Holographic Algorithms with Matchgates Capture Precisely Tractable Planar #CSP
HHolographic Algorithms with Matchgates CapturePrecisely Tractable Planar
Jin-Yi Cai ∗ Pinyan Lu † Mingji Xia ‡ Abstract
Valiant introduced matchgate computation and holographic algorithms. A number of seemingly exponentialtime problems can be solved by this novel algorithmic paradigm in polynomial time. We show that, in avery strong sense, matchgate computations and holographic algorithms based on them provide a universalmethodology to a broad class of counting problems studied in statistical physics community for decades. Theycapture precisely those problems which are every problem in this class belongs to precisely three categories: (1) those which are tractable(i.e., polynomial time computable) on general graphs, or (2) those which are
Given a set of functions F , the Counting Constraint Satisfaction Problem F ) is the following problem: Aninput instance consists of a set of variables X = { x , x , . . . , x n } and a set of constraints where each constraint is afunction f ∈ F applied to some variables in X . The output is the sum, over all assignments to X , of the productsof these function evaluations. This sum-of-product evaluation is called the partition function . In the special casewhere f ∈ F outputs values in { , } it counts the number of satisfying assignments. But constraint functionstaking real or complex values are also interesting, called (real or complex) weighted F consists ofreal or complex valued functions in general. There is a deeper reason for allowing this generality: The theory of holographic reductions is a powerful tool which operates naturally over C , even if the original problem has only0-1 valued functions.A closely related framework for locally constrained counting problems is called Holant Problems [14, 16]. Thisframework is inspired by the introduction of Holographic Algorithms by L. Valiant [40, 39]. In two ground-breakingpapers [38, 40] Valiant introduced matchgates and holographic algorithms based on matchgates to solve a numberof problems in polynomial time, which appear to require exponential time. At the heart of these exotic algorithmsis a tensor transformation from a given problem to the problem of counting (complex) weighted perfect matchingsover planar graphs. The latter problem has a remarkable P-time algorithm (FKT-algorithm) [33, 24, 25]. Planarityis crucial, as counting perfect matchings over general graphs is symmetric function with at most 3 Boolean variables on a matchgate.This work has been extended in [11]. In particular we have obtained a complete characterization of all realizablesymmetric functions by matchgates over the complex field C .The study of “tractable ∗ University of Wisconsin-Madison. [email protected] . Supported by NSF CCF-0830488 and CCF-0511679. † Microsoft Research Asia. [email protected] ‡ Institute of Software, Chinese Academy of Sciences. [email protected] . Supported by the Grand Challenge Program “NetworkAlgorithms and Digital Information” of the Institute of Software, CAS. Partially supported by NSFC 60970003. a r X i v : . [ c s . CC ] A ug o systems with polynomial time computable partition functions. This is captured completely by the computerscience notion of “tractable or are there other more exotic algorithmic paradigms yet undiscovered? A suspicion (and perhapsan audacious proposition) is that they have indeed captured all tractable planar counting problems. If so it wouldprovide a universal methodology to a broad class of counting problems studied in statistical physics and beyond.The results of this paper can be viewed as an affirmation of that suspicion. Within the framework of weightedBoolean all symmetric real valued functions.While G = ( V, E ) is given,where each v ∈ V is attached a function f v ∈ F , mapping { , } deg( v ) → R or C . We consider all edge assignments σ : E → { , } . For each σ , f v takes its input bits from the incident edges E ( v ) at v , and evaluates to f v ( σ | E ( v ) ).The counting problem on instance G is to compute Holant G = (cid:80) σ (cid:81) v ∈ V f v ( σ | E ( v ) ). In effect, in a Holantproblem, edges are variables and vertices represent constraint functions. This framework is very natural; e.g., theproblem of Perfect Matching corresponds to attaching the
Exact-One function at each vertex, taking 0-1inputs. The class of all Holant problems with function set F is denoted by Holant( F ).Every k : { , } k → { , } the Equality function of arity k , which is 1 on 0 k and 1 k , and is 0 elsewhere. Then we canturn the v on LHS by= deg( v ) . In fact, F ) is exactly the same as Holant( F ∪ { = k | k ≥ } ). Thus, Equality functions are available for free, or assumed to be present. However,when we wish to discuss some restricted classes of counting problems, e.g., for 3-regular graphs, the framework ofHolant problems is the more natural one. And as it turns out, the main technical breakthrough for our dichotomytheorem for planar X . For a symmetric function on k variables, we denoteit as [ f , f , . . . , f k ], where f i is the value of f on inputs of Hamming weight i . E.g., (= ) = [1 , , (= ) = [1 , , ) = [1 , , ,
1] etc. When we relax Holant problems by allowing all
Equality functions for free, we obtain = [1 ,
0] and = [0 ,
1] denote the constant 0 and 1 unary(arity 1) functions. Then Holant c is the natural class of Holant problems where and are free. This amountsto computing Holant on input graphs where we can set 0 or 1 to some dangling edges (one end has degree 1).Another class of Holant problems is called Holant ∗ problems where we assume all unary functions [ u , u ] arefree.In [16] we obtained a dichotomy theorem for (complex) Holant ∗ problems and (real) Holant c problems. Thedichotomy criterion for Holant ∗ problems is still valid for planar graphs . The proof of dichotomy theorems in thispaper starts from there.In Section 4, we prove that for any real-valued symmetric function set F , the planar Holant c ( F ) problemis tractable (i.e., computable in P) iff either it is tractable over general graphs (for which we already have aneffective dichotomy theorem [16]), or it is tractable because every function in F is realizable by a matchgate, inwhich case the planar Holant c ( F ) problem is computable by matchgates in P-time using FKT. In all other cases the problem is ∗ A crucial ingredient of the proof is a crossover construction whose validity is proved ∗ Strictly speaking, we must only consider F where functions take computable real numbers; this will be assumed implicitly. F , the planar F ) problem is tractable iff either it is tractable as F ) withoutthe planarity restriction (for which we have an effective dichotomy theorem [16]), or it is tractable becauseevery function in F is realizable by a matchgate under a specific holographic basis transformation. Thus planar F ) is solvable by a holographic algorithm in the second case. For all other F the problem is c in Section 4.Our third result is a dichotomy theorem for planar 2-3 regular bipartite Holant problems (Section 6). (Thistheorem deals with Holant problems without assuming unary and .) This includes Holant problems for 3-regular graphs as a special case. The tractability criterion is the same: Either it is tractable for general graphs(for which we also have an effective dichotomy theorem [9]), or it is tractable by a suitable holographic algorithm,which is a holographic reduction to FKT using matchgates. In all other cases the problem is Our functions take values in C by default. The framework of Holant problems is defined for functions mappingany [ q ] k → C for a finite q . Our results in this paper are for the Boolean case q = 2. So we give the followingdefinitions only for q = 2 for notational simplicity.A signature grid Ω = ( H, F , π ) consists of a graph H = ( V, E ), and a labeling π which labels each vertex witha function f v ∈ F . The Holant problem on instance Ω is to compute Holant Ω = (cid:80) σ (cid:81) v ∈ V f v ( σ | E ( v ) ), a sum overall edge assignments σ : E → { , } . A function f v can be represented as a vector of length 2 deg( v ) , or a tensorin ( C ) ⊗ deg( v ) . A function f ∈ F is also called a signature . We denote by = k the Equality signature of arity k .A symmetric function f on k Boolean variables can be expressed by [ f , f , . . . , f k ], where f i is the value of f oninputs of Hamming weight i . Thus, (= k ) = [1 , , . . . , ,
1] (with k − Definition 2.1.
Given a set of signatures F , we define a counting problem Holant( F ) :Input: A signature grid Ω = ( G, F , π ) ;Output: Holant Ω . Planar Holant problems are Holant problems on planar graphs.
Definition 2.2.
Given a set of signatures F , we define a counting problem Pl-Holant( F ) :Input: A signature grid Ω = ( G, F , π ) , where G is a planar graph;Output: Holant Ω . We would like to characterize the complexity of Holant problems in terms of its signature sets. † For some F , it is possible that Holant( F ) is F ) is tractable. These new tractable cases makedichotomies for planar Holant problems more challenging. This is also the focus of this work. Some specialfamilies of Holant problems have already been widely studied. For example, if F contains all Equality signatures { = , = , = , . . . } , then this is exactly the weighted Definition 2.3.
Let U denote the set of all unary signatures. Given a set of signatures F , we use Holant ∗ ( F ) (or Pl-Holant ∗ ( F ) respectively) to denote Holant(
F ∪ U ) (or Pl-Holant(
F ∪ U ) respectively). † Usually our set of signatures F is a finite set, and the assertion of either Holant( F ) is tractable or F to be infinite, e.g., to include { = , = , = , . . . } or allunary signatures. Holant( F ) is tractable means that it is computable in P even when we include the description of the signatures inthe input Ω in the input size. Holant( F ) is F for which the problem is efinition 2.4. Given a set of signatures F , we use Holant c ( F ) (or Pl-Holant c ( F ) respectively) to denote Holant(
F ∪ { [1 , , [0 , } ) ( or Pl-Holant(
F ∪ { [1 , , [0 , } ) respectively). Replacing a signature f ∈ F by a constant multiple cf , where c (cid:54) = 0, does not change the complexity ofHolant( F ). So we view f and cf as the same signature. An important property of a signature is whether it isdegenerate. Definition 2.5.
