Lemma 2.12, (cid:99) f (cid:48)(cid:48) ∈ { (cid:98) f } (cid:98) B(cid:54) = = (cid:99) F .2. Then, by Lemma 6.3, (cid:99) f (cid:48)(cid:48) = (cid:98) Q (cid:98) f (cid:48) for some (cid:98) Q = (cid:104) e − i δ e i δ (cid:105) ∈ (cid:99) O where 0 (cid:54) δ < π/ 2, and thebinary signature (cid:98) b = ( e i δ , , , e − i δ ) is realizable from (cid:98) f (cid:48) where (cid:98) f (cid:48) is realizable from (cid:98) f . Thus, (cid:98) b is realizable from (cid:98) F . If e i δ (cid:54) = ± ± i , then (cid:98) b / ∈ (cid:98) B ⊗ . Since (cid:98) F is non- (cid:98) B hard, we get (cid:54) δ < π/ e i δ = 1 or i and then, δ = 0 or π/ 4. If δ = 0, then e i δ = e − i δ = 1 and (cid:99) f (cid:48)(cid:48) = (cid:98) Q (cid:98) f (cid:48) = (cid:98) f (cid:48) . Thus, (cid:98) f (cid:48) ∈ (cid:99) F . If δ = π/ 4, then (cid:98) f (cid:48) = (cid:98) Q − (cid:99) f (cid:48)(cid:48) where (cid:98) Q − = (cid:104) e i π/ e − i π/ (cid:105) = (cid:98) H. Since (cid:99) f (cid:48)(cid:48) ∈ (cid:99) F , (cid:98) f (cid:48) = (cid:98) H (cid:99) f (cid:48)(cid:48) ∈ (cid:98) H (cid:99) F = (cid:100) F H . 3. Finally, by Lemma 6.1, (cid:98) f (cid:48) is realized by extending variables of (cid:98) f with nonzero binary sig-natures realized from (cid:98) ∂ (cid:98) f . If we can realize a nonzero binary signature that is not in (cid:98) B ⊗ from (cid:98) ∂ (cid:98) f by factorization, then since (cid:98) F is non- (cid:98) B hard, we get (cid:98) ∂ (cid:98) f are in (cid:98) B ⊗ .Then, (cid:98) f (cid:48) is realized by extending variables of (cid:98) f with nonzero binary signatures in (cid:98) B ⊗ . Thus, (cid:98) f (cid:48) ∈ { (cid:98) f } (cid:98) B(cid:54) = . By Lemma 2.12, (cid:98) f ∈ { (cid:98) f (cid:48) } (cid:98) B(cid:54) = . Since (cid:98) f (cid:48) ∈ (cid:99) F or (cid:100) F H , (cid:98) f ∈ (cid:99) F or (cid:100) F H . Contradiction.Thus, Holant( (cid:54) = | (cid:98) F ) is (cid:98) F contains a nonzero 6-ary signature (cid:98) f / ∈ (cid:98) B ⊗ ∪ (cid:99) F ∪ (cid:100) F H .We go back to real-valued Holant problems under the Z -transformation. Consider the problemHolant( f , F ) where f = Z (cid:98) f = χ S · ( − x + x + x + x x + x x + x x + x x + x x + x x and S = S ( f ) = E . The signature f has a quite similar matrix form to (cid:98) f . M , ( f ) = − − − − − − − − − − − − − − − − . Since (cid:99) f H = (cid:98) H (cid:98) f = (cid:100) Hf , f H = Z (cid:99) f H = Hf . Also, since (cid:99) F = { (cid:98) f } (cid:98) B(cid:54) = , F = Z (cid:99) F = { f } B = is the set of signatures realizable by extending variables of f with binary signatures in B using= . Similarly, since (cid:100) F H = { (cid:99) f H } (cid:98) B(cid:54) = , F H = Z (cid:100) F H = { f } B = is the set of signatures realizable byextending variables of f H with binary signatures in B using = . Notice that f ∈ A and B ⊆ A .Thus, F ⊆ A . Also, the binary signature (1 , , − , 1) with a signature matrix H is in A . Thus, f H ∈ A and then F H ⊆ A . Also, S ( f ) = E and one can check that S ( f H ) = O . Thus, forevery f ∈ F ∪ F H , S ( f ) = E or O . Since f and f H satisfy , one can easily checkthat every f ∈ F ∪ F H satisfies .We want to show that Holant( f , F ) ≡ T Holant( (cid:54) = | (cid:98) f , (cid:98) F ) is F thatdoes not satisfy condition (T). By Lemma 6.6, { f } ∪ F is non- B hard. We restate Lemmas 6.8and 6.9 in the setting of Holant( F ) for non- B hard F . Corollary 6.10. Suppose that F is non- B hard. Then, Holant( F ) is F contains anonzero signature f of arity at most where f / ∈ B ⊗ ∪ F ∪ F H . emark: Notice that B ⊗ ∪ F ∪ F H ⊆ A . Thus, for any non- B hard set F , Holant( F ) is F contains a nonzero signature f of arity at most 6 where f / ∈ A .Now, we show that all four binary signatures in B are realizable from f . Lemma 6.11. Holant( B , f , F ) (cid:54) Holant( f , F ) .Proof. Consider ∂ f . Notice that (cid:20) f f (cid:21) = (cid:20) − − − − − − − − (cid:21) . Thus, ∂ f ( x , x , x , x ) = f + f has the truth table (0 , , , , , , , , , , − , , − , , , . In other words, ∂ f (0011) = 1, ∂ f (0101) = 1, ∂ f (1010) = − ∂ f (1100) = − 1, and ∂ f = 0 elsewhere. Then, S ( ∂ f ) = { ( x , x , x , x ) ∈ Z | x (cid:54) = x ∧ x (cid:54) = x } , and ∂ f ( x , x , x , x ) = ( (cid:54) = − )( x , x ) ⊗ ( (cid:54) = )( x , x ) . Thus, by factorization we can realize (cid:54) = − and (cid:54) = . Then connecting a variable of (cid:54) = − with a variableof (cid:54) = (using = ), we will get = − . Thus, B is realizable from f .We define the problem Holant b ( F ) to be Holant( B ∪ F ) . For all { i, j } and every b ∈ B , considersignatures ∂ bij f (i.e., ∂ + ij f , ∂ (cid:98) + ij f , ∂ − ij f and ∂ (cid:98) − ij f ) realized by merging variables x i and x j of f using the binary signature b . If there were one that is not in B ⊗ , then by Corollary 6.10, we wouldbe done. However, it is observed in [15] that f satisfies the following Bell property. Definition 6.12 (Bell property) . An irreducible signature f satisfies the Bell property if for allpairs of indices { i, j } and every b ∈ B , ∂ bij f ∈ B ⊗ . It can be directly checked that Lemma 6.13. Every signature in F ∪ F H satisfies the Bell property. Now consider all possible gadget constructions. If we could realize a signature of arity at most6 that is not in B ⊗ ∪ F ∪ F H from B and f by any possible gadget, then by Corollary 6.10 therewould be a somewhat more straightforward proof to our dichotomy theorem for the case of arity 6.However, after many failed attempts, we believe there is a more intrinsic reason why this approachcannot work. The following conjecture formulates this difficulty. This truly makes f the Lord ofIntransigence at arity 6. Conjecture 6.14. All nonzero signatures of arity at most realizable from B ∪{ f } are in B ⊗ ∪F .Also, all signatures of arity at most realizable from B ∪ { f H } are in B ⊗ ∪ F H . So to prove the b ( f , F ), we have to make additional use of F . Inparticular, we need to use a non-affine signature in F .33 The Holant b ( F ) In this section, we prove that for all real-valued non- B hard set F that does not satisfy condition(T), Holant b ( F ) is F that does not satisfycondition (T), the set { f } ∪ F is non- B hard, and since B is realizable from f , Holant( f , F ) is (cid:54) = | (cid:98) F ) is (cid:98) F contains a 6-ary signature that is not in (cid:98) O ⊗ (Lemma 7.21).Since F does not satisfy condition (T), F (cid:54)⊆ A . Thus, it contains a signature f of arity 2 n thatis not in A . In the following, we will prove the b ( F ) where F is non- B hardby induction on 2 n (cid:62) 2. For the base cases 2 n (cid:54) 6, by Corollary 6.10 and the Remark after that,Holant b ( F ) is n (cid:62) A , we wantto realize a signature of lower arity 2 k (cid:54) n − A , or else we get f / ∈ A be a nonzero signature of arity 2 n (cid:62) 8. We first show that if f does not have parity,then we get f has parity. If f is reducible, since f has even arity (as we assumed so starting from Section 3), it is a tensor product of two signatures ofodd arity, or a tensor product of two signatures of even arity which are not both in A since f / ∈ A .Thus, by factorization, we can realize a nonzero signature of odd arity and we get A . Thus, we mayassume that f is irreducible. Then by Lemma 4.4 and the Remark after Definition 4.1 we mayassume f satisfies .Consider signatures ∂ bij f (i.e., ∂ + ij f , ∂ − ij f , ∂ (cid:98) + ij f and ∂ (cid:98) − ij f ) realized by merging variables x i and x j of f using b ∈ B for all pairs of indices { i, j } and every b ∈ B . If there is one signature that is not in A , then we have realized a signature of arity 2 n − A . Otherwise, ∂ bij f ∈ A for all { i, j } and every b ∈ B . We denote this property by f ∈ (cid:82) B A . Now, assuming that f has parity, f satisfies and f ∈ (cid:82) B A , we would like to reach a contradiction by showing that this wouldforce f itself to belong to A . However, quite amazingly, there do exist non-affine signatures thatsatisfy these stringent conditions. We will show how they are discovered and handled (Lemmas 7.9,7.16 and 7.18).In this section, all signatures are real-valued. When we say an entry of a signature has norm a , we mean it takes value ± a . Since B is available in Holant b ( F ), if a signature f is realizable inHolant b ( F ), then we can realize all signatures in { f } B = that are realizable by extending f with B ⊗ (using = ). If we extend the variable x i of f with (cid:54) = , then we will get a signature g where g i = f i and g i = f i . This is a flipping operation on the variable x i . If we extend the variable x i of f with = − , then we will get a signature g where g i = f i and g i = − f i . We call this a negatingoperation on the variable x i . In the following, once f is realizable in Holant b ( F ), we may modifyit by flipping or negating. This will not change the complexity of the problem. We first show that if F contains a signature that does not have parity, then we can get Lemma 7.1. Suppose that F is non- B hard and F contains a signature f of arity n . If f doesnot have parity, then Holant b ( F ) is We prove this lemma by induction on 2 n . We first consider the base case that 2 n = 2. Since f has no parity, f / ∈ B . Since F is non- B hard, Holant b ( F ) is b ( F ) is n = 2 k (cid:62) 2. Consider the case that 2 n =2 k + 2 (cid:62) 4. We will show that we can realize a signature g of arity 2 k with no parity from f , i.e.,Holant b ( g, F ) (cid:54) T Holant b ( F ). Then by the induction hypothesis, we have Holant b ( F ) is n = 2 k + 2 . Since f has no parity, f (cid:54)≡ 0. It has at least a nonzero entry. By flipping variables of f , we mayassume that f ( (cid:126) n ) = x (cid:54) = 0. We denote (cid:126) n by α = 000 δ where δ = (cid:126) n − . Since f has no parityand f ( (cid:126) n ) (cid:54) = 0, there exists an input α (cid:48) with wt( α (cid:48) ) ≡ f ( α (cid:48) ) = x (cid:48) (cid:54) = 0. Since2 n (cid:62) 4, we can find three bits of α (cid:48) such that on these three bits, the values of α (cid:48) are the same. Byrenaming variables of f which gives a permutation of α (cid:48) , without loss of generality, we may assumethat these are the first three bits, i.e, α (cid:48) = α (cid:48) = α (cid:48) .We first consider the case that α (cid:48) α (cid:48) α (cid:48) = 000. Then, α (cid:48) = 000 δ (cid:48) for some δ (cid:48) ∈ Z n − wherewt( δ (cid:48) ) = wt(000 δ (cid:48) ) = wt( α (cid:48) ) ≡ . We consider the following six entries of f . x = f (000 δ ) , x (cid:48) = f (000 δ (cid:48) ) , y = f (011 δ ) , y (cid:48) = f (011 δ (cid:48) ) , z = f (101 δ ) , z (cid:48) = f (101 δ (cid:48) ) . Consider signatures ∂ +23 f and ∂ − f realized by connecting variables x and x of f using = +2 and = − respectively. Clearly, ∂ +23 f and ∂ − f have arity 2 n − 2. If one of them has no parity, thenwe are done. Thus, we may assume that ∂ +23 f and ∂ − f both have parity. Note that x + y and x (cid:48) + y (cid:48) are entries of the signature ∂ +23 f on inputs 0 δ and 0 δ (cid:48) respectively. Clearly, wt(0 δ ) = 0 andwt(0 δ (cid:48) ) ≡ . Since ∂ +23 f has parity, at least one of x + y and x (cid:48) + y (cid:48) is zero. Thus, we have( x + y )( x (cid:48) + y (cid:48) ) = 0. Also, note that x − y and x (cid:48) − y (cid:48) are entries of the signature ∂ − f on inputs0 δ and 0 δ (cid:48) respectively. Then, since ∂ − f has parity, similarly we have ( x − y )( x (cid:48) − y (cid:48) ) = 0. Thus,( x + y )( x (cid:48) + y (cid:48) ) + ( x − y )( x (cid:48) − y (cid:48) ) = 2( xx (cid:48) + yy (cid:48) ) = 0 . (7.1)Consider signatures ∂ +13 f and ∂ − f realized by connecting variables x and x of f using = and= − respectively. Again if one of them has no parity, then we are done. Suppose that ∂ +13 f and ∂ − f both have parity. Then, ( x + z )( x (cid:48) + z (cid:48) ) = 0 since x + z and x (cid:48) + z (cid:48) are entries of ∂ +13 f on inputs0 δ and 0 δ (cid:48) respectively. Similarly, ( x − z )( x (cid:48) − z (cid:48) ) = 0. Thus,( x + z )( x (cid:48) + z (cid:48) ) + ( x − z )( x (cid:48) − z (cid:48) ) = 2( xx (cid:48) + zz (cid:48) ) = 0 . (7.2)Consider signatures ∂ (cid:98) +12 f and ∂ (cid:98) − f realized by connecting variables x and x of f using (cid:54) = and (cid:54) = − respectively. Again if one of them has no parity, then we are done. Suppose that ∂ (cid:98) +12 f and ∂ (cid:98) − f both have parity. Then, ( y + z )( y (cid:48) + z (cid:48) ) = 0 since y + z and y (cid:48) + z (cid:48) are entries of ∂ (cid:98) +12 f on inputs 1 δ and 1 δ (cid:48) respectively, and wt(1 δ ) = 1 and wt(1 δ (cid:48) ) ≡ 0( mod 2). Similarly, ( y − z )( y (cid:48) − z (cid:48) ) = 0. Thus,( y + z )( y (cid:48) + z (cid:48) ) + ( y − z )( y (cid:48) − z (cid:48) ) = 2( yy (cid:48) + zz (cid:48) ) = 0 . (7.3)Then, consider (7.1) + (7.2) − (7.3). We have xx (cid:48) = 0 . However, since x = f ( (cid:126) n ) (cid:54) = 0 and x (cid:48) = f ( α (cid:48) ) (cid:54) = 0, xx (cid:48) (cid:54) = 0. Contradiction.For the case that α (cid:48) α (cid:48) α (cid:48) = 111, we have α (cid:48) = 111 δ (cid:48) for some δ (cid:48) ∈ Z n − where wt( δ (cid:48) ) =wt(111 δ (cid:48) ) − α (cid:48) ) − ≡ . We consider the following six entries of f . x = f (000 δ ) , x (cid:48) = f (111 δ (cid:48) ) , y = f (011 δ ) , y (cid:48) = f (100 δ (cid:48) ) , z = f (101 δ ) , z (cid:48) = f (010 δ (cid:48) ) . We still consider signatures ∂ +23 f , ∂ − f , ∂ +13 f , ∂ − f , ∂ (cid:98) +12 f and ∂ (cid:98) − f and suppose that they all haveparity. Then, similar to the above proof of the case α (cid:48) α (cid:48) α (cid:48) = 000, we can show that xx (cid:48) = 0.Contradiction.Thus, among ∂ +23 f , ∂ − f , ∂ +13 f , ∂ − f , ∂ (cid:98) +12 f and ∂ (cid:98) − f , at least one does not have parity. Thus,we realized a 2 k -ary signature with no parity. By our induction hypothesis, we are done.35 .2 Norm condition Under the assumptions that f has parity, f satisfies and f ∈ (cid:82) B A , we consider whetherall nonzero entries of f have the same norm. In Lemma 7.9, we will show that the answer is yes,but only for signatures of arity 2 n (cid:62) 10 (this lemma does not require F to be non- B hard). For asignature f of arity 2 n = 8, we show that either all nonzero entries of f have the same norm, or oneof the following signatures g or g (cid:48) is realizable. These two signatures are defined by g = χ S − · f and g (cid:48) = q − · f , where S = S ( q ) = E , q = χ S ( − (cid:80) (cid:54) i Define the graphs G n and H n as follows. The vertex set V ( G n ) is the set E n of all even weighted points in Z n . The vertex set V ( H n ) is the set O n of all odd weighted pointsin Z n . Two points u, v ∈ E n (or O n ) are connected by an edge iff wt( u ⊕ v ) = 2 . Let α ( G n ) be the independence number of G n i.e, the size of a maximum independent set of G n , and α ( H n ) be the independence number of H n . Let S ⊆ [2 n ]. We define ϕ S be a mappingthat flips the values on bits in S for all u ∈ E n . In other words, suppose that u (cid:48) = ϕ S ( u ). Then, u (cid:48) i = u i if i ∈ S and u (cid:48) i = u i if i / ∈ S where u (cid:48) i and u i are values of u and u on bit i respectively.For all S , clearly wt( u ⊕ v ) = 2 iff wt( ϕ S ( u ) ⊕ ϕ S ( v )) = 2 . When | S | is odd, ϕ S ( E n ) = O n .One can easily check that ϕ S gives an isomorphism between G n and H n . When | S | is even, ϕ S ( E n ) = E n . Then, ϕ S gives an automorphism of G n . Also, by permuting these 2 n bits, wecan get an automorphism of G n . In fact, the automorphism group of G n is generated by theseoperations. Lemma 7.3. Let n (cid:62) . Every automorphism ψ of G n is a product ϕ S ◦ π for some automorphism π induced by a permutation of n bits, and an automorphism ϕ S given by flipping the values onsome bits in a set S of even cardinality.Proof. Let ψ be an arbitrary automorphism of G n . Suppose ψ ( (cid:126) n ) = u . Let S ⊆ [2 n ] be theindex set where u i = 1. Then | S | = wt( u ) is even, and ψ (cid:48) = ϕ S ◦ ψ fixes (cid:126) n . Consider ψ (cid:48) ( v ) forall v ∈ E n of wt( v ) = 2. Since ψ (cid:48) is an automorphism fixing (cid:126) n , ψ (cid:48) ( v ) has weight 2. We denoteby e ij the 2 n -bit string with wt( e ij ) = 2 having 1’s on bits i and j , for 1 (cid:54) i < j (cid:54) n . Then, e = 11 (cid:126) n − . By a suitable permutation π of the variables, we have π ◦ ψ (cid:48) ( e ) = e , while stillfixing (cid:126) n . We will show that π ◦ ψ (cid:48) = π ◦ ϕ S ◦ ψ is the identity mapping, i.e., π ◦ ϕ S ◦ ψ = 1 G n . Then, ψ = ϕ − S ◦ π − . We are done. 36or simplicity of notations, we reuse ψ to denote π ◦ ψ (cid:48) in the following and we show that ψ = 1 G n . Consider e i , for 3 (cid:54) i (cid:54) n . Note that ψ ( e i ) is some e st and must have Hammingdistance 2 to e . It is easy to see that the only possibilities are s ∈ { , } and t > 2, i.e., from e we flip exactly one bit in { , } and another bit in { , . . . , n } . Suppose there are i (cid:54) = i (cid:48) ( i, i (cid:48) (cid:62) ψ ( e i ) = e t and ψ ( e i (cid:48) ) = e t (cid:48) . Since wt( e i ⊕ e i (cid:48) ) = 2, we must have t = t (cid:48) . Since2 n (cid:62) 6, we can pick another i (cid:48)(cid:48) (cid:62) i (cid:48)(cid:48) (cid:54) = i and i (cid:48) . Then, this leads to a contradiction since e i (cid:48)(cid:48) must either be mapped to e t if ψ ( e i (cid:48)(cid:48) ) = e t (cid:48)(cid:48) , or be mapped to e t if ψ ( e i (cid:48)(cid:48) ) = e t (cid:48)(cid:48) ; neitheris possible. Thus either ψ ( e i ) is some e t for all 3 (cid:54) i (cid:62) n , or is some e t for all 3 (cid:54) i (cid:62) n . Bya permutation of { , } (which maintains the property that ψ fixes (cid:126) n and e ) we may assume itis the former. Then the mapping i (cid:55)→ t given by ψ ( e i ) = e t for 3 (cid:54) i (cid:62) n defines a permutationof the variables for 3 (cid:54) i (cid:62) n (which again maintains the property that ψ fixes (cid:126) n and e )and, after a permutation of the variables we may now assume that ψ fixes (cid:126) n and all e i . Forany 1 (cid:54) i < j (cid:54) n , we have wt( ψ ( e ij )) = 2 and ψ ( e ij ) has distance 2 from both ψ ( e i ) = e i and ψ ( e j ) = e j . Then ψ ( e ij ) must be obtained from e i by flipping exactly one bit in { , i } andanother bit out of { , i } . However, it cannot flip bit i which would result in some e t for some t > 2, because ψ already fixed e t . Thus, it flips bit 1 but not bit i . Similarly in view of e j , wemust flip bit 1 but not bit j . Hence ψ ( e ij ) = e ij , and therefore ψ fixes all v with Hamming weightwt( v ) (cid:54) ψ fixes all v of wt( v ) (cid:54) k , for some k (cid:62) 1. If k < n we prove that ψ alsofixes all v of wt( v ) = 2 k + 2. For notational simplicity we may assume v = (cid:126) k +2 (cid:126) n − k − . As2 k + 2 (cid:62) 4, we can choose u = (cid:126) k (cid:126) n − k − and w = 00 (cid:126) k (cid:126) n − k − , and the two 00 in u and w among the first 2 k + 2 bits are in disjoint bit positions. Clearly wt( ψ ( v )) (cid:62) k + 2 since all stringsof wt (cid:54) k are fixed. Also since ψ ( v ) has Hamming distance 2 from ψ ( u ) = u and ψ ( w ) = w , it hasweight exactly 2 k + 2, and is obtained from u by flipping two bits from 00 to 11 in positions > k ,as well as obtained from w by flipping two bits from 00 to 11 in positions in { , } ∪ { t | t > k + 2 } .In particular, it is 1 in positions 1 to 2 k (in view of u ), and it is also 1 in positions 3 to 2 k + 2. Buttogether these positions cover all bits 1 to 2 k + 2. Thus ψ ( v ) = v . This completes the induction,and proves the lemma for all 2 n (cid:62) Remark: The condition 2 n (cid:62) n = 4: ψ fixes 0000 and 1111, and it maps α to α for all α ∈ { , } with wt( α ) = 2. If ψ were to be expressibleas ϕ S ◦ π , then since ψ (0000) = 0000, we have S = ∅ . Then by ψ (0011) = 1100 and ψ (0101) = 1010,the permutation π must map variable x to x . However this violates ψ (1001) = 0110. Lemma 7.4. Let { G n } be the sequence of graphs defined above. • If n = 8 , then α ( G ) = | E | = 2 , and the maximum independent set I of G is unique upto an automorphism, where I = { , , , , , , , , , , , , , , , } . • If n (cid:62) , then α ( G n ) < | E n | = 2 n − .Proof. We first consider the case 2 n = 6. One can check that the following set I = { , , , } 37s an independent set of G . Thus, α ( G ) (cid:62) 4. Next, we show that α ( G ) = 4 and I is the uniquemaximum independent set of α ( G ) up to an automorphism.Let J be an maximum independent set of G . Clearly, | J | (cid:62) 4. After an automorphism of G by flipping some bits, we may assume that (cid:126) ∈ J . Then for any u ∈ E with wt( u ) = 2, u / ∈ J . If (cid:126) ∈ J , then for any u ∈ E with wt( u ) = 4, u / ∈ J . Thus, J is maximal with | J | = 2 < 4, a contradiction. Thus, we have (cid:126) / ∈ J . Then all vertices in J , except (cid:126) havehamming weight 4. After an automorphism by permuting bits (this will not change (cid:126) ), we mayassume that u = 001111 ∈ J . Consider some other v ∈ J with wt( v ) = 4. If v v = 01 or 10,then wt( v v v v ) = 3. Thus, wt( u ⊕ v ) = wt(00 ⊕ v v ) + wt(1111 ⊕ v v v v ) = 1 + 1 = 2.Contradiction. The only v ∈ J with wt( v ) = 4, and v v = 00 is v = 001111 = u . Thus, v v = 11,i.e., both bits of v are 1 where u is 00. After an automorphism by permuting bits in { , , , } (this will not change (cid:126) and u ), we may assume that v = 110011 ∈ J . For any other w ∈ J withwt( w ) = 4, we must have w w = 11 (by the same proof for the pair ( u, v ) applied to ( u, w )), andalso w w = 11 (by the same proof for the pair ( u, v ) applied to ( v, w )). Thus, w = 111100. Then, J = { (cid:126) , u, v, w } = I is maximal. Thus, α ( G ) = 4 and I is the unique maximum independentset of α ( G ) up to an automorphism.Consider 2 n = 8. One can check that I is an independent set of G . Thus, α ( G ) (cid:62) 16. Weuse G ab to denote the subgraph of G induced by vertices { u ∈ E | u u = ab } for ( a, b ) ∈ Z .Clearly, G and G are isomorphic to G , and G and G are isomorphic to H . Since H isisomorphic to G , G and G are also isomorphic to G . Let J be a maximum independent setof G . Clearly, | J | (cid:62) | I | = 16. Also, we use J ab to denote the subset { u ∈ J | u u = ab } for ( a, b ) ∈ Z . Similarly, we can define I ab . Since J is an independent set of G , clearly, forevery ( a, b ) ∈ Z , J ab is an independent set of G ab . Since G ab is isomorphic to G and α ( G ) = 4,thus | J ab | (cid:54) 4. Then | J | (cid:54) 16. Thus, | J | = 16, and | J ab | = 4 for every ( a, b ) ∈ Z . Since themaximum independent set of G is unique up to an automorphism of G , which can be extendedto an automorphism of G by fixing the first two bits, we may assume that J = I = { , , , } after an automorphism of G .Then, consider J . We show that for any u ∈ J , u (cid:54) = u , u (cid:54) = u and u (cid:54) = u . Otherwise,by switching the pair of bits { , } with { , } or { , } (this will not change J ), we may assumethat u = u . Then wt( u u u u ) is odd. Since wt( u ) is even, wt( u u u u ) is odd. Thus, either u = u or u = u . Still by switching the pair { , } with { , } (again this will not change J ), we may assume that u = u . Then since wt( u u u u ) is odd, we have u (cid:54) = u . Then,one can check that there exists some v ∈ J such that v v v v = u u u u . Since v = v and u (cid:54) = u , wt( u u ⊕ v v ) = 1. Also since v = v and u (cid:54) = u , wt( u u ⊕ v v ) = 1. Thus,wt( u ⊕ v ) = wt( u u ⊕ v v ) + wt( u u ⊕ v v ) = 2 . This means that the vertices u and v areconnected in the graph G , a contradiction. Thus, for any u ∈ J , u (cid:54) = u , u (cid:54) = u and u (cid:54) = u .By permuting bit 3 with bit 4, bit 5 with bit 6, and bit 7 with bit 8 (this will not change J ),we may assume that 01010101 ∈ J . Consider some other w ∈ J . Since w i +1 (cid:54) = w i +2 for any i = 1 , w i +1 w i +2 = 01 or 10. Among these three pairs, let k denote the number ofpairs that are 01. If k = 3, then w = 01010101. Contraction. If k = 2, then wt(01010101 ⊕ w ) = 2.Contradiction. If k = 0, then w = 01101010. One can check that { , } is alreadya maximal independent set of G and it has size 2 < 4. Contradiction. Thus, k = 1. Then, w can38ake (cid:0) (cid:1) possible values. Thus, J ⊆ I = { , , , } . Since, | J | = 4, J = I . Consider some u ∈ J . Similar to the proof of J , we can show that u (cid:54) = u , u (cid:54) = u and u (cid:54) = u . Thus, u can take 2 possible values. Moreover, for any 01 u (cid:48) ∈ J , 10 u (cid:48) / ∈ J . Thus,there are only four remaining values that u can take. Then, J ⊆ I = { , , , } . Since | J | = 4, J = I .Finally, consider J . We show that for any u ∈ J , u = u , u = u and u = u . Otherwise,by permuting the pair of bits { , } with { , } or { , } (one can check that this will not change J and J ), we may assume that u (cid:54) = u . Since wt( u ) is even, between wt( u u ) and wt( u u ),exactly one is even and the other is odd. By permuting the pair of bits { , } with { , } , we mayfurther assume that u (cid:54) = u and u = u . Then, one can check that there exists some v ∈ J suchthat u u u u = v v v v . Since u = u and v (cid:54) = v , wt( u u ⊕ v v ) = 1. Also since u = u and v (cid:54) = v , wt( u u ⊕ v v ) = 1. Thus, wt( u ⊕ v ) = wt( u u ⊕ v v ) + wt( u u ⊕ v v ) = 2 . Contradiction. Thus, for any u ∈ J , it can take 2 possible values. Moreover, for any 00 u (cid:48) ∈ J ,we have 11 u (cid:48) / ∈ J . Thus, there are only four remaining values that u can take. Then, J ⊆ I = { , , , } . Thus, after an automorphism, J = I . In other words, I is the unique maximum independent setof G up to an automorphism.Now, we consider the case 2 n (cid:62) 10. For every ( a, b ) ∈ Z , we define G ab n to be the subgraph of G n induced by { u ∈ G n | u u = ab } , and it is isomorphic to G n − . Thus, α ( G n ) (cid:54) α ( G n ) + α ( G n ) + α ( G n ) + α ( G n ) = 4 α ( G n − ) . Then, α ( G n − ) < (2 n − − will imply that α ( G n ) < n − . Thus, in order to prove α ( G n ) < n − for all 2 n (cid:62) 10, it suffices to prove that α ( G ) < − . For a contradiction, supposethat α ( G ) (cid:62) − . Let I be a maximum independent set of G . Then, | I | (cid:62) . We define I ab = { u ∈ I | u u = ab } for every ( a, b ) ∈ Z . Since G ab is isomorphic to G and α ( G ) = 2 , | I ab | (cid:54) for every ( a, b ) ∈ Z . Then, | I | (cid:54) · . Thus, | I | = 2 and | I ab | = 2 for every( a, b ) ∈ Z . Since the maximum independent set of G is unique up to an automorphism of G which can be extended to an automorphism of G by fixing the first two bits, we may assume that I = { u | u ∈ I } .Consider I . Since | I | (cid:54) = 0, there exists some 01 v ∈ I . Since wt( v ) is odd, among wt( v v ),wt( v v ), wt( v v ) and wt( v v ), there is an odd number (either one or three) of pairs such thatwt( v i +1 v i +2 ) (1 (cid:54) i (cid:54) 4) is odd, i.e., v i +1 (cid:54) = v i +2 . In other words, there are exactly three pairsamong v v , v v , v v and v v such that the values inside each pair are all equal with each otheror all distinct with each other. By permuting these pairs of bits { , } , { , } , { , } and { , } (this will not change I ), we may assume that either v = v , v = v , v = v and v (cid:54) = v ,or v (cid:54) = v , v (cid:54) = v , v (cid:54) = v and v = v . In both cases, one can check that there exists some00 u ∈ I such that u i = v i for i ∈ { , . . . , } . Moreover, u = u if v (cid:54) = v , and u (cid:54) = u if v = v . Then, wt(00 u ⊕ v ) = wt(00 ⊕ 01) + wt( u u ⊕ v v ) = 2. This contradiction provesthat α ( G ) < − , and the lemma is proved. 39 emark: We remark that I = S ( f ). Later, we will see that the signature f , this Queen of theNight, and its support S ( f ) have even more extraordinary properties.We consider a particular gadget construction that will be used in our proof. Let h be a 4-arysignature with signature matrix M , ( h ) = H = (cid:20) − − (cid:21) . Notice that H H T = H H = 2 I ,and h is an affine signature. The following is called an H gadget construction on f , denoted by H ij f . This is the signature obtained by connecting variables x and x of h with variables x i and x j of f using = , respectively. Note that H ij f is not necessarily realizable from f since h maynot be available. However, we will analyze the structure of f by analyzing H ij f . For convenience,we consider ( i, j ) = (1 , 2) and we use (cid:101) f to denote H f . The following results (Lemmas 7.5 and7.6) about (cid:101) f = H f hold for all H ij f by replacing { , } with { i, j } . Note that (cid:101) f has the followingsignature matrix M ( (cid:101) f ) = (cid:101) f (cid:101) f (cid:101) f (cid:101) f = H M ( f ) = f + f f + f f − f f − f = ∂ +12 f ∂ (cid:98) +12 f ∂ (cid:98) − f ∂ − f . We give the following relations between f and (cid:101) f . Lemma 7.5. 1. If f has even parity then (cid:101) f also has even parity.2. If f ∈ A , then (cid:101) f ∈ A .3. If M ( m f ) = λI for some real λ (cid:54) = 0 , then M ( m (cid:101) f ) = 2 λI .4. If ∂ +12 f, ∂ − f, ∂ (cid:98) +12 f, ∂ (cid:98) − f ∈ A , then (cid:101) f , (cid:101) f , (cid:101) f , (cid:101) f ∈ A .5. For { u, v } disjoint with { , } and b ∈ B , if ∂ buv f ∈ A , then ∂ buv (cid:101) f ∈ A .Proof. Since h has even parity and h ∈ A , the first and second propositions hold.If M ( m f ) = λI , then M ( m (cid:101) f ) = M ( (cid:101) f ) M T ( (cid:101) f ) = H M ( f ) M T ( f ) H T = λH I H T =2 λI . The third proposition holds.By the matrix form M ( (cid:101) f ), the fourth proposition holds.Since the H gadget construction only touches variables x and x of f , it commutes withmerging gadgets on variables other than x and x . Thus ∂ bij (cid:101) f = (cid:103) ∂ bij f . For all b ∈ B and all { i, j } disjoint with { , } , if ∂ bij f ∈ A where ∂ bij f are signatures realized by connecting variables x i and x j of f using b , then ∂ bij (cid:101) f = (cid:103) ∂ bij f ∈ A . The last proposition holds.Clearly, if f ∈ (cid:82) B A , then (cid:101) f , (cid:101) f , (cid:101) f , (cid:101) f ∈ A . Thus, for every ( a, b ) ∈ Z , if (cid:101) f ab (cid:54)≡ n ab . Let n ab = 0 if (cid:101) f ab ≡ 0. We have thefollowing results regarding these norms n ab . Lemma 7.6. Let f be an irreducible signature of arity n (cid:62) . Suppose that f has even parity, f satisfies and f ∈ (cid:82) B A .1. For any ( a, b ) , ( c, d ) ∈ Z , there exists some k ∈ Z such that n ab = √ k n cd (cid:54) = 0 , and n ab = n cd iff | S ( (cid:101) f ab ) | = | S ( (cid:101) f cd ) | . . Furthermore, if (cid:101) f ( (cid:126) n − ) (cid:54) = 0 and n > n , then S ( (cid:101) f ) = E n − and n ab = n or n for every ( a, b ) ∈ Z ; in particular, n = 2 n . Symmetrically, if (cid:101) f ( (cid:126) n − ) (cid:54) = 0 and n < n , then S ( (cid:101) f ) = E n − and n ab = n or n for every ( a, b ) ∈ Z , and n = 2 n .Proof. Since f satisfies , M ( m f ) = λI for some real λ (cid:54) = 0. Then, by Lemma 7.5, M ( m (cid:101) f ) = 2 λI (cid:54) = 0. Thus, | (cid:101) f ab | = 2 λ (cid:54) = 0 for every ( a, b ) ∈ Z . Also, since f ∈ (cid:82) B A , byLemma 7.5, for every ( a, b ) ∈ Z , (cid:101) f ab ∈ A . Thus, S ( (cid:101) f ab ) is affine and | S ( (cid:101) f ab ) | = 2 k ab for someinteger k ab (cid:62) 0. Note that | (cid:101) f ab | = n ab · | S ( (cid:101) f ab ) | = n ab · k ab (cid:54) = 0 . Thus, for any ( a, b ) , ( c, d ) ∈ Z , n ab · k ab = n cd · k cd (cid:54) = 0. Then, n ab = √ k n cd (cid:54) = 0 where k = k cd − k ab ∈ Z . Clearly, k = 0 iff | S ( (cid:101) f ab ) | = 2 k ab = 2 k cd = | S ( (cid:101) f cd ) | . Now we prove the second part of this lemma. We give the proof for the case that (cid:101) f ( (cid:126) n − ) (cid:54) = 0and n > n . The proof of the case that (cid:101) f ( (cid:126) n − ) (cid:54) = 0 and n < n is symmetric. We firstshow that S ( (cid:101) f ) = E n − . For a contradiction, suppose that S ( (cid:101) f ) (cid:54) = E n − . Since f has evenparity, by Lemma 7.5, (cid:101) f has even parity. Then, (cid:101) f also has even parity. Thus, S ( (cid:101) f ) (cid:40) E n − .There exists θ ∈ E n − such that θ / ∈ S ( (cid:101) f ). Also, since n (cid:54) = 0, (cid:101) f (cid:54)≡ 0. Then, S ( (cid:101) f ) (cid:54) = ∅ and there exists δ ∈ E n − such that δ ∈ S ( (cid:101) f ). Then, we can find a pair α, β ∈ E n − wherewt( α ⊕ β ) = 2 such that one is in S ( (cid:101) f ) and the other one is not in S ( (cid:101) f ). • If wt( α ) (cid:54) = wt( β ), then clearly the difference between their Hamming weights is 2 sincewt( α ⊕ β ) = 2. Thus, α and β differ in two bits i, j on which one takes value 00 and the othertakes value 11. • If wt( α ) = wt( β ), then they differ in two bits i, j on which one takes value 01 and the othertakes value 10. Without loss of generality, we assume that α i α j = 01 and β i β j = 10. Theytake the same value on other bits. Since α, β ∈ E n − and 2 n (cid:62) 6, they have even Hammingweight and length at least 4. Thus, there is another bit k such that on this bit, α k = β k = 1.Consider γ ∈ E n − where γ i γ j γ k = 000 and γ takes the same value as α and β on other bits.Clearly, wt( γ ) + 2 = wt( α ) = wt( β ). If γ ∈ S ( (cid:101) f ), then between α and β , we pick the onethat is not in S ( (cid:101) f ). Otherwise, we pick the one that is in S ( (cid:101) f ). In both cases, we canget a pair of inputs in E n − such that one is in S ( (cid:101) f ) and the other is not in S ( (cid:101) f ), andthey have Hamming distance 2 as well as different Hamming weights.Thus, we can always find a pair α, β ∈ E n − where wt( α ⊕ β ) = 2 and α, β differ in two bits i, j on which one takes value 00 and the other takes value 11, such that one is in S ( (cid:101) f ) and theother is not in S ( (cid:101) f ). Clearly, { i, j } is disjoint with { , } .Consider signatures ∂ + ij (cid:101) f and ∂ − ij (cid:101) f . Then, (cid:101) f (11 α ) + (cid:101) f (11 β ) is an entry of ∂ + ij (cid:101) f , and (cid:101) f (11 α ) − (cid:101) f (11 β ) is an entry of ∂ − ij (cid:101) f . Since between (cid:101) f (11 α ) and (cid:101) f (11 β ), exactly one is nonzero and it hasnorm n , we have | (cid:101) f (11 α ) + (cid:101) f (11 β ) | = | (cid:101) f (11 α ) − (cid:101) f (11 β ) | = n . Thus, both ∂ + ij (cid:101) f and ∂ − ij (cid:101) f have an entry with norm n . Let δ ∈ E n where δ i δ j = 11 and δ takesvalue 0 on other bits. Then, clearly, (cid:101) f ( (cid:126) n ) + (cid:101) f ( δ ) is an entry of ∂ + ij (cid:101) f , and (cid:101) f ( (cid:126) n ) − (cid:101) f ( δ ) is an entryof ∂ − ij (cid:101) f . Here, E n − = { ( x , . . . , x ) ∈ Z | x + · · · + x = 0 } . When context is clear, we do not specify the variables of E n − and also O n − . If (cid:101) f ( δ ) (cid:54) = 0, then | (cid:101) f ( δ ) | = n since δ δ = 00. Since (cid:101) f ( (cid:126) n ) (cid:54) = 0, | (cid:101) f ( (cid:126) n ) | = n . Thus,between (cid:101) f ( (cid:126) n ) + (cid:101) f ( δ ) and (cid:101) f ( (cid:126) n ) − (cid:101) f ( δ ), one has norm 2 n and the other is zero. Therefore,between ∂ + ij (cid:101) f and ∂ − ij (cid:101) f , one signature has an entry with norm 2 n . Remember that both ∂ + ij (cid:101) f and ∂ − ij (cid:101) f have an entry with norm n . Clearly, 2 n > n . Thus, between ∂ + ij (cid:101) f and ∂ − ij (cid:101) f , there is a signature that has two entries with different norms. Clearly, such a signatureis not in A . However, since f ∈ (cid:82) B A , by Lemma 7.5, ∂ + ij (cid:101) f , ∂ − ij (cid:101) f ∈ A . Contradiction. • If (cid:101) f ( δ ) = 0, then | (cid:101) f ( (cid:126) n ) + (cid:101) f ( δ ) | = | (cid:101) f ( (cid:126) n ) | = n > n . Thus, ∂ + ij (cid:101) f has two nonzero entrieswith different norms n and n . Then, ∂ + ij (cid:101) f / ∈ A . Contradiction.Thus, S ( (cid:101) f ) = E n − .Then, we show that n ab = n or 2 n for any ( a, b ) ∈ Z . Clearly, we may assume that ( a, b ) (cid:54) =(1 , n ab (cid:54) = n and 2 n . First, we show that | S ( (cid:101) f ab ) | < n − and n ab > n . Since f has parity, (cid:101) f ab also has parity (either even or odd depending on whetherwt( ab ) = 0 or 1). Thus | S ( (cid:101) f ab ) | (cid:54) | E n − | = | O n − | = 2 n − . If the equality holds, then n ab = n since n ab | S ( (cid:101) f ab ) | = n | S ( (cid:101) f ) | and | S ( (cid:101) f ) | = 2 n − . Contradiction. Thus, | S ( (cid:101) f ab ) | < n − and also n ab > n . Depending on whether f ab has even parity or odd parity, we can pick a pair of inputs α, β withwt( α ⊕ β ) = 2 from E n − or O n − such that exactly one is in S ( f ab ) and the other is not in S ( f ab ). Suppose that α and β differ in bits i, j . Depending on whether α i = α j or α i (cid:54) = α j , wecan connect variables x i and x j of (cid:101) f using = +2 and = − , or (cid:54) = +2 and (cid:54) = − . We will get two signatures ∂ + ij (cid:101) f and ∂ − ij (cid:101) f , or ∂ (cid:98) + ij (cid:101) f and ∂ (cid:98) − ij (cid:101) f . We consider the case that α i = α j . For the case that α i (cid:54) = α j , theanalysis is the same by replacing ∂ + ij (cid:101) f and ∂ − ij (cid:101) f with ∂ (cid:98) + ij (cid:101) f and ∂ (cid:98) − ij (cid:101) f respectively.Consider signatures ∂ + ij (cid:101) f and ∂ − ij (cid:101) f . Then, (cid:101) f ( abα ) + (cid:101) f ( abβ ) is an entry of ∂ + ij (cid:101) f , and (cid:101) f ( abα ) − (cid:101) f ( abβ ) is an entry of ∂ − ij (cid:101) f . Since between α and β , exactly one is in S ( f ab ), between (cid:101) f ( abα ) and (cid:101) f ( abβ ), exactly one is nonzero and it has norm n ab . Thus, | (cid:101) f ( abα ) + (cid:101) f ( abβ ) | = | (cid:101) f ( abα ) − (cid:101) f ( abβ ) | = n ab . Both ∂ + ij (cid:101) f and ∂ − ij (cid:101) f have an entry with norm n ab .Let α (cid:48) , β (cid:48) ∈ E n − where α (cid:48) i α (cid:48) j = α i α j , α (cid:48) k = α (cid:48) i ⊕ α (cid:48) j for some k (cid:54) = i, j and α (cid:48) takes value0 on other bits, and β (cid:48) i β (cid:48) j = β i β j , β (cid:48) k = β (cid:48) i ⊕ β (cid:48) j for the same k (cid:54) = i, j and β (cid:48) takes value 0 onother bits. Clearly, α (cid:48) and β (cid:48) differ in bits i and j and they differ in the same way as α and β . Then, (cid:101) f (11 α (cid:48) ) + (cid:101) f (11 β (cid:48) ) is an entry of ∂ + ij (cid:101) f , and (cid:101) f (11 α (cid:48) ) − (cid:101) f (11 β (cid:48) ) is an entry of ∂ − ij (cid:101) f . Since S ( (cid:101) f ) = E n − , both (cid:101) f (11 α (cid:48) ) and (cid:101) f (11 β (cid:48) ) are nonzero and they have norm n . Thus, between (cid:101) f (11 α (cid:48) ) + (cid:101) f (11 β (cid:48) ) and (cid:101) f (11 α (cid:48) ) − (cid:101) f (11 β (cid:48) ), exactly one is zero and the other has norm 2 n . Thus,between signatures ∂ ij (cid:101) f and ∂ − ij (cid:101) f , there is a signature that has two entries with different norms 2 n and n ab . Such a signature is not in A . However, since f ∈ (cid:82) B A , by Lemma 7.5, ∂ ij (cid:101) f , ∂ − ij (cid:101) f ∈ A .Contradiction. Thus, n ab = n or 2 n for any ( a, b ) ∈ Z .We also give the following results about multilinear polynomials F ( x , . . . , x n ) ∈ Z [ x , . . . , x n ].We use d ( F ) to denote the total degree of F . For { i, j } ⊆ { , . . . , n } = [ n ], we use F abij ∈ Z [ { x , . . . , x n }\{ x i , x j } ] to denote the polynomial obtained by setting ( x i , x j ) = ( a, b ) in F . Since 2 n − (cid:62) 4, such a k exists. Here, α (cid:48) k = 0 since α i = α j in this case under discussion. For the case that α i (cid:54) = α j , we have α (cid:48) k = 1. efinition 7.7. Let F ( x , . . . , x n ) ∈ Z [ x , . . . , x n ] be a multilinear polynomial. We say F is acomplete quadratic polynomial if d ( F ) = 2 and for all { i, j } ⊆ [ n ] , the quadratic term x i x j appearsin F . We say F is a complete cubic polynomial if d ( F ) = 3 and for all { i, j, k } ⊆ [ n ] , the cubicterm x i x j x k appears in F . Lemma 7.8. Let F ( x , . . . , x n ) ∈ Z [ x , . . . , x n ] be a multilinear polynomial.1. If for all { i, j } ⊆ [ n ] , F ij + F ij ≡ or , and F ij + F ij ≡ or , then d ( F ) (cid:54) . Moreover,if d ( F ) = 2 , then F is a complete quadratic polynomial.2. If for all { i, j } ⊆ [ n ] , d ( F ij + F ij ) (cid:54) , and d ( F ij + F ij ) (cid:54) , then d ( F ) (cid:54) . Moreover, if d ( F ) = 3 , then F is a complete cubic polynomial.Proof. We prove the first part. The proof for the second part is similar which we omit here.For all { i, j } ⊆ [ n ], we write F ∈ Z [ x , . . . , x n ] as a polynomial of variables x i and x j . F = X ij x i x j + Y ij x i + Z ij x j + W ij where X ij , Y ij , Z ij , W ij ∈ Z [ { x , . . . , x n }\{ x i , x j } ]. Then, F ij = W ij and F ij = X ij + Y ij + Z ij + W ij . Thus, X ij + Y ij + Z ij = F ij + F ij ≡ . Also, F ij = Z ij + W ij and F ij = Y ij + W ij . Thus, Y ij + Z ij = F ij + F ij ≡ . Then, X ij ≡ { i, j } . Thus, d ( F ) (cid:54) d ( F ) = 2. then F has at least a quadratic term x u x v ( u (cid:54) = v ). Without loss ofgenerality, we assume that the term x x appears in F . We first show that for all 2 (cid:54) j (cid:54) n , thequadratic term x x j appears in F . Since x x is already in F , we may assume that 3 (cid:54) j . Wewrite F as a polynomial of variables x and x j . F = X j x x j + Y j x + Z j x j + W j , where X j , Y j , Z j , W j do not involve x and x j . Since x x appears in F , x appears in Y j . Aswe have proved above, Y j + Z j ≡ x also appears in Z j , which means that x x j appears in F . Then, for all 2 (cid:54) j (cid:54) n , x x j appears in F .Then, for all 2 (cid:54) i < j (cid:54) n , we write F as a polynomial of variables x and x i . F = X i x x i + Y i x + Z i x i + W i , where X i , Y i , Z i , W i do not involve x and x i . Since x x j appears in F , x j appears in Y i . Since Y i + Z i ≡ x j also appears in Z i . Thus, x i x j appears in F . Then, for all 2 (cid:54) i < j (cid:54) n ,the quadratic term x i x j appears in F . Thus, for all { i, j } ⊆ [ n ], x i x j appears in F .Now, we are ready to take a major step towards Theorem 7.19. Lemma 7.9. Let n (cid:62) and let f ∈ F be a n -ary irreducible signature with parity. Then, • Holant b ( F ) is • there is a signature g / ∈ A of arity k < n that is realizable from f and B , or • after normalization, f ( α ) = ± for all α ∈ S ( f ) . roof. Since f is irreducible, we may assume that it satisfies . Otherwise, we get f ∈ (cid:82) B A . Otherwise, we can realize a signatureof arity 2 n − A by merging f using some b ∈ B .For any four entries x, y, z, w of f on inputs α, β, γ, δ ∈ Z n written in the form of a 2-by-2matrix [ x yz w ] = (cid:104) f ( α ) f ( β ) f ( γ ) f ( δ ) (cid:105) , we say that such a matrix is a distance-2 square if there exist four bits i, j, k, (cid:96) such that α i α j = β i β j = γ i γ j = δ i δ j , α k α (cid:96) = γ k γ (cid:96) = β k β (cid:96) = δ k δ (cid:96) and α , β , γ and δ take thesame values on other bits. An equivalent description is that δ = α ⊕ β ⊕ γ, wt( α ⊕ β ) = 2 , wt( α ⊕ γ ) = 2 and wt( α ⊕ δ ) = 4 . (7.5)Indeed (7.5) is clearly satisfied by any distance-2 square. Conversely, suppose (7.5) holds. If weflip any bit i in all α, β, γ and δ , both (7.5) and the bitwise description are invariant, and thus wemay assume α = (cid:126) n . By wt( α ⊕ γ ) = 2, there exist two bits i, j such that γ i γ j = 11, and γ takes0 on other bits. By wt( α ⊕ β ) = 2, there exits two bits k, (cid:96) such that β k β (cid:96) = 11, and β takes 0 onother bits. Since δ = α ⊕ β ⊕ γ , wt( β ⊕ γ ) = wt( α ⊕ δ ) = 4 . Thus, the bits i, j, k, (cid:96) are distinct fourbits. Then, δ i δ j δ k δ (cid:96) = 1111 and δ takes 0 on other bits. Thus, α , β , γ and δ satisfy the bitwisedescription of distance-2 squares.We give an example of such a distance-2 square. Let (cid:20) x yz w (cid:21) = (cid:20) f ( α ) f ( β ) f ( γ ) f ( δ ) (cid:21) = (cid:20) f (0001 θ ) f (0010 θ ) f (1101 θ ) f (1110 θ ) (cid:21) where θ ∈ Z n − is an arbitrary binary string of length 2 n − 4. In this example, ( i, j ) = (1 , 2) and( k, (cid:96) ) = (3 , x yz w ] has the property described in(7.6) ∼ (7.9).By connecting variables x and x of f using = +2 and = − respectively, we get signatures ∂ +12 f and ∂ − f . By our assumption, ∂ +12 f and ∂ − f are affine signatures. Note that, x + z and y + w areentries of ∂ +12 f on inputs 01 θ and 10 θ ∈ Z n − . Since ∂ +12 f ∈ A , if x + z and y + w are both nonzero,then they have the same norm. Thus, we have ( x + z )( y + w ) = 0 or ( x + z ) = ( y + w ) . Similarly, x − z and y − w are entries of ∂ − f ∈ A . Thus, we have ( x − z )( y − w ) = 0 or ( x − z ) = ( y − w ) .Also, by connecting variables x and x of f using (cid:54) = and (cid:54) = − respectively, we get signatures ∂ (cid:98) +34 f and ∂ (cid:98) − f that are affine signatures. Note that, x + y and z + w are entries of ∂ (cid:98) +34 f on inputs00 θ and 11 θ . Since ∂ (cid:98) +34 f ∈ A , we have ( x + y )( z + w ) = 0 or ( x + y ) = ( z + w ) . Similarly, x − y and z − w are entries of (cid:98) ∂ − f . Then, we have ( x − y )( z − w ) = 0 or ( x − y ) = ( z − w ) .Now, consider an arbitrary distance-2 square [ x yz w ] = (cid:104) f ( α ) f ( β ) f ( γ ) f ( δ ) (cid:105) . Depending on whether α i = α j or α i (cid:54) = α j , we can use = +2 and = − , or (cid:54) = +2 and (cid:54) = − respectively, to connect variables x i and x j of f to produce two signatures ∂ + ij f and ∂ − ij f , or ∂ (cid:98) + ij f and ∂ (cid:98) − ij f in either case, such that x ± z and y ± w are both entries of the resulting two signatures. Since the two resulting signatures arein affine, we have ( x + z )( y + w ) = 0 or ( x + z ) = ( y + w ) , (7.6)and ( x − z )( y − w ) = 0 or ( x − z ) = ( y − w ) . (7.7)Similarly, by connecting variables x k and x (cid:96) of f using either = ± or (cid:54) = ± , we have( x + y )( z + w ) = 0 or ( x + y ) = ( z + w ) (7.8)44nd ( x − y )( z − w ) = 0 or ( x − y ) = ( z − w ) . (7.9)Now, we show that by solving equations (7.6) ∼ (7.9), every distance-2 square has one of thefollowing forms (after normalization, row or column permutation, multiplying a − (cid:20) (cid:21) , (cid:20) (cid:21) , (cid:20) (cid:21) , (cid:20) (cid:21) , (cid:20) (cid:21) , (cid:20) − (cid:21) , (cid:124) (cid:123)(cid:122) (cid:125) type I (cid:20) aa (cid:21) ( a > , (cid:124) (cid:123)(cid:122) (cid:125) type II or (cid:20) − (cid:21)(cid:124) (cid:123)(cid:122) (cid:125) type III . We say that the first six forms are type I, and the other two are type II and type III respectively.These forms listed above are canonical forms of each type.Let [ x yz w ] be a distance-2 square. Consider p = ( x + y )( z + w )( x + z )( y + w )( x − y )( z − w )( x − z )( y − w ) . • If p = 0, then among its eight factors (four sums and four differences), at least one factor iszero. By taking transpose and row permutation, we may assume that x + y = 0 or x − y = 0.If x + y = 0, then by multiplying the column [ yw ] with − 1, we can modify this distance-2square to get x − y = 0. Thus, we may assume that x − y = 0. If x = y = 0, then by (7.6),we have z = 0 or w = 0, or z = ± w . Thus, after normalizing operations of row and columnpermutation and multiplication by − 1, we reach the following canonical forms [ ], [ ] or[ ] . Otherwise, x = y (cid:54) = 0. Consider q = ( x + z )( y + w )( x − z )( y − w ). – If q = 0, then among its four factors (two sums and two differences), at least one is zero.By column permutation on the matrix [ x yz w ] and multiplying the row ( z, w ) with − x and y ), we may assume that x − z = 0. Thus, x = y = z (cid:54) = 0. We normalize their values to 1. Then by (7.6), 1 + w = 0 or 1 + w = ± w = − , − 3. If w = ± 1, then [ x yz w ] has the canonical form [ ] or (cid:2) − (cid:3) . If w = − 3, then [ x yz w ] = (cid:2) − (cid:3) which has the canonical form (cid:2) − (cid:3) (Type III). – If q (cid:54) = 0, then ( x + z )( y + w ) (cid:54) = 0 and ( x − z )( y − w ) (cid:54) = 0. By equations (7.6) and (7.7),( x + z ) = ( y + w ) and ( x − z ) = ( y − w ) . Thus, xz = yw . Since x = y (cid:54) = 0, z = w .If z = w = 0, then this gives the canonical form [ ]. Otherwise, z = w (cid:54) = 0. Then z + w (cid:54) = 0 and hence by (7.8), z + w = ± ( x + y ). Since z = w and x = y , we get z = ± x .Thus, x + z = 0 or x − z = 0. Contradiction. • If p (cid:54) = 0, then all its eight factors are nonzero. Thus by (7.6) ∼ (7.9), ( x + z ) = ( y + w ) ,( x − z ) = ( y − w ) , ( x + y ) = ( z + w ) and ( x − y ) = ( z − w ) . By solving these equations,we have x = w , y = z , and xy = zw . If x = y = z = w = 0, then it gives the canonicalform [ ]. Otherwise, by permuting rows and columns, we may assume that x (cid:54) = 0 and | x | is the smallest among the norms of nonzero entries in [ x yz w ]. We normalize x to 1. Since x = w , we get w = ± 1. By multiplying the row ( z, w ) with − xy = zw ), we may assume that w = 1. Then, xy = zw implies that y = z . If y = z = 0,then [ x yz w ] has the canonical form [ ]. Otherwise, since | x | = 1 is the smallest norm amongnonzero entries, y = z = ± a where a (cid:62) . If a = 1 (i.e., y = z = ± x yz w ] has thecanonical form [ ]. If a > 1, then [ x yz w ] has the canonical form of Type II.Thus, every distance-2 square has a canonical form of Type I, II or III.Note that given a particular distance-2 square of f , by normalization, and renaming or flippingor negating variables of f , we can always modify this distance-2 square to get its canonical form.45learly, for signatures of arity at least 4, distance-2 squares exist. We consider the following twocases according to which types of distance-2 squares appear in f . Case 1. All distance-2 squares in f are of type I.We show that (after normalization) f ( α ) = ± α ∈ S ( f ). Since f (cid:54)≡ 0, it has at leastone nonzero entry. By normalization, we may assume that 1 is the smallest norm of all nonzeroentries of f . Then by flipping variables of f , we may assume that f ( (cid:126) n ) = 1. For a contradiction,suppose that there is some β ∈ S ( f ) such that f ( β ) (cid:54) = ± 1. Then by our assumption that 1 isthe smallest norm and | f ( β ) | (cid:54) = 1, we have | f ( β ) | > 1. Also, since f has parity and (cid:126) n ∈ S ( f ), f has even parity. Thus, wt( β ) ≡ . By renaming variables of f , we may assume that β = (cid:126) m (cid:126) n − m , for some m (cid:62) 1. (This does not affect the normalization f ( (cid:126) n ) = 1). Then, weshow that for all α = δ(cid:126) n − m where δ ∈ Z m , f ( α ) = ± 1. We prove this by induction on wt( δ ).This will lead to a contradiction when wt( δ ) = 2 m , since | f ( β ) | = | f ( (cid:126) m (cid:126) n − m ) | (cid:54) = 1.Since f ( (cid:126) n ) = 1, we may assume wt( δ ) (cid:62) 2. We first consider the base case that wt( δ ) = 2. Byrenaming the first 2 m variables, without loss of generality, we may assume that δ = 11 (cid:126) m − andthen α = 11 (cid:126) n − = 1100 (cid:126) n − . This renaming will not change β . Consider the following distance-2square (cid:20) x yz w (cid:21) = (cid:20) f (0000 (cid:126) n − ) f (1100 (cid:126) n − ) f (0011 (cid:126) n − ) f (1111 (cid:126) n − ) (cid:21) . Recall our assumption that every distance-2 square is of type I. Here x = f ( (cid:126) n ), and y = f ( α ).Since x = 1, [ x yz w ] being of type I implies that y = 0 or ± − | y | = 1; for a contradiction, supposethat y = 0. We consider the following two extra entries of f , where δ = 00 (cid:126) m − .x (cid:48) = f ( δ(cid:126) n − m ) = f (00 (cid:126) m − (cid:126) n − m ) and y (cid:48) = f ( β ) = f (11 (cid:126) m − (cid:126) n − m ) . By connecting variables x and x of f using = and = − , we get signatures ∂ f and ∂ − f re-spectively. Note that both x + y and x (cid:48) + y (cid:48) are entries of ∂ f . Since ∂ f ∈ A , we have( x + y )( x (cid:48) + y (cid:48) ) = 0 or ( x + y ) = ( x (cid:48) + y (cid:48) ) . We can also consider ∂ − f and get ( x − y )( x (cid:48) − y (cid:48) ) = 0or ( x − y ) = ( x (cid:48) − y (cid:48) ) . Since x = 1 and y = 0, we have (cid:104) x (cid:48) + y (cid:48) = 0 or ( x (cid:48) + y (cid:48) ) = 1 (cid:105) and (cid:104) x (cid:48) − y (cid:48) = 0 or ( x (cid:48) − y (cid:48) ) = 1 (cid:105) . Recall that | y (cid:48) | = | f ( β ) | > 1. Clearly x (cid:48) + y (cid:48) = 0 and x (cid:48) − y (cid:48) = 0 cannot be both true, otherwise y (cid:48) = 0. Suppose one of them is true, then x (cid:48) = ± y (cid:48) . And at least one of ( x (cid:48) + y (cid:48) ) = 1 or ( x (cid:48) − y (cid:48) ) = 1holds. So either | x (cid:48) + y (cid:48) | = 1 or | x (cid:48) − y (cid:48) | = 1. Substituting x (cid:48) = ± y (cid:48) we reach a contradiction to | y (cid:48) | > 1. So neither x (cid:48) + y (cid:48) = 0 nor x (cid:48) − y (cid:48) = 0 holds. Then ( x (cid:48) + y (cid:48) ) = 1 and ( x (cid:48) − y (cid:48) ) = 1.Subtracting them, we get x (cid:48) y (cid:48) = 0, and since y (cid:48) (cid:54) = 0, we get x (cid:48) = 0. But then this contradicts | y (cid:48) | > x (cid:48) + y (cid:48) ) = 1. Therefore, y (cid:54) = 0. Then, y = ± 1. Thus, y = f ( δ(cid:126) n − m ) = ± δ with wt( δ ) = 2 . If 2 m = 2, then the induction is finished. Otherwise, 2 m > 2. Inductively for some 2 k (cid:62) f ( θ(cid:126) n − m ) = ± θ ∈ Z m with wt( θ ) (cid:54) k < m . Let δ be such thatwt( δ ) = 2 k + 2 (cid:54) m and we show that f ( δ(cid:126) n − m ) = ± 1. Since wt( δ ) = 2 k + 2 (cid:62) 4, we canfind four bits of δ such that the values of δ are 1 on these four bits. Without loss of generality, weassume that they are the first four bits, i.e. δ = 1111 δ (cid:48) where δ (cid:48) ∈ Z m − . Consider the following46istance-2 square (cid:20) x yz w (cid:21) = (cid:20) f (0000 δ (cid:48) (cid:126) n − m ) f (0011 δ (cid:48) (cid:126) n − m ) f (1100 δ (cid:48) (cid:126) n − m ) f (1111 δ (cid:48) (cid:126) n − m ) (cid:21) . Clearly, three entries in this distance-2 square have input strings of weight at most 2 k , namelywt(0000 δ (cid:48) (cid:126) n − m ) = 2 k − 2, and wt(0011 δ (cid:48) (cid:126) n − m ) = wt(1100 δ (cid:48) (cid:126) n − m ) = 2 k . By our inductionhypothesis, x, y, z ∈ { , − } . Then, since the distance-2 square [ x yz w ] is of type I, we have w = f ( δ(cid:126) n − m ) = ± 1. The induction is complete. This finishes the proof of Case 1. Case 2. There is a type II or type III distance-2 square in f .This is the case where signatures g and g (cid:48) appear. We handle this case in two steps. Step 1. We show that after flipping variables of f , S ( f ) = E n , and after normalization f ( α ) = ± ± α ∈ S ( f ). Let S ( f ) = { α ∈ S ( f ) | f ( α ) = ± } . We also show that | S ( f ) | = 2 n − = | S ( f ) | , and for any distinct α, β ∈ S ( f ), wt( α ⊕ β ) (cid:62) f . We show that the onlypossible Type II distance-2 square in f has the canonical form [ ]. Suppose that a distance-2square of Type II appears in f . By flipping and negating variables, we modify f such that thisdistance-2 square is in its canonical form [ aa ] ( a > . Also, by flipping variables and renamingvariables, we may assume that this distance-2 square appears on inputs α , β , γ and δ where (cid:20) f ( α ) f ( β ) f ( γ ) f ( δ ) (cid:21) = (cid:20) f (0000 (cid:126) n − ) f (0011 (cid:126) n − ) f (1100 (cid:126) n − ) f (1111 (cid:126) n − ) (cid:21) = (cid:20) aa (cid:21) . Then, we consider the entries of (cid:101) f on inputs α , β , γ and δ . We have (cid:34) (cid:101) f ( α ) (cid:101) f ( β ) (cid:101) f ( γ ) (cid:101) f ( δ ) (cid:35) = (cid:20) f ( α ) + f ( γ ) f ( β ) + f ( δ ) f ( α ) − f ( γ ) f ( β ) − f ( δ ) (cid:21) = (cid:20) a a − a a − (cid:21) . Since a > 1, clearly 1 + a (cid:54) = 0, 1 − a (cid:54) = 0 and | a | > | − a | . Since f has parity and f ( (cid:126) n ) = 1, f has even parity. By Lemma 7.6(2), S ( (cid:101) f ) = E n − and | a | = 2 | − a | . Since a > 1, we have1 + a = 2( a − a = 3. Thus, the only possible Type II distance-2 square in f has thecanonical form [ ].Under the assumption that a Type II distance-2 square appears in f and (cid:104) f ( α ) f ( β ) f ( γ ) f ( δ ) (cid:105) = [ ], wehave (cid:104) (cid:101) f ( α ) (cid:101) f ( β ) (cid:101) f ( γ ) (cid:101) f ( δ ) (cid:105) = (cid:2) − (cid:3) . As showed above, by Lemma 7.6(2), S ( (cid:101) f ) = E n − and n , n = 2or 4. We first prove Claim 1. S ( f ) = S ( f ) = E n − , f ( θ ) , f ( θ ) = ± or ± for all θ ∈ E n − , and | S ( f ) | + | S ( f ) | = 2 n − . Remember that (cid:101) f , (cid:101) f ∈ A . For any of them, its nonzero entries have the same norm. Since (cid:101) f ( α ) = (cid:101) f (00 (cid:126) n − ) = 1 + 3 = 4 and S ( (cid:101) f ) ⊆ E n − , for every θ ∈ E n − , (cid:101) f (00 θ ) = ± (cid:101) f ( γ ) = (cid:101) f (11 (cid:126) n − ) = 1 − − 2, and S ( (cid:101) f ) = E n − , for every θ ∈ E n − , (cid:101) f (11 θ ) = ± f (00 θ ) = (cid:101) f (00 θ ) + (cid:101) f (11 θ )2 = ( ± 4) + ( ± ± . f (00 θ ) = ± ± θ ∈ E n − . Also, f (11 θ ) = (cid:101) f (00 θ ) − (cid:101) f (11 θ )2 = ( ± − ( ± − ( ± . Thus, f (11 θ ) = ± ± θ ∈ E n − . Additionally note that, for any θ ∈ E n − if (cid:101) f (00 θ ) = ± 4, then of the two values f (00 θ ) and f (11 θ ), exactly one is ± ± (cid:101) f (00 θ ) = 0, then f (00 θ ) = ± f (11 θ ) = ± 1. Since | (cid:101) f | = 4 · | S ( (cid:101) f ) | = | (cid:101) f | = 2 · | S ( (cid:101) f ) | = 2 · | E n − | , we have | S ( (cid:101) f ) | = | E n − | / n − . Thus, there are exactly 2 n − entries of (cid:101) f having value ± 4, which give arise to exactly 2 n − many entries of value ± f and f .Claim 1 has been proved.Next, we prove Claim 2. S ( f ) = S ( f ) = O n − , f ( θ ) , f ( θ ) = ± or ± for all θ ∈ O n − ,and | S ( f ) | + | S ( f ) | = 2 n − . We have (cid:101) f ( (cid:126) n ) = 4. We have n = 4 and n = 2. Also recall that we have showed that n , n = 2 or 4, by Lemma 7.6(2). There are three cases. • n = n = 2. Since n = n = 2 and | (cid:101) f | = n · | S ( (cid:101) f ) | = n · | S ( (cid:101) f ) | = | (cid:101) f | , we have | S ( (cid:101) f ) | = | S ( (cid:101) f ) | = | E n − | = 2 n − . Since (cid:101) f has even parity, S ( (cid:101) f ) ⊆ O n − . As | O n − | = 2 n − , we get S ( (cid:101) f ) = O n − .Similarly, we can show that S ( (cid:101) f ) = O n − . Let ζ = 0110 (cid:126) n − and η = 1010 (cid:126) n − . Then, (cid:101) f ( ζ ) = ± (cid:101) f ( η ) = ± . Note that f ( ζ ) = (cid:101) f ( ζ ) + (cid:101) f ( η )2 and f ( η ) = (cid:101) f ( ζ ) − (cid:101) f ( η )2 . If (cid:101) f ( ζ ) = (cid:101) f ( η ), then f ( ζ ) = ± f ( η ) = 0. If (cid:101) f ( ζ ) = − (cid:101) f ( η ), then f ( ζ ) = 0 and f ( η ) = ± f ( ζ ) = ± 2. Let ξ = 1001 (cid:126) n − . Consider the followingdistance-2 square. (cid:20) f ( α ) f ( ζ ) f ( ξ ) f ( δ ) (cid:21) = (cid:20) f (0000 (cid:126) n − ) f (0110 (cid:126) n − ) f (1001 (cid:126) n − ) f (1111 (cid:126) n − ) (cid:21) = (cid:20) ± ∗ (cid:21) . Clearly, it is not of type I nor type III. Also, it is not of type II with the canonical form [ ].Contradiction. If f ( η ) = ± 2, then similarly by considering the distance-2 square (cid:104) f ( α ) f ( η ) f ( τ ) f ( δ ) (cid:105) where τ = 0101 (cid:126) n − , we get a contradiction.48 n = n = 4. We still consider f ( ζ ) = (cid:101) f ( ζ ) + (cid:101) f ( η )2 and f ( η ) = (cid:101) f ( ζ ) − (cid:101) f ( η )2 , where ζ = 0110 (cid:126) n − and η = 1010 (cid:126) n − . Then, as ζ has leading bits 01 and η has leading bits 10, f ( ζ ) = ( ± 4) + ( ± , ( ± 4) + 02 or 0 + 02 and f ( η ) = ( ± − ( ± , ± ( ± − 02 or 0 − . Thus, f ( ζ ) , f ( η ) = ± , ± f ( ζ ) or f ( η ) = ± , ± (cid:104) f ( α ) f ( ζ ) f ( ξ ) f ( δ ) (cid:105) or (cid:104) f ( α ) f ( η ) f ( τ ) f ( δ ) (cid:105) , we still get a contradiction. Thus we have f ( ζ ) = f ( η ) = 0.Then, consider the signature H f , denoted by (cid:101) f (cid:48) . Since f has even parity, f satisfies and f ∈ (cid:82) B A , (cid:101) f (cid:48) has even parity, (cid:101) f (cid:48) , (cid:101) f (cid:48) , (cid:101) f (cid:48) , (cid:101) f (cid:48) ∈ A . Let n (cid:48) , n (cid:48) , n (cid:48) and n (cid:48) denote the norms of nonzero entries in (cid:101) f (cid:48) , (cid:101) f (cid:48) , (cid:101) f (cid:48) , and (cid:101) f (cid:48) respectively. Notice that (cid:101) f (cid:48) ( α ) = (cid:101) f (cid:48) ( (cid:126) n ) = f (0000 (cid:126) n − ) + f (0110 (cid:126) n − ) = f ( α ) + f ( ζ ) = 1 + 0 = 1 . Thus, n (cid:48) = 1. Also, notice that (cid:101) f (cid:48) ( γ ) = (cid:101) f (cid:48) (1100 (cid:126) n − ) = f (1010 (cid:126) n − ) − f (1100 (cid:126) n − ) = f ( η ) − f ( γ ) = 0 − − . Thus, n (cid:48) = 3. But by Lemma 7.6(1), n (cid:48) = √ k n (cid:48) for some k ∈ Z . However, clearly,3 (cid:54) = √ k for any k ∈ Z . Contradiction. • Therefore exactly one of n and n is 2 and the other is 4. Let ( a, b ) = (0 , 1) or (1 , 0) besuch that n ab = 2. Since n = 2 and | S ( (cid:101) f ) | = | E n − | =2 n − , we have | S ( (cid:101) f ab ) | = 2 n − .Since (cid:101) f has even parity, (cid:101) f ab has odd parity, thus S ( (cid:101) f ab ) = O n − . Then, similar to the proofof f and f , we can show that for every θ ∈ O n − , f ( θ ) , f ( θ ) = ± ± 1. Also,among f and f , exactly 2 n − many entries are ± S ( f ) = E n , f ( α ) = ± ± α ∈ S ( f ), and | S ( f ) | = 2 n − = | S ( f ) | . Also remember that by our assumption, f ( (cid:126) n ) = 1.Now, we show that for any distinct α, β ∈ S ( f ), wt( α ⊕ β ) (cid:62) 4. For a contradiction, supposethat α, β ∈ S ( f ) and wt( α ⊕ β ) = 2, and they differ at bits i and j . By renaming variables,without loss of generality, we may assume that { i, j } = { , } . This renaming does not changethe value of f ( (cid:126) n ) = 1. Since f (11 (cid:126) n − ) = ± ± 3, of the values f (00 (cid:126) n − ) + f (11 (cid:126) n − )and f (00 (cid:126) n − ) − f (11 (cid:126) n − ), which are respectively an entry of (cid:101) f and an entry of (cid:101) f , at leastone has norm 2. Thus, among n and n , at least one is 2. Since f ( α ) = ± f ( β ) = ± f ( α ) + f ( β ) and f ( α ) − f ( β ), exactly one has norm 6 and the other has norm 0. Clearly, f ( α ) + f ( β ) and f ( α ) − f ( β ) are entries of (cid:101) f since α and β differ at bits 1 and 2. Thus, among n , n , n and n , one has norm 6. By Lemma 7.6(1), 2 = √ k · k ∈ N . Contradiction.This proves that for any distinct α, β ∈ S ( f ), wt( α ⊕ β ) (cid:62) f .Finally, within Step 1 of Case 2, we consider the case that a type III distance-2 square appearsin f . By flipping and negating variables, we modify f such that this distance-2 square is in its49anonical form (cid:2) − (cid:3) . Also, by flipping variables and renaming variables, still we may assume thatthis distance-2 square appears on inputs α , β , γ and δ where (cid:20) f ( α ) f ( β ) f ( γ ) f ( δ ) (cid:21) = (cid:20) f (0000 (cid:126) n − ) f (0011 (cid:126) n − ) f (1100 (cid:126) n − ) f (1111 (cid:126) n − ) (cid:21) = (cid:20) − (cid:21) . Then, we consider the entries of (cid:101) f on inputs α , β , γ and δ . We have (cid:34) (cid:101) f ( α ) (cid:101) f ( β ) (cid:101) f ( γ ) (cid:101) f ( δ ) (cid:35) = (cid:20) f ( α ) + f ( γ ) f ( β ) + f ( δ ) f ( α ) − f ( γ ) f ( β ) − f ( δ ) (cid:21) = (cid:20) − (cid:21) . Then exactly in the same way as the above proof when (cid:104) (cid:101) f ( α ) (cid:101) f ( β ) (cid:101) f ( γ ) (cid:101) f ( δ ) (cid:105) = (cid:2) − (cid:3) , we can show thatthe same result holds. Thus, S ( f ) = E n , f ( α ) = ± ± α ∈ S ( f ), | S ( f ) | = 2 n − = | S ( f ) | , and for any distinct α, β ∈ S ( f ) with wt( α ⊕ β ) = 2, α and β cannot be both in S ( f ).This finishes the proof of Step 1 of Case 2. Step 2. Now we show that either g or g (cid:48) is realizable from f . We will show that they areboth irreducible and do not satisfy , which gives G n with vertex set E n , and there is an edge between α and β if wt( α ⊕ β ) = 2.I.e., we view every α ∈ E n as a vertex, and the edges are distance 2 neighbors in Hamming distance.Then, S ( f ) is an independent set of G n . Remember that 2 n (cid:62) n (cid:62) 10, then by Lemma 7.4, | S ( f ) | < | S ( f ) | . Contradiction. Thus, 2 n = 8. After renamingand flipping variables, we may assume that S ( f ) = I = S ( f ). For brevity of notation, let S = E and T = S ( f ). We can pick ( x , . . . , x ) as a set of free variables of S = E . Then,there exists a multilinear polynomial F ( x , . . . , x ) ∈ Z [ x , . . . , x ], and a multilinear polynomial G ( x , . . . , x ) ∈ Z [ x , . . . , x ] that is viewed as a representative for its image in the quotient algebra Z [ x , . . . , x ] / ( P , P , P , P ) where P , P , P , P are the four linear polynomials in (7.4) such that T is decided by P = P = P = P = 0, such that f = χ S ( − F ( x ,...,x ) + 4 χ T ( − G ( x ,...,x ) . We note that such multilinear polynomials F ( x , . . . , x ) and G ( x , . . . , x ) exist: For any pointin S \ T we can choose a unique value s ∈ Z which represents the ± f as ( − s , andfor any point in T ⊆ S we can choose unique values t ∈ Z and s (cid:48) ∈ Z such that ( − s (cid:48) + 4( − t represents the ± f .For { i, j } ⊆ [7] = { , . . . , } , remember that F abij ∈ Z [ { x , . . . , x }\{ x i , x j } ] is the functionobtained by setting ( x i , x j ) = ( a, b ) in F . Similarly, we can define G abij with respect to P = P = P = P = 0 (any assignment of ( x i , x j ) = ( a, b ) is consistent with P = P = P = P = 0 whichdefines T ). We make the following claim about F abij . Claim 3. For all { i, j } ⊆ [7] , F ij + F ij ≡ or , and also F ij + F ij ≡ or . We first show how this claim will let us realize g or g (cid:48) , and lead to d ( F ) (cid:54) If d ( F ) (cid:54) 1, then F is an affine linear combination of variables x , . . . , x , i.e., F = λ + (cid:80) i =1 λ i x i where λ i ∈ Z for 0 (cid:54) i (cid:54) 7. Notice that if we negate the variable x i of f , wewill get a signature f (cid:48) ( x , . . . , x ) = ( − x i f ( x , . . . , x ) . For every x i appearing in F (i.e., λ i = 1), we negate the variable x i of f . Also, if λ = 1, then we normalize f by a scalar − f (cid:48) = χ S · χ T ( − G (cid:48) ( x ,...,x ) . This will not change the support of f and also norms of entries of f . Thus, f (cid:48) ( α ) = ± ± α ∈ S ( f (cid:48) ) = E . Then, for every α ∈ T , f (cid:48) ( α ) = 1 + 4( − G (cid:48) ( α ) = ± 3, which impliesthat ( − G (cid:48) ( α ) = − f (cid:48) ( α ) = − 3, because 1 + 4 = 5 cannot be an entry of f (cid:48) . Therefore, f (cid:48) = χ S − χ T = g . Thus, g is realizable from f .By merging variables x and x of g using = , we can get a 6-ary signature h . We renamevariables x , x , x to x , x , x and variables x , x , x to x , x , x (The choice of merging x and x is just for a simple renaming of variables). Then after normalization by a scalar1 / h has the following signature matrix M , ( h ) = A = − − − − − − − − . Consider the inner product (cid:104) h , h (cid:105) . One can check that (cid:104) h , h (cid:105) = (cid:88) (cid:54) i,j (cid:54) A i,j · A i +4 ,j +4 = 8 (cid:54) = 0 . (This is the sum of pairwise products of every entry in the upper left 4 × A with the corresponding entry of the lower right 4 × A .) In fact, notice that h ( α ) = h ( α ) = h ( α ). By considering the representative matrix M r ( h ) of h (see Table 4), wehave M r ( h ) = − − − − . Then, (cid:104) h , h (cid:105) = 2(perm( M r ( h ) [1 , ) + perm( M r ( h ) [3 , )) = 2(2 + 2) = 8 (cid:54) = 0 . Also, since S ( h ) = E , it is easy to see that h is irreducible. Since h does not satisfy , we get • If d ( F ) = 2, then by Lemma 7.8, for all { i, j } ⊆ [7], x i x j appears in F . Then, F = (cid:80) (cid:54) i 1, we get a signature f (cid:48) = χ S ( − (cid:80) (cid:54) i 3, of norm 3 for f (cid:48) . The other choice would give 1 + 4 = 5 to be an entry of f (cid:48) , acontradiction. Therefore, f (cid:48) ( α ) = q − χ T = g (cid:48) . Thus, g (cid:48) is realizable from f .By merging variables x and x of g (cid:48) using = − , we can get a 6-ary signature h (cid:48) . Afterrenaming variables (same as we did for h ) and normalization by a scalar − / 2, we have M , ( h (cid:48) ) = B = − − 10 1 − − 11 0 0 1 0 − − − − 11 0 0 − − − − − − − − . Consider the inner product (cid:104) h (cid:48) , h (cid:48) (cid:105) . One can check that (cid:104) h (cid:48) , h (cid:48) (cid:105) = (cid:88) (cid:54) i,j (cid:54) B i,j · B i +4 ,j +4 = − (cid:54) = 0 . Also, since S ( h (cid:48) ) = E , it is easy to see that h (cid:48) is irreducible. Since h (cid:48) does not satisfy , we get { i, j } ⊆ [7], F ij + F ij ≡ F ij + F ij ≡ { i, j } = { , } . The proof for arbitrary { i, j } is the sameby replacing { , } by { i, j } . Since f ∈ (cid:82) B A , (cid:101) f , (cid:101) f , (cid:101) f , (cid:101) f ∈ A . Remember all nonzero entriesin (cid:101) f ab have the same norm, denoted by n ab . We first show that between (cid:101) f and (cid:101) f , exactly onehas support E n − and its nonzero entries have norm 2 and the other has nonzero entries of norm4, and between (cid:101) f and (cid:101) f , exactly one has support O n − and its nonzero entries have norm 2and the other has nonzero entries of norm 4. (This is not what we have proved in Step 1 where { , } is a pair of particularly chosen indices. Here { , } means an arbitrary pair { i, j } .)Consider f ( (cid:126) ) and f ( (cid:126) ). By Step 1 of Case 2 and Lemma 7.4, we may assume that S ( f ) = S ( f ) (after flipping and renaming variables). We have 00 (cid:126) ∈ S ( f ) and 11 (cid:126) / ∈ S ( f ).Thus, f ( (cid:126) ) = ± f ( (cid:126) ) = ± 1. (This is true when replacing { , } by an arbitrary pair ofindices { i, j } .) Thus, between (cid:101) f ( (cid:126) ) = f ( (cid:126) ) + f ( (cid:126) ) and (cid:101) f ( (cid:126) ) = f ( (cid:126) ) − f ( (cid:126) ) , one has norm 2 and the other has norm 4. They are both nonzero. Then, between n and n ,one is 2 and the other is 4. By Lemma 7.6(2), between (cid:101) f and (cid:101) f , the one whose nonzero entries52ave norm 2 has support E , and moreover n and n = 2 or 4. Since there exists ( a, b ) = (0 , , 1) such that | (cid:101) f ab | = n ab · | S ( (cid:101) f ab ) | = 2 · | E | , for (cid:101) f cd where ( c, d ) = (0 , 1) or (1 , n cd = 2, then | S ( (cid:101) f cd ) | = | E | = | O | . Since (cid:101) f cd has oddparity, S ( (cid:101) f cd ) ⊆ O . Thus, | S ( (cid:101) f cd ) | = 2 n − implies that S ( (cid:101) f cd ) = O . • If n = n = 2, then S ( (cid:101) f ) = S ( (cid:101) f ) = O . For an arbitrary θ ∈ O , f (01 θ ) = (cid:101) f (01 θ ) + (cid:101) f (10 θ )2 = ( ± 2) + ( ± f (10 θ ) = (cid:101) f (01 θ ) − (cid:101) f (10 θ )2 = ( ± − ( ± . Thus, between f (01 θ ) and f (10 θ ), exactly one has norm 2 and the other has norm 0. Thisgives a contradiction since every nonzero entry of f has norm 1 or 3. • If n = n = 4, then still consider f (01 θ ) and f (10 θ ) for an arbitrary θ ∈ O . We knowthat f (01 θ ) , f (10 θ ) = ± , ± f (01 θ ) = 0 or f (10 θ ) = 0 cannot occursince S ( f ) = E n and clearly, 01 θ, θ ∈ E n . Thus, f (01 θ ) , f (10 θ ) = ± , ± 2. Still, we get acontradiction since every nonzero entry of f has norm 1 or 3. • Thus, between n and n , one is 2 and the other is 4.Then, between (cid:101) f and (cid:101) f , exactly one has support O and its nonzero entries have norm 2, andthe other has nonzero entries of norm 4.Now, we show that F + F ≡ (cid:101) f and (cid:101) f , (cid:101) f = f − f is the signature whose support is E and nonzero entries have norm 2; thecase where it is (cid:101) f will be addressed shortly. Let S be the subspace in Z obtained by setting x = x = 0 in S = S ( f ) = E , and S be the subspace in Z obtained by setting x = x = 1.Similarly, we can define T and T , replacing S in the definition by T = S ( f ) = I . Clearly, S = S = { ( x , . . . , x ) ∈ Z | x + · · · x = 0 } = E . Also, one can check that T is disjoint with T . Then f = χ S ( − F ( x ,...,x ) + 4 χ T ( − G ( x ,...,x ) , and f = χ S ( − F ( x ,...,x ) + 4 χ T ( − G ( x ,...,x ) . Thus, (cid:101) f = χ E (( − F ( x ,...,x ) − ( − F ( x ,...,x ) ) + 4 χ T ( − G ( x ,...,x ) − χ T ( − G ( x ,...,x ) . Since S ( (cid:101) f ) = E and n = 2, (cid:101) f ( θ ) = ± θ ∈ E . If θ / ∈ T ∪ T , then (cid:101) f ( θ ) = ( − F ( θ ) − ( − F ( θ ) = ± . If θ ∈ T ∪ T , then it belongs to exactly one of T or T , (cid:101) f ( θ ) = ( − F ( θ ) − ( − F ( θ ) + 4 a = ± , where a = ± 1. In this case, the sum of the first two terms is still ( − F ( θ ) − ( − F ( θ ) = ± ± − ( ± 1) is 0 and then we would have 4 a = ± 2, acontradiction. Thus, for every ( x , . . . , x ) ∈ Z which decides every ( x , . . . , x ) ∈ E by x = x + · · · + x , ( − F ( x ,...,x ) − ( − F ( x ,...,x ) = ± . − F ( x ,...,x ) = − ( − F ( x ,...,x ) . Thus, ( − F ( x ,...,x )+ F ( x ,...,x ) = − . Then, F + F ≡ (cid:101) f and (cid:101) f ) it is (cid:101) f = f + f the signature whosesupport is E and nonzero entries have norm 2. Then similarly for every ( x , . . . , x ) ∈ Z , whichdetermines every ( x , . . . , x ) ∈ E ,( − F ( x ,...,x ) + ( − F ( x ,...,x ) = ± . This implies that ( − F ( x ,...,x ) = ( − F ( x ,...,x ) . Thus, ( − F ( x ,...,x )+ F ( x ,...,x ) = 1Then, F + F ≡ . We have proved that, F + F ≡ (cid:101) f and (cid:101) f . One of them is a signature whose support is O n − and nonzeroentries have norm 2. Then similarly, for every ( x , . . . , x ) ∈ Z which decides every ( x , . . . , x ) ∈ O by x = 1 + x + · · · + x ,( − F ( x ,...,x ) + ( − F ( x ,...,x ) = ± , or ( − F ( x ,...,x ) − ( − F ( x ,...,x ) = ± . Then, F + F ≡ F + F ≡ 1. The above proof holds for all { i, j } ⊆ [7]. Thus, for all { i, j } ⊆ [7], F ij + F ij ≡ F ij + F ij ≡ Remark: The above proof does not require F to be non- B hard. Then, by further assuming that nonzero entries of f have the same norm, we show that f has affinesupport or we can get the B hard set F (Lemma 7.16). Here, we do require F to be non- B hard.We first give one more result about (cid:101) f . Remember that if f ∈ (cid:82) B A , then (cid:101) f , (cid:101) f , (cid:101) f , (cid:101) f ∈ A , and n ab denotes the norm of nonzero entries of (cid:101) f ab . Let (cid:101) B = (cid:110) (cid:102) = +2 , (cid:102) = − , (cid:102) (cid:54) = +2 , (cid:102) (cid:54) = − (cid:111) where (cid:102) = +2 = (2 , , , (cid:102) = − = (0 , , , (cid:102) (cid:54) = +2 = (0 , , , 0) and (cid:102) (cid:54) = − = (0 , , , (cid:101) B areobtained by performing the H gadget construction on binary signatures in B . Lemma 7.10. Let f be an irreducible signature of arity n (cid:62) with the following properties.1. f has even parity, f satisfies , and f ∈ (cid:82) B A ; . for all { i, j } disjoint with { , } and every b ∈ B , either M ( m ( ∂ bij f )) = λ bij I for some real λ bij (cid:54) = 0 , or there exists a nonzero binary signature g bij ∈ B such that g bij ( x , x ) | ∂ bij f .If S ( (cid:101) f ) = S ( (cid:101) f ) , n > n > , then S ( (cid:101) f ) = O n − .Proof. We first analyze the second property of f , i.e., the property about ∂ bij f . • If M ( m ( ∂ bij f )) = λ bij I , by Lemma 7.5, then M ( m ( (cid:93) ∂ bij f )) = 2 λ bij I . Since { i, j } is disjointwith { , } , the H gadget on variables x and x commutes with the merging gadget onvariables x i and x j . Thus, (cid:103) ∂ bij f = ∂ bij (cid:101) f . Let ( ∂ bij (cid:101) f ) ab be the signature obtained by settingvariables x and x of ∂ bij (cid:101) f to a and b , and ∂ bij ( (cid:101) f ab ) be the signature obtained by mergingvariables x i and x j of (cid:101) f ab . Again, since { , } and { i, j } are disjoint, ( ∂ bij (cid:101) f ) ab = ∂ bij ( (cid:101) f ab ). Wedenote them by ∂ bij (cid:101) f ab . Then, since M ( m ( (cid:93) ∂ bij f )) = M ( m ( ∂ bij (cid:101) f )) = 2 λ bij I , | ∂ bij (cid:101) f | = | ∂ bij (cid:101) f | = | ∂ bij (cid:101) f | = | ∂ bij (cid:101) f | = 2 λ bij (cid:54) = 0 . • If g bij ( x , x ) | ∂ bij f , i.e, ∂ bij f = g bij ( x , x ) ⊗ h , then (cid:103) ∂ bij f = ∂ bij (cid:101) f = (cid:102) g bij ( x , x ) ⊗ h . Since g bij ∈ B , (cid:102) g bij ∈ (cid:101) B . By the form of signatures in (cid:101) B , among ∂ bij (cid:101) f , ∂ bij (cid:101) f , ∂ bij (cid:101) f and ∂ bij (cid:101) f , atmost one is a nonzero signature.Combining the above two cases we have that, among ∂ bij (cid:101) f , ∂ bij (cid:101) f , ∂ bij (cid:101) f and ∂ bij (cid:101) f , if at leasttwo of them are nonzero signatures then they are all nonzero signatures.Now, we show that S ( (cid:101) f ) = O n − . Since f has even parity, (cid:101) f also has even parity. Then, (cid:101) f has odd parity, i.e., S ( (cid:101) f ) ⊆ O n − . For a contradiction, suppose that S ( (cid:101) f ) (cid:40) O n − . Since n > S ( (cid:101) f ) (cid:54) = ∅ . Then, we can pick a pair of inputs α, β ∈ O n − with wt( α ⊕ β ) = 2 suchthat α ∈ S ( (cid:101) f ) and β / ∈ S ( (cid:101) f ) . Also, since S ( (cid:101) f ) = S ( (cid:101) f ), α ∈ S ( (cid:101) f ) and β / ∈ S ( (cid:101) f ) . Thus, | (cid:101) f ( α ) | = n and | (cid:101) f ( β ) | = 0, and | (cid:101) f ( α ) | = n and | (cid:101) f ( β ) | = 0 . Suppose that α and β differ in bits i and j . Clearly, { i, j } is disjoint with { , } . Depending whether α i = α j or α i (cid:54) = α j ,we connect variables x i and x j of (cid:101) f using = +2 or (cid:54) = +2 . We get signatures ∂ + ij (cid:101) f or ∂ (cid:98) + ij (cid:101) f respectively.We consider the case that α i = α j ; in this case { α i α j , β i β j } = { , } . For the case that α i (cid:54) = α j ,the analysis is the same by replacing ∂ + ij (cid:101) f with ∂ (cid:98) + ij (cid:101) f .Consider ∂ + ij (cid:101) f . Then, because { α i α j , β i β j } = { , } , (cid:101) f ( α ) + (cid:101) f ( β ) and (cid:101) f ( α ) + (cid:101) f ( β ) areentries of ∂ + ij (cid:101) f ; more precisely, they are entries of ∂ + ij (cid:101) f and ∂ + ij (cid:101) f respectively. Since (cid:101) f ( β ) = (cid:101) f ( β ) = 0, we have | (cid:101) f ( α ) + (cid:101) f ( β ) | = | (cid:101) f ( α ) | = n (cid:54) = 0 , and | (cid:101) f ( α ) + (cid:101) f ( β ) | = | (cid:101) f ( α ) | = n (cid:54) = 0 . Thus, ∂ + ij (cid:101) f has a nonzero entry with norm n , and then ∂ + ij (cid:101) f (cid:54)≡ 0. Also, we have ∂ + ij (cid:101) f (cid:54)≡ ∂ + ij (cid:101) f , ∂ + ij (cid:101) f , ∂ + ij (cid:101) f and ∂ + ij (cid:101) f are nonzero, it follows that all of themare nonzero signatures.Then ∂ + ij (cid:101) f (cid:54)≡ 0. Let ∂ + ij (cid:101) f ( γ ) be a nonzero entry of ∂ + ij (cid:101) f . Then, ∂ + ij (cid:101) f ( γ ) = (cid:101) f ij ( γ ) + (cid:101) f ij ( γ ) (cid:54) = 0. Clearly, (cid:101) f ij ( γ ) and (cid:101) f ij ( γ ) are entries of (cid:101) f , and they have norm n or 0.Thus, ∂ + ij (cid:101) f ( γ ) has norm 2 n or n . Also, ∂ + ij (cid:101) f ( γ ) is an entry of ∂ + ij (cid:101) f on the input 00 γ . Thus, For the case that α i (cid:54) = α j , ∂ + ij (cid:101) f ( γ ) = (cid:101) f ij ( γ ) + (cid:101) f ij ( γ ) will be replced by ∂ (cid:98) + ij (cid:101) f ( γ ) = (cid:101) f ij ( γ ) + (cid:101) f ij ( γ ). + ij (cid:101) f has a nonzero entry with norm 2 n or n . Since n > n , both 2 n and n are not equalto n . Thus, ∂ + ij (cid:101) f has two nonzero entries with different norms. Such a signature is not in A .However, since f ∈ (cid:82) B A , by Lemma 7.5, ∂ + ij (cid:101) f ∈ A . Contradiction. Thus, S ( (cid:101) f ) = O n − .We also give a result about the edge partition of complete graphs into two complete tripartitesubgraphs. This result should also be of independent interest. We say a graph G = ( V, E ) istripartite if V = V (cid:116) V (cid:116) V and every e ∈ E is between distinct V i and V j . Here (cid:116) denotes disjointunion. The parts V i are allowed to be empty. It is a complete tripartite graph if every pair betweendistinct V i and V j is an edge. Definition 7.11. Let K n be the complete graph on n vertices. We say K n has a tripartite 2-partition if there exist complete tripartite subgraphs T and T such that { E ( T ) , E ( T ) } is a par-tition of E ( K n ) , i.e., E ( K n ) = E ( T ) (cid:116) E ( T ) . We say T and T are witnesses of a tripartite2-partition of K n . Lemma 7.12. K n has a tripartite 2-partition iff n (cid:54) . For n = 5 , up to an automorphism of K ,there is a unique tripartite 2-partition where T is a triangle on { v , v , v } and T is the completetripartite graph with parts { v , v , v } , { v } and { v } .Proof. Let T be a complete tripartite graph. Let G , be the union of K and an isolated vertex.We first prove the following two claims. Claim 1. G , is not an induced subgraph of T . For a contradiction, suppose G , = ( V, E ) is an induced subgraph of T , where V = { v , v , v } ,and E = { ( v , v ) } . Then, v and v belong to distinct parts of T . Since ( v , v ) , ( v , v ) / ∈ E ( T ), v and v belong to the same part of T , and so are v and v . Thus, v and v belong to the samepart of T . This contradiction proves Claim 1. Claim 2. K is not an induced subgraph of T . For a contradiction, suppose K on V = { v , v , v , v } is an induced subgraph of T . Then, forany two distinct vertices v i , v j ∈ V , the edge ( v i , v j ) ∈ K shows that v i and v j belong to distinctparts in T . But T has at most three distinct nonempty parts. This contradiction proves Claim 2.Now, we prove this lemma. The cases n = 1 , , n = 4, we have the followingtwo tripartite 2-partitions of K , with V ( T ) = { v }(cid:116){ v }(cid:116){ v } and V ( T ) = { v , v , v }(cid:116){ v }(cid:116)∅ ,or alternatively with V ( T (cid:48) ) = { v } (cid:116) { v } (cid:116) ∅ and V ( T (cid:48) ) = { v , v } (cid:116) { v } (cid:116) { x } .We consider n (cid:62) 5. Suppose K n has a tripartite 2-partition with complete tripartite subgraphs T = ( V , E ) and T = ( V , E ). We write ( A i , B i , C i ) for the three parts of T i , i = 1 , V = V ∪ V , as all vertices of V must appear in either T or T , for otherwise any edgeincident to v ∈ V \ ( V ∪ V ) is not in E ∪ E . If all parts of both T and T have size at most 1,then | E (cid:116) E | (cid:54) < | K | (cid:54) | K n | , a contradiction. So at least one part, say A , has size | A | (cid:62) a, a (cid:48) ∈ A . Then, ( a, a (cid:48) ) / ∈ E . Thus, ( a, a (cid:48) ) ∈ E and a, a (cid:48) ∈ V .We show that ( V \ A ) ∩ ( V \ A ) = ∅ . Otherwise, there exists v ∈ ( V \ A ) ∩ ( V \ A ) . Then,edges ( v, a ) , ( v, a (cid:48) ) ∈ E . Thus, among edges ( v, a ) , ( v, a (cid:48) ) and ( a, a (cid:48) ) of K n , ( a, a (cid:48) ) is the only onein T . Since v, a, a (cid:48) ∈ V , G , is an induced subgraph of T . A violation of Claim 1.If both V \ A and V \ A are nonempty, then an edge in K n between u ∈ V \ A and v ∈ V \ A is in neither E nor E , since u (cid:54)∈ V and v (cid:54)∈ V . This is a contradiction. If V \ A = ∅ , then56 = ∅ , and then all edges of K n belong to T , which violates Claim 2. So V \ A = ∅ . Since V = V ∪ V , V \ A = ∅ implies that V = V .Clearly V \ A = B (cid:116) C . If | B | (cid:62) 2, then there exists some { u, v } ⊆ B ⊆ V \ A , whichis disjoint from V . Thus { u, v } (cid:54)∈ E (cid:116) E , a contradiction. Hence | B | (cid:54) 1. Similarly | C | (cid:54) | A | (cid:62) 4, then there is a K inside A which must be an induced subgraph of T , aviolation of Claim 2. Thus | A | (cid:54) 3. It follows that n (cid:54) V = V = A (cid:116) B (cid:116) C . If n = 5,then | A | = 3 and | B | = | C | = 1. After relabeling vertices, we may assume that A = { v , v , v } , B = { v } and C = { v } . Then, we have A = { v } , B = { v } and C = { v } . This gives theunique tripartite 2-partition of K .We will apply Lemma 7.12 to multilinear Z -polynomials. Remember that we take the reductionof polynomials in Z [ x , . . . , x n ] modulo the ideal generated by { x i − x i | i ∈ [ n ] } replacing any F by its unique multilinear representative. Definition 7.13. Let F ( x , . . . , x n ) ∈ Z [ x , . . . , x n ] be a complete quadratic polynomial. We say F has a twice-linear 2-partition if there exist L , L , L , L ∈ Z [ x , . . . , x n ] where d ( L ) = d ( L ) = d ( L ) = d ( L ) (cid:54) such that F = L · L + L · L . Lemma 7.12 gives the following result about multilinear Z -polynomials. Lemma 7.14. Let F ( x , . . . , x n ) ∈ Z [ x , . . . , x n ] be a complete quadratic polynomial. For n (cid:62) , F does not have a twice-linear 2-partition. For n = 5 , F has a twice-linear 2-partition F = L · L + L · L iff (after renaming variables) the cross terms of L · L and L · L correspondto the unique tripartite 2-partition of K , and we have L · L = ( x + x + a )( x + x + b ) and L · L = ( x + x + x + x + c )( x + x + x + x + d ) for some a, b, c, d ∈ Z .Proof. We first analyze the quadratic terms that appear in a product of two linear polynomials.We use x i ∈ L to denote that a linear term x i appears in a linear polynomial L . Let L and L betwo linear polynomials.Let U = { x i | x i ∈ L , x i / ∈ L } , U = { x i | x i ∈ L , x i ∈ L } , and U = { x i | x i / ∈ L , x i ∈ L } . Then, L = (cid:88) x i ∈ U x i + (cid:88) x j ∈ U x j + a, and L = (cid:88) x j ∈ U x j + (cid:88) x k ∈ U x k + b for some a, b ∈ Z . The quadratic terms in L · L are from( (cid:88) x i ∈ U x i + (cid:88) x j ∈ U x j ) · ( (cid:88) x j ∈ U x j + (cid:88) x k ∈ U x k )which are enumerated in (cid:88) x i ∈ U ,x j ∈ U x i x j + (cid:88) x i ∈ U ,x k ∈ U x i x k + (cid:88) x j ∈ U ,x k ∈ U x j x k . Note that each term x i for i ∈ U is replaced by x i (thus no longer counted as a quadratic term)as we calculate modulo the ideal generated by { x i − x i | i ∈ [ n ] } , and every pairwise cross productterm x i x j for i, j ∈ U and i (cid:54) = j disappears since it appears exactly twice.If we view variables x , . . . , x n as n vertices and each quadratic term x i x j as an edge betweenvertices x i and x j , then the quadratic terms in L · L are the edges of a complete tripartite subgraph57 of K n (the parts of a tripartite graph could be empty) and V ( T ) = U (cid:116) U (cid:116) U . Therefore, L · L is one of the two terms of a twice-linear 2-partition of a complete quadratic polynomialover n variables iff T is one tripartite complete graph in a tripartite 2-partition of the completegraph K n . By Lemma 7.12, a tripartite 2-partition does not exist for K n when n (cid:62) 6. Thus, F does not have twice-linear partition when n (cid:62) 6. When n = 5, the tripartite 2-partition of K isunique up to relabeling. One tripartite complete graph of this tripartite 2-partition is a triangle,and we may assume it is on { x , x , x } . Then, we take L · L = ( x + x + a )( x + x + b ) forsome a, b ∈ Z , and L · L = ( x + x + x + x + c )( x + x + x + x + d ) for some c, d ∈ Z .Thus, a complete quadratic polynomial F ( x , . . . , x ) over 5 variables has a twice-linear 2-partitioniff (after renaming variables) F = L · L + L · L .Now, we are ready to make a further major step towards Theorem 7.19. We first give apreliminary result. Lemma 7.15. Let f be a n -ary signature, where n (cid:62) . If f ∈ (cid:82) B A and | f ( α ) | = 1 for all α ∈ S ( f ) , then for all { i, j } ⊆ [2 n ] , S ( f ij ) = S ( f ij ) or S ( f ij ) ∩ S ( f ij ) = ∅ , and S ( f ij ) = S ( f ij ) or S ( f ij ) ∩ S ( f ij ) = ∅ .Proof. We first prove that for all { i, j } ⊆ [2 n ], S ( f ij ) = S ( f ij ) or S ( f ij ) ∩ S ( f ij ) = ∅ .For a contradiction, suppose that there exist α, β ∈ Z n − such that α ∈ S ( f ij ) ∩ S ( f ij ) and β ∈ S ( f ij )∆ S ( f ij ), where ∆ denotes the symmetric difference between two sets. Considersignatures ∂ + ij f and ∂ − ij f . Then, f ij ( α ) + f ij ( α ) and f ij ( β ) + f ij ( β ) are entries of ∂ + ij f , and f ij ( α ) − f ij ( α ) and f ij ( β ) − f ij ( β ) are entries of ∂ − ij f . Since α ∈ S ( f ij ) ∩ S ( f ij ), f ij ( α ) = ± f ij ( α ) = ± 1. Then between f ij ( α ) + f ij ( α ) and f ij ( α ) − f ij ( α ), exactly one has norm 2 andthe other is 0. However, since β ∈ S ( f ij )∆ S ( f ij ), between f ij ( β ) and f ij ( β ), exactly one is 0and the other has norm 1. Thus, | f ij ( β ) + f ij ( β ) | = | f ij ( β ) − f ij ( β ) | = 1 . Then, between ∂ + ij f and ∂ − ij f , there is a signature that has an entry of norm 1 and an entry of norm 2. Clearly, such asignature is not in A . However, since f ∈ (cid:82) B A , we have ∂ + ij f , ∂ − ij f ∈ A . Contradiction.By considering signatures ∂ (cid:98) + ij f and ∂ (cid:98) − ij f , similarly we can show that S ( f ij ) = S ( f ij ) or S ( f ij ) ∩ S ( f ij ) = ∅ .The next lemma is a major step. Lemma 7.16. Suppose that F is non- B hard. Let f ∈ F be an irreducible n -ary (2 n (cid:62) signature with parity. Then, • Holant b ( F ) is • there is a signature g / ∈ A of arity k < n that is realizable from f and B , or • f has affine support.Proof. Again, we may assume that f satisfies and f ∈ (cid:82) B A . Also, by Lemma 7.9, wemay assume that f ( α ) = ± α ∈ S ( f ) after normalization.For any four distinct binary strings α, β, γ, δ ∈ Z n with α ⊕ β ⊕ γ = δ , we define a score T ( α, β, γ, δ ) = ( t , t , t ) which are the values of wt( α ⊕ β ) = wt( γ ⊕ δ ) , wt( α ⊕ γ ) = wt( β ⊕ δ ) andwt( α ⊕ δ ) = wt( β ⊕ γ ) ordered from the smallest to the largest. We order the scores lexicographically,i.e., we say T = ( t , t , t ) < T (cid:48) = ( t (cid:48) , t (cid:48) , t (cid:48) ) if t < t (cid:48) , or t < t (cid:48) when t = t (cid:48) , or t < t (cid:48) when t = t (cid:48) and t = t (cid:48) . Note that since α, β, γ, δ are distinct, the smallest value of the score T is582 , , α, β, γ, δ ) where α ⊕ β ⊕ γ = δ forms a non-affine quadrilateral of f if exactlythree of them are in S ( f ) and the fourth is not.For a contradiction, suppose that S ( f ) is not affine. Then, f has at least a non-affine quadrilat-eral. Among all non-affine quadrilaterals of f , we pick the one ( α, β, γ, δ ) with the minimum score T min = T ( α, β, γ, δ ) = ( t , t , t ) . Without loss of generality, we may assume that among α, β, γ and δ , δ is the one that is not in S ( f ).We first consider the case that (2 , , < T min . We prove that we can realize a non-affinesignature from f by merging. Depending on the values of T min , there are three cases. • t (cid:62) 4. Without loss of generality, we may assume that t = wt( α ⊕ β ). (Note that eventhough we have named δ as the one not belonging to S ( f ), since α ⊕ β ⊕ γ ⊕ δ = 0, we canname them so that t = wt( α ⊕ β ).) Then, there are at least four bits on which α and β differ. Among these four bits, there are at least two bits on which γ is identical to α or β .Without loss of generality, we assume that these are the first two bits and γ γ = α α . Wehave β β = α α , and as δ = α ⊕ β ⊕ γ , we have δ δ = α α . Also by flipping variables,we may assume that α = (cid:126) n = 00 (cid:126) n − . Then, β = 11 β ∗ , γ = 00 γ ∗ and δ = 11 δ ∗ where β ∗ , γ ∗ , δ ∗ ∈ Z n − and δ ∗ = β ∗ ⊕ γ ∗ . We consider the following eight inputs of f . α = 00 α ∗ α (cid:48) = 11 α ∗ β (cid:48) = 00 β ∗ β = 11 β ∗ γ = 00 γ ∗ γ (cid:48) = 11 γ ∗ δ (cid:48) = 00 δ ∗ δ = 11 δ ∗ Note that γ (cid:48) = α ⊕ α (cid:48) ⊕ γ , and wt( α ⊕ α (cid:48) ) = 2 < t . Then, T ( α, α (cid:48) , γ, γ (cid:48) ) < T ( α, β, γ, δ ) . By our assumption that T ( α, β, γ, δ ) is the minimum score among all non-affine quadrilateralsof f , ( α, α (cid:48) , γ, γ (cid:48) ) is not a non-affine quadrilateral of f . Since α, γ ∈ S ( f ), α (cid:48) and γ (cid:48) areeither