Dichotomy Result on 3-Regular Bipartite Non-negative Functions
DDichotomy Result on 3-Regular BipartiteNon-negative Functions
Austen Z. Fan ∗ [email protected] Jin-Yi Cai † [email protected] Abstract
We prove a complexity dichotomy theorem for a class of Holant problems on 3-regular bi-partite graphs. Given an arbitrary nonnegative weighted symmetric constraint function f =[ x , x , x , x ], we prove that the bipartite Holant problem Holant ( f | (= )) is either computablein polynomial time or f is explicit. Holant problems are also called edge-coloring models. They can express a broad class of countingproblems, such as counting matchings ( ), perfect matchings ( ), edge-colorings,cycle coverings, and a host of counting orientation problems such as counting Eulerian orientations.Given an input graph G = ( V, E ), we identify each edge e ∈ E as a variable over some finitedomain D , and identify each vertex v as a constraint function f v . Then the partition function isthe following sum of product (cid:80) σ : E → D (cid:81) v ∈ V f v ( σ | E ( v ) ), where E ( v ) denotes the edges incident to v and f v ( σ | E ( v ) ) is the evaluation of f v on the restriction of σ on E ( v ). For example, and are counting problems specified by the constraint function At-Most-One , respectively
Exact-One , which outputs value 1 if the input bits have at most one 1, respectively exactly one1, and outputs 0 otherwise. Thus, every term (cid:81) v ∈ V f v ( σ | E ( v ) ) evaluates to 1 or 0, and is 1 iff theassignment σ is a mathching, respectively a perfect mathching. The framework of Holant problemsis intimately related to Valiant’s holographic algorithms [16]. Holant problems can encompass allcounting constraint satisfaction problems ( ∗ Department of Computer Sciences, University of Wisconsin-Madison. † Department of Computer Sciences, University of Wisconsin-Madison. Supported by NSF CCF-1714275. a r X i v : . [ c s . CC ] N ov n this paper we consider the Boolean domain D = { , } . Significant knowledge has been gainedabout the complexity of Holant problems [1, 2, 4, 8, 13, 14]. However, there has been very limitedprogress on bipartite Holant problems where the input graph G = ( U, V, E ) is bipartite, and everyconstraint function on U and V comes from two separately specified sets of constraint functions.This is not an oversight; the reason is a serious technical obstacle. When the graph is bipartiteand, say, r -regular, there is a curious number theoretic limitation as to what types of subgraphfragments, called gadgets, one can possibly construct. It turns out that every constructible gadgetmust have a rigid arity restriction; e.g., if the gadget represents a constraint function that can beused for a vertices in U or in V , the arity (the number of input variables) of the function must becongruent to 0 modulo r .We initiate in this paper the study of Holant problems on bipartite graphs. To be specific,we classify Holant problems on 3-regular bipartite graphs where vertices of one side are labeledwith a nonnegative weighted symmetric constraint function f = [ x , x , x , x ], which takes value x , x , x , x respectively when the input has Hamming weight 0, 1, 2, 3; the vertices of the other sideare labeled by the ternary equality function (= ). Such graphs can be viewed as incidence graphsof hypergraphs. Thus one can interpret problems of this type as on 3-uniform (every hyperedge hassize 3) 3-regular (every vertex appears in 3 hyperedges) hypergraphs. In this view, e.g., countingperfect matchings on 3-uniform hypergraphs (the number of subsets of hyperedges that cover everyvertex exactly once) corresponds to the Exact-One function [0 , , , f | (= )). We prove that for all f , this problem is either computablein polynomial time or x , x , x , x .Suppose ( X, S ) is a 3-uniform set system where every x ∈ X appears in 3 sets in S . In the 0-1case, this dichotomy completely classifies the complexity of counting the number of ways to choose S (cid:48) ⊆ S while satisfying some local constraint specified by f . In the more general nonnegativeweighted case, this is to compute a weighted sum of products.As mentioned earlier the main technical obstacle to this regular bipartite Holant dichotomy is thenumber theoretic arity restriction. We overcome this obstacle by considering straddled constraintfunctions, i.e., those functions which have some input variables that must be connected to one sideof the bipartite graph while some other variables must be connected to the other side. Then weintroduce a lemma that let’s interpolate a degenerate constraint function by iterating the straddledfunction construction. Using a Vandermonde system we can succeed in this interpolation. Typicallyin proving a Holant dichotomy, getting a degenerate constraint function is useless and signifiesfailure. But here we turn the table, and transform this “failure” to a “success” by “peel” off aconstraint function which is a tensor factor, whereby to break the number theoretic arity restriction.This paper is a mere starting point for understanding bipartite Holant problems. Almost everygeneralization is an open problem at this point, including more than one constraint function oneither side, other regularity parameter r , real or complex valued constraint functions which allowcancellations, etc. The bigger picture is to gain a systematic understanding of all such countingproblems in a classification program, a theme seems second to none in its centrality to countingcomplexity theory, short of proving PF (cid:54) = Preliminaries
In this paper we consider the following subclass of Holant problems. An input 3-regular bipartitegraph G = ( U, V, E ) is given, where each vertex on V is assigned the Equality of arities 3 (= )and each vertex on U is assigned a ternary symmetric constraint function (also called a signature) f with nonnegative values. The problem is to compute the quantityHolant ( G ) = (cid:88) σ : E →{ , } (cid:89) u ∈ U f (cid:0) σ | E ( u ) (cid:1) (cid:89) v ∈ V (= ) (cid:0) σ | E ( v ) (cid:1) Equivalently, this can be stated as a weighted counting constraint satisfaction problem onBoolean variables defined on 3-regular bipartite graphs: The input is a 3-regular bipartite G =( U, V, E ), where every v ∈ V is a Boolean variable and every u ∈ U represents the nonnegativevalued constraint function f . An edge ( u, v ) ∈ E indicates that v appears in the constraint at u . Being 3-regular means that every constraint has 3 variables and every variable appears in 3constraints.We adopt the notation f = [ x , x , x , x ] to represent the ternary symmetric signature f where f (0 , ,
0) = x , f (0 , ,
1) = f (0 , ,
0) = f (1 , ,
0) = x , f (0 , ,
1) = f (1 , ,
1) = f (1 , ,
0) = x and f (1 , ,
1) = x . The Equality of arities 3 is (= ) = [1 , , , V are on the right hand side (RHS) and vertices in U are on the left hand side (LHS). We denotethis problem Holant ( f | (= )).A gadget, such as those illustrated in Figure 1, is a bipartite graph G = ( U, V, E in , E out ) withinternal edges E in and dangling edges E out . There can be m dangling edges internally incident tovertices from U and n dangling edges internally incident to vertices from V . These m + n danglingedges correspond to Boolean variables x , . . . , x m , y , . . . , y n and the gadget defines a signature f ( x , . . . , x m , y , . . . , y n ) = (cid:88) σ : E in →{ , } (cid:89) u ∈ U f (cid:0)(cid:98) σ | E ( u ) (cid:1) (cid:89) v ∈ V (= ) (cid:0)(cid:98) σ | E ( v ) (cid:1) , where (cid:98) σ denotes the extension of σ by the assignment on the dangling edges.To preserve the bipartite structure, we must be careful in any gadget construction how eachexternal wire is to be connected, i.e., as an input variable whether it is on the LHS (like those of f which can be used to connect to (= ) on the RHS), or it is on the RHS (like those of (= ) whichcan be used to connect to f on