A signature is degenerate iff it is a tensor product of unary signatures. In particular, a symmetricsignature in F is degenerate iff it can be expressed as λ [ x, y ] ⊗ k . F -Gate and Matchgate A signature from F is a basic function which can be used at a vertex in an input graph. Instead of a single vertex,we can use graph fragments to generalize this notion. An F -gate Γ is a tuple ( H, F , π ), where H = ( V, E, D )is a graph where the edge set consists of regular edges E and dangling edges D . Some nodes of degree 1 aredesignated as external nodes, and all other nodes are internal nodes; a dangling edge connects an internal nodeto an external node, while a regular edge connects two internal nodes. The labeling π assigns a function from F to each internal node. The dangling edges define variables for the F -gate. (See Figure 1 for one example.) WeFigure 1: An F -gate with 5 dangling edges.denote the regular edges in E by 1 , , . . . , m , and denote the dangling edges in D by m + 1 , m + 2 , . . . , m + n .Then we can define a function for this F -gate Γ = ( H, F , π ),Γ( y , y , . . . , y n ) = (cid:88) x ,x ,...x m H ( x , x , . . . , x m , y , y , . . . y n ) , where ( y , y , . . . , y n ) ∈ { , } n denotes an assignment on the dangling edges and H ( x , x , . . . , x m , y , y , . . . , y n )denotes the value of the signature grid on an assignment of all edges. We will also call this function the signatureof the F -gate Γ. An F -gate can be used in a signature grid as if it is just a single node with the particularsignature.Using the idea of F -gates, we can reduce one Holant problem to another. Let g be the signature of some F -gateΓ. Then Holant( F ∪ { g } ) ≤ T Holant( F ). The reduction is quite simple. Given an instance of Holant( F ∪ { g } ),by replacing every appearance of g by an F -gate Γ, we get an instance of Holant( F ). Since the signature of Γ is g , the values for these two signature grids are identical.We note that even for a very simple signature set F , the signatures for all F -gates can be quite complicatedand expressive. Matchgate signatures are an example. Matchgate is introduced by Valiant [38 , , F -gates, where F contains Exact-One functions of all arities and weight functions ([1 , , w ], w ∈ C ) on edges. (Formally, we replaceeach matchgate edge of weight w by a path of length 2, and the new node on the path is assigned weight function[1 , , w ].) The signature function Γ defined above for a matchgate is called a matchgate signature, or a standardsignature. A signature function is realizable by a matchgate if it is the standard signature of that matchgate.(After a holographic transformation, a signature function is realizable under a basis if it is the transformedsignature of a matchgate; see below.) 4 .3 Holographic Reduction To introduce the idea of holographic reductions, it is convenient to consider bipartite graphs. This is without lossof generality. For any general graph, we can make it bipartite by replacing each edge by a path of length two,and giving each new vertex the
Equality function = on 2 inputs. (This is just the incident graph.)We use Holant( G|R ) to denote all counting problems, expressed as Holant problems on bipartite graphs H = ( U, V, E ), where each signature for a vertex in U or V is from G or R , respectively. An input instance forthe bipartite Holant problem is a bipartite signature grid and is denoted as Ω = ( H, G|R , π ). Signatures in G aredenoted by column vectors (or contravariant tensors); signatures in R are denoted by row vectors (or covarianttensors) [18].One can perform (contravariant and covariant) tensor transformations on the signatures. We will definea simple version of holographic reductions, which are invertible. They are called holographic because theymay produce exponential cancellations in the tensor space. Suppose Holant( G|R ) and Holant( G (cid:48) |R (cid:48) ) are twoHolant problems defined for the same family of graphs, and T ∈ GL ( C ). We say that there is an (invertible)holographic reduction from Holant( G|R ) to Holant( G (cid:48) |R (cid:48) ), and T is the basis transformation, if the contravariant transformation G (cid:48) = T ⊗ g G and the covariant transformation R = R (cid:48) T ⊗ r map G ∈ G to G (cid:48) ∈ G (cid:48) and R ∈ R to R (cid:48) ∈ R (cid:48) , and vice versa, where G and R have arity g and r respectively. (Notice the reversal of directions whenthe transformation T ⊗ n is applied. This is the meaning of contravariance and covariance .) Theorem 2.6 (Valiant’s Holant Theorem [40]) . Suppose there is a holographic reduction from
G|R to G (cid:48) |R (cid:48) mapping signature grid Ω to Ω (cid:48) , then Holant Ω = Holant Ω (cid:48) . In particular, for invertible holographic reductions from Holant(
G|R ) to Holant( G (cid:48) |R (cid:48) ), one problem is in Piff the other one is, and similarly one problem is T ⊗ n F or T F , we viewthe signature or signatures as column vectors (or contravariant tensors); whenever we write a transformation as F T ⊗ n or F T , we view the signature or signatures as row vectors (or covariant tensors). In this subsection, we state some known dichotomy theorems. We first review three dichotomy theorems from[16].
Theorem 2.7.
Let F be a set of symmetric signatures over C . Then Holant ∗ ( F ) is computable in polynomialtime in the following three cases. In all other cases, Holant ∗ ( F ) is F is of arity no more than two;2. There exist two constants a and b (not both zero, depending only on F ), such that for every signature [ x , x , . . . , x n ] ∈ F one of the two conditions is satisfied: (1) for every k = 0 , , . . . , n − , we have ax k + bx k +1 − ax k +2 = 0 ; (2) n = 2 and the signature [ x , x , x ] is of form [2 aλ, bλ, − aλ ] .3. For every signature [ x , x , . . . , x n ] ∈ F , one of the two conditions is satisfied: (1) For every k = 0 , , . . . , n − , we have x k + x k +2 = 0 ; (2) n = 2 and the signature [ x , x , x ] is of form [ λ, , λ ] .The same dichotomy also holds for Pl-Holant ∗ ( F ) . Theorem 2.8.
Let F be a set of real symmetric signatures, and let F , F and F be three families of signaturesdefined as F = { λ ([1 , ⊗ k + i r [0 , ⊗ k ) | λ ∈ C , k = 1 , , . . . , r = 0 , , , } ; F = { λ ([1 , ⊗ k + i r [1 , − ⊗ k ) | λ ∈ C , k = 1 , , . . . , r = 0 , , , } ; F = { λ ([1 , i ] ⊗ k + i r [1 , − i ] ⊗ k ) | λ ∈ C , k = 1 , , . . . , r = 0 , , , } . Then
Holant c ( F ) is computable in polynomial time if (1) After removing unary signatures from F , it falls in oneof the three Classes of Theorem 2.7 (this implies Holant ∗ ( F ) is computable in polynomial time) or (2) (Withoutremoving any unary signature) F ⊆ F ∪ F ∪ F . Otherwise, Holant c ( F ) is F ∪ F ∪ F , up to an arbitrary scalar factor:1. ( F ): [1 , , , . . . , , − ,
2. ( F ): [1 , , , , . . . , ,
3. ( F ): [0 , , , , . . . , ,
4. ( F ): [1 , , − , , , , − , , . . . , − ,
5. ( F ): [0 , , , − , , , , − , . . . , − ,
6. ( F ): [1 , , − , − , , , − , − , . . . , − ,
7. ( F ): [1 , − , − , , , − , − , , . . . , − . Definition 2.9. A k -ary function f ( x , . . . , x k ) is affine if it has the form χ [ AX =0] i (cid:80) nj =1 (cid:104) α j ,X (cid:105) where X = ( x , x , . . . , x k , , and χ is a 0-1 indicator function such that χ [ AX =0] is 1 iff AX = 0 . Note that theinner product (cid:104) α, X (cid:105) is calculated over F , while the summation over j on the exponent of i = √− is over F .We use A to denote the set of all affine functions.We use P to denote the set of functions which can be expressed as a product of unary functions, binary equalityfunctions ( [1 , , on some two variables) and binary disequality functions ( [0 , , on some two variables). Theorem 2.10.
Suppose F is a set of functions mapping Boolean inputs to complex numbers. If F ⊆ A or F ⊆ P , then F ) is computable in polynomial time. Otherwise, F ) is As we mentioned in [16], the class A is a natural generalization of the symmetric signatures family F ∪F ∪F .It is easy to show that the set of symmetric signatures in A is exactly F ∪ F ∪ F .The following dichotomy for 2-3 regular graphs is from [27]. Theorem 2.11. ([27])
The problem Holant ([ y , y , y ] | [1 , , , is y , y , y ∈ C except in thefollowing cases, for which the problem is in P: (1) y = y y ; (2) y = y and y y = − y ( y (cid:54) = 0 ) ; (3) y = 0 ; (4) y = y = 0 . If we restrict the input to planar graphs, then these four categories are tractable in P,as well as a fifth category y = y , and the problem remains A matchgate is called even (respectively odd) if it has an even (respectively odd) number of vertices. The followingtwo lemmas are from [8].
Lemma 2.12.
A symmetric signature [ z , . . . , z m ] is the standard signature of some even matchgate iff for allodd i , z i = 0 , and there exist r and r not both zero, such that for every even ≤ k ≤ m , r z k − = r z k . Lemma 2.13.
A symmetric signature [ z , . . . , z m ] is the standard signature of some odd matchgate iff for alleven i , z i = 0 , and there exist r and r not both zero, such that for every odd ≤ k ≤ m , r z k − = r z k . In [13], we characterized all symmetric signatures realizable by matchgates under a given basis. Here we statethe theorem for a particular basis (cid:20) − (cid:21) , which will be used in Theorem 5.1. Theorem 2.14.