both in S ( f ) or both not in S ( f ). Also, note that γ (cid:48) = α (cid:48) ⊕ β ⊕ δ , and wt( α (cid:48) ⊕ β ) =wt( α ⊕ β ) − t − < t . Then, T ( α (cid:48) , β, γ (cid:48) , δ ) < T ( α, β, γ, δ ) . Again since T ( α, β, γ, δ ) is the minimum score among all non-affine quadrilaterals of f ,( α (cid:48) , β, γ (cid:48) , δ ) is not a non-affine quadrilateral. Since β ∈ S ( f ) and δ / ∈ S ( f ), α (cid:48) and γ (cid:48) are not both in S ( f ). Thus, α (cid:48) , γ (cid:48) / ∈ S ( f ). Similarly, ( β (cid:48) , β, δ (cid:48) , δ ) and ( α, β (cid:48) , γ, δ (cid:48) ) are notnon-affine quadrilaterals of f , since their scores are less than T ( α, β, γ, δ ). Since β ∈ S ( f )and δ / ∈ S ( f ), we cannot have both β (cid:48) , δ (cid:48) ∈ S ( f ) from considering ( β (cid:48) , β, δ (cid:48) , δ ), and thenfrom ( α, β (cid:48) , γ, δ (cid:48) ), we cannot have exactly one of β (cid:48) , δ (cid:48) is in S ( f ). Thus, both β (cid:48) , δ (cid:48) / ∈ S ( f ).In other words, we have f ( α (cid:48) ) = f ( β (cid:48) ) = f ( γ (cid:48) ) = f ( δ (cid:48) ) = 0.Consider the signature ∂ f . Then, f ( α ) + f ( α (cid:48) ), f ( β ) + f ( β (cid:48) ), f ( γ ) + f ( γ (cid:48) ) and f ( δ ) + f ( δ (cid:48) )are entries of ∂ f on inputs α ∗ , β ∗ , γ ∗ and δ ∗ respectively. Since f ( α ) + f ( α (cid:48) ) = f ( α ) (cid:54) = 0, f ( β ) + f ( β (cid:48) ) = f ( β ) (cid:54) = 0 and f ( γ ) + f ( γ (cid:48) ) = f ( γ ) (cid:54) = 0, α ∗ , β ∗ , γ ∗ ∈ S ( ∂ f ) . Meanwhilewe have f ( δ ) + f ( δ (cid:48) ) = 0 + 0 = 0, thus δ ∗ / ∈ S ( ∂ f ). Thus, ( α ∗ , β ∗ , γ ∗ , δ ∗ ) is a non-affinequadrilateral of ∂ f . Then, ∂ f is a non-affine signature of arity 2 n − 2. Contradiction. • t = 2 and t (cid:62) 4. Without loss of generality, we assume that wt( α ⊕ γ ) = 2 and wt( α ⊕ β ) = t (cid:62) 4. (Again, using α ⊕ β ⊕ γ ⊕ δ = 0, a moment reflection shows that this is indeed withoutloss of generality, even though we have named δ (cid:54)∈ S ( f ).) Again by flipping variables, wemay assume that α = (cid:126) n . Then, wt( γ ) = 2 and wt( β ) (cid:62) 4. Take four bits where β i = 1, for59t most two of these we can have γ i = 1, thus there exist two other bits of these four bits(we may assume that they are the first two bits) such that γ γ = 00 and β β = 11. Then, α = 00 α ∗ , β = 11 β ∗ , γ = 00 γ ∗ , and δ = 11 δ ∗ by δ = α ⊕ β ⊕ γ , where β ∗ , γ ∗ , δ ∗ ∈ Z n − ,wt( β ∗ ) (cid:62) 2, wt( γ ∗ ) = 2 and δ ∗ = β ∗ ⊕ γ ∗ . Still, we consider the following eight inputs of f . α = 00 α ∗ α (cid:48) = 11 α ∗ β (cid:48) = 00 β ∗ β = 11 β ∗ γ = 00 γ ∗ γ (cid:48) = 11 γ ∗ δ (cid:48) = 00 δ ∗ δ = 11 δ ∗ Note that wt( α ⊕ γ ) = 2 and wt( α ⊕ α (cid:48) ) = 2 < t . Then, T ( α, α (cid:48) , γ, γ (cid:48) ) < T ( α, β, γ, δ ) . Then similarly since T ( α, β, γ, δ ) is the minimum, ( α, α (cid:48) , γ, γ (cid:48) ) is not a non-affine quadrilateral.Since α, γ ∈ S ( f ), α (cid:48) and γ (cid:48) are either both in S ( f ) or both not in it. Also, note thatwt( α (cid:48) ⊕ γ (cid:48) ) = 2 and wt( α (cid:48) ⊕ β ) = wt( α ⊕ β ) − t − < t . Then, T ( α (cid:48) , β, γ (cid:48) , δ ) < T ( α, β, γ, δ ) . Thus, ( α (cid:48) , β, γ (cid:48) , δ ) is not a non-affine quadrilateral. Since β ∈ S ( f ) and δ / ∈ S ( f ), α (cid:48) and γ (cid:48) are not both in S ( f ). Thus, α (cid:48) , γ (cid:48) / ∈ S ( f ). Similarly, by considering ( β (cid:48) , β, δ (cid:48) , δ ) and( α, β (cid:48) , γ, δ (cid:48) ), we know that they are not non-affine quadrilaterals. Thus, β (cid:48) , δ (cid:48) / ∈ S ( f ). Inother words, we have f ( α (cid:48) ) = f ( β (cid:48) ) = f ( γ (cid:48) ) = f ( δ (cid:48) ) = 0. Still consider the signature ∂ f .We have ∂ f / ∈ A . Contradiction. • t = 2, t = 2 and t = 4. In this case, by the definition of distance-2 squares (equation (7.5)), (cid:104) f ( α ) f ( β ) f ( γ ) f ( δ ) (cid:105) forms a distance-2 square. Clearly, it is not of type I, II or III since exactly oneentry of this square is zero. As proved in Lemma 7.9, since f has a distance-2 square that isnot type I, II or III, then we can realize a non-affine signature by merging. Contradiction.Now, we consider the case that T min = (2 , , | S ( f ) | = 2 n − . We consider the non-affine quadrilateral ( α, β, γ, δ ) withscore T = (2 , , (cid:20) α βγ δ (cid:21) = (cid:20) (cid:126) n − (cid:126) n − (cid:126) n − (cid:126) n − (cid:21) , and δ is the only one among four that is not in S ( f ). By normalization, we may assume that f ( α ) = 1. If f ( γ ) = − 1, then we negate the variable x of f . This keeps f unchanged butchanges f to − f , so this does not change the value of f ( α ), but changes the value of f ( γ ) to1. Thus, without loss of generality, we may assume that f ( α ) = f ( γ ) = 1. Clearly, f has evenparity. Consider the signature (cid:101) f by the H gadget applied on variables x and x of f . We have (cid:101) f ( (cid:126) n − ) = f ( α ) + f ( γ ) = 2 and (cid:101) f (1 (cid:126) n − ) = f ( β ) + f ( δ ) = f ( β ) since f ( δ ) = 0. Rememberthat since f ∈ (cid:82) B A , by Lemma 7.5, for all ( a, b ) ∈ Z , (cid:101) f ab ∈ A and we use n ab to denote thenorm of its nonzero entries. Thus, n = 2 and n = 1. Also, we have f ( β ) (cid:54) = 0 which is thesame as 1 (cid:126) n − ∈ S ( f ), and f ( δ ) = 0 which is the same as 1 (cid:126) n − / ∈ S ( f ). By Lemma 7.15, S ( f ) ∩ S ( f ) = ∅ . Remember that (cid:101) f = f + f and (cid:101) f = f − f . Then, S ( (cid:101) f ) = S ( f ) ∪ S ( f ) = S ( (cid:101) f ) . ∂ bij f for all { i, j } disjoint with { , } and every b ∈ B . By Lemma 4.3 andits remark, we may assume that either M ( m ( ∂ bij f )) = λ bij I for some real λ bij (cid:54) = 0, or there existsa nonzero binary signature g bij ∈ O such that g bij ( x , x ) | ∂ bij f . Otherwise, we get g bij ( x , x ) | ∂ bij f . If ∂ bij f ≡ 0, then we can let g bij ∈ B since a zerosignature can be divided by any nonzero binary signature. If ∂ bij f (cid:54)≡ 0, we can realize g bij byfactorization. If g bij / ∈ B ⊗ , then we get F is non- B hard. Thus, we may assumethat g bij ∈ B after normalization. Therefore, for all { i, j } disjoint with { , } and every b ∈ B ,we may assume that either M ( m ( ∂ bij f )) = λ bij I for some real λ bij (cid:54) = 0, or there exists a nonzerobinary signature g bij ∈ B such that g bij ( x , x ) | ∂ bij f . Then, by Lemma 7.10, S ( (cid:101) f ) = O n − .Thus, | S ( (cid:101) f ) | = 2 n − .Now consider again the signature f . Since f satisfies , and all its nonzero entrieshave norm 1, for any ( a, b ) ∈ Z , | f ab | = | S ( f ab ) | . Then, | S ( f ) | = | S ( f ) | = | S ( f ) | = | S ( f ) | . Remember that S ( f ) ∩ S ( f ) = ∅ , and S ( (cid:101) f ) = S ( f ) ∪ S ( f ). Then, S ( f ) and S ( f )form an equal size partition of S ( (cid:101) f ). Thus, | S ( f ) | = | S ( f ) | = | S ( (cid:101) f ) | = 2 n − . Also, | S ( f ) | = | S ( f ) | = 2 n − . Therefore, | S ( f ) | = | S ( f ) | + | S ( f ) | + | S ( f ) | + | S ( f ) | = 4 · n − = 2 n − . Since all nonzero entries of f have norm 1, | f | = | S ( f ) | = 2 n − . Also, since f satisfies ,for all { i, j } ∈ [2 n ] and all ( a, b ) ∈ Z , | f abij | = | f | = 2 n − .We denote S ( f ) by S . Since f has even parity, for every ( x , . . . , x n ) ∈ S , x + · · · + x n = 0,i.e., S ⊆ E n . Let F ( x , . . . , x n − ) ∈ Z [ x , . . . , x n − ] be the multilinear polynomial such that F ( x , . . . , x n − ) = (cid:40) , ( x , . . . , x n − , x n ) ∈ S , ( x , . . . , x n − , x n ) / ∈ S where x n = n − (cid:88) i =1 x i . Then, S = { ( x , . . . , x n ) ∈ E n | F ( x , . . . , x n − ) = 1 } . Now, we show that for all { i, j } ⊆ [2 n − F ij + F ij ≡ F ij + F ij ≡ { i, j } = { , } . The proof for arbitrary { i, j } is thesame by replacing { , } by { i, j } . Consider S = S ( f ) = { ( x , . . . , x n ) ∈ E n − | F ( x , . . . , x n − ) = 1 } , and S = S ( f ) = { ( x , . . . , x n ) ∈ E n − | F ( x , . . . , x n − ) = 1 } . Then, S ∩ S = { ( x , . . . , x n ) ∈ E n − | F · F = 1 } , and S ∪ S = { ( x , . . . , x n ) ∈ E n − | F + F + F · F = 1 } . By Lemma 7.15, S = S or S ∩ S = 0. 61 If S = S , then for every ( x , . . . , x n − ) ∈ Z n − which decides every ( x , . . . , x n ) ∈ E n − by x n = x + · · · + x n − , F ( x , . . . , x n − ) = F ( x , . . . , x n − ) . Thus, F + F ≡ • If S ∩ S = ∅ , then since | S | = | S | = 2 n − (which is still true when replacing { , } by anarbitrary { i, j } ), | S ∪ S | = | S | + | S | = 2 n − . Since S ∪ S ⊆ E n − and | E n − | = 2 n − , S ∪ S = E n − . Thus, for every ( x , . . . , x n − ) ∈ Z n − which decides every ( x , . . . , x n ) ∈ E n − by x n = x + · · · + x n − , F ( x , . . . , x n − ) · F ( x , . . . , x n − ) = 0 , and F ( x , . . . , x n − ) + F ( x , . . . , x n − ) + F · F ( x , . . . , x n − ) = 1 . Thus, F + F ≡ F + F ≡ { i, j } ⊆ [2 n − F ij + F ij ≡ F ij + F ij ≡ d ( F ) (cid:54) d ( F ) (cid:54) 1, then clearly, S = { ( x , . . . , x n ) ∈ E n | F ( x , . . . , x n − ) = 1 } is an affine linearspace. Thus, f has affine support. Otherwise, d ( F ) = 2. By Lemma 7.8, F is a complete quadraticpolynomial. Consider signatures f and f . Remember that f (000 (cid:126) n − ) = f (110 (cid:126) n − ) = 1.Thus, (cid:126) n − ∈ S ∩ S (cid:54) = ∅ . Then, S = S . Let S + = { α ∈ S | f ( α ) = f ( α ) } and S − = { α ∈ S | f ( α ) = − f ( α ) } . Then, as f takes ± S + = S ( ∂ +12 f ) and S − = S ( ∂ − f ) . Since ∂ +12 f, ∂ − f ∈ A , S + and S − are affine linear subspaces of E n − . Also, by , (cid:104) f , f (cid:105) = | S + | − | S − | = 0.Thus, | S + | = | S − | = | S | = 2 n − . Since | E n − | = 2 n − , S + is a an affine linear subspaces of E n − decided by two affine linear constraints L +1 = 1 and L +2 = 1. (Here both L +1 and L +2 are affine linear forms which may have nonzero constant terms, but we write the constraints as L +1 = 1and L +2 = 1.) In other words, S + = { ( x , . . . , x n ) ∈ E n − | L +1 = 1 and L +2 = 1 } = { ( x , . . . , x n ) ∈ E n − | L +1 · L +2 = 1 } . Since for every ( x , . . . , x n ) ∈ E n − , x + · · · + x n = 0, we may substitute the appearance of x n in L +1 and L +2 by x + · + x n − . Thus, we may assume that L +1 , L +2 ∈ Z [ x , . . . , x n − ], and d ( L +1 ) = d ( L +2 ) = 1. Similarly, there exist L − , L − ∈ Z [ x , . . . , x n − ] with d ( L − ) = d ( L − ) = 1such that S − = { ( x , . . . , x n ) ∈ E n − | L − = 1 and L − = 1 } = { ( x , . . . , x n ) ∈ E n − | L − · L − = 1 } . Clearly, S + ∩ S − = ∅ . Then S + ∪ S − = { ( x , . . . , x n ) ∈ E n − | L +1 · L +2 + L − · L − = 1 } . Remember that S = S + ∪ S − = { ( x , . . . , x n ) ∈ E n − | F = 1 } . L +1 · L +2 + L − · L − = F . Since for all 1 (cid:54) i < j (cid:54) n − 1, the quadratic term x i x j appears in F , for all 3 (cid:54) i < j (cid:54) n − 1, the quadratic term x i x j appears in F . Thus, F ∈ Z [ x , . . . , x n − ]is a complete quadratic polynomial over 2 n − n (cid:62) 8, 2 n − (cid:62) 5. By Lemma 7.14, we have 2 n − F = ( x + x + a )( x + x + b ) + ( x + x + x + x + c )( x + x + x + x + d )where a, b, c, d ∈ Z . Without loss of generality, we may assume that L +1 · L +2 = ( x + x + a )( x + x + b ) . Then, S + = S ( ∂ +12 f ) = { ( x , . . . , x ) ∈ E n − | x = x + a and x = x + b } , for some a, b ∈ Z .Clearly ∂ +12 f is a 6-ary signature and | S ( ∂ +12 f ) | = 2 − = 2 . We show that ∂ +12 f / ∈ B ⊗ ∪ F ∪F H . Then, by Corollary 6.10, we get F ∪ F H is either E or O whose sizes are both 2 . Thus, ∂ +12 f / ∈ F ∪ F H . For any 6-ary signature g in B ⊗ , its 6 variables can be divided into three independent pairs such that on the support S ( g ), thevalues of variables inside each pair do not rely on the values of variables of other pairs. Thus, if wepick any three variables in S ( g ), the degree of freedom of them is at least 2; more precisely, thereare at least 4 assignments on these three variables which can be extended to an input in S ( g ).However, in S ( ∂ +12 f ), the degree of freedom of variables x , x , x is only 1, namely there are onlytwo assignments on x , x , x that can be extended to an input in S ( ∂ +12 f ). Thus, ∂ +12 f / ∈ B ⊗ .This completes the proof of Lemma 7.16. Finally, by further assuming that f has affine support, we consider whether f itself is an affinesignature. We prove that this is true only for signature of arity 2 n (cid:62) 10. For signature f of arity2 n = 8, we show that either f ∈ A or the following signature is realizable. h = χ T · ( − x x x + x x x + x x x + x x x , where T = S ( h ) = S ( f ) . Note that in the support S ( f ) (see its definition (7.4) for this Queen of the Night f ), by tak-ing x , x , x , x as free variables, the remaining 4 variables are mod 2 sums of (cid:0) (cid:1) subsets of { x , x , x , x } . Clearly, h is not affine, but it has affine support and all its nonzero entries havethe same norm. One can check that h satisfies and h ∈ (cid:82) B A . But fortunately, weshow that by merging h , we can realize a 6-ary signature that is not in B ⊗ ∪ F ∪ F H . By Corollary6.10, we are done.After we give one more result about multilinear boolean polynomials, we make our final steptowards Theorem 7.19. Lemma 7.17. Let F ( x , . . . , x n ) ∈ Z [ x , . . . , x n ] be a complete cubic polynomial, L ( x , . . . , x n ) ∈ Z [ x , . . . , x n ] and d ( L ) (cid:54) . If we substitute x by x n +1 + L ( x , . . . , x n ) in F to get F (cid:48) , and suppose F (cid:48) ( x , . . . , x n +1 ) = F ( x n +1 + L, x , . . . , x n ) ∈ Z [ x , . . . , x n +1 ] is also a complete cubic polynomial,then • If n (cid:62) , then L must be a constant (cid:15) = 0 or . • If n = 4 , then L must be either (cid:15) , or of the form x i + x j + (cid:15) , for some (cid:15) = 0 or , for some { i, j } ∈ { , , } . roof. Since F ( x , . . . , x n ) is a complete cubic polynomial, we can write it as F ( x , . . . , x n ) = x · (cid:88) (cid:54) i 2. Then, F (cid:48) ( x , . . . , x n , x n +1 ) = ( x n +1 + L ) · (cid:88) (cid:54) i 2. Then,there is no cubic term in G (cid:48) ( x , . . . , x n , x n +1 ). Since F (cid:48) ( x , . . . , x n , x n +1 ) is a complete cubic poly-nomial over variables ( x , . . . , x n , x n +1 ) and x n +1 · (cid:80) (cid:54) i 4, we have Q (cid:54)≡ 0. For every x i x j ∈ Q , since x ∈ L , the cubic term x x i x j will appear in L · (cid:80) (cid:54) i 5, then x x , x x , x x ∈ Q . Thus, exactly one between x and x is in L , exactlyone between x and x is in L , and exactly one between x and x is in L . Clearly, this is acontradiction.If n = 4, then Q = x x . Either x or x appears in L . Thus, L is a sum of two variablesamong { x , x , x } plus a constant 0 or 1. Lemma 7.18. Let F be non- B hard. Let f ∈ F be an irreducible n -ary (2 n (cid:62) signature withparity. Then, • Holant b ( F ) is • there is a signature g / ∈ A of arity k < n that is realizable from f and B , or • f ∈ A .Proof. Again, we may assume that f satisfies and f ∈ (cid:82) B A . Also by Lemmas 7.9 and7.16, we may assume that f ( α ) = ± α ∈ S ( f ) and S ( f ) is an affine linear space. Let { x , . . . , x m } be a set of free variables of S ( f ). Then, on the support S ( f ), every variable x i (1 (cid:54) i (cid:54) n ) is expressible as a unique affine linear combination over Z of these free variables,i.e., x i = L i ( x , . . . , x m ) = λ i + λ i x + . . . + λ mi x m , where λ i , . . . , λ mi ∈ Z . Clearly, for 1 (cid:54) i (cid:54) m , L ( x i ) = x i . Then, S ( f ) = { ( x , . . . , x n ) ∈ Z n | x = L , . . . , x n = L n } = { ( x , . . . , x n ) ∈ Z n | x m +1 = L m +1 , . . . , x n = L n } . Also, let I ( x i ) = { (cid:54) k (cid:54) m | λ ki = 1 } . Clearly, for 1 (cid:54) i (cid:54) m , I ( x i ) = { i } . For m + 1 (cid:54) i (cid:54) n ,we show that | I x i | (cid:62) 2. For a contradiction, suppose that there exists m + 1 (cid:54) i (cid:54) n such that | I x i | = 0 or 1. If | I x i | = 0, then x i takes a constant value in S . Then, among f i and f i , one is azero signature. Thus, f is reducible. Contradiction. If | I x i | = 1, then x i = x k or x k + 1 for somefree variable x k . Then, among f ik , f ik , f ik and f ik , two are zero signatures. Thus, f does notsatisfy . Contradiction. 64ince f ( α ) = ± α ∈ S ( f ) and each α ∈ S ( f ) can be uniquely decided by its valueon the first m free variables, there exists a unique multilinear boolean polynomial F ( x , . . . , x m ) ∈ Z [ x , . . . , x m ] such that f ( x , . . . , x m , . . . , x n ) = χ S ( − F ( x ,...,x m ) where S = S ( f ). If d ( F ) (cid:54) 2, then clearly f ∈ A . We are done. Thus, we may assume that d ( F ) > m > 2. Remember that F abij denotes the polynomial obtained by settingvariables ( x i , x j ) of F to ( a, b ) ∈ Z . Then, f abij = ( − F abij on S ( f ). We will show that for all i, j ∈ [ m ], d ( F ij + F ij ) (cid:54) d ( F ij + F ij ) (cid:54) 1. For brevity of notation, we prove this for { i, j } = { , } . The proof for arbitrary { i, j } is the same by replacing { , } with { i, j } . We firstshow that d ( F ij + F ij ) (cid:54) 1. We use S to denote S ( f ij ) and S to denote S ( f ij ) . By Lemma 7.15,there are two cases, S = S or S ∩ S = ∅ . • Suppose that S = S . For convenience, we use L i to denote ( L i ) and L i to denote ( L i ) .Then, S = { ( x , . . . , x n ) ∈ Z n − | x m +1 = L m +1 , . . . , x n = L n } S = { ( x , . . . , x n ) ∈ Z n − | x m +1 = L m +1 , . . . , x n = L n } . So L i ≡ L i for all i (cid:62) m + 1. Thus, either { , } ⊆ I ( x i ) or { , } ∩ I ( x i ) = ∅ for i (cid:62) m + 1.Let S + = { α ∈ S | f ij ( α ) = f ij ( α ) } and S − = { α ∈ S | f ij ( α ) = − f ij ( α ) } . Then, (cid:104) f ij , f ij (cid:105) = 1 · | S + | − · | S − | = 0. Since S = S + ∪ S − , | S + | = | S − | = | S | . Note that S ( ∂ f ) = S + and S ( ∂ − f ) = S − . By our assumption that f ∈ (cid:82) B A , ∂ f, ∂ − f ∈ A .Thus, both S + and S − are affine linear subspaces of S = S . Since | S + | = | S − | = | S | / L ( x , . . . , x n ) such that S + = { ( x , . . . , x n ) ∈ S | L ( x , . . . , x n ) = 0 } , and S − = { ( x , . . . , x n ) ∈ S | L ( x , . . . , x n ) = 1 } . For ( x , . . . , x n ) ∈ S , and i (cid:62) m + 1, we can substitute the variable x i that appears in L ( x , . . . , x n ) with L i ≡ L i . Then, we get an (affine) linear polynomial L (cid:48) ( x , . . . , x m ) ∈ Z [ x , . . . , x m ] such that L (cid:48) ( x , . . . , x m ) = L ( x , . . . , x m , x m +1 , . . . , x n ) for ( x , . . . , x n ) ∈ S .Thus, S + = { ( x , . . . , x n ) ∈ S | L (cid:48) ( x , . . . , x m ) = 0 } , and S − = { ( x , . . . , x n ) ∈ S | L (cid:48) ( x , . . . , x m ) = 1 } . Note that as | S + | = | S − | > 0, the affine linear polynomial L (cid:48) is non-constant, i.e., d ( L (cid:48) ) = 1.Then, for every ( x , . . . , x m ) ∈ Z m − ,( − F ( x ,...,x m ) = ( − F ( x ,...,x m ) if L (cid:48) ( x , . . . , x m ) = 0and ( − F ( x ,...,x m ) = − ( − F ( x ,...,x m ) if L (cid:48) ( x , . . . , x m ) = 1 . Thus, ( − F ( x ,...,x m )+ F ( x ,...,x m ) = ( − L (cid:48) ( x ,...,x m ) . Therefore, F ( x , . . . , x m ) + F ( x , . . . , x m ) ≡ L (cid:48) ( x , . . . , x m ) . Then, d ( F + F ) = 1.65 Suppose that S ∩ S = ∅ . Then, there exists a variable x i where i (cid:62) m + 1 such that between { , } , exactly one index is in I ( x i ). Without loss of generality, we may assume that i = m +1,1 ∈ I ( x m +1 ) and 2 / ∈ I ( x m +1 ). Then, x m +1 = x + K ( x , . . . , x m ) where K ∈ Z [ x , . . . , x m ]is an (affine) linear polynomial. Consider S . S = { ( x , . . . , x n ) ∈ Z n − | x = x = 0 , x m +1 = x + K, x m +2 = L m +2 . . . , x n = L n } . Since x = x on S , for every i (cid:62) m + 2, if x or x appear in L i , we substitute each one ofthem with x m +1 + K . We get a linear polynomial K i ∈ Z [ x , . . . , x m , x m +1 ]. Then, for every( x , . . . , x n ) ∈ S , L i = K i . Thus, S = { ( x , . . . , x n ) ∈ Z n − | x m +1 + K = 0 , x m +2 = K m +2 . . . , x n = K n } . Similarly, we have S = { ( x , . . . , x n ) ∈ Z n − | x m +1 + K = 1 , x m +2 = K m +2 . . . , x n = K n } . Let S ∪ = S ∪ S . Then, S ∪ = { ( x , . . . , x n ) ∈ Z n − | x m +2 = K m +2 . . . , x n = K n } . Thus, we can pick x , . . . , x m , x m +1 as a set of free variables of S ∪ .Consider g = ∂ f . Clearly, S ( g ) = S ∪ since S ∩ S = ∅ . Then, there exists a uniquemultilinear boolean polynomial G ( x , . . . , x m +1 ) ∈ Z [ x , . . . , x m +1 ] such that g ( x , . . . , x n ) = χ S ∪ · ( − G ( x ,...,x m +1 ) . For every ( x , . . . , x n ) ∈ S that is uniquely decided by (0 , , x , . . . , x m ) ∈ { (0 , } × Z m − , x m +1 = K ( x , . . . , x m ) and f ( x , . . . , x n ) = g ( x , . . . , x n ). Thus, for every ( x , . . . , x m ) ∈ Z m − , ( − F ( x ,...,x m ) = ( − G ( x ,...,x m ,K ) . Also, for every ( x , . . . , x n ) ∈ S that is uniquely decided by (1 , , x , . . . , x m ) ∈ { (1 , } × Z m − , x m +1 = K ( x , . . . , x m ) + 1, and f ( x , . . . , x n ) = g ( x , . . . , x n ). Thus, for every( x , . . . , x m ) ∈ Z m − , ( − F ( x ,...,x m ) = ( − G ( x ,...,x m ,K +1) . Thus, F ( x , . . . , x m ) ≡ G ( x , . . . , x m , K ) and F ( x , . . . , x m ) ≡ G ( x , . . . , x m , K + 1).Since f ∈ (cid:82) B A , g = ∂ f ∈ A . Thus, g (cid:48) ( x , . . . , x m , x m +1 ) = ( − G ( x ,...,x m ,x m +1 ) is also in ∈ A . Let y = x m +1 + K ( x , . . . , x m ) ∈ Z [ x , . . . , x m +1 ] be an affine linear combina-tion of variables x , . . . , x m +1 . Since g ∈ A , by Lemma 2.19, d [ G ( x , . . . , x m , K ) + G ( x , . . . , x m , K + 1)] (cid:54) . Thus, d ( F + F ) (cid:54) 1. Also if d ( G ) = 1, then by Lemma 2.19 d ( F + F ) = 0 , i.e., F + F ≡ . (7.10)66imilarly, we can show that d ( F + F ) (cid:54) 1. Thus, for all i, j ∈ [ m ], d ( F ij + F ij ) (cid:54) d ( F ij + F ij ) (cid:54) 1. By Lemma 7.8, d ( F ) (cid:54) d ( F ) (cid:54) 2, then clearly f ∈ A . We are done. Otherwise, d ( F ) = 3 and by Lemma 7.8, F is acomplete cubic multilinear polynomial over m variables. If we pick another set X of m free variablesand substitute variables of F by variables in X , then we will get a cubic multilinear polynomial F (cid:48) over variables in X . Same as the analysis of F , F (cid:48) is also a complete cubic polynomial. Inparticular, consider the variable x m +1 . Recall that | I ( x m +1 ) | (cid:62) 2. Without loss of generality, weassume that 1 ∈ I ( x m +1 ). Then, x m +1 = x + L ( x , . . . , x m ) where L ( x , . . . , x m ) is an affine linearcombination of variables x , . . . , x m . We substitute x in F by x m +1 + L , and we get a completecubic multilinear polynomial F (cid:48) ( x , . . . , x m +1 ) ∈ Z [ x , . . . , x m +1 ] . By Lemma 7.17, if m (cid:62) 5, then x m +1 = x or x m +1 = x . Thus, I ( x m +1 ) = { } . This contradicts with | I ( x m =1 ) | (cid:62) 2. Thus, m (cid:54) m = 4, then by Lemma 7.17, x = x + (cid:15) , or x = x + x i + x j + (cid:15) , where (cid:15) = 0 