A symmetric signature [ x , x , . . . , x n ] is realizable under the basis (cid:20) − (cid:21) iff it takes one ofthe following forms: Form 1: there exist constants λ, s, t and (cid:15) where (cid:15) = ± , such that for all i, ≤ i ≤ n , x i = λ [( s + t ) n − i ( s − t ) i + (cid:15) ( s − t ) n − i ( s + t ) i ] . • Form 2: there exist a constant λ , such that for all i, ≤ i ≤ n , x i = λ [( n − i )( − i + i ( − i − ] . • Form 3: there exist a constant λ , such that for all i, ≤ i ≤ n , x i = λ [( n − i ] . In this section, we discuss the interpolation method we will use in this paper. Polynomial interpolation is apowerful tool in the study of counting problems initiated by Valiant [36] and further developed by Vadhan, Dyerand Greenhill [34, 20] and others. The method we use here is essentially the same as Vadhan [34].For some set of signatures F , suppose we want to show that for all unary signatures f = [ x, y ], we haveHolant( F ∪ { [ x, y ] } ) ≤ T Holant( F ). Let Ω = ( G, F ∪ { [ x, y ] } , π ). We want to compute Holant Ω in polynomialtime using an oracle for Holant( F ).Let V f be the subset of vertices in G assigned f in Ω. Suppose | V f | = n . We can classify all 0-1 assignments σ in the Holant sum according to how many vertices in V f whose incident edge is assigned a 0 or a 1. Then theHolant value can be expressed as Holant Ω = (cid:88) ≤ i ≤ n c i x i y n − i , (1)where c i is the sum over all edge assignments σ , of products of evaluations at all v ∈ V ( G ) − V f , where σ is suchthat exactly i vertices in V f have their incident edges assigned 0 (and n − i have their incident edges assigned 1.)If we can evaluate these c i , we can evaluate Holant Ω .Now suppose { G s } is a sequence of F -gates, and each G s has one dangling edge. Denote the signature of G s by f s = [ x s , y s ], for s = 0 , , . . . . If we replace each occurrence of f by f s in Ω we get a new signature grid Ω s ,which is an instance of Holant( F ), with Holant Ω s = (cid:88) ≤ i ≤ n c i x is y n − is . (2)One can evaluate Holant Ω s by oracle access to Holant( F ). Note that the same set of values c i occurs. We cantreat c i in (2) as a set of unknowns in a linear system. The idea of interpolation is to find a suitable sequence { f s } such that the evaluation of Holant Ω s gives a linear system (2) of full rank, from which we can solve all c i .In this paper, the sequence { G s } will be constructed recursively using suitable gadgetry. There are two gadgetsin a recursive construction: one gadget has arity 1, giving the initial signature g = [ x , y ]; the other has arity 2,giving the recursive iteration. It is more convenient to use a 2 × A to denote it. So we can recursivelyconnect them as in Figure 2 and get { G s } .Figure 2: Recursive construction.The signatures of { G s } have the following relation, (cid:20) x s y s (cid:21) = (cid:20) a a a a (cid:21) (cid:20) x s − y s − (cid:21) , (3)7here A = (cid:20) a a a a (cid:21) and g = (cid:20) x y (cid:21) .We call this gadget pair ( A, g ) a recursive construction. It follows from Lemma 6.1 in [34] that
Lemma 3.1.
Let α, β be the two eigenvalues of A . If the following three conditions are satisfied1. det( A ) (cid:54) = 0 ;2. g is not a column eigenvector of A (nor the zero vector);3. α/β is not a root of unity;then the recursive construction ( A, g ) can be used to interpolate all unary signatures. A similar interpolation method also works for signatures with larger arity but have two degrees of freedom.For example, all signatures of form [0 , x, , y ]. This is used in the proof of Lemma 4.9. c Problems
Before presenting the main dichotomy theorem for planar Holant c problems, we prove the following theorem,which plays a crucial role in the proof of the main theorem. Theorem 4.1.
Let a, b ∈ R . • If ab (cid:54) = 1 then Pl-Holant c ([ a, , , , b ]) is • If ab = 1 then Pl-Holant c ([ a, , , , b ]) is solvable in P. We first prove three lemmas which will be used in the proof of this theorem.
Lemma 4.2.
Let a, b, x ∈ R , ab (cid:54) = 0 and x (cid:54) = ± . Then Pl-Holant c ( { [ a, , , , b ] , [0 , , , x ] } ) is Firstly, we show how to realize (= ) = [1 , , , , , ,
1] by [ a, , , , b ]. [ a, , , , b ] can be attached to avertex of degree 4. We can connect 3 pairs of edges of two copies of [ a, , , , b ] to realize the binary function[ a , , b ].If a = b , then we connect one pair of edges from two copies of [ a, , , , b ] to get [ a , , , , , , b ]. This isthe same as (= ) = [1 , , , , , ,
1] after factoring out the non-zero factor a = b .If a (cid:54) = b , then we connect [ a, , , , b ] with a chain of [ a , , b ] of length i to get [ a i +1 , , , , b i +1 ]. Becausefor any i (cid:54) = j , a i +1 /b i +1 (cid:54) = a j +1 /b j +1 , we can realize (= ) = [1 , , , ,
1] using polynomial interpolation, asfollows. Consider any signature grid on a planar graph G with n occurrences of = together with some othersignatures. Let x k,(cid:96) be the sum, over all 0-1 edge assignments σ , of the products of all other vertex functionvalues in G except at n vertices with = , where k, (cid:96) ≥ k + (cid:96) = n , and in σ exactly k occurrences of = haveinput 0, and exactly (cid:96) occurrences of = have input 1. The Holant value is (cid:80) k + (cid:96) = n x k,(cid:96) . Now substitute eachoccurrence of = by [ a i +1 , , , , b i +1 ]. The new signature grid has Holant value (cid:80) k + (cid:96) = n x k,(cid:96) ( a k b (cid:96) ) i +1 . Thisgives a Vandermonde system from which we solve for x k,(cid:96) . Now we have = . Then we connect two copies of = on one pair of edges to get = .Take a vertex of degree 6 in a planar graph attached with = , where the 6 incident edges are its variables.We will bundle two adjacent variables to form 3 bundles of 2 edges each. Then if the inputs are restricted to { (0 , , (1 , } on each bundle, then the function takes value 1 on ((0 , , (0 , , (0 , , , (1 , , (1 , { (0 , , (1 , } , it is the ternary Equality function = .Let F = [0 , , , x ] and let H ( x , x , y , y ) = (cid:80) z =0 , F ( x , y , z ) F ( x , y , z ). This H is realizable by connect-ing one pair of edges of two copies of F . (See Figure 3.) We will consider H as a function in ( x , x ) and ( y , y ).However we will only connect H externally by connecting ( x , x ) and ( y , y ) to some bundle of two adjacentedges of some = . Since = enforces the values on the bundle to be either (0 ,
0) or (1 , H to the domain { (0 , , (1 , } . On this domain, H is a symmetric function of arity 2, andcan be denoted as [1 , , x ]. (Note that H is not a symmetric function of arity 4 on { , } , as H (0 , , ,
1) = x .)Now we have reduced Pl-Holant c ( { [1 , , , , [1 , , x ] } ) to Pl-Holant c ( { [ a, , , , b ] , [0 , , , x ] } ).8igure 3: The gadget for function H and [1 , , x k ].Using (= ) = [1 , , , Equality function = k of any arity k ≥
3. Then we can realize[1 , , x k ], for all k ≥
1. (See Figure 3.) If x = 0, then we already have [1 , , x (cid:54) = 0. Because x (cid:54) = 1 and being a positive real number, we can realize [1 , ,
0] by interpolation. Nowwe have reduced the problem Pl-Holant([1 , , , | [1 , , c ( { [1 , , , , [1 , , x ] } ). The bipartiteproblem Pl-Holant([1 , , , | [1 , , Vertex Covers on planar 3-regulargraphs [42].The following lemma handles a special case of Theorem 4.1. The proof uses Lemma 4.2.
Lemma 4.3.
Pl-Holant c ([0 , , , , is We construct a reduction from Pl-Holant c ([1 , , , , , [0 , , , c ([0 , , , , F = [0 , , , , F ) realizing the following sequence offunctions: H ( x , x , y , y ) = (cid:88) x ,x =0 , F ( x , x , x , x ) F ( y , y , x , x ) , and for i ≥ H i +2 ( x , x , y , y ) = (cid:88) x ,x =0 , H i ( x , x , x , x ) H ( y , y , x , x ) . The gadget for H i is composed of 2 i functions F . As an example, the gadget for H is shown in Figure 4.Figure 4: The gadget for H .By calculation, H i (0 , , ,
0) = H i (1 , , ,
1) = 1, and H i (0 , , ,
1) = H i (0 , , ,
0) = H i (1 , , ,
1) = H i (1 , , ,
0) = 2 i − , and H i is zero on other inputs. Again we will consider the inputs to H i as bundled into( x , x ) and ( y , y ).Given a planar graph G as an instance of Pl-Holant c ([1 , , , , , [0 , , , n vertices in G attached with the function (= ) = [1 , , , , i = 1 , , . . . , n + 1, we construct an instance G i ofPl-Holant c ([0 , , , , by a copy of H i , and replace each occurrenceof [0 , , ,
0] by [0 , , , ,
0] connected with a [0 , , , , with H i , we have bundled two adjacent edges together (in the planar embedding) for each vertex attached with= .Let x a,b denote the summation, over all 0-1 edge assignments σ , of the products of all other vertex functionvalues in G except at those n vertices with = , where a, b ≥ a + b = n , and in σ exactly a occurrences of= have inputs { , } , and exactly b occurrences of = have inputs { , , , } .Note that the Holant value on G i is (cid:88) a + b = n x ab a (2 i − ) b .
9n the other hand, the value of Pl-Holant c ([1 , , , , , [0 , , , G is exactly x n, .When we take 1 ≤ i ≤ n + 1, we get a system of linear equations in x ab , whose coefficient matrix is a fullranked Vandermonde matrix. Solving this Vandermonde system we obtain the value x n, .The following result can be proved by interpolation as well. Lemma 4.4.