or 1,for some 2 (cid:54) i < j (cid:54) 4. Since | I ( x ) | (cid:62) 2, the case that x = x + (cid:15) is impossible. Similarly,for i (cid:62) m + 2, the variable x i is a sum of three variables in { x , x , x , x } plus a constant 0 or1. If there exist x i and x j for 5 (cid:54) i < j (cid:54) n such that I ( x i ) = I ( x j ). Then, x i = x j or x j .Thus, among f ij , f ij , f ij and f ij , two are zero signatures. Thus, f does not satisfy .Contradiction. Thus, I ( x i ) (cid:54) = I ( x j ) for any 5 (cid:54) i < j (cid:54) n . There are only (cid:0) (cid:1) = 4 ways topick three variables from { x , x , x , x } . Thus, 2 n (cid:54) n (cid:62) n = 8. Clearly, | S ( f ) | = 2 = 16. Due to , for all { i, j } ∈ [8], | S ( f ij ) | = | S ( f ij ) | = | S ( f ij ) | = | S ( f ij ) | = 4. • If there exists { i, j } such that S ( f ij ) = S ( f ij ), then for any point α in S ( f ij ) = S ( f ij ),regardless whether f ij ( α ) = f ij ( α ) or f ij ( α ) = − f ij ( α ), either α ∈ S ( ∂ + ij f ) or α ∈ S ( ∂ − ij f ).Thus, S ( ∂ + ij f ) ∪ S ( ∂ − ij f ) = S ( f ij ) = S ( f ij ) . Also, by , (cid:104) f ij , f ij (cid:105) = | S ( ∂ − ij f ) | − | S ( ∂ + ij f ) | = 0 . Thus, | S ( ∂ + ij f ) | = | S ( ∂ − ij f ) | = 2. Note that every 6-ary signature in B ⊗ has support of size8, and every signature in F and F H has support of size 32. Thus, ∂ + ij f / ∈ B ∪ F ∪ F H .Then, by Corollary 6.10, we get { i, j } such that S ( f ij ) = S ( f ij ), then we have | S ( ∂ (cid:98) + ij f ) | = | S ( ∂ (cid:98) − ij f ) | = 2. Thus, ∂ (cid:98) + ij f / ∈ B ⊗ ∪ F ∪ F H .Again, we get • Otherwise, for all { i, j } ∈ [8], S ( f ij ) ∩ S ( f ij ) = ∅ and S ( f ij ) ∩ S ( f ij ) = ∅ . Then, S ( ∂ + ij f ) = S ( f ij ) ∪ S ( f ij ). Thus, | S ( ∂ + ij f ) | = 8. Clearly, ∂ + ij f / ∈ F ∪ F H . If ∂ + ij f / ∈ B ⊗ ,then we get ∂ + ij f ∈ B ⊗ . Then, ∂ + ij f = χ S ( ∂ + ij f ) ( − G + ij where d ( G + ij ) = 1 . As we proved above in equation (7.10), F ij + F ij ≡ . Similarly, suppose ∂ (cid:98) + ij f ∈ B ⊗ ,and we can show that F ij + F ij ≡ . Thus, for all { i, j } ⊆ [8], F ij + F ij ≡ F ij + F ij ≡ . Then, by Lemma 7.8, d ( F ) (cid:54) 2. Contradiction.Suppose that m = 3. Remember that for 4 (cid:54) i (cid:54) n , | I ( x i ) | (cid:62) 2. Thus, x i is a sum of at leasttwo variables in { x , x , x } plus a constant 0 or 1. Again, if there exist x i and x j for 4 (cid:54) i < j (cid:54) n I ( x i ) = I ( x j ), then among f ij , f ij , f ij and f ij , two are zero signatures. Contradiction.Thus, I ( x i ) (cid:54) = I ( x j ) for any 4 (cid:54) i < j (cid:54) n . There are (cid:0) (cid:1) + (cid:0) (cid:1) = 4 different ways to pick at leasttwo variables from { x , x , x } . Thus, 2 n (cid:54) Theorem 7.19. Suppose that F is non- B hard. Then, Holant b ( F ) is Since F does not satisfy condition (T), F contains a signature f / ∈ A . Suppose that f hasarity 2 n . We prove this theorem by induction on 2 n .If 2 n = 2 , b ( F ) is k (cid:62) 6, Holant b ( F ) is n (cid:54) k . We consider thecase that 2 n = 2 k + 2 (cid:62) 8. First, suppose that f is reducible. If it is a tensor product of twosignatures of odd arity, then we can realize a signature of odd arity by factorization. We get A since f / ∈ A . Then, we can realize a non-affine signature of arity 2 m (cid:54) k byfactorization. By our induction hypothesis, we get f isirreducible. If f has no parity, then we get f has parity. Then by Lemma 7.18, Holant b ( F ) is m (cid:54) k . By our induction hypothesis, we get B is realizable from f and { f } ∪ F is non- B hard for any real-valued F that does notsatisfy condition (T), we have the following result. Lemma 7.20. Holant b ( f , F ) is Combining Theorem 6.5 and Lemma 7.20, we have the following result. This concludes Sections6 and 7, and we are done with the arity 6 case. Lemma 7.21. If (cid:98) F contains a signature (cid:98) f of arity and (cid:98) f / ∈ (cid:98) O ⊗ , then Holant( (cid:54) = | (cid:98) F ) is We have seen some extraordinary properties of the signature f . Now, we formally analyze it.Remember that f = χ T where T = S ( f ) = { ( x , x , . . . , x ) ∈ Z | x + x + x + x = 0 , x + x + x + x = 0 ,x + x + x + x = 0 , x + x + x + x = 0 } . = { , , , , , , , , , , , , , , , } . (8.1)One can see that S ( f ) has the following structure: the sums of the first four variables, and thelast four variables are both even; the assignment of the first four variables are either identicalto, or complement of the assignment of the last four variables. Another interesting description of S ( f ) is as follows: One can take 4 variables, called them y , y , y , y . Then on the support theremaining 4 variables are mod 2 sums of (cid:0) (cid:1) subsets of { y , y , y , y } , and y , y , y , y are freevariables. (However, the 4 variables ( y , y , y , y ) cannot be taken as ( x , x , x , x ) in the abovedescription (8.1). But one can take ( y , y , y , y ) = ( x , x , x , x ). More specifically, one can takeany 3 variables x i , x j , x k from { x , . . . , x } first as free variables, which excludes one unique other68 (cid:96) from the remainder set X (cid:48) = { x , . . . , x } \ { x i , x j , x k } , and then one can take any one variable x r ∈ X (cid:48) as the 4th free variable. Then the remaining 4 variables are the mod 2 sums of (cid:0) (cid:1) subsetsof the 4 free variables { x i , x j , x k , x r } , and in particular x (cid:96) = x i + x j + x k , on S ( f ).) We give thefollowing Figure 2 to visualize the signature matrix M ( f ). A block with orange color denotesan entry +1. Other blank blocks are zeros. Figure 2: A visualization of f , which happens to be the same as (cid:98) f = Z − f One can check that f satisfies both and f ∈ (cid:82) O ⊗ . Also, f is unchanged underthe holographic transformation by Z − , i.e., (cid:98) f = Z − f = f . (cid:98) f In this subsection, we show how this extraordinary signature (cid:98) f was discovered. We use the notation (cid:98) f since we consider the problem Holant( (cid:54) = | (cid:98) F ) for complex-valued (cid:98) F satisfying ars . We provethat if (cid:98) F contains an 8-ary signature (cid:98) f where (cid:98) f / ∈ (cid:98) O ⊗ , then Holant( (cid:54) = | (cid:98) F ) is (cid:98) f isrealizable from (cid:98) f (Theorem 8.5).Remember that D = {(cid:54) = } . Then D ⊗ = { λ · ( (cid:54) = ) ⊗ n | λ ∈ R \{ } , n (cid:62) } is the set of tensorproducts of binary disequalities (cid:54) = up to a nonzero real scalar. If for all pairs of indices { i, j } , (cid:98) ∂ ij (cid:98) f ∈ D ⊗ , then we say (cid:98) f ∈ (cid:98)(cid:82) D ⊗ . Clearly, if (cid:98) f ∈ D ⊗ and (cid:98) f has arity greater than 2, then (cid:98) f ∈ (cid:98)(cid:82) D ⊗ . We first show the following result for signatures of arity at least 8. Lemma 8.1. Let (cid:98) f / ∈ (cid:98) O ⊗ be a signature of arity n (cid:62) in (cid:98) F . Then, • Holant( (cid:54) = | (cid:98) F ) is • there is a signature (cid:98) g / ∈ (cid:98) O ⊗ of arity k (cid:54) n − that is realizable from (cid:98) f , or • there is an irreducible signature (cid:99) f ∗ ∈ (cid:98)(cid:82) D ⊗ of arity n that is realizable from (cid:98) f .Proof. Since (cid:98) f / ∈ (cid:98) O ⊗ , (cid:98) f (cid:54)≡ 0. Again, we may assume that (cid:98) f is irreducible. Otherwise, by factoriza-tion, we can realize a nonzero signature of odd arity and we get (cid:98) O ⊗ and we are done. Under the assumptionthat (cid:98) f is irreducible, we may further assume that (cid:98) f satisfies by Lemma 4.4. Considersignatures (cid:98) ∂ ij (cid:98) f for all pairs of indices { i, j } . If there exists a pair { i, j } such that (cid:98) ∂ ij (cid:98) f / ∈ (cid:98) O ⊗ , thenlet (cid:98) g = (cid:98) ∂ ij (cid:98) f , and we are done. Thus, we may also assume that (cid:98) f ∈ (cid:98)(cid:82) (cid:98) O ⊗ .69f for all pairs of indices { i, j } , we have (cid:98) ∂ ij (cid:98) f ≡ 0. Then, by Lemma 2.10, (cid:98) f ( α ) = 0 for all α withwt( α ) (cid:54) = 0 or 2 n . Since f (cid:54)≡ ars , | (cid:98) f ( (cid:126) n ) | = | (cid:98) f ( (cid:126) n ) | (cid:54) = 0. Clearly, such a signature doesnot satisfy . Contradiction. Thus, without loss of generality, we assume that (cid:98) ∂ (cid:98) f (cid:54)≡ (cid:98) ∂ (cid:98) f ∈ (cid:98) O ⊗ , without loss of generality, we may assume that in the UPF of (cid:98) ∂ (cid:98) f , variables x and x appear in one binary signature b ( x , x ), x and x appear in one binary signature b ( x , x ) and so on. Thus, we have (cid:98) ∂ (cid:98) f = (cid:98) b ( x , x ) ⊗ (cid:98) b ( x , x ) ⊗ (cid:98) b ( x , x ) ⊗ . . . ⊗ (cid:100) b n − ( x n − , x n ) . By Lemma 2.7, all these binary signatures (cid:98) b , (cid:98) b , . . . , (cid:100) b n − are realizable from f by factorization.Note that for nonzero binary signatures (cid:98) b i ( x i +1 , x i +2 ) (1 (cid:54) i (cid:54) n − x i +1 of two copies of (cid:98) b i ( x i +1 , x i +2 ) using (cid:54) = (mating two binary signatures), then we get (cid:54) = up to a scalar. We consider the following gadget construction on (cid:98) f . Recall that in the setting ofHolant( (cid:54) = | (cid:98) F ), variables are connected using (cid:54) = . For 1 (cid:54) i (cid:54) n − 1, by a slight abuse of namesof variables, we connect the variable x i +1 of (cid:98) f with the variable x i +1 of (cid:98) b i ( x i +1 , x i +2 ) using (cid:54) = . We get a signature (cid:98) f (cid:48) of arity 2 n . (Note that, as a complexity reduction using factorization(Lemma 2.7), we can only apply it a constant number of times. However, the arity 2 n of (cid:98) f isconsidered a constant, as (cid:98) f ∈ (cid:98) F , which is independent of the input size of a signature grid to theproblem Holant( (cid:54) = | (cid:98) F ).) We denote this gadget construction by G and we write (cid:98) f (cid:48) as G ◦ (cid:98) f . G is constructed by extending variables of (cid:98) f using binary signatures realized from (cid:98) ∂ (cid:98) f . It doesnot change the irreducibility of (cid:98) f . Thus, (cid:98) f (cid:48) is irreducible since (cid:98) f is irreducible. Similarly, we mayassume that (cid:98) f (cid:48) ∈ (cid:98)(cid:82) (cid:98) O ⊗ . Otherwise, we are done.Consider the signature (cid:98) ∂ (cid:98) f (cid:48) . Since the above gadget construction G does not touch variables x and x of f , G commutes with the merging gadget (cid:98) ∂ . (Succinctly, the commutativity can beexpressed as (cid:98) ∂ (cid:98) f (cid:48) = (cid:98) ∂ ( G ◦ (cid:98) f ) = G ◦ (cid:98) ∂ (cid:98) f . ) Thus, (cid:98) ∂ (cid:98) f (cid:48) can be realized by performing the gadgetconstruction G on (cid:98) ∂ (cid:98) f , which connects each binary signature (cid:98) b i ( x i +1 , x i +2 ) in the UPF of (cid:98) ∂ (cid:98) f with another copy of (cid:98) b i ( x i +1 , x i +2 ) (in the mating fashion). Thus, each binary signature (cid:98) b i in (cid:98) ∂ (cid:98) f is changed to (cid:54) = up to a nonzero scalar after this gadget construction G . After normalizationand renaming variables, we have (cid:98) ∂ (cid:98) f (cid:48) = ( (cid:54) = )( x , x ) ⊗ ( (cid:54) = )( x , x ) ⊗ ( (cid:54) = )( x , x ) ⊗ . . . ⊗ ( (cid:54) = )( x n − , x n ) . (8.2)Thus, (cid:98) ∂ (cid:98) f (cid:48) ∈ D ⊗ . Moreover, for all pairs of indices { i, j } disjoint with { , } , we have (cid:98) ∂ ( ij )(12) (cid:98) f (cid:48) ∈ D ⊗ , and hence (cid:98) ∂ ( ij )(12) (cid:98) f (cid:48) (cid:54)≡ . (8.3)A fortiori, for all pairs of indices { i, j } disjoint with { , } , (cid:98) ∂ ij (cid:98) f (cid:48) (cid:54)≡ (cid:99) f ∗ of arity 2 n from (cid:98) f (cid:48) such that (cid:99) f ∗ ∈ (cid:98)(cid:82) D ⊗ . We first prove the following claim. Claim. Let (cid:98) h ∈ (cid:98)(cid:82) (cid:98) O ⊗ be a signature of arity n (cid:62) . If (cid:98) ∂ ij (cid:98) h ∈ D ⊗ for all { i, j } disjointwith { , } , then (cid:98) h ∈ (cid:98)(cid:82) D ⊗ . Clearly, we only need to show that (cid:98) ∂ k (cid:98) h ∈ D ⊗ for all 2 (cid:54) k (cid:54) n . Then, by symmetry we alsohave (cid:98) ∂ k (cid:98) h ∈ D ⊗ for k = 1 and all 3 (cid:54) k (cid:54) n . This will prove (cid:98) h ∈ (cid:98)(cid:82) D ⊗ . Consider (cid:98) ∂ k (cid:98) h for an70rbitrary 2 (cid:54) k (cid:54) n . Since for all { i, j } disjoint with { , } , we have (cid:98) ∂ ij (cid:98) h ∈ D ⊗ , a fortiori for all { i, j } disjoint with { , } ∪ { k } , (cid:98) ∂ (1 k )( ij ) (cid:98) h ∈ D ⊗ . (8.4)Since (cid:98) h has arity 2 n (cid:62) 8, we can pick a pair of indices { i, j } disjoint with { , } ∪ { k } . Since (cid:98) ∂ (1 k )( ij ) (cid:98) h ∈ D ⊗ , which is nonzero, a fortiori we have (cid:98) ∂ k (cid:98) h (cid:54)≡ 0. So we may consider the UPF of (cid:98) ∂ k (cid:98) h , which is known to be in (cid:98) O ⊗ . For a contradiction, suppose that there is a binary signature (cid:98) b (as a factor of (cid:98) ∂ k (cid:98) h ) such that (cid:98) b is not an associate of (cid:54) = . Among the two variables in the scopeof (cid:98) b , at least one is not x . We pick such a variable x s where x s (cid:54) = x . Then, we consider anotherbinary signature (cid:98) b in the UPF of (cid:98) ∂ k (cid:98) h . • If (cid:98) b = λ · (cid:54) = , for some nonzero scalar λ , then we pick a variable x t in the scope of (cid:98) b thatis not x . Consider (cid:98) ∂ ( st )(1 k ) (cid:98) h . When merging variables x s and x t of (cid:98) ∂ k (cid:98) h , we connect thevariable x s of (cid:98) b with the variable x t of λ · (cid:54) = , and the resulting binary signature is just λ · (cid:98) b ,which is not an associate of (cid:54) = . Thus, we have (cid:98) ∂ ( st )(1 k ) (cid:98) h / ∈ D ⊗ . • Otherwise, (cid:98) b is not an associate of (cid:54) = . Since (cid:98) ∂ k (cid:98) h has arity 2 n − (cid:62) 6, we can find anotherbinary signature (cid:98) b in the UPF of (cid:98) ∂ k (cid:98) h . We pick a variable x t in the scope of (cid:98) b that is not x . Consider (cid:98) ∂ ( st )(1 k ) (cid:98) h . Now, when merging variables x s and x t of (cid:98) ∂ k (cid:98) h , the binary signature (cid:98) b is untouched. Thus, we have (cid:98) b | (cid:98) ∂ ( st )(1 k ) (cid:98) h , which implies that (cid:98) ∂ ( st )(1 k ) (cid:98) h / ∈ D ⊗ .Note that in both cases, { s, t }∩ ( { , }∪{ k } ) = ∅ . Therefore the two cases above both contradict(8.4) by picking { i, j } = { s, t } . Thus, (cid:98) ∂ k (cid:98) h ∈ D ⊗ for all 2 (cid:54) k (cid:54) n . Then similarly, we can showthat (cid:98) ∂ k (cid:98) h ∈ D ⊗ for all 3 (cid:54) k (cid:54) n . This finishes the proof of our Claim.Remember that (cid:98) ∂ ij (cid:98) f (cid:48) (cid:54)≡ { i, j } disjoint with { , } . We consider the UPF of (cid:98) ∂ ij (cid:98) f (cid:48) . Since (cid:98) f (cid:48) ∈ (cid:98)(cid:82) (cid:98) O ⊗ , there are two cases depending on whether variables x and x appear in one binarysignature or two distinct binary signatures. Case 1. For every { i, j } disjoint with { , } , in the UPF of (cid:98) ∂ ij (cid:98) f (cid:48) , x and x appear in one nonzerobinary signature (cid:99) b ij ( x , x ) ∈ (cid:98) O . In other words, for every { i, j } disjoint with { , } , (cid:98) ∂ ij (cid:98) f (cid:48) = (cid:99) b ij ( x , x ) ⊗ (cid:99) g ij , for some (cid:99) g ij (cid:54)≡ . (These factors (cid:99) b ij and (cid:99) g ij are nonzero since (cid:98) ∂ ij (cid:98) f (cid:48) (cid:54)≡ (cid:99) g ij ∼ (cid:98) ∂ (12)( ij ) (cid:98) f (cid:48) , and by (8.3), we have (cid:99) g ij ∈ D ⊗ . Also for { k, (cid:96) } disjoint with both { i, j } and { , } , (cid:98) ∂ ( k(cid:96) )( ij ) (cid:98) f (cid:48) (cid:54)≡ (cid:98) ∂ (12)( k(cid:96) )( ij ) (cid:98) f (cid:48) = (cid:98) ∂ ( ij )( k(cid:96) )(12) (cid:98) f (cid:48) (cid:54)≡ { i, j } (cid:54) = { k, (cid:96) } that are both disjoint with { , } , (cid:99) b ij ( x , x ) ∼ (cid:99) b k(cid:96) ( x , x ). If { i, j } is disjoint with { k, (cid:96) } , then (cid:99) b ij ( x , x ) | (cid:98) ∂ ( k(cid:96) )( ij ) (cid:98) f (cid:48) and (cid:99) b k(cid:96) ( x , x ) | (cid:98) ∂ ( ij )( k(cid:96) ) (cid:98) f (cid:48) .Since (cid:98) ∂ ( k(cid:96) )( ij ) (cid:98) f (cid:48) = (cid:98) ∂ ( ij )( k(cid:96) ) (cid:98) f (cid:48) (cid:54)≡ 0, by Lemma 2.5, we have (cid:99) b ij ( x , x ) ∼ (cid:99) b k(cid:96) ( x , x ). Otherwise, { i, j } and { k, (cid:96) } are not disjoint. Since (cid:98) f (cid:48) has arity (cid:62) 8, we can find another pair of indices { s, t } such that it is disjoint with { , } ∪ { i, j } ∪ { k, (cid:96) } . Then, by the above argument, we have (cid:99) b ij ( x , x ) ∼ (cid:99) b st ( x , x ) , and (cid:99) b st ( x , x ) ∼ (cid:99) b k(cid:96) ( x , x ) . Thus, (cid:99) b ij ( x , x ) ∼ (cid:99) b k(cid:96) ( x , x ) . We can usea binary signature (cid:98) b ( x , x ) to denote these binary signature (cid:99) b ij ( x , x ) for all { i, j } disjoint with { , } . Then, (cid:98) b ( x , x ) | (cid:98) ∂ ij (cid:98) f (cid:48) for all { i, j } disjoint with { , } . Also, (cid:98) b ( x , x ) is realizable from (cid:98) f (cid:48) by merging and factorization.Then, we consider the following gadget construction G on (cid:98) f (cid:48) . By a slight abuse of variablenames, we connect the variable x of (cid:98) f (cid:48) with the variable x of (cid:98) b ( x , x ) and we get a signature71 f ∗ . Clearly, G is constructed by extending variables of (cid:98) f (cid:48) . It does not change the irreducibilityof (cid:98) f (cid:48) . Thus, (cid:99) f ∗ is irreducible. Again, we may assume that (cid:99) f ∗ ∈ (cid:98)(cid:82) (cid:98) O ⊗ . Consider (cid:98) ∂ ij (cid:99) f ∗ for all { i, j } disjoint with { , } . Since the above gadget construction G only touches the variable x of f (cid:48) , itcommutes with the merging operation (cid:98) ∂ ij . Thus, (cid:98) ∂ ij (cid:99) f ∗ can be realized by performing the gadgetconstruction G on (cid:98) ∂ ij (cid:98) f (cid:48) , i.e., connecting the binary signature (cid:98) b ( x , x ) in the UPF of (cid:98) ∂ ij (cid:98) f (cid:48) withitself (in the mating fashion), which changes (cid:98) b ( x , x ) to (cid:54) = up to some nonzero scalar λ ij . Thus,for all { i, j } disjoint with { , } , after renaming variables, we have (cid:98) ∂ ij (cid:99) f ∗ = λ ij · ( (cid:54) = )( x , x ) ⊗ (cid:99) g ij ∈ D ⊗ . Thus, (cid:98) ∂ ij (cid:99) f ∗ ∈ D ⊗ for all { i, j } disjoint with { , } . By our Claim, (cid:99) f ∗ ∈ (cid:98)(cid:82) D ⊗ . We are done withCase 1. Case 2. There is a pair of indices { i, j } disjoint with { , } such that x and x appear in twodistinct nonzero binary signatures (cid:98) b (cid:48) ( x , x u ) and (cid:98) b (cid:48) ( x , x v ) in the UPF of (cid:98) ∂ ij (cid:98) f (cid:48) . In other words,there exits { i, j } such that (cid:98) ∂ ij (cid:98) f (cid:48) = (cid:98) b (cid:48) ( x , x u ) ⊗ (cid:98) b (cid:48) ( x , x v ) ⊗ (cid:99) h ij , for some (cid:99) h ij (cid:54)≡ . (8.5)Since (cid:99) h ij | (cid:98) ∂ (12)( ij ) (cid:98) f (cid:48) and (cid:98) ∂ (12)( ij ) (cid:98) f (cid:48) ∈ D ⊗ , we have (cid:99) h ij ∈ D ⊗ . Also, after merging variables x and x (using (cid:54) = ) in (cid:98) ∂ ij (cid:98) f (cid:48) , variables x u and x v form a binary disequality up to a nonzero scalar (this binarysignature on x u and x v must be a binary disequality because we already know (cid:98) ∂ (12)( ij ) (cid:98) f (cid:48) ∈ D ⊗ ).In other words, by connecting the variable x of (cid:98) b (cid:48) ( x , x u ) and the variable x of (cid:98) b (cid:48) ( x , x v ) (using (cid:54) = ), we get λ · (cid:54) = ( x u , x v ) for some λ (cid:54) = 0. By Lemma 2.13, we have (cid:98) b (cid:48) ∼ (cid:98) b (cid:48) . Also, connecting thevariable x u of (cid:98) b (cid:48) and the variable x v of (cid:98) b (cid:48) (using (cid:54) = ) will give the binary signature λ · (cid:54) = ( x , x )as well.We consider the following gadget construction G on (cid:98) f (cid:48) . By a slight abuse of variable names,we connect variables x and x of (cid:98) f (cid:48) with the variable x of (cid:98) b (cid:48) and x of (cid:98) b (cid:48) using (cid:54) = respectively.We get a signature (cid:99) f ∗ . Again, (cid:99) f ∗ is irreducible since the gadget construction G does not changethe irreducibility of (cid:98) f (cid:48) . Also, we may assume that (cid:99) f ∗ ∈ (cid:98)(cid:82) (cid:98) O ⊗ . Otherwise, we are done. Consider (cid:98) ∂ ij (cid:99) f ∗ . Similarly, by the commutitivity of the gadget construction G and the merging gadget (cid:98) ∂ ij , (cid:98) ∂ ij (cid:99) f ∗ can be realized by connecting variables x and x of (cid:98) ∂ ij (cid:98) f (cid:48) with the variable x of (cid:98) b (cid:48) and thevariable x of (cid:98) b (cid:48) respectively. After renaming variables, we have (cid:98) ∂ ij (cid:99) f ∗ = λ ij · ( (cid:54) = )( x , x u ) ⊗ ( (cid:54) = ) ( x , x v ) ⊗ (cid:98) h ij ∈ D ⊗ . (8.6)We now show that (cid:98) ∂ (cid:99) f ∗ ∈ D ⊗ . Note that it is realized in the following way; we first connectvariables x and x of (cid:98) f (cid:48) with the variable x of (cid:98) b (cid:48) ( x , x u ) and the variable x of (cid:98) b (cid:48) ( x , x v ) respec-tively (using (cid:54) = ) to get (cid:99) f ∗ , and then after renaming variables x u and x v to x and x respectively,we merge them using (cid:54) = (see Figure 3(a)). By associativity of gadget constructions, we can changethe order; we first connect the variable x u of (cid:98) b (cid:48) ( x , x u ) with the variable x v of (cid:98) b (cid:48) ( x , x v ) (using (cid:54) = ), and then we use the resulting binary signature to connect variables x and x of (cid:98) f (cid:48) (edgesare connected using (cid:54) = ). Note that connecting x u of (cid:98) b (cid:48) ( x , x u ) with x v of (cid:98) b (cid:48) ( x , x v ) gives λ · (cid:54) = up to a nonzero scalar λ , and λ · (cid:54) = is unchanged by extending both of its two variables with (cid:54) = (cid:98) ∂ (cid:99) f ∗ and (cid:98) ∂ (cid:98) f (cid:48) (see Figure 3(b)). Thus, (cid:98) ∂ (cid:99) f ∗ is actually realized by merging x and x of (cid:98) f (cid:48) (using (cid:54) = ) up to anonzero scalar. Thus, we have (cid:98) ∂ (cid:99) f ∗ ∼ (cid:98) ∂ (cid:98) f (cid:48) , and hence (cid:98) ∂ (cid:99) f ∗ ∈ D ⊗ , by the form (8.2) of (cid:98) ∂ (cid:98) f (cid:48) .Then, we show that (cid:98) ∂ st (cid:99) f ∗ ∈ D ⊗ for all pairs of indices { s, t } disjoint with { , , i, j } and { s, t } (cid:54) = { u, v } where u and v are named in (8.6). Clearly, (cid:98) ∂ st (cid:99) f ∗ (cid:54)≡ (cid:98) ∂ ( st )(12) (cid:99) f ∗ ∈ D ⊗ . Wefirst show that in the UPF of (cid:98) ∂ st (cid:99) f ∗ , x and x appear in two distinct nonzero binary signatures.Otherwise, for a contradiction, suppose that there is a nonzero binary signature (cid:98) b ∗ ( x , x ) suchthat (cid:98) b ∗ ( x , x ) | (cid:98) ∂ st (cid:99) f ∗ . Then, (cid:98) b ∗ ( x , x ) | (cid:98) ∂ ( ij )( st ) (cid:99) f ∗ = (cid:98) ∂ ( st )( ij ) (cid:99) f ∗ (cid:54)≡ 0. By the form (8.6) of (cid:98) ∂ ij (cid:99) f ∗ ,the only way that x and x can form a nonzero binary signature in (cid:98) ∂ ( st )( ij ) (cid:99) f ∗ is that the merginggadget is actually merging x u and x v . Thus, { s, t } = { u, v } . Contradiction. Therefore, for some i (cid:48) and j (cid:48) , we have (cid:98) ∂ st (cid:99) f ∗ = (cid:100) b ∗ st ( x , x i (cid:48) ) ⊗ (cid:100) b ∗ st ( x , x j (cid:48) ) ⊗ (cid:99) h st , (8.7)for some (cid:100) b ∗ st ( x , x i (cid:48) ) , (cid:100) b ∗ st ( x , x j (cid:48) ) , (cid:99) h st (cid:54)≡ (cid:98) ∂ st (cid:99) f ∗ (cid:54)≡ 0. Since (cid:99) h st | (cid:98) ∂ (12)( st ) (cid:99) f ∗ and (cid:98) ∂ (12)( st ) (cid:99) f ∗ ∈D ⊗ , we have (cid:99) h st ∈ D ⊗ . Also, by Lemma 2.13, (cid:100) b ∗ st ∼ (cid:100) b ∗ st . For a contradiction, suppose that (cid:98) ∂ st (cid:99) f ∗ / ∈ D ⊗ , then (cid:100) b ∗ st ( x , x i (cid:48) ) (cid:54)∼ ( (cid:54) = ), and (cid:100) b ∗ st ( x , x j (cid:48) ) (cid:54)∼ ( (cid:54) = ). Consider the signature (cid:98) ∂ ( st )( ij ) (cid:99) f ∗ .Since { s, t } (cid:54) = { u, v } , by the form (8.6) of (cid:98) ∂ ij (cid:99) f ∗ , x and x appear in two binary signatures inthe UPF of (cid:98) ∂ ( st )( ij ) (cid:99) f ∗ . Remember that (cid:98) ∂ ( st )( ij ) (cid:99) f ∗ = (cid:98) ∂ ( ij )( st ) (cid:99) f ∗ . By the form (8.7) of (cid:98) ∂ st (cid:99) f ∗ , if { i (cid:48) , j (cid:48) } = { i, j } , then, after merging x i and x j of (cid:98) ∂ st (cid:99) f ∗ , x and x will form a new binary signaturein (cid:98) ∂ ( ij )( st ) (cid:99) f ∗ . Contradiction. Thus, { i (cid:48) , j (cid:48) } (cid:54) = { i, j } . Then, when merging x i and x j of (cid:98) ∂ st (cid:99) f ∗ , among (cid:100) b ∗ st ( x , x i (cid:48) ) and (cid:100) b ∗ st ( x , x j (cid:48) ), at least one binary signature is untouched. Thus, (cid:98) ∂ ( ij )( st ) (cid:99) f ∗ has afactor that is not an associate of (cid:54) = . A contradiction with (cid:98) ∂ ( ij )( st ) (cid:99) f ∗ ∈ D ⊗ , which is a consequenceof (8.6). Thus, (cid:98) ∂ st (cid:99) f ∗ ∈ D ⊗ .Then, we show that (cid:98) ∂ uv (cid:99) f ∗ ∈ D ⊗ . Recall the form (8.6) of (cid:98) ∂ ij (cid:99) f ∗ . Clearly, { u, v } is disjoint with { , , i, j } . Also, (cid:98) ∂ uv (cid:99) f ∗ (cid:54)≡ (cid:98) ∂ ( ij )( uv ) (cid:99) f ∗ ∈ D ⊗ . Consider the UPF of (cid:98) ∂ uv (cid:99) f ∗ . • If x and x appear in one nonzero binary signature (cid:99) b ∗ uv ( x , x ), then (cid:98) ∂ uv (cid:99) f ∗ = (cid:99) b ∗ uv ( x , x ) ⊗ (cid:99) g uv for some (cid:99) g uv (cid:54)≡ . Then, we have (cid:99) g uv ∼ (cid:98) ∂ (12)( uv ) (cid:99) f ∗ ∈ D ⊗ since (cid:98) ∂ (cid:99) f ∗ ∈ D ⊗ . Also, since (cid:99) b ∗ uv ( x , x ) | (cid:98) ∂ ( ij )( uv ) (cid:99) f ∗ ∈D ⊗ , we have (cid:99) b ∗ uv ( x , x ) ∈ D ⊗ . Hence, (cid:98) ∂ uv (cid:99) f ∗ ∈ D ⊗ .73 If x and x appear in two distinct nonzero binary signatures (cid:100) b ∗ uv ( x , x i (cid:48) ) and (cid:100) b ∗ uv ( x , x j (cid:48) ),then (cid:98) ∂ uv (cid:99) f ∗ = (cid:100) b ∗ uv ( x , x i (cid:48) ) ⊗ (cid:100) b ∗ uv ( x , x j (cid:48) ) ⊗ (cid:100) h uv for some (cid:100) h uv (cid:54)≡ . Then, we have (cid:100) h uv ∈ D ⊗ since (cid:98) ∂ (12)( uv ) (cid:99) f ∗ ∈ D ⊗ . By the form (8.6) of (cid:98) ∂ ij (cid:99) f ∗ , after mergingvariables x u and x v of (cid:98) ∂ ij (cid:99) f ∗ , variables x and x form a binary (cid:54) = in (cid:98) ∂ ( uv )( ij ) (cid:99) f ∗ = (cid:98) ∂ ( ij )( uv ) (cid:99) f ∗ .On the other hand, by the form of (cid:98) ∂ uv (cid:99) f ∗ , the only way that x and x form a binary aftermerging two variables in (cid:98) ∂ uv (cid:99) f ∗ is to merge x i (cid:48) and x j (cid:48) . Thus, we have { i (cid:48) , j (cid:48) } = { i, j } . Since (cid:99) f ∗ has arity 2 n (cid:62) 8, we can find another pair of indices { s, t } disjoint with { , , i, j, u, v } .When merging variables x s and x t in (cid:98) ∂ uv (cid:99) f ∗ , binary signatures (cid:100) b ∗ uv ( x , x i (cid:48) ) and (cid:100) b ∗ uv ( x , x j (cid:48) )are untouched. Thus, we have (cid:100) b ∗ uv ( x , x i (cid:48) ) ⊗ (cid:100) b ∗ uv ( x , x j (cid:48) ) | (cid:98) ∂ ( st )( uv ) (cid:99) f ∗ . As showed above, wehave (cid:98) ∂ st (cid:99) f ∗ ∈ D ⊗ and then (cid:98) ∂ ( st )( uv ) (cid:99) f ∗ ∈ D ⊗ . Thus, (cid:100) b ∗ uv ( x , x i (cid:48) ) ⊗ (cid:100) b ∗ uv ( x , x j (cid:48) ) ∈ D ⊗ and then (cid:98) ∂ uv (cid:99) f ∗ ∈ D ⊗ .So far, we have shown that (cid:98) ∂ (cid:99) f ∗ ∈ D ⊗ , (cid:98) ∂ ij (cid:99) f ∗ ∈ D ⊗ and (cid:98) ∂ st (cid:99) f ∗ ∈ D ⊗ for all { s, t } disjoint with { , , i, j } . If we can further show that (cid:98) ∂ ik (cid:99) f ∗ ∈ D ⊗ for all k (cid:54) = 1 , , i, j , and then symmetrically (cid:98) ∂ jk (cid:99) f ∗ ∈ D ⊗ for all k (cid:54) = 1 , , i, j , then (cid:98) ∂ st (cid:99) f ∗ ∈ D ⊗ for all { s, t } disjoint with { , } . Thus, by ourClaim, (cid:99) f ∗ ∈ (cid:98)(cid:82) D ⊗ . This will finish the proof of Case 2.Now we prove (cid:98) ∂ ik (cid:99) f ∗ ∈ D ⊗ for all k (cid:54) = 1 , , i, j . Since (cid:98) ∂ ( ik )(12) (cid:99) f ∗ ∈ D ⊗ , we have (cid:98) ∂ ik (cid:99) f ∗ (cid:54)≡ 0. Sowe can consider the UPF of (cid:98) ∂ ik (cid:99) f ∗ . • If x and x appear in one nonzero binary signature, then (cid:98) ∂ ik (cid:99) f ∗ = (cid:99) b ∗ ik ( x , x ) ⊗ (cid:99) g ik for some (cid:99) g ik ∈ D ⊗ . Here, (cid:99) g ik ∈ D ⊗ since (cid:98) ∂ ( ik )(12) (cid:99) f ∗ ∈ D ⊗ . Since (cid:99) f ∗ has arity 2 n (cid:62) 8, we can pick a pair of indices { s, t } disjoint with { , , i, j, k } , and merge variables x s and x t of (cid:98) ∂ ik (cid:99) f ∗ . Then, (cid:99) b ∗ ik ( x , x ) | (cid:98) ∂ ( st )( ik ) (cid:99) f ∗ . Since (cid:98) ∂ st (cid:99) f ∗ ∈ D ⊗ , (cid:98) ∂ ( st )( ik ) (cid:99) f ∗ = (cid:98) ∂ ( ik )( st ) (cid:99) f ∗ ∈ D ⊗ . Thus, (cid:99) b ∗ ik ( x , x ) ∈ D ⊗ and then (cid:98) ∂ ik (cid:99) f ∗ ∈ D ⊗ . • If x and x appear in two nonzero distinct binary signatures, then (cid:98) ∂ ik (cid:99) f ∗ = (cid:100) b ∗ ik ( x , x p ) ⊗ (cid:100) b ∗ ik ( x , x q ) ⊗ (cid:99) h ik for some (cid:99) h ik ∈ D ⊗ . Again, here (cid:99) h ik ∈ D ⊗ since (cid:98) ∂ ( ik )(12) (cid:99) f ∗ ∈ D ⊗ . By connecting variables x and x of (cid:98) ∂ ik (cid:99) f ∗ , x p and x q will form a binary disequality up to a nonzero scalar (this binary signature isdisequality because we know that (cid:98) ∂ ( ik )(12) (cid:99) f ∗ ∈ D ⊗ ). By Lemma 2.13, as the type of binarysignatures, (cid:100) b ∗ ik ∼ (cid:100) b ∗ ik . Between x p and x q , at least one of them is not x j ; suppose that it is x p . We pick a variable x r in the scope of (cid:99) h ik that is also not x j (such a variable x r exists as2 n (cid:62) x p and x r of (cid:98) ∂ ik (cid:99) f ∗ , the binary signature (cid:100) b ∗ ik ( x , x q ) is untouched.Since { p, r } is disjoint with { , , i, j } , we have (cid:100) b ∗ ik ( x , x q ) | (cid:98) ∂ ( ik )( pr ) (cid:99) f ∗ ∈ D ⊗ . Thus, we have (cid:100) b ∗ ik ( x , x q ) ∈ D ⊗ and so does (cid:100) b ∗ ik ( x , x p ), since we have shown that they are associates asthe type of binary signatures. Thus, (cid:98) ∂ ik (cid:99) f ∗ ∈ D ⊗ .As remarked earlier, by symmetry, we also have (cid:98) ∂ jk (cid:99) f ∗ ∈ D ⊗ for all k (cid:54) = 1 , , i, j . Thus, we aredone with Case 2.Thus, an irreducible signature (cid:99) f ∗ ∈ (cid:98)(cid:82) D ⊗ of arity 2 n is realized from (cid:98) f .74 emark: Since (cid:99) f ∗ is realized from (cid:98) f by gadget construction, (cid:99) f ∗ satisfies ars as (cid:98) f does.We first give a condition (Lemma 8.3) in which we can quite straightforwardly get the (cid:54) = | (cid:98) f , (cid:98) F ) by given (cid:98) f ∈ (cid:98)(cid:82) D ⊗ is an irreducible 8-ary signature. Lemma 8.2. Let (cid:98) f = a (1 , ⊗ n + ¯ a (0 , ⊗ n + ( (cid:54) = )( x i , x j ) ⊗ (cid:98) g h be an irreducible n -ary signature,where n (cid:62) and (cid:98) g h is a nonzero EO signature (i.e., with half-weighted support) of arity n − .Then, (cid:98) f does not satisfy .Proof. By renaming variables, without loss of generality, we may assume that { i, j } = { , } .For any input 00 β (cid:54) = (cid:126) n of (cid:98) f , we have (cid:98) f (00 β ) = ( (cid:54) = )(0 , · (cid:98) g h ( β ) = 0. Thus, | (cid:98) f | = (cid:88) β ∈ Z n − | (cid:98) f (00 β ) | = | (cid:98) f ( (cid:126) n ) | . On the other hand, since both ( (cid:54) = )( x , x ) and (cid:98) g h are nonzero EO signatures, ( (cid:54) = )( x , x ) ⊗ (cid:98) g h is a nonzero EO signature. Then, we can pick an input 01 γ ∈ Z n with wt(01 γ ) = n such that (cid:98) f (01 γ ) = ( (cid:54) = )(0 , · (cid:98) g h ( γ ) (cid:54) = 0 . Since γ ∈ Z n − , and wt( γ ) = n − (cid:62) 1, there exists a bit γ i in γ such that γ i = 0. Without loss of generality, we may assume that 01 γ = 010 γ (cid:48) . Then, | (cid:98) f | (cid:62) | (cid:98) f ( (cid:126) n ) | + | (cid:98) f (010 γ (cid:48) ) | > | (cid:98) f ( (cid:126) n ) | = | (cid:98) f | . Note that the constant λ for the norm squares must be the same for all index pairs { i, j } ⊆ [2 n ] inorder to satisfy in Definition 4.1. Thus, (cid:98) f does not satisfy . Lemma 8.3. Let (cid:98) f ∈ (cid:98)(cid:82) D ⊗ be an irreducible 8-ary signature in (cid:98) F . If there exists a binary disequality ( (cid:54) = )( x i , x j ) and two pairs of indices { u, v } and { s, t } where { u, v }∩{ s, t } (cid:54) = ∅ such that ( (cid:54) = )( x i , x j ) | (cid:98) ∂ uv (cid:98) f and ( (cid:54) = )( x i , x j ) | (cid:98) ∂ st (cid:98) f , then Holant( (cid:54) = | (cid:98) F ) is For all pairs of indices { i, j } , since (cid:98) ∂ ij (cid:98) f ∈ D ⊗ , S ( (cid:98) ∂ ij (cid:98) f ) is on half-weight. By Lemma 2.10,we have (cid:98) f ( α ) = 0 for all wt( α ) (cid:54) = 0 , , 8. Suppose that (cid:98) f ( (cid:126) ) = a and by ars (cid:98) f ( (cid:126) ) = ¯ a . We canwrite (cid:98) f in the following form (cid:98) f = a (1 , ⊗ + ¯ a (0 , ⊗ + (cid:98) f h , where (cid:98) f h is an EO signature of arity 8.Clearly, (cid:98) ∂ ij (cid:98) f = (cid:98) ∂ ij (cid:98) f h for all { i, j } . Then, (cid:98) f h ∈ (cid:98)(cid:82) D ⊗ since (cid:98) f ∈ (cid:98)(cid:82) D ⊗ . In addition, since there existsa binary disequality ( (cid:54) = )( x i , x j ) and two pairs of indices { u, v } and { s, t } where { u, v } ∩ { s, t } (cid:54) = ∅ such that ( (cid:54) = )( x i , x j ) | (cid:98) ∂ uv (cid:98) f h , (cid:98) ∂ st (cid:98) f h , by Lemma 2.11, (cid:98) f h ∈ D ⊗ and ( (cid:54) = )( x i , x j ) | (cid:98) f h . Thus, (cid:98) f = a (1 , ⊗ + ¯ a (0 , ⊗ + ( (cid:54) = )( x i , x j ) ⊗ (cid:98) g h , where (cid:98) g h ∈ D ⊗ is a nonzero EO signature or arity 6 since (cid:98) f h ∈ D ⊗ . By Lemma 8.2, (cid:98) f does notsatisfy . Thus, Holant( (cid:54) = | (cid:98) F ) is D ⊗ , we give the following property. Now we adopt the following notation forbrevity. We use ( i, j ) to denote the binary disequality ( (cid:54) = )( x i , x j ) on variables x i and x j . Lemma 8.4. Let (cid:98) f ∈ D ⊗ be a signature of arity at least . If there exist { u, v } (cid:54) = { s, t } such that ( i, j ) | (cid:98) ∂ uv (cid:98) f and ( i, j ) | (cid:98) ∂ st (cid:98) f , then ( i, j ) | (cid:98) f . roof. For a contradiction, suppose that ( i, j ) (cid:45) (cid:98) f . Thus x i and x j appear in two separate disequal-ities in the UPF of (cid:98) f . Since (cid:98) f ∈ D ⊗ , there exists { (cid:96), k } such that ( i, (cid:96) ) ⊗ ( j, k ) | (cid:98) f . By merging twovariables of (cid:98) f , the only way to make x i and x j to form a binary disequality is by merging x (cid:96) and x k . By the hypothesis of the lemma, { (cid:96), k } = { u, v } = { s, t } . Contradiction. Theorem 8.5. Let (cid:98) f / ∈ (cid:98) O ⊗ be a signature of arity in (cid:98) F . Then • Holant( (cid:54) = | (cid:98) F ) is • there exists some (cid:98) Q ∈ (cid:99) O such that Holant( (cid:54) = | (cid:98) f , (cid:98) Q (cid:98) F ) (cid:54) T Holant( (cid:54) = | (cid:98) F ) . Proof. By Lemma 8.1, we may assume that an irreducible signature (cid:99) f ∗ of arity 8 where (cid:99) f ∗ ∈ (cid:98)(cid:82) D ⊗ is realizable from (cid:98) f , and (cid:99) f ∗ also satisfies ars . Otherwise, Holant( (cid:54) = | (cid:98) F ) is (cid:98) g / ∈ (cid:98) O ⊗ of arity 2 , (cid:98) f is realizable from (cid:99) f ∗ , or otherwise we get (cid:99) f ∗ by (cid:98) f . We first show that after renaming variables by applying a suitablepermutation to { , , . . . , } , for all { i, j } ⊆ { , , , } , ( (cid:96), k ) | (cid:98) ∂ ij (cid:98) f where { (cid:96), k } = { , , , }\{ i, j } .Furthermore, we show that either Holant( (cid:54) = | (cid:98) F ) is , | (cid:98) ∂ (cid:98) f , (5 , | (cid:98) ∂ (cid:98) f , (6 , | (cid:98) ∂ (cid:98) f , and (1 , | (cid:98) ∂ (cid:98) f or (1 , | (cid:98) ∂ (cid:98) f . (8.8)Consider (cid:98) ∂ (cid:98) f . Since (cid:98) f ∈ (cid:98)(cid:82) D ⊗ , (cid:98) ∂ (cid:98) f ∈ D ⊗ . By renaming variables, without loss of generality,we may assume that (cid:98) ∂ (cid:98) f = λ · (3 , ⊗ (5 , ⊗ (7 , , (8.9)for some λ ∈ R \ { } . Then, consider (cid:98) ∂ (cid:98) f . (cid:98) ∂ (cid:98) f , and (cid:98) ∂ (cid:98) f . There are two cases. • Case 1. (1 , | (cid:98) ∂ (cid:98) f , (cid:98) ∂ (cid:98) f and (cid:98) ∂ (cid:98) f . Then we can write (cid:98) ∂ (cid:98) f = (1 , ⊗ (cid:98) h for some (cid:98) h ∈ D ⊗ .Clearly, (cid:98) h ∼ (cid:98) ∂ (12)(56) (cid:98) f . By the form (8.9) and commutativity, (cid:98) ∂ (12)(56) (cid:98) f ∼ (3 , ⊗ (7 , . Thus, (cid:98) h ∼ (3 , ⊗ (7 , λ ∈ R \ { } , (cid:98) ∂ (cid:98) f = λ · (1 , ⊗ (3 , ⊗ (7 , . (8.10)Similarly, we have (cid:98) ∂ (cid:98) f = λ · (1 , ⊗ (3 , ⊗ (5 , , and (cid:98) ∂ (cid:98) f = λ · (1 , ⊗ (5 , ⊗ (7 , , for some λ , λ ∈ R \ { } .Let (cid:98) g = (1 , ⊗ (3 , { i, j } ⊆ { , , , } and { (cid:96), k } = { , , , }\{ i, j } . If we mergevariables x i and x j of (cid:98) g , i.e., if we form (cid:98) ∂ ij (cid:98) g , then clearly variables x (cid:96) and x k will form adisequality. Thus, for all { i, j } ⊆ { , , , } , ( (cid:96), k ) | (cid:98) ∂ ij (cid:98) g . Then, ( (cid:96), k ) | (cid:98) ∂ ij (cid:98) g ⊗ (7 , ∼ (cid:98) ∂ ( ij )(56) (cid:98) f by (8.10), and similarly ( (cid:96), k ) | (cid:98) ∂ ij (cid:98) g ⊗ (5 , ∼ (cid:98) ∂ ( ij )(78) (cid:98) f . By Lemma 8.4, ( (cid:96), k ) | (cid:98) ∂ ij (cid:98) f . • Case 2. Among (cid:98) ∂ (cid:98) f , (cid:98) ∂ (cid:98) f , and (cid:98) ∂ (cid:98) f , there is at least one signature that is not divisible by(1 , , (cid:45) (cid:98) ∂ (cid:98) f . Since (cid:98) ∂ (cid:98) f ∈ D ⊗ , there exists { u, v } disjoint from { , , , } such that (1 , u ) ⊗ (2 , v ) | (cid:98) ∂ (cid:98) f . Then, by merging variables x and x of (cid:98) ∂ (cid:98) f , we have ( u, v ) | (cid:98) ∂ (12)(56) (cid:98) f ; comparing it to (cid:98) ∂ (56)(12) (cid:98) f using the form of (8.9)and by unique factorization we get { u, v } = { , } or { , } . Without loss of generality (i.e.,this is still within the freedom of our naming variables subject to the choices made so far),76e may assume that { u, v } = { , } and furthermore, u = 3 and v = 4. Then, for some λ (cid:48) ∈ R \ { } , (cid:98) ∂ (cid:98) f = λ (cid:48) · (1 , ⊗ (2 , ⊗ (7 , . (8.11)Then, consider (cid:98) ∂ (cid:98) f . We show that (5 , | (cid:98) ∂ (cid:98) f . Otherwise, there exists { s, t } disjoint from { , , , } such that (5 , s ) ⊗ (6 , t ) | (cid:98) ∂ (cid:98) f . By merging two variables of (cid:98) ∂ (cid:98) f , the only wayto make x and x form a binary disequality is to merge x s and x t . By the form (8.9),(5 , | (cid:98) ∂ (12)(78) (cid:98) f . Thus, { s, t } = { , } . From (5 , s ) ⊗ (6 , t ) | (cid:98) ∂ (cid:98) f , and { s, t } = { , } we knowthat x and x will form a binary disequality in (cid:98) ∂ (56)(78) (cid:98) f . Thus, (1 , | (cid:98) ∂ (56)(78) (cid:98) f . However,by (8.11) (cid:98) ∂ (56)(78) (cid:98) f ∼ (1 , ⊗ (2 , (cid:98) ∂ (cid:98) f = (5 , ⊗ (cid:98) g (cid:48) and (cid:98) g (cid:48) ∼ (cid:98) ∂ (56)(78) (cid:98) f ∼ (1 , ⊗ (2 , . Then, for some λ (cid:48) ∈ R \ { } , (cid:98) ∂ (cid:98) f = λ (cid:48) · (1 , ⊗ (2 , ⊗ (5 , . (8.12)Let { i, j } ⊆ { , , , } and { (cid:96), k } = { , , , }\{ i, j } . If we merge variables x i and x j of (cid:98) g (cid:48) ,which is an associate of (1 , ⊗ (2 , x (cid:96) and x k will form a disequality.Thus, for all { i, j } ⊆ { , , , } , ( (cid:96), k ) ∼ (cid:98) ∂ ij (cid:98) g (cid:48) . Then, ( (cid:96), k ) | (cid:98) ∂ ij (cid:98) g (cid:48) ⊗ (7 , ∼ (cid:98) ∂ ( ij )(56) (cid:98) f (by(8.11)) and ( (cid:96), k ) | (cid:98) ∂ ij (cid:98) g (cid:48) ⊗ (5 , ∼ (cid:98) ∂ ( ij )(78) (cid:98) f (by (8.12)). By Lemma 8.4, ( (cid:96), k ) | (cid:98) ∂ ij (cid:98) f .Thus, in both cases, we have ( (cid:96), k ) | (cid:98) ∂ ij (cid:98) f where { i, j } (cid:116) { (cid:96), k } = { , , , } is an arbitrarydisjoint union of two pairs. Now, we show that in both cases, (with possibly switching the names x and x , which we are still free to do), we can have(5 , | (cid:98) ∂ (cid:98) f , (5 , | ∂ (cid:98) f , (6 , | (cid:98) ∂ (cid:98) f . (8.13)Clearly, by the form (8.9), we have (5 , | (cid:98) ∂ (cid:98) f . Consider ∂ (cid:98) f . We already know that (2 , | ∂ (cid:98) f (in both cases). If (5 , | (cid:98) ∂ (cid:98) f , then since (5 , | (cid:98) ∂ (cid:98) f and { , } ∩ { , } (cid:54) = ∅ , by Lemma 8.3,Holant( (cid:54) = | (cid:98) F ) is , | (cid:98) ∂ (cid:98) f or (5 , | (cid:98) ∂ (cid:98) f . By renaming variables x and x ,we may assume that in both cases (cid:98) ∂ (cid:98) f = (2 , ⊗ (5 , ⊗ (6 , . (8.14)This renaming will not change any of the above forms of (cid:99) ∂ ij (cid:98) f . Consider (cid:98) ∂ (cid:98) f . We already have(1 , | (cid:98) ∂ (cid:98) f . We know (cid:98) ∂ (cid:98) f ∈ D ⊗ , and so in its UPF, (6 , r ) | (cid:98) ∂ (cid:98) f , for some r ∈ [8] \ { , , , , } .If (5 , | (cid:98) ∂ (cid:98) f , then since (5 , | (cid:98) ∂ (cid:98) f and { , } ∩ { , } (cid:54) = ∅ , by Lemma 8.3, we get , | (cid:98) ∂ (cid:98) f , then since (6 , | (cid:98) ∂ (cid:98) f by (8.14) and { , } ∩ { , } (cid:54) = ∅ , again by Lemma 8.3, we get r = 7 and (6 , | (cid:98) ∂ (cid:98) f . Therefore, we have established(8.13) in both cases. Furthermore, in Case 1, we have (1 , | (cid:98) ∂ (cid:98) f by form (8.10), and in Case 2,we have (1 , | (cid:98) ∂ (cid:98) f by form (8.11).Now, we show that for any α ∈ Z with wt( α ) = 1, (cid:98) f α ≡ 0. Since (3 , | (cid:98) ∂ (cid:98) f , ( (cid:98) ∂ (cid:98) f ) ≡ { , } is disjoint with { , } ,( (cid:98) ∂ (cid:98) f ) = (cid:98) ∂ ( (cid:98) f ) = (cid:98) f + (cid:98) f ≡ . (8.15)Since (1 , | (cid:98) ∂ (cid:98) f , ( (cid:98) ∂ (cid:98) f ) = (cid:98) ∂ ( (cid:98) f ) = (cid:98) f + (cid:98) f ≡ . (8.16)Since (1 , | (cid:98) ∂ (cid:98) f , ( (cid:98) ∂ (cid:98) f ) = (cid:98) ∂ ( (cid:98) f ) = (cid:98) f + (cid:98) f ≡ . (8.17)77omparing (8.15), (8.16) and (8.17), we have (cid:98) f = (cid:98) f = (cid:98) f ≡ . Since (2 , | (cid:98) ∂ (cid:98) f , ( (cid:98) ∂ (cid:98) f ) = (cid:98) ∂ ( (cid:98) f ) = (cid:98) f + (cid:98) f ≡ . Plug in (cid:98) f ≡ 0, we have (cid:98) f ≡ 0. Thus for any α ∈ Z with wt( α ) = 1, we have (cid:98) f α ≡ α ∈ Z with wt( α ) = 3 and any β ∈ Z , by ars we have, (cid:98) f α ( β ) = (cid:98) f α ( β ) = 0since wt( α ) = 1 . Thus, for any α ∈ Z with wt( α ) = 3, we also have (cid:98) f α ≡ α ∈ Z be an assignment of the first four variables of f , and β ∈ Z be an assignment of thelast four variables of f . Thus, for any α, β ∈ Z , (cid:98) f ( αβ ) = 0 if wt( α ) = 1 or 3. Also, since (cid:98) f ∈ (cid:98)(cid:82) D ⊗ ,by Lemma 2.10, (cid:98) f ( αβ ) = 0 if wt( α ) + wt( β ) (cid:54) = 0 , αβ ∈ S ( (cid:98) f ), | (cid:98) f ( αβ ) | = | (cid:98) f ( αβ ) | = | (cid:98) f ( αβ ) | = | (cid:98) f ( αβ ) | . By ars , | (cid:98) f ( αβ ) | = | (cid:98) f ( αβ ) | and | (cid:98) f ( αβ ) | = | (cid:98) f ( αβ ) | . So, we only need to show that | (cid:98) f ( αβ ) | = | (cid:98) f ( αβ ) | . (8.18)Pick an arbitrary { i, j } ⊆ { , , , } and an arbitrary { u, v } ⊆ { , , , } . Let { (cid:96), k } = { , , , }\{ i, j } and { s, t } = { , , , }\{ u, v } . Since (cid:98) f satisfies , by equation (4.6),we have | (cid:98) f ijuv | = | (cid:98) f ijuv | . Since (cid:98) f ( αβ ) = 0 if wt( α ) = 1 or 3, or wt( α ) + wt( β ) (cid:54) = 0 , | (cid:98) f ij(cid:96)kuvst | + | (cid:98) f ij(cid:96)kuvst | = | (cid:98) f ij(cid:96)kuvst | + | (cid:98) f ij(cid:96)kuvst | . (8.19)Note that for | (cid:98) f ijuv | , since we set x i x j = 00, the only possible nonzero terms are for x (cid:96) x k = 00 or11; furthermore, as we also set x u x v = 00, then x s x t = 00 if x (cid:96) x k = 00, and x s x t = 11 if x (cid:96) x k = 11.The situation is similar for | (cid:98) f ijuv | .Also, by considering | (cid:98) f ijst | = | (cid:98) f ijst | , we have | (cid:98) f ij(cid:96)kuvst | + | (cid:98) f ij(cid:96)kuvst | = | (cid:98) f ij(cid:96)kuvst | + | (cid:98) f ij(cid:96)kuvst | . (8.20)Comparing equations (8.19) and (8.20), we have | (cid:98) f ij(cid:96)kuvst | = | (cid:98) f ij(cid:96)kuvst | , and | (cid:98) f ij(cid:96)kuvst | = | (cid:98) f ij(cid:96)kuvst | . Also, by ars , | (cid:98) f ij(cid:96)kuvst | = | (cid:98) f ij(cid:96)kuvst | . As ( i, j, k, (cid:96) ) is an arbitrary permutation of (1 , , , 4) and ( u, v, s, t ) is an arbitrary permutation of(5 , , , (cid:98) f ( αβ ) vanishes if wt( α ) + wt( β ) (cid:54) = 0 , α, β ∈ Z . Hence, for all α, β ∈ Z , | (cid:98) f ( αβ ) | = | (cid:98) f ( αβ ) | = | (cid:98) f ( αβ ) | = | (cid:98) f ( αβ ) | . x x x x x x x α = 0110 (Col 1) α = 1010 (Col 2) α = 1100 (Col 3) α = 0110 (Row 1) m = (cid:98) f m = (cid:98) f m = (cid:98) f α = 1010 (Row 2) m = (cid:98) f m = (cid:98) f m = (cid:98) f α = 1100 (Row 3) m = (cid:98) f m = (cid:98) f m = (cid:98) f Table 6: Representative entries of (cid:98) f in terms of normsNote that (cid:98) f has at most 4 + (cid:0) (cid:1) · (cid:0) (cid:1) = 40 many possibly non-zero entries. In terms of norms,these 40 entries can be represented by (cid:98) f (cid:126) and the following 9 entries in Table 6. In other words,for every αβ ∈ Z where wt( α ) ≡ wt( β ) ≡ α ) + wt( β ) ≡ (cid:98) f ( αβ ), (cid:98) f ( αβ ), (cid:98) f ( αβ ) and (cid:98) f ( αβ ) appears in Table 6. We also view these 9 entries inTable 6 as a 3-by-3 matrix denoted by M = ( m ij ) i,j =1 .Let (cid:98) f (cid:126) = a . First we show that | m i, | + | m i, | + | m i, | = | a | , for i = 1 , , . (8.21)and | m ,j | + | m ,j | + | m ,j | = | a | , for j = 1 , , . (8.22)Let ( i, j, k ) be an arbitrary permutation of (1 , , | (cid:98) f ijk | = | (cid:98) f ijk | .Then, we have | (cid:98) f ijk | + | (cid:98) f ijk | + | (cid:98) f ijk | = | (cid:98) f ijk | = | a | . By taking ( i, j, k ) = (1 , , , (2 , , 3) and (3 , , i = 1 , , | (cid:98) f ijk | = | (cid:98) f ijk | where ( i, j, k ) is an arbitrary permutation of (5 , , , | (cid:98) ∂ (cid:98) f , we have (cid:98) ∂ (cid:98) f ( x , . . . , x ) = 0 if x = x . Notice that m + m = (cid:98) f + (cid:98) f is an entry of (cid:98) ∂ (cid:98) f on