Let a (cid:54)∈ {− , , } be a real number. Then we can interpolate all [ x, , y, and [0 , y, , x ] for x, y ∈ C starting from either [0 , , , a ] or [ a, , , .Proof. The recursive construction is depicted by Figure 5. By a simple parity argument, every F -gate N i has aFigure 5: The recursive construction. The signature of every vertex in the gadget is [0 , , , a ].signature of the form [0 , x i , , y i ]. After some calculation, we see that they satisfy the following recursive relation: (cid:20) x i +1 y i +1 (cid:21) = (cid:20) a + 1) a + a a + a ) a + 1 (cid:21) (cid:20) x i y i (cid:21) . The signatures we want to interpolate are of arity 3. But since all of them take the form [0 , x i , , y i ] with twodegrees of freedom, we can use the interpolation method in Section 3. Now we verify that the conditions ofthat theorem are satisfied. Let A = (cid:20) a + 1) ( a + a )3( a + a ) a + 1 (cid:21) , then ( A, [1 , a ] T ) forms a recursive construction. Sincedet( A ) = 3( a − (cid:54) = 0, the first condition holds. Its characteristic equation is X − ( a +3 a +4) X +3( a − = 0.For this quadratic equation, the discriminant ∆ = ( a − a − + 12( a + a ) >
0. So A has two distinct realeigenvalues. The sum of the two eigenvalues is tr A = a + 3 a + 4 >
0. So they are not opposite to each other.Therefore, the ratio of these two eigenvalues is not a root of unity and the third condition holds. Consider thesecond condition: if the initial vector [1 , a ] T is a column eigenvector of A , then we have A (cid:20) a (cid:21) = λ (cid:20) a (cid:21) , where λ is an eigenvalue of A . From this, we will conclude that a ( a − a −
1) = 0, which can not happen given a (cid:54)∈ {− , , } . To sum up, this recursive relation satisfies all three conditions of Lemma 3.1 and can be used tointerpolate all signatures of the form [0 , x, , y ]. This completes the proof. Proof of Theorem 4.1 If ab = 1, then [ a, , , , b ] is realizable by some matchgate, by Lemma 2.12. Thisrealizability also applies to the unary functions [1 ,
0] and [0 , c ([ a, , , , b ]) canbe solved in polynomial time by matchgate computation via the FKT method [33, 24, 25]. In the following weassume that ab (cid:54) = 1 and prove that the problem is a = b = 0 is proved in Lemma 4.3. Nowwe can assume at least one of a and b is non-zero, and by symmetry we assume a (cid:54) = 0.We know from our dichotomy for Holant c problems [16] that Holant c ([ a, , , , b ]) for general graphs is a = b = 1 or a = b = −
1, in which cases it is tractable. Both of these tractable cases are also includedin the tractable cases ( ab = 1) here. Therefore, if we can realize a cross function X with a planar gadget when10 b (cid:54) = 1, we can reduce Holant c ([ a, , , , b ]) for general graphs to Pl-Holant c ([ a, , , , b ]) and finish the proof.Here a cross function X has 4 input bits, and satisfies X = X = X = X = 1 and X α = 0 for allother inputs α ∈ { , } .If { a, b } (cid:54)⊂ {− , , } , we can use Lemma 4.4 to interpolate all [ x, , y, x, y ∈ C . If { a, b } ⊂ {− , , } ,then there are only four cases: [1 , , , , − , , , , − , , , ,
1] and [ − , , , , c , , c ,
0] where c c (cid:54) = 0 and c (cid:54) = ± c using thegadget in Figure 6. ( For [1 , , , , − , , ,
0] by using [1 ,
0] in the gadget; for [1 , , , , , , ,
0] by using [1 , − , , , , , , ,
8] by using [0 , − , , , , , , , , c (cid:48) , , , c (cid:48) ∈ R and c (cid:48) (cid:54)∈ { , ± } . As aresult, we can also interpolate all [ x, , y, x, y ∈ C .Figure 6: The signature of the degree 1 vertex in the gadget is [1 ,
0] or [0 , x, , y, x, y ∈ C , to build new gadgets. We alsohave all [ x, , y ] by connecting [ x, , y,
0] to a [1 , (cid:112) t/a, , (cid:112) a/t ] to each edge of the signature[ a, , , , b ], we get [ t, , , , ct ] for all t (cid:54) = 0, where c = ab (cid:54) = 1. Using all these, we will build a planar gadgetin Figure 7 to realize the cross function X . In the equations below x, y, t are three variables we can set to anycomplex numbers, with t (cid:54) = 0. The parameter c is given and not equal to 1.(Of course we presumably could not build a cross function X if c = 1; this is exactly when the problem is in P,and this is also exactly when our construction of X fails. If a cross function X were to exist when c = 1 then P = c = 1 is exactly when our constructionfails. This failure condition is by no means obvious from the equations below.)Figure 7: This gadget is to realize the Cross function. The signature for the center vertex (black and square) is[ t, , , , ct ]. The signature for the vertexes in the four corners (red and circle) is [ x, , , y, , , X = x y t + t + 4 x y + 4 x + 4 x y + 2 cx tX = 2 y t + 12 y + 2 ctX = X = 2 xy t + 4 x y + 4 + 4 xy + 2 cxtX = X = X = X = x y t + yt + 3 x y + 3 + 6 xy + 2 cxt . Here we prove that for any c (cid:54) = 1, we can assign suitable complex values to x, y and t , where t (cid:54) = 0, such that A = B = C (cid:54) = 0, and D = 0, where A , B , C and D denote respectively the four functions of x, y and t listed inthe four lines above. Claim 1.
For any c (cid:54) = 1 , ( x − = 16 c − has a solution x (cid:54)∈ { , +1 , − } . This x satisfies (cid:18) − x ( x + 3) x − (cid:19) (cid:18) x + 3 x − (cid:19) + cx + 6 = 0 . (4) Proof.
Clearly x = 1 is not a solution to ( x − = c − . Also the equation has two distinct roots. When c = 17there is a solution x = 2 (cid:54)∈ { , +1 , − } . When c (cid:54) = 17, we can verify x = 0 is not a solution. Hence the equationalways has a solution other than 0 , ± x − − x − x )( x + 3) + ( cx + 6)( x − x + 1)= − ( x + 4 x + 5 x + 6) + cx + (6 − c ) x + ( −
12 + c ) x + 6= ( c − x − c − x + ( c − x = ( c − x [( x − − / ( c − . Now we fix x (cid:54)∈ { , +1 , − } satisfying (4) for any given c (cid:54) = 1. Claim 2.
For any c (cid:54) = 1 , we can pick z (cid:54) = ± such that z (1 + z ) = x ( x + 3) x − . (5) Proof.
We are given x (cid:54) = 0 , ±
1. If x = −
3, we can pick z = 0. Now suppose x (cid:54) = −
3. Consider the quadraticequation in z z ( x −
1) = x ( x + 3)(1 + z ) . This is quadratic since x ( x + 3) (cid:54) = 0. We can check that z = +1 (and − x = − z (cid:54) = − x (cid:54) = 1 is equivalent to (5).Hence we have a solution z (cid:54) = ± z (cid:54) = ± y = z/x such that xy (cid:54) = ±
1, for any c (cid:54) = 1. Claim 3.
For any c (cid:54) = 1 , there exist x (cid:54)∈ { , +1 , − } and y such that xy (cid:54) = ± satisfying x y )(1 + xy ) · x + 3 x − cx + 6 = 0 . (6)12 roof. x y )(1 + xy ) · x + 3 x − cx + 6= 2 (cid:18) − z (1 + z ) (cid:19) · x + 3 x − cx + 6= (cid:18) − x ( x + 3) x − (cid:19) · x + 3 x − cx + 6= 0 . Here we used (5) and (4).Now we will set t = 4 / (1 + xy ) . Clearly t (cid:54) = 0. We next verify that D = 0.By (5) and (6) we get 8 y (1 + x y )(1 + xy ) + cx + 6 = 0 . Then t y (1 + x y ) + 2 cx + 3 t (1 + xy ) = 0 . Thus D = yt (1 + x y ) + 3(1 + xy ) + 2 cxt = 0 . Next we show that C = − xy ) − x (cid:54) = 0.By D = 0, we have C = 2 xy xy ) + 4(1 + xy ) − xy + [ − yt (1 + x y ) − xy ) ] . Hence C = 8 xy (1 + xy ) + (1 + xy ) − xy − y x y )(1 + xy ) = 4 y (1 + xy ) (cid:2) xy − − x y (cid:3) + (1 − xy ) = (cid:18) − y (1 + xy ) + 1 (cid:19) (1 − xy ) = 4(1 − xy ) − x (cid:54) = 0 , using (5).The next task is to show B = C .We have C = 4(1 − xy ) + xB. Hence B = 1 x (cid:20) − xy ) − x − − xy ) (cid:21) = 4(1 − xy ) x (cid:20) − x − (cid:21) = 4(1 − xy ) − x = C. Finally we verify A = C as well. A = ( x y + 1) t + x [ C − xy t ] = C + ( x − C − x y t + ( x y + 1) t = C − − xy ) + t ( x y − = C. Now we come to the main dichotomy theorem for Pl-Holant c problems.13 heorem 4.5. Let F be a set of real symmetric signatures. Pl-Holant c ( F ) is F satisfies one ofthe following conditions, in which case it is tractable:1. Holant c ( F ) is tractable (for which we have an effective dichotomy [16]); or2. Every signature in F is realizable by some matchgate (for which we have a complete characterization [8]). Before we give the proof, we do some normalization of the signature set F . Since any degenerate signature[ x, y ] ⊗ k can be replaced by the corresponding unary signature [ x, y ] without changing the complexity of theproblem, we always assume that all the signatures in F , whose arity is greater than 1, are non-degenerate. Since[1 ,
0] and [0 ,
1] are freely available, we can construct any sub-signature of an original signatures as well as anysignature realizable by some F -gate.The main idea of the proof is to interpolate all unary functions. If we can do that, we can reduce the problemPl-Holant ∗ ( F ) to Pl-Holant c ( F ) and finish the proof. We note that our dichotomy in [16] for Holant ∗ ( F ) alsoholds for planar graphs. In some cases, we cannot interpolate all unary functions, then we prove the theoremseparately, mainly using Lemma 4.2 and Theorem 4.1. The following lemma is for interpolation of unary functions. Lemma 4.6.
If we can construct from F a gadget with signature [ a, b, c ] , where b (cid:54) = ac , b (cid:54) = 0 and a + c (cid:54) = 0 ,then we can interpolate all unary functions. (Hence the conclusions of Theorem 4.5 hold.)Proof. we use the interpolation method as described in Section 3. We consider two recursive constructions( (cid:20) a bb c (cid:21) , (cid:20) (cid:21) ) and ( (cid:20) a bb c (cid:21) , (cid:20) (cid:21) ), and argue that at least one of them will succeed given the conditions on a, b, c .We use A to denote (cid:20) a bb c (cid:21) . Since b (cid:54) = ac , A is non-degenerate, the first condition of Lemma 3.1 is satisfied forboth recursive constructions. If both [1 ,
0] and [0 ,
1] are column eigenvectors of A , then b = 0, a contradiction.So at least for one of the two recursive constructions, the second condition of Lemma 3.1 is satisfied. Since A isa real symmetric matrix, both its eigenvalues are real. If the ratio of two real numbers is a root of unity, theymust be the same or opposite to each other. If the two eigenvalues are the same, we have b = 0 and a = c , acontradiction. If the two eigenvalues are opposite to each other, then we have a + c = 0, also a contradiction.Therefore, the third condition of Lemma 3.1 is also satisfied for both recursive constructions. To sum up, at leastone of the two recursive constructions satisfies all the conditions of Lemma 3.1. This completes the proof.If we can construct from F a gadget with a binary symmetric signature [ a, b, c ], which satisfies all the conditionsin Lemma 4.6, then we are done. For most cases, we prove the theorem by interpolating all unary signatures.However, in some more delicate cases, we are not able to do that. For example, if all signatures from F havethe parity condition, which includes a proper superset of matchgate signatures, then all unary signatures we canrealize have form [ a,
0] or [0 , a ], so we can not interpolate all unary signatures. For these cases, our starting pointis Theorem 4.1.We define some families of symmetric signatures, which will be used in our proof. G = { [ a, , , · · · , , b ] | ab (cid:54) = 0 }G = { [ x , x , · · · , x k ] | ∀ i is even , x i = 0 or ∀ i is odd , x i = 0 }G = { [ x , x , · · · , x k ] | ∀ i, x i + x i +2 = 0 }M = { f | f is realizable by some matchgate } . We note that G , G and G are supersets of F , F and F respectively. Furthermore (the real part of) F ⊆M ⊆ G . The conditions in G are called parity conditions. The following several lemmas all have the form “If F (cid:54)⊆ A , then the conclusions of Theorem 4.5 hold.” After proving each lemma, in subsequent lemmas, we onlyneed to consider the case that
F ⊆ A . Lemma 4.7. If F (cid:54)⊆ G ∪ G ∪ G , then the conclusions of Theorem 4.5 hold.Proof. Since
F (cid:54)⊆ G ∪ G ∪ G , there exists an f ∈ F and f (cid:54)∈ G ∪ G ∪ G . Since all unary signatures are in G ,the arity of f is greater than 1 and f is non-degenerate. There are two cases according to whether f has a zeroentry or not. 141) f has some zero entries. If there exists a sub-signature of f of the form [0 , a, b ] or [ a, b, ab (cid:54) = 0,then we are done by Lemma 4.6. Otherwise, we can conclude that there are no two successive non-zero entries.So the signature f has this form [0 i x i x i · · · x k i k ], where k ≥ x j (cid:54) = 0 and for all 1 ≤ j ≤ k − i j ≥ ≤ j ≤ k − i j is odd, (including k = 1), then f ∈ G , a contradiction. Otherwise there exists asub-signature of form [ x, , , · · · , , y ], where xy (cid:54) = 0 and there are a positive even number of 0s between x and y . If this is the entire f , then f ∈ G , a contradiction. So there is one 0 before x or after y . By symmetry, weassume there is a 0 before x , so we have a sub-signature [0 , x, , , · · · , , y ], whose arity is even and at least 4.We label its dangling edges 1 , , · · · , k . Then for every i = 1 , , · · · , k −
1, we connect dangling edges 2 i + 1 and2 i + 2 together to form a regular edge. After that, we have an F -gate with arity 2, and its signature is [0 , x, y ].Then we are done by Lemma 4.6.(2) f has no zero entry. We only need to prove that we can construct a function [ a (cid:48) , b (cid:48) , c (cid:48) ] satisfying the threeconditions in Lemma 4.6. Suppose all sub-signatures of f with arity 2 do not satisfy all the three conditions. Foreach sub-signature [ a (cid:48) , b (cid:48) , c (cid:48) ], either a (cid:48) + c (cid:48) = 0, or b (cid:48) = a (cid:48) c (cid:48) . If all of them satisfy a (cid:48) + c (cid:48) = 0, then f ∈ G . Acontradiction. If all of them satisfy b (cid:48) = a (cid:48) c (cid:48) , then f is degenerate. A contradiction. W.l.o.g., we can assume thereis a sub-signature [ a, b, c, d ] of f , such that a + c = 0, b + d (cid:54) = 0, and c = bd . We get this sub-signature [ a, b, c, d ] by[1 ,
0] and [0 , a, b, c, d ], we can get a function [ a (cid:48) , b (cid:48) , c (cid:48) ] = [ a +2 b + c , ab +2 bc + cd, b +2 c + d ] =[2( b + c ) , c ( b + d ) , ( b + d ) ]. Then b (cid:48) = c ( b + d ) (cid:54) = 0. a (cid:48) + c (cid:48) >
0. And a (cid:48) c (cid:48) − b (cid:48) = ( b + d ) (2 b + c ) >
0. We aredone by Lemma 4.6.The following lemma uses Theorem 4.1 in an essential way, which in turns depends on the crossover.
Lemma 4.8. If F (cid:54)⊆ G ∪ M ∪ G , then the conclusions of Theorem 4.5 hold.Proof. If F (cid:54)⊆ G ∪G ∪G , then by Lemma 4.7, we are done. Otherwise, there exists a signature f ∈ F ⊆ G ∪G ∪G and f (cid:54)∈ G ∪ M ∪ G . Then it must be the case that f ∈ G . Note that every signature with arity at most 3 in G (this is called the parity condition) is also contained in M , so f is of arity greater than 3. Let f = [ x , x , · · · , x n ],for some n ≥
4. Suppose there exists some i ∈ [2 , , · · · , n −
2] such that x i (cid:54) = 0. If x i − x i +2 (cid:54) = x i , then we canget [ x i − , , x i , , x i +2 ] by [1 ,
0] and [0 ,
1] which restrict the signature to a sub-signature. Then the problem is x i − x i +2 = x i (cid:54) = 0. Then starting from x i − (cid:54) = 0and if i − ∈ [2 , , · · · , n − x i − x i = x i − (cid:54) = 0. Similarly we can start with x i +2 . A signaturesatisfying the parity condition and is a geometric series on the alternate entries is realizable by a matchgate [7, 8],a contradiction.Now we may assume x i = 0 for all i ∈ [2 , , · · · , n − f ∈ G − ( M ∪ G ), we know that there areonly three possible subcases: (1) n is odd, n ≥ x x n − (cid:54) = 0 and x = x n = 0; (2) n is odd, n ≥ x x n (cid:54) = 0and x = x n − = 0; (3) n ≥ x x n − (cid:54) = 0 and x = x n = 0. This uses the theory of matchgaterealizability [7, 8]. Crucially, if n is even and n <
6, then n = 4 and the case x x n − (cid:54) = 0, x = x n = 0 belongsto M . The subcases (1) and (2) are reversals of each other and (3) contains a signature in form (1) and (2). Soafter normalizing (and connecting pairs of edges together if n > , , , , , x ] where x (cid:54) = 0. So we have both sub-signature [0 , , ,
0] and [1 , , , , x ]. As we proved in Lemma 4.2, the problem is Lemma 4.9. If [0 , , , x ] ∈ F (or [1 , , x, ∈ F ) where x ∈ R , x (cid:54) = ± , then the conclusions of Theorem 4.5hold.Proof. If x (cid:54) = 0, we can use Lemma 4.4 to interpolate [0 , , , , , ,
0] from F . If F (cid:54)⊆ G ∪ M ∪ G , then by Lemma 4.8, we are done. If F ⊆ M , then the problem is tractable and we are done.Otherwise, there exists a signature f ∈ F ⊆ G ∪ M ∪ G and f (cid:54)∈ M . That is f ∈ ( G ∪ G − M ).If f has arity ≥ x , x , − x , − x , x · · · ] ∈ G − M , then we will have x x (cid:54) = 0. Otherwisewe would have f ∈ M , a contradiction. Connecting one unary signature [ x , x ] to [0 , , , x , x , f = [1 , , , · · · , , y ] ∈ G − M , where y (cid:54) = 0. Since f (cid:54)∈ M , its arity n is greater than 2. If n is odd, we can connect its edges except one to get a unary signature [1 , y ]. Then we can use a similar argument asabove and we are done. If n is even, then it is at least 4, since f (cid:54)∈ M . After connecting its edges except four, wecan get [1 , , , , y ]. Together with [0 , , , emma 4.10. If F (cid:54)⊆ G ∪ F ∪ G , then the conclusions of Theorem 4.5 hold.Proof. If F (cid:54)⊆ G ∪G ∪G , then by Lemma 4.7, we are done. Otherwise, there exists a signature f ∈ F ⊆ G ∪G ∪G and f (cid:54)∈ G ∪ F ∪ G . Then it must be the case that f ∈ G . Note that every signature with arity less than 3 in G is also contained in G ∪ G , so f is of arity greater than 2. Since f (cid:54)∈ G , there is some non-zero in the middleof the signature f , after normalization, we can assume there is a sub-signature of form [0 , , , x ] (or [ x, , , x (cid:54) = ±
1, then by Lemma 4.9, we are done. Otherwise, for every such pattern, we have x = ±
1. Since f (cid:54)∈ F ,then there is some sub-signature [0 , , , −
1] and because f (cid:54)∈ G , there is some sub-signature [0 , , , , , , , −
1] of f . Then by Theorem 4.1, we know that the problem is Lemma 4.11. If F (cid:54)⊆ G ∪ G , then the conclusions of Theorem 4.5 hold.Proof. If F (cid:54)⊆ G ∪ F ∪ G , then by Lemma 4.10, we are done. Otherwise, there exists a signature f ∈ F ⊆G ∪ F ∪ G and f (cid:54)∈ G ∪ G . Then it must be the case that f ∈ F . Note that every signature with arity lessthan 3 in F is also contained in G ∪ G , so f is of arity at least 3. Then f has a sub-signature [1 , , ,
0] or[0 , , , , , , F ⊆ F ∪ F ∪ F , then Theorem 4.5 trivially holds andthere is nothing to prove. If not, there exists a signature g ∈ F − F ∪ F ∪ F . By F ⊆ G ∪ F ∪ G , either g ∈ G − F ∪ F ∪ F ( ⊆ G − F ) or g ∈ G − F ∪ F ∪ F ( ⊆ G − F ).For the first case, g ∈ ( G − F ), after a scale, g is of form [1 , , , · · · , b ], where b (cid:54)∈ {− , , } . If the arityof g is odd, we can realize [1 , b ]. (We connect every two adjacent dangling edges into one edge and leave onedangling edge.) Then connecting this unary signature to one dangling edge of [1 , , , , b, g is even, we can realize [1 , , b ] (leavetwo dangling edges). By connecting one of its dangling edge to one dangling edge of [1 , , , , , b, g ∈ ( G − F ), g has a sub-signature of form [1 , b ], where b (cid:54)∈ {− , , } . By the sameargument as above, Theorem 4.5 holds. This completes the proof. Lemma 4.12. If F (cid:54)⊆ G ∪ F , then the conclusions of Theorem 4.5 hold.Proof. If F (cid:54)⊆ G ∪ G , then by Lemma 4.11, we are done. Otherwise, there exists a signature f ∈ F ⊆ G ∪ G and f (cid:54)∈ G ∪ F . Then it must be the case that f ∈ G , and f has a sub-signature of form [1 , a, − a (cid:54)∈ {− , , } .If F ⊆ { [1 , , }∪G , then Holant ∗ ( F ) is polynomial time computable by Theorem 2.7 and as a result Theorem4.5 trivially holds and we are done.If not, there exists a signature g ∈ F ⊆ G ∪ G and g (cid:54)∈ { [1 , , } ∪ G . Then it must be the case that g ∈ G .The arity of g is greater than 1, as g (cid:54)∈ G .If the arity of g is 2, then g is of form [1 , , b ], where b (cid:54)∈ {− , , } . Connecting two signatures [1 , , b ] to bothsides of one binary signature [1 , a, − , ab, − b ]. It satisfies all the conditionsof Lemma 4.6, and we are done. If the arity of g is greater than 2, then we can always realize a signature [1 , , , b ],where b (cid:54) = 0. (We connect the unary signature [1 , a ] to all its dangling edges except the three ones.) Then wecan use an F -gate in Figure 8. Its signature is [1 , a b, b ], and by Lemma 4.6, we are done. This completes theFigure 8: The function on degree 2 nodes is [1 , a, − , , , b ].proof.By the above lemmas, the only case left we have to handle is that F ⊆ G ∪ F . This is done by the followinglemma, which completes the proof of Theorem 4.5. 16 emma 4.13. If F ⊆ G ∪ F , then the conclusions of Theorem 4.5 hold.Proof. If F ⊆ F ∪ F , then by Theorem 2.8 part (2), Holant c ( F ) is computable in polynomial time. Similarly,if F ⊆ U ∪ F ∪ { [1 , , } , then by Theorem 2.8 part (1), and then by Theorem 2.7 part (3), Holant c ( F ) iscomputable in polynomial time. Hence in these two cases, Theorem 4.5 holds. Now suppose F (cid:54)⊆ F ∪ F and F (cid:54)⊆ U ∪ F ∪ { [1 , , } .There exists f ∈ F − F ∪ F . Since F ⊆ G ∪ F , such an f ∈ G .Now there are two cases. The first case is that we have such an f (cid:54)∈ U , and so, f ∈ F ∩ G − ( F ∪ F ∪ U ).The arity of f is greater than 1. By connecting its dangling edges together except two or three depends on theparity of the arity of f , we can assume f has form [1 , , a ] or [1 , , , a ], where a (cid:54)∈ {− , , } .The second case is every f ∈ F ∩ G − ( F ∪ F ) is also in U . By F (cid:54)⊆ U ∪ F ∪ { [1 , , } , there exists f ∈ F − ( U ∪ F ∪ { [1 , , } ). Since F ⊆ G ∪ F , and f (cid:54)∈ F , we get f ∈ G . If f (cid:54)∈ F , we could use this f as the f above, namely f ∈ F ∩ G − ( F ∪ F ∪ U ). A contradiction. Thus f ∈ F . Also we have some f ∈ F − ( F ∪ F ). So f ∈ G , since F ⊆ G ∪ F . Also since we are in this second case, certainly f ∈ U .So we have f , f ∈ F ∩ G such that f ∈ F but f (cid:54)∈ U ∪ F ∪ { [1 , , } , and f ∈ U but f (cid:54)∈ F . Thearity of f is at least 2. We claim it is greater than 2. Otherwise, f being from F and not [1 , , f = [1 , , − ∈ F , a contradiction. So f has form [1 , , , . . . , ±
1] of arity at least 3. f is of form [1 , a (cid:48) ], where a (cid:48) (cid:54)∈ {− , , } ; this follows from f ∈ U ∩ G − F . By connecting all the dangling edges of f except two with f , we can construct an F -gate with signature of form [1 , , a ], where a (cid:54)∈ {− , , } . This is one of the abovetwo forms after the first case. To sum up, in both cases, we have some f of the form [1 , , a ] or [1 , , , a ], where a (cid:54)∈ {− , , } .If F ⊆ G ∪ { [0 , , } ∪ U , then by Theorem 2.8 part (1), and then by Theorem 2.7 part (2) (with a = 0and b = 1), Holant c ( F ) is computable in polynomial time and Theorem 4.5 holds. Otherwise, there exists g ∈ F ⊆ G ∪ F , and g (cid:54)∈ G ∪ { [0 , , } ∪ U . Then g must be in F , and have one of the following sub-signatures: [1 , , − , [1 , − , − , [1 , , − , , [0 , , , − F . By symmetry (taking the reversal of both f and g ), we only need to consider the cases f = [1 , , a ] or[1 , , , a ], where a (cid:54)∈ {− , , } , and g = [1 , , −
1] or [1 , , − , f and g , we have four cases. If f = [1 , , a ] and g = [1 , , − f gf , we can realize [1 , a, − a ]. By Lemma 4.6, we are done. If f = [1 , , a ] and g = [1 , , − , g , we extend it by one copy of f . Then we can realize [1 , , − a , f = [1 , , , a ] and g = [1 , , − ,
1] (sub-signature of g ) to onedangling edge of f , and realize a binary signature f = [1 , , a ]. This reduces it to the first case, which has beenproved. If f = [1 , , , a ] and g = [1 , , − , , a ] from f by connecting two ofits dangling edges together, and then connect this unary signature to one dangling edge of g to realize [1 , − a, − , − a, − (cid:54)∈ G ∪ F , by Lemma 4.12, we are done. In this section, we prove a dichotomy for planar real weighted (cid:20) − (cid:21) (this can be computed by the characterization in [11]).Now we present the dichotomy theorem for planar weighted Theorem 5.1.
Let F be a set of real symmetric functions. Pl - CSP( F ) is F satisfies one ofthe following conditions, in which case it is tractable:1. F ) is tractable (for which we have an effective dichotomy [16]); or2. Every function in F is realizable by some matchgate under basis (cid:20) − (cid:21) (for which we have a completecharacterization [8]). The main proof idea is to reduce Pl-Holant c problems to Pl- F ) is exactly the sameas planar Holant with all the Equality functions, i.e., Pl-Holant(
F ∪ { [1 , , [1 , , , [1 , , , , [1 , , , , , . . . } ).17e can use a holographic reduction under the basis H = (cid:20) − (cid:21) . Under this transformation, the problem istransformed to, and hence has the same complexity as Pl-Holant( H F ∪ { [1 , , [1 , , , [1 , , , , [1 , , , , , . . . } ).Since this holographic reduction gives us [1 ,
0] (from [1 , , c problem and apply Theorem 4.5 to H F ∪ { [1 , , , [1 , , , , [1 , , , , , . . . } toget a proof of Theorem 5.1. In the following, we show how to realize (or interpolate) [0 , , ,
1] in some cases is difficult. The following lemma says thatit is also sufficient if we can realize (or interpolate) [0 , , , ,
1] can be viewed as two copies of [0 , , ,
1] = [0 , ⊗ [0 , , ,
1] to replace two occurrences of [0 , ,
1] used in theinput instance; the second difficulty, which is more subtle, is that we have to pair up two copies of [0 ,
1] whilemaintaining planarity of the instance.
Lemma 5.2.
Pl-Holant ( F ∪ { [1 , , [0 , , , [1 , , , [1 , , , , [1 , , , , , . . . } ) is c ( F ∪ { [0 , , , [1 , , , [1 , , , , [1 , , , , , . . . } ) is There is one more function [0 ,
1] in the second signature set than the first, so obviously the first one canbe reduced to the second one. Hence if the second problem is in P, so is the first. We have already proved adichotomy theorem for Pl-Holant c problems. So now we may assume the second problem is , , , | [1 , , , , , , , ,
0] (respectively counting
Vertex Cover , Matching for planar 3-regular graphs, or
PerfectMatching for general 3-regular graphs) is reduced to it by a chain of reductions. There are only three reductionmethods in this reduction chain, direct gadget construction, polynomial interpolation, and holographic reduction.Given an instance G of Pl-Holant([1 , , , | [1 , , , , , , , , G ∪ G , which denotes the disjoint union of two copies of G .Notice that the value of Pl-Holant([1 , , , | [1 , , , , , , , ,
0] on the instance G is a non-negative integer, and the value on G ∪ G is its square. So we can compute the value on G uniquelyfrom its square. Suppose the reduction chain on the instance G produced instances G , G , . . . , G m of the secondproblem. The same reduction applied to G ∪ G produces instances of the form G ∪ G , G ∪ G , . . . , G m (cid:48) ∪ G m (cid:48) .(We note that the reduction on G ∪ G may produce polynomially more instances than on G because of polynomialinterpolation.)Now we only need to show how to transform instances G ∪ G , G ∪ G , . . . , G m (cid:48) ∪ G m (cid:48) in the second problem,to instances of the first problem with the same values (replacing all occurrences of the signature [0 ,
1] by some[0 , , G i ∪ G i is a planar graph with zero or more vertices of degree one attached with the function [0 , , ,
1] to replace one pair of [0 , G i . Let the outer face be the root. Choose an arbitrary leaf ofthis tree, which corresponds to a face C of G i . Suppose C (cid:48) is the face corresponding to the parent of C in thetree. If there are an even number of vertices of degree one attached with [0 ,
1] in face C , we can perfectly matchthem and realize them using [0 , ,
1] while maintaining planarity in this face. This can be done by matching thesedangling vertices of degree one in a clockwise fashion on this face C . If there are an odd number of [0 ,
1] in face C , we choose one edge e between C and C (cid:48) , and add a new vertex v e on e , and connect two new vertices of degreeone to v e . The two new vertices are attached [0 , v e has degree 4 and is attached [1 , , , , , , , ,
1] connected by two [0 ,
1] is the same as the function [1 , , e itself. We put one new vertex with [0 ,
1] in face C , and the other one in face C (cid:48) . Now, there are an even numberof [0 ,
1] in face C , and we can replace them by [0 , ,
1] in C , as before. We may repeat this process, until we reachthe root in the dual graph of G i . If we do the same for the two G i in G i ∪ G i , we will have an even number of[0 ,
1] in the common outer face and can at last perfectly match the [0 ,
1] vertices and realize them by [0 , , emma 5.3. If we can realize (or interpolate) [0 , or [0 , , from H F∪{ [1 , , [1 , , , [1 , , , , [1 , , , , , . . . } ,then the conclusion of Theorem 5.1 holds. Next we give two lemmas which give a general condition to realize or interpolate [0 ,
1] or [0 , , Lemma 5.4.
Let a ∈ R . If a (cid:54)∈ { , , − } , then we can interpolate [0 , from ( { [1 , a ] , [1 , , [1 , , , [1 , , , , . . . } ) .Proof. For every j ≥
1, we can take a function F j +1 = [1 , , , , , . . . ] of arity j + 1, and connect j functions [1 , a ]to it. The row vector form of the function (i.e., a listing of its values) of arity j composed of j copies of [1 , a ] is(1 , a ) ⊗ j .The column vector form of F j +1 is 1 / (cid:18) (cid:19) ⊗ ( j +1) + 1 / (cid:18) − (cid:19) ⊗ ( j +1) . The 2 j × F j +1 is1 / (cid:18) (cid:19) ⊗ j ⊗ (1 ,
1) + 1 / (cid:18) − (cid:19) ⊗ j ⊗ (1 , − , a ) ⊗ j (cid:34) / (cid:18) (cid:19) ⊗ j ⊗ (1 ,
1) + 1 / (cid:18) − (cid:19) ⊗ j ⊗ (1 , − (cid:35) = (1 + a ) j ,
1) + (1 − a ) j , − a ∈ R and a (cid:54)∈ { , , − } , (1 + a ) / (1 − a ) is well defined and is neither zero nor a root of unity. Wecan interpolate any unary function x (1 ,
1) + y (1 , − , Lemma 5.5.
Let a ∈ R . If a (cid:54)∈ { , , − } , then we can interpolate [0 , , from [1 , , a ] .Proof. The function of a chain of length j composed of [1 , , a ] is [1 , , a j ]. Since the real number a (cid:54)∈ { , , − } ,we can interpolate all [ x, , y ], and in particular [0 , , Proof of Theorem 5.1.
In this proof, we augment the class F ∪ F ∪ F to include those degenerate signatureswhich can be obtained from tensor products from unary signatures in F ∪ F ∪ F .If H F ⊆ F ∪ F ∪ F , then the problem is tractable even for general graphs and the conclusion of the theoremholds. Now we assume that there exists an f ∈ H F − ( F ∪ F ∪ F ). In the following, we will prove that we canrealize (or interpolate) [0 ,
1] or [0 , ,
1] from f and { [1 , , [1 , , , [1 , , , , [1 , , , , , . . . } .The general thrust of the proof is to squeeze all possible f into several standardized forms, and either prove , , , f = [ f , f , . . . , f n ]. Since we have [1 , f we already have as the signature of a realizable function. Given a function g with arity r >
1, we often use thegadget composed of two copies of g such that r − g . We separate two cases according to whether f = 0, or f (cid:54) = 0 which we normalize to f = 1.1. f = 0.As the constant 0 function is in F ∪ F ∪ F , f is not identically 0, and thus for some i ≥ f i (cid:54) = 0. If f = 0 and f (cid:54) = 0, then we can connect n − ,
0] to f to get [0 , f ], which is [0 ,
1] up to a factor.So we have f = f = 0, then n ≥
2. If f (cid:54) = 0, then we can connect n − ,
0] to f to get[0 , , f ], which is [0 , ,
1] up to a factor.So we have f = f = f = 0, then n ≥
3. Let m ≤ n be the first nonzero, f = f = f = · · · = f m − = 0, f m (cid:54) = 0, then we can connect a function [1 , ,
1] to two dangling edges of f to get a function whose firstnonzero entry is f m at index m −
2. We can repeat this process until exactly one or two zeros are left atindex 0 or at index 0 and 1, and we reach one of the two scenarios above.2. f = 1.By Lemma 5.4, we only need to consider f ∈ { , , − } . Otherwise, we are done.19a) f = 1 and f = ± n = 1, then f = [ f , f ] = [1 , ± ∈ F ∪ F ∪ F , a contradiction. Therefore we have n ≥
2, we cantake its initial part [1 , f , f ]. Connecting one edge to [1 , f ], we get [1 + f , f + f f ] = [2 , ± (1 + f )].By Lemma 5.4, we only need to consider f ∈ { , − , − } .We can construct another gadget which connects two inputs of [1 , , ,
0] by [1 , f , f ]. This produces aunary signature [1 + f , f ]. It follows that f (cid:54) = −
1, since otherwise we have [0 ,
1] after normalization.Next we rule out f = −
3. The double gadget of [ f , f , f ] = [1 , ± , f ] has signature [2 , − ,
10] and[2 , , , ± ,
5] and 5 (cid:54)∈ { , − , − } . Hence, we may assume f = 1.Our goal in this case 2.(a) of f = 1 and f = ± , ± , , . . . ]. Assume we haveproved that f j = 1 (respectively f j = ( − j ) for j = 0 , , . . . , m ( m ≥ f (cid:54)∈ F ∪ F ∪ F , thearity n > m . We can connect [ f , f ] to [ f , f , . . . , f m +1 ], to get a function [ f + f , f f + f f , f f + f f , . . . ] of arity m , which is [2 , , . . . , f m +1 ] (respectively [2 , − , . . . , f m − − f m , f m − f m +1 ]). Bywhat has been proved inductively, f m +1 = 1 (respectively f m +1 = f m − ). So in this case we showedthat either f ∈ F ∪ F ∪ F , which is a contradiction, or f = 1 and f = 0.Since [1 , ∈ F , and f (cid:54)∈ F , we have n >
1. If f were degenerate it would be [1 , ⊗ n = [1 , , . . . , F ∪ F ∪ F . But f does not. So f is non-degenerate,in particular there is some nonzero entry other than f . Suppose f m is the first nonzero f i , for i > , ,
1] to [ f , , . . . , , f m ] to get [1 , f m ] or [1 , , f m ], we have f m = ± f (cid:54)∈ F we have n > m .We prove m is an even number. Otherwise, we can get [1 , f m , f m +1 ] by connecting some [1 , , f m +1 = 1. We can also get [1 , , , f m , f m +1 ], since m >
1, whosedouble gadget has the signature [1 + f m , f m f m +1 , f m + f m +1 ] = [2 , ± , , ± m must be even. Next we show that in fact m = 2. Otherwise, m ≥ , , , , f m , f m +1 ], whose double gadget has the signature [1 + f m , f m f m +1 , f m + f m +1 ] =[2 , ± f m +1 , f m +1 ]. By what has been proved so far this also leads to m = 2 and reached [1 , , ± , f ], whose double gadget has the signature [2 , ± f , f ],so f = 0.Again since f (cid:54)∈ F ∪ F ∪ F we have n >
3. Hence we have [1 , , ± , , f ]. For [1 , , − , , f ], byconnecting two edges with [1 , , , , − f ], and we must have f = 1, or else we havethe signature [0 , , , , , , f ], by connecting two edges with [1 , , , , f ], and it follows from Lemma 5.5 that f ∈ { , − , − } . Connecting three edges of[1 , , , , f ] to three edges of [1 , , , , , , f ], which rules out f = −
1, by Lemma 5.5again. The double gadget of [1 , , , , f ] gives [4 , , f ], which rules out f = −
3. To sum up, weget f = 1.We have reached [1 , , ± , , , . . . ]. The rest of the proof is similar to the induction proof for the case2.(a) but by skipping all entries with an odd index. Assume we have proved that f j are of the properform, for j = 0 , , . . . , m . More precisely, f j = 0 for all odd j ≤ m , and, either f j = 1 for all j ≤ m/ f j = ( − j for all j ≤ m/
2. Since f (cid:54)∈ F ∪ F ∪ F , the arity n > m . We connect [ f , f , f ] to[ f , f , . . . , f m +1 ], to get a function g of arity m −
1. If m is even, f m − = 0, and f m +1 is added to orsubtracted from f m − , namely f m − ± f m +1 , to form the last entry in g at index m −
1. This entryshould be zero by induction, so f m +1 = 0. If m is odd, we can repeat the proof in the case 2.(a), butwe ignore all zero entries at odd indexed locations, then the induction can be completed as before.This completes the proof. In this section we prove a dichotomy for Holant on planar 2-3 regular graphs. This setting is very interesting forat least two reasons. From dichotomy theorem point of view, this is the simplest nontrivial setting and always20erves as the starting point of more general dichotomy theorems as in [16, 9]. This was also a focus of severalprevious work [14, 26, 15, 27], whose result is the starting point of this theorem. From the holographic algorithmspoint of view, most of the known holographic algorithms [40, 39] are essentially for planar 2-3 regular graphs.The dichotomy theorem here explains the reason why they are special and why many variations of them are c and Pl- Theorem 6.1.
Let [ y , y , y ] and [ x , x , x , x ] be two complex symmetric signatures with arity 2 and 3 respec-tively. Then Pl-Holant([ y , y , y ] | [ x , x , x , x ]) is [ y , y , y ] and [ x , x , x , x ] satisfy one ofthe following conditions, in which case it is tractable:1. Holant([ y , y , y ] | [ x , x , x , x ]) is tractable (for which we have an effective dichotomy [9]); or2. There exists a basis T such that both [ y , y , y ]( T − ) ⊗ and T ⊗ [ x , x , x , x ] are realizable by some match-gates (for which we have a complete characterization [11]).Proof. If [ x , x , x , x ] or [ y , y , y ] is degenerate, the problem is tractable, even for the non-planar case,and so this falls in condition . Now we assume that they are both non-degenerate. As proved in [16], wecan choose an invertible T such that [ x , x , x , x ] (or its reversal, which is similar and we omit that case)can be written as T ⊗ [1 , , ,
1] or T ⊗ [1 , , , z , z , z ] | [1 , , , z , z , z ] | [1 , , , z , z , z ] | [1 , , , z = z . This condition is exactly the same as the condition that there existsa basis T such that both [ y , y , y ]( T − ) ⊗ and T ⊗ [1 , , ,
1] are realizable by some matchgates. This provesTheorem 6.1 for case (1).Now we consider Pl-Holant([ z , z , z ] | [1 , , , z = 0, the problem is trivially tractable even for generalgraphs. This can be seen by a simple counting argument: in a bipartite graph the LHS vertices all have thesignature [0 , z , z ] and thus at least half the edges must be 1, while the RHS vertices all have the signature[1 , , ,
0] and thus less than half the edges are 1. This is also the only case where the problem is not , , ,
0] by [9]. Now we assume z (cid:54) = 0. Then it is sufficient to prove that eitherthe problem is T such that [1 , , , T ⊗ and ( T − ) ⊗ [ z , z , z ]are realizable by some matchgates. Let T = (cid:20) √ z z / √ z (cid:112) ( z z − ( z ) ) /z (cid:21) . Note that T is well defined andinvertible since z (cid:54) = 0 and [ z , z , z ] is non-degenerate ( i.e., z z − ( z ) (cid:54) = 0). Then we can verify that[1 , , , T ⊗ = [ √ z ( z + 3 z ) , (cid:112) z ( z z − ( z ) ) , ,
0] and ( T − ) ⊗ [ z , z , z ] = [1 , , . We note that (cid:112) z ( z z − ( z ) ) (cid:54) = 0. If √ z ( z + 3 z ) = 0, then both [ √ z ( z + 3 z ) , (cid:112) z ( z z − ( z ) ) , , , ,
1] can be realized by matchgates and the problem for planar graphs is tractable. We denote v = √ z ( z +3 z ) √ z ( z z − ( z ) ) (cid:54) = 0. Then the problem is equivalent to (non-bipartite) Pl-Holant([ v, , , Claim:
Let v (cid:54) = 0 be a complex number. Then Pl-Holant([ v, , , v + 3 v, v + 1 , v,
1] by connecting 3 copies of [ v, , , v + 3 v, v + 1 , v, v + 3 v, v + 1 , v, T = 12 (cid:32)(cid:20) v + 11 (cid:21) ⊗ + (cid:20) v − (cid:21) ⊗ (cid:33) . Then the following reduction chain holds:Pl-Holant([ v + 3 v, v + 1 , v, ≡ T Pl-Holant([1 , , | [ v + 3 v, v + 1 , v, ≡ T Pl-Holant([ v + 2 v + 2 , v , v − v + 2] | [1 , , , , v, , , (cid:20) v + 1 v −
11 1 (cid:21) . This transforms the problem to ourfirst case where the RHS all have [1 , , , v + 2 v + 2) =( v − v + 2) . Since ( v + 2 v + 2) − ( v − v + 2) = 4 v (3 v + 16 v + 12) and v (cid:54) = 0, we will have proved theclaim as long as 3 v + 16 v + 12 (cid:54) = 0. There are four roots for the equation 3 v + 16 v + 12 = 0, and for thesefour exceptional values of v , we prove it separately as follows.In addition to the gadget in Figure 9, we can construct a gadget in Figure 10 with a binary signature[ v + 2 , v, v + 2 , v, | [ v + 3 v, v + 1 , v, v, , , (cid:20) v + 1 v −
11 1 (cid:21) , we will get an equivalent problem Pl-Holant([
X, Y, Z ] | [1 , , , X =( v + 2)( v + 2 v + 1) + 2 v ( v + 1) + 1, Y = ( v + 2)( v −
1) + 2 v + 1, and Z = ( v + 2)( v − v + 1) + 2 v ( v −
1) + 1.Again this transforms the problem to our first case, and, it is easy to verify that any root of 3 v + 16 v + 12 = 0is not a tractable case here. This completes the proof of the claim and also the proof of the theorem. Appendix: Some Connections to Statistical Physics
In this section we describe some background and connections from Statistical Physics. Our discussion is necessarilya superficial one, both due to our limited knowledge and limitation on space. The purpose is to illustrate that,even at such a superficial level, a strong connection exists, and that our complexity results may shed some lighton the venerable question from physics: Exactly what “systems” can be solved “exactly” and what “systems” are“difficult”.The Ising model was named after Ernst Ising [22]. Wilhelm Lenz invented this model and gave it to his studentIsing to work on it. The model consists of a discrete set of variables, called spins, that can be assigned one of twovalues (states). These spins are usually placed on a lattice structure or a graph, and each spin interacts with itsnearest neighbors.Denoting the values each spin i can take as σ i = +1 and −
1, the energy (the Hamiltonian) of the Isingmodel is H ( σ ) = − (cid:80) edge { i,j } J i,j σ i σ j . The interaction between spins i and j is called ferromagnetic if J i,j > J i,j <
0, and noninteracting if J i,j = 0. E.g., if all the spins are placed on a one-dimensionallattice, then the antiferromagnetic one-dimensional Ising model (with the same value J i,j = J <
0) has the energyfunction H = (cid:80) i σ i σ i +1 , after normalization. The ferromagnetic two-dimensional Ising model on a square lattice(with the same value J i,j = J >
0) has energy H = − (cid:80) i,j ( σ i,j σ i,j +1 + σ i,j σ i +1 ,j ). The Ising model may bemodified by magnetic fields which amounts to a unary function at each spin H = − (cid:80) edge { i,j } J i,j σ i σ j − (cid:80) i h i σ i .The model is a statistical model. The central premise of statistical physics is that the probability of each22onfiguration σ is given by the Boltzmann distribution, e − H ( σ ) /kT / (cid:80) σ e − H ( σ ) /kT , where k is Boltzmann constantand T is the (absolute) temperature. This focuses attention on the partition function Z = (cid:88) σ e − H ( σ ) /kT . Note that the exponential e − H ( σ ) /kT turns this into a sum-of-product functions exactly as we discussed in Exact-One function. Freedman, Lov´asz and Schrijver [21] recently proved that this partitionfunction cannot be expressed as a graph homomorphism function, where the vertices are variables as in the Isingmodel. However in the framework of Holant problems we can find a unity for all these problems.We note the following. In the paper [11] we gave a complete characterization of matchgate realizable symmetricsignatures. The following lemma is proved [11]:
Lemma 6.2.
The set of bases under which the signature [ x , x , x ] is realizable as a signature by some matchgateis (cid:26)(cid:20)(cid:18) n n (cid:19) , (cid:18) p p (cid:19)(cid:21) ∈ GL ( C ) (cid:12)(cid:12)(cid:12)(cid:12) x p − x p n + x n = 0 , x p − x p n + x n = 0or x p p − x ( n p + n p ) + x n n = 0 (cid:27) . This has the consequence that under the basis (cid:20)(cid:18) n n (cid:19) , (cid:18) p p (cid:19)(cid:21) = (cid:20)(cid:18) (cid:19) , (cid:18) − (cid:19)(cid:21) , the signature [ x, y, x ] isrealizable by a matchgate, for all values x and y . In terms of the Ising model, when two interacting spins i and j take the same assignment value σ i = σ j = ±
1, the contribution to the Hamiltonian is − J i,j , and when theytake the opposite assignment σ i = − σ j = ±
1, the contribution is J i,j . Translating this to the contributions tothe partition function we get exactly the local constraint evaluation x = e J i,j /kT when inputs are 00 or 11, and y = e − J i,j /kT when inputs are 01 and 10.Then, the theory of Holographic Algorithms tells us that for planar graphs, this Ising model is exactly solvableby a holographic reduction to the FKT algorithm.The present paper, especially Theorem 5.1, tells us why this is exactly where physicists stopped, and attemptsto generalize this to non-planar systems have not been successful in the past 85 years.Sorin Istrail [23] showed that computing the free energy of an arbitrary subgraph of an Ising model on a latticeof dimension three or more is NP-hard; see a nice article by Barry Cipra in the SIAM News [17]. A very partiallist of a great deal of research on this and related models, from a computational complexity perspective, can befound in [2, 3, 4, 7, 13, 16, 15, 19, 26, 5, 6, 12, 10, 11]. Acknowledgments
We thank the following colleagues for their interests and helpful comments: Xi Chen, Martin Dyer, Alan Frieze,Sean Hallgren, Leslie Goldberg, Sorin Istrail, Richard Lipton, Jason Morton, Dana Randall, Leslie Valiant andSantosh Vempala.
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