the input 101100. Thus, m + m = 0. Also, since (5 , | (cid:98) ∂ (cid:98) f , we have m + m = 0 . Since (6 , | (cid:98) ∂ (cid:98) f , we have m + m = 0 . Let x = | m | = | m | , y = | m | = | m | , and z = | m | = | m | . Plug x , y , z into equations (8.21)and (8.22). We have | m | + y + x = | m | + z + z = z + | m | + x = y + | m | + y = z + y + | m | = x + x + | m | . x = y = z and | m | = | m | = | m | . Consider m + m = (cid:98) f + (cid:98) f and m + m = (cid:98) f + (cid:98) f . They are entries of (cid:98) ∂ (cid:98) f on inputs 100110 and 101010. By form (8.9) of (cid:98) ∂ (cid:98) f , we have m + m = m + m ∈ R \{ } . Remember that we also have (1 , | (cid:98) ∂ (cid:98) f or (1 , | (cid:98) ∂ (cid:98) f . We first consider the case that (1 , | (cid:98) ∂ (cid:98) f . Then m + m = (cid:98) f + (cid:98) f = 0 . Thus, m + m = m − m ∈ R \{ } . Since | m | = | m | , | m | = | m | and m + m = 0 , | m | = | m | = | m | = | m | . Thus, m = m and m = − m . Let Re ( x ) the real part of a number x . Then, Re ( m ) + Re ( m ) = 2 Re ( m ) = Re ( m ) − Re ( m ) = − Re ( m ) . Thus, Re ( m ) = 0. Then, Re ( m ) = Re ( m ) = 0. Thus, m + m / ∈ R \{ } since Re ( m + m ) = 0 . Contradiction.Now, we consider the case that (1 , | (cid:98) ∂ (cid:98) f . Then m + m = (cid:98) f + (cid:98) f = 0 . Since m + m = 0 and m + m = 0, we have m = − m . Thus, we have m + m = m + m = m − m ∈ R \{ } . Taking the imaginary part, Im ( m ) + Im ( m ) = Im ( m ) − Im ( m ) = 0. Adding the two, weget Im ( m ) + Im ( m ) = 0, and thus, m + m ∈ R . Since | m | = | m | , m = m . Then, Re ( m ) = Re ( m ) . Also, since m + m = m − m ∈ R \{ } , Re ( m ) + Re ( m ) = Re ( m ) − Re ( m ) = Re ( m ) − Re ( m ) (cid:54) = 0 . Thus, Re ( m ) = 0, and Re ( m ) (cid:54) = 0. Suppose that m = d i for some d ∈ R . Then there exists c ∈ R \{ } such that m = c − d i and then m = c + d i . Remember that m + m = 0. Thus, m = − d i . Consider m + m = (cid:98) f + (cid:98) f = c − d i . It is an entry of the signature (cid:98) ∂ (cid:98) f . Since (cid:98) ∂ (cid:98) f ∈ D ⊗ , c − d i ∈ R . Thus, d = 0. Then, m = 0and m ∈ R . Thus, x = | m | = | m | = y = | m | = | m | = z = | m | = | m | = 0 , and | m | = | m | = | m | = | a | = | (cid:98) f ( (cid:126) | . (cid:98) f (cid:54)≡ a (cid:54) = 0. Thus, S ( (cid:98) f ) = { δδ, δδ, δδ, δδ ∈ Z | δ = 0000 , α , α , α } , where α , α , α are named in Table 6. It is easy to see that S ( (cid:98) f ) = S ( (cid:98) f ). Since m ∈ R , and | m | = | a | (cid:54) = 0, we can normalize it to 1. Since, (cid:98) ∂ (cid:98) f ∈ D ⊗ , we have1 = (cid:98) f ( α α ) + (cid:98) f ( α α ) = (cid:98) f ( α α ) + (cid:98) f ( α α ) = (cid:98) f ( α α ) + (cid:98) f ( α α ) = (cid:98) f ( α α ) + (cid:98) f ( α α ) . Since, (cid:98) f ( α α ) = (cid:98) f ( α α ) = (cid:98) f ( α α ) = (cid:98) f ( α α ) = 0, (cid:98) f ( α α ) = (cid:98) f ( α α ) = (cid:98) f ( α α ) = (cid:98) f ( α α ) = 1 . Similarly, since (cid:98) ∂ (cid:98) f ∈ D ⊗ , (cid:98) f ( α α ) = (cid:98) f ( α α ) = (cid:98) f ( α α ) = (cid:98) f ( α α ) = 1 . By ars , we have1 = (cid:98) f ( α α ) = (cid:98) f ( α α ) = (cid:98) f ( α α ) = (cid:98) f ( α α ) = (cid:98) f ( α α ) = (cid:98) f ( α α ) = (cid:98) f ( α α ) . Also, since (cid:98) ∂ (cid:98) f ∈ D ⊗ ,1 = (cid:98) f ( α α ) = (cid:98) f + (cid:98) f = (cid:98) f + (cid:98) f = (cid:98) f . Then, by ars , (cid:98) f = (cid:98) f = 1 . Thus, (cid:98) f ( γ ) = 1 for any γ ∈ S ( (cid:98) f ) with wt( γ ) = 4.Remember that (cid:98) f ( (cid:126) ) = a where | a | = 1. Then, (cid:98) f ( (cid:126) ) = a by ars . Suppose that a = e i θ . Let (cid:98) Q = (cid:104) ρ ρ (cid:105) ∈ (cid:99) O where ρ = e − i θ/ . Consider the holographic transformation by (cid:98) Q . (cid:98) Q does notchange the entries of (cid:98) f on half-weighed inputs, but change the values of (cid:98) f ( (cid:126) ) and (cid:98) f ( (cid:126) ) to 1. Thus, (cid:98) Q (cid:98) f = (cid:98) f . Then, Holant( (cid:54) = | (cid:98) f , (cid:98) Q (cid:98) F ) (cid:54) T Holant( (cid:54) = | (cid:98) F ) . Now, we want to show that Holant( (cid:54) = | (cid:98) f , (cid:98) Q (cid:98) F ) is (cid:98) Q ∈ (cid:99) O and all (cid:98) F where F = Z (cid:98) F is a real-valued signature set that does not satisfy condition (T). If so, then we are done.Recall that for all (cid:98) Q ∈ (cid:99) O , (cid:98) Q (cid:98) F = (cid:100) Q F for some Q ∈ O . Moreover, for all Q ∈ O , and allreal-valued F that does not satisfy condition (T), Q F is also a real-valued signature set that doesnot satisfy condition (T). Thus, it suffices for us to show that Holant( (cid:54) = | (cid:98) f , (cid:98) F ) is F that does not satisfy condition (T).The following Lemma shows that (cid:98) f gives non- (cid:98) B hardness (Definition 6.7). Lemma 8.6. Holant( (cid:54) = | (cid:98) f , (cid:98) F ) is (cid:98) F contains a nonzero binary signature (cid:98) b / ∈ (cid:98) B ⊗ .Equivalently, Holant( f , F ) is F contains a nonzero binary signature b / ∈ B ⊗ .Proof. We prove this lemma in the setting of Holant( (cid:54) = | (cid:98) f , (cid:98) F ). If (cid:98) b / ∈ (cid:98) O ⊗ , then by Lemma 5.1,we get (cid:98) b ∈ (cid:98) O ⊗ . Then, (cid:98) b has parity. We first consider thecase that (cid:98) b has even parity, i.e., (cid:98) b = ( a, , , ¯ a ). Since (cid:98) b (cid:54)≡ a (cid:54) = 0. We can normalize a to e i θ where0 (cid:54) θ < π . Then ¯ a = e − i θ . Since (cid:98) b / ∈ (cid:98) B , a (cid:54) = ± a (cid:54) = ± i . Thus, θ (cid:54) = 0 and θ (cid:54) = π .81e connect variables x and x of (cid:98) f with the two variables of (cid:98) b (using (cid:54) = ), and we get a 6-arysignature denoted by (cid:98) g . We rename variables x , x , x of (cid:98) g to x , x , x and variables x , x , x to x , x , x . Then, (cid:98) g has the following signature matrix M , ( (cid:98) g ) = e − i θ e i θ e i θ e − i θ e i θ e − i θ e − i θ 00 0 0 0 0 0 0 e i θ . Now, we show that (cid:98) g / ∈ (cid:98) O ⊗ . For a contradiction, suppose that (cid:98) g ∈ (cid:98) O ⊗ . Notice that S ( (cid:98) g ) = { ( x , . . . , x ) ∈ Z | x = x , x = x and x = x } . Then, we can write (cid:98) g as (cid:98) g = (cid:98) b ( x , x ) ⊗ (cid:98) b ( x , x ) ⊗ (cid:98) b ( x , x ) , where (cid:98) b = ( e i θ , , , e − i θ ), (cid:98) b = ( e i θ , , , e − i θ ) and (cid:98) b = ( e i θ , , , e − i θ ) . Then notice that (cid:98) g = e − i θ = (cid:98) b (0 , · (cid:98) b (0 , · (cid:98) b (0 , 0) = e i ( θ + θ + θ ) , and (cid:98) g = e − i θ = (cid:98) b (0 , · (cid:98) b (1 , · (cid:98) b (1 , 1) = e i ( θ − θ − θ ) . By multiplying the above two equations, we have e − i θ = e i ( θ + θ + θ ) · e i ( θ − θ − θ ) = e i θ . Also, notice that (cid:98) g = e i θ = (cid:98) b (0 , · (cid:98) b (0 , · (cid:98) b (1 , 1) = e i ( θ + θ − θ ) , and (cid:98) g = e i θ = (cid:98) b (0 , · (cid:98) b (1 , · (cid:98) b (0 , 0) = e i ( θ − θ + θ ) . By multiplying them, we have e i θ = e i ( θ + θ − θ ) · e i ( θ − θ + θ ) = e i θ . Thus, e i θ = e − i θ . Then, e i θ = 1 . Since, θ ∈ [0 , π ), θ = 0 or π . Contradiction. Thus, (cid:98) g / ∈ (cid:98) O ⊗ . ByLemma 7.21, we get (cid:98) b has odd parity, i.e., (cid:98) b ( y , y ) = (0 , e i θ , e − i θ , 0) where θ ∈ [0 , π ) afternormalization. We still consider the 6-ary signature (cid:98) g (cid:48) that is realized by connecting variables x and x of (cid:98) f with the two variables y and y of (cid:98) b (using (cid:54) = ). Then, after renaming variables, (cid:98) g (cid:48) has the following signature matrix M , ( (cid:98) g (cid:48) ) = e − i θ e i θ 00 0 0 0 0 e i θ e − i θ e i θ e − i θ e − i θ e i θ . (cid:98) g (cid:48) / ∈ (cid:98) O ⊗ . Thus, by Lemma 7.21, we get Z -transformation. Consider the problemHolant( f , F ). Remember that f = (cid:98) f . We observe that, by Lemma 8.6 the set { f } ∪ F is non- B hard, according to Definition 6.7. Then if we apply Theorem 7.19 to the set { f } ∪ F we see thatHolant b ( f , F ) is if we were able to show that B is realizable from f then we wouldbe done, since by Theorem 8.5, we either already have the F ), or we canrealize f from F , and thus the following reduction chain holdsHolant b ( f , F ) (cid:54) T Holant( f , F ) (cid:54) T Holant( F ) . Thus we get the F ) in either way.However, since f has even parity and all its entries are non-negative, all gadgets realizable from f have even parity and have non-negative entries. Thus, = − , (cid:54) = and (cid:54) = − cannot be realized from f by gadget construction. In fact, it is observed in [15] that f satisfies the following strong Bellproperty. Definition 8.7. A signature f satisfies the strong Bell property if for all pairs of indices { i, j } ,and every b ∈ B , the signature ∂ bij f realized by merging x i and x j of f using b is in { b } ⊗ . In this subsection, not using gadget construction but critically based on the strong Bell property of f , we prove that Holant b ( f , F ) (cid:54) T Holant( f , F ) in a novel way. We define the following Holantproblems with limited appearance. Definition 8.8. Let F be a signature set containing a signature f . The problem Holant( f (cid:54) k , F ) contains all instances of Holant( F ) where the signature f appears at most k times. Lemma 8.9. For any b ∈ B , Holant( b, f , F ) (cid:54) T Holant( b (cid:54) , f , F ) . Proof. Consider an instance Ω of Holant( b, f , F ). Suppose that b appears n times in Ω. If n (cid:54) b (cid:54) , f , F ). Otherwise, n (cid:62) 3. Consider the gadget ∂ bij f realized by connecting two variables x i and x j of f using b . (This gadget uses b only once.) Since f satisfies the strong Bell property, ∂ bij f = b ⊗ . Thus, by replacing three occurrences of b in Ω by ∂ bij f , we can reduce the number of occurrences of b by 2. We carry out this replacement a linearnumber of times to obtain an equivalent instance of Holant( b (cid:54) , f , F ), of size linear in Ω.Now, we are ready to prove the reduction Holant b ( f , F ) (cid:54) T Holant( f , F ). Note that ifHolant( f , F ) is b ∈ B , if we connect avariable of b with a variable of another copy of b using = , we get ± (= ). Also, for any b , b ∈ B where b (cid:54) = b if we connect the two variables of b with the two variables of b , we get a value 0. Lemma 8.10. Holant b ( f , F ) (cid:54) T Holant( f , F ) . Proof. We prove this reduction in two steps. Step 1. There exists a signature b ∈ B\{ = } such that Holant( b , f , F ) (cid:54) T Holant( f , F ) . We consider all binary and 4-ary signatures realizable by gadget constructions from { f } ∪ F . Ifa binary signature g / ∈ B is realizable from { f } ∪ F , then by Lemma 8.6, Holant( f , F ) is g ∈ B\{ = } is realizable from { f } ∪ F , then we are done bychoosing b = g . So we may assume that all binary signatures g realizable from { f } ∪ F are = (upto a scalar) or the zero binary signature, i.e., g = µ · (= ) for some µ ∈ R . Similarly, if a nonzero4-ary signature h / ∈ B ⊗ is realizable, then we have Holant( f , F ) is { f } ∪ F is non- B hard. If a nonzero 4-ary signature h ∈ B ⊗ \{ = } ⊗ isrealizable, then we can realize a binary signature b ∈ B\{ = } by factorization, and we are done.Thus, we may assume that all 4-ary signatures h realizable from { f } ∪ F are (= ) ⊗ or the 4-aryzero signature, i.e., h = λ · (= ) ⊗ for some λ ∈ R .Now, let b be a signature in B\{ = } . We show that Holant( b (cid:54) , f , F ) (cid:54) T Holant( f , F ) . Consider an instance Ω of Holant( b (cid:54) , f , F ). • If b does not appear in Ω, then Ω is already an instance of Holant( f , F ). • If b appears exactly once in Ω (we may assume it does connect to itself), then we mayconsider the rest of Ω that connects to b as a gadget realized from { f } ∪ F , which musthave signature λ · (= ), for some λ ∈ R . Connecting the two variables of b by (= ) for every b ∈ B\{ = } will always gives 0. Thus, Holant(Ω) = 0. • Suppose b appears exactly twice in Ω. It is easy to handle when the two copies of b form agadget of arity 0 or 2 to the rest of Ω. We may assume they are connected to the rest of Ω insuch a way that the rest of Ω forms a 4-ary gadget h realized from { f } ∪ F . We can namethe four dangling edges of h in any specific ordering as ( x , x , x , x ). Then h ( x , x , x , x ) = λ · (= )( x , x j ) ⊗ (= )( x k , x (cid:96) )for some partition { , , , } = { , j } (cid:116) { k, (cid:96) } , and some λ ∈ R . (Note that while wehave named four specific dangling edges as ( x , x , x , x ), the specific partition { , , , } = { , j } (cid:116) { k, (cid:96) } and the value λ are unknown at this point.) We consider the following threeinstances Ω , Ω , and Ω , where Ω s ( s ∈ { , , } ) is the instance formed by mergingvariables x and x s of h using = , and merging the other two variables of h using = (seeFigure 4 where h = h = (= ) and h = λ · h ⊗ h ). Since h is a gadget realized from { f } ∪ F , Ω , Ω , and Ω are instances of Holant( f , F ). Note that Holant(Ω s ) = 4 λ when s = j and Holant(Ω s ) = 2 λ otherwise. Thus, by computing Holant(Ω s ) for s ∈ { , , } , wecan get λ , and if λ (cid:54) = 0 the partition { , j } (cid:116) { k, (cid:96) } of the four variables. Thus we can get theexact structure of the 4-ary gadget h . In either case (whether λ = 0 or not), we can computethe value of Holant(Ω).Thus, Holant( b (cid:54) , f , F ) (cid:54) T Holant( f , F ) . By Lemma 8.9, Holant( b , f , F ) (cid:54) T Holant( f , F ) . Figure 4: Instances Ω j , Ω k and Ω (cid:96) tep 2. For any b ∈ B\{ = } , we have Holant b ( f , F ) (cid:54) T Holant( b , f , F ) . We show that we can get another b ∈ B\{ = , b } , i.e., for some binary signature b ∈ B\{ = , b } we have the reduction Holant( b , b , f , F ) (cid:54) T Holant( b , f , F ) . Then, by connecting one variableof b and one variable of b using = , we get the third signature in B\{ b , b } . Then, the lemmais proved. The proof is similar to the proof in Step 1. We consider all binary and 4-ary gadgetsrealizable from { b , f } ∪ F . Still, we may assume that all realizable binary signatures are of theform µ · (= ) or µ · b for some µ ∈ R , and all realizable 4-ary signatures are of form λ · (= ) ⊗ , λ · b ⊗ or λ · (= ) ⊗ b for some λ ∈ R . Otherwise, we can show that Holant( b , f , F ) is b ∈ B\{ = , b } directly by gadget construction.Then, let b be an arbitrary signature in B\{ = , b } . We show thatHolant( b (cid:54) , b , f , F ) (cid:54) T Holant( b , f , F ) . Consider an instance Ω of Holant( b (cid:54) , b , f , F ). If b does not appear in Ω, then Ω is alreadyan instance of Holant( b , f , F ). If b appears exactly once in Ω, then it is connected with abinary gadget g where g = µ · (= ) or g = µ · b . In both cases, the evaluation is 0. Thus,Holant(Ω) = 0. Suppose b appears exactly twice in Ω. Again it is easy to handle the case ifthe rest of Ω forms a gadget of arity 0 or 2 to the two occurrences of b . So we may assumethe two occurrences of b are connected to a 4-ary gadget h = λ · (= ) ⊗ , λ · b ⊗ or λ · (= ) ⊗ b . We denote the four variables of h by ( x , x , x , x ), by an arbitrary ordering of the four danglingedges. Then h ( x , x , x , x ) = λ · h ( x , x j ) ⊗ h ( x k , x (cid:96) ) where h , h ∈ { = , b } , for some λ and { j, k, (cid:96) } = { , , } . (Note that at the moment the values λ and j, k, (cid:96) are unknown.) We considerthe following three instances Ω , Ω and Ω , where Ω s ( s ∈ { , , } ) is the instance formed byconnecting variables x and x s of h using = , and connecting the other two variables of h using= (again see Figure 4). Clearly, Ω , Ω and Ω are instances of Holant( b , f , F ). Consider theevaluations of these instances. We have three cases. • If h = h = (= ), then Holant(Ω s ) = 4 λ when s = j and Holant(Ω s ) = 2 λ when s (cid:54) = j . • If h = h = b , then Holant(Ω s ) = 0 when s = j . If M ( b ) is the 2 by 2 matrix form for thebinary signature b where we list its first variable as row index and second variable as columnindex, then we have Holant(Ω k ) = λ · tr( M ( b ) M ( b ) T ), and Holant(Ω (cid:96) ) = λ · tr( M ( b ) ),where tr denotes trace. For b = (= − ) or ( (cid:54) = +2 ), the matrix M ( b ) is symmetric, and the valueHolant(Ω s ) = 2 λ in both cases s = k or s = (cid:96) . For b = ( (cid:54) = − ), M ( b ) T = − M ( b ), and wehave Holant(Ω k ) = 2 λ , and Holant(Ω (cid:96) ) = − λ . • If one of h and h is = and the other is b , then Holant(Ω s ) = 0 for all s ∈ { j, k, (cid:96) } . Thus, if the values of Holant(Ω s ) for s ∈ { , , } are not all zero, then λ (cid:54) = 0 and the third case isimpossible, and we can tell whether h is in the form λ · (= ) ⊗ or λ · ( b ) ⊗ . Moreover we can getthe exact structure of h , i.e., the value λ and the decomposition form of h and h . Otherwise, thevalues of Holant(Ω s ) for s ∈ { , , } are all zero. Then we can write h = λ · (= )( x , x j ) ⊗ b ( x k , x (cid:96) )or h = λ · b ( x , x j ) ⊗ (= )( x k , x (cid:96) ), including possibly λ = 0, which means h ≡ 0. (Note that if λ (cid:54) = 0, this uniquely identifies that we are in the third case; if λ = 0 then this form is still formallyvalid, even though we cannot say this uniquely identifies the third case. But when λ = 0 all threecases are the same, i.e., h ≡ λ and thedecomposition form of h .We further consider the following three instances Ω (cid:48) , Ω (cid:48) and Ω (cid:48) , where Ω (cid:48) s ( s ∈ { , , } ) isthe instance formed by connecting variables x and x s of h using b , and connecting the other twovariables of h using = . (In other words, we replace the labeling = of the edge that is connected85o the variable x in each instance illustrated in Figure 4 by b .) It is easy to see that Ω (cid:48) , Ω (cid:48) andΩ (cid:48) are instances of Holant( b , f , F ). Consider the evaluations of these instances. • If h = (= )( x , x j ), then Holant(Ω (cid:48) s ) = 0 when s = j . Also we have Holant(Ω k ) = λ · tr( M ( b ) ), and Holant(Ω (cid:96) ) = λ · tr( M ( b ) M ( b ) T ). For b = (= − ) or (cid:54) = +2 , the matrix M ( b ) is symmetric, and the value Holant(Ω s ) = 2 λ in both cases s = k or s = (cid:96) . For b = ( (cid:54) = − ), M ( b ) T = − M ( b ), and we have Holant(Ω k ) = − λ , and Holant(Ω (cid:96) ) = 2 λ . • If h = b ( x , x j ), then Holant(Ω (cid:48) s ) = 4 λ when s = j and Holant(Ω (cid:48) s ) = 2 λ when s (cid:54) = j .Thus, by further computing Holant(Ω (cid:48) s ) for s ∈ { , , } , we can get the exact structure of h .Therefore, by querying Holant( b , f , F ) at most 6 times, we can compute h exactly. Then, wecan compute Holant(Ω) easily. Thus, Holant( b (cid:54) , b , f , F ) (cid:54) T Holant( b , f , F ) . By Lemma 8.9,Holant( b , b , f , F ) (cid:54) T Holant( b , f , F ) . The other signature in B\{ = , b , b } can be realized byconnecting b and b . Thus, Holant b ( f , F ) (cid:54) T Holant( b , f , F ) . Therefore, Holant b ( f , F ) (cid:54) T Holant( f , F ) . Since Holant b ( f , F ) (cid:54) T Holant( f , F ) and { f } ∪ F is non- B hard for any real-valued F thatdoes not satisfy condition (T), by Theorem 7.19, we have the following result. Lemma 8.11. Holant( f , F ) is Combining Theorem 8.5 and Lemma 8.11, we have the following result. Lemma 8.12. If (cid:98) F contains a signature (cid:98) f of arity and (cid:98) f / ∈ (cid:98) O ⊗ , then Holant( (cid:54) = | (cid:98) F ) is n (cid:62) Now, we show that our induction framework works for signatures of arity 2 n (cid:62) Lemma 9.1. If (cid:98) F contains a signature (cid:98) f of arity n (cid:62) and (cid:98) f / ∈ (cid:98) O ⊗ , then, • Holant( (cid:54) = | (cid:98) F ) is P-hard, or • a signature (cid:98) g / ∈ (cid:98) O ⊗ of arity k (cid:54) n − is realizable from (cid:98) f .Proof. By Lemma 8.1, we may assume that an irreducible signature (cid:99) f ∗ of arity 2 n (cid:62) 10 where (cid:99) f ∗ ∈ (cid:98)(cid:82) D ⊗ is realizable, and (cid:99) f ∗ satisfies ars . We show that (cid:99) f ∗ does not satisfy , andhence we get { i, j } , since (cid:98) ∂ ij (cid:99) f ∗ ∈ D ⊗ , S ( (cid:98) ∂ ij (cid:99) f ∗ ) is on half-weight. By Lemma 2.10,we have (cid:99) f ∗ ( α ) = 0 for all wt( α ) (cid:54) = 0 , n, n . Suppose that (cid:99) f ∗ ( (cid:126) n ) = a and (cid:99) f ∗ ( (cid:126) n ) = ¯ a by ars . Wecan write (cid:99) f ∗ in the following form (cid:99) f ∗ = a (1 , ⊗ n + ¯ a (0 , ⊗ n + (cid:99) f ∗ h . where (cid:99) f ∗ h is an EO signature of arity 2 n (cid:62) ∂ ij (cid:99) f ∗ = ∂ ij (cid:99) f ∗ h for all { i, j } . Then, (cid:99) f ∗ h ∈ (cid:98)(cid:82) D ⊗ since (cid:99) f ∗ ∈ (cid:98)(cid:82) D ⊗ . Since (cid:99) f ∗ h is an EOsignature of arity at least 10 and (cid:99) f ∗ h ∈ (cid:98)(cid:82) D ⊗ , by Lemma 2.11, we have (cid:99) f ∗ h ∈ D ⊗ . Recall that allsignatures in D ⊗ are nonzero by definition. Pick some { i, j } such that ( (cid:54) = )( x i , x j ) | (cid:99) f ∗ h . Then, (cid:99) f ∗ = a (1 , ⊗ n + ¯ a (0 , ⊗ n + (cid:98) b ∗ ( x i , x j ) ⊗ (cid:98) g ∗ h , where (cid:98) g ∗ h ∈ D ⊗ is a nonzero EO signature since (cid:99) f ∗ h ∈ D ⊗ . By Lemma 8.2, (cid:99) f ∗ does not satisfy . Thus, Holant( (cid:54) = | (cid:98) F ) is emark: Indeed, following from our proof, we can also show that there is no irreducible signature (cid:98) f of arity 2 n (cid:62) 10 that satisfies both and (cid:98) f ∈ (cid:98)(cid:82) (cid:98) O ⊗ .Finally, we give the proof of Theorem 1.2. We restate it here. Theorem 9.2. Let F be a set of real-valued signatures. If F satisfies the tractability condition (T) in Theorem 2.22, then Holant( F ) is polynomial-time computable; otherwise, Holant( F ) is By Theorem 2.22, if F satisfies condition (T), then Holant( F ) is P-time computable. Sup-pose that F does not satisfy condition (T). If F contains a nonzero signature of odd arity, then byTheorem 2.25, Holant( F ) is (cid:54) = | (cid:98) F ) ≡ T Holant( F ) is F is a set of signatures of even arity. Since F does not satisfy condition (T), (cid:98) F (cid:54)⊆ T . Since (cid:98) O ⊗ ⊆ T ,there is a signature (cid:98) f ∈ (cid:98) F of arity 2 n such that (cid:98) f / ∈ (cid:98) O ⊗ . We prove this theorem by induction on2 n . When 2 n (cid:54) 8, by Lemmas 5.1, 5.2, 7.21, 8.12, Holant( (cid:54) = | (cid:98) F ) is k (cid:62) 8, if 2 n (cid:54) k , then Holant( (cid:54) = | (cid:98) F ) is n = 2 k + 2 (cid:62) 10. By Lemma 9.1, Holant( (cid:54) = | (cid:98) F ) is (cid:54) = | (cid:98) g, (cid:98) F ) (cid:54) T Holant( (cid:54) = | (cid:98) F )for some (cid:98) g / ∈ (cid:98) O ⊗ of arity (cid:54) k . By the induction hypothesis, Holant( (cid:54) = | (cid:98) g, (cid:98) F ) is (cid:54) = | (cid:98) F ) is Acknowledgement We would like to thank Professor Mingji Xia for pointing out that the strong Bell property canbe used to prove Lemma 8.9, which led to the inspiration to prove Lemma 8.10. Without thisinspiration this paper may languish for much more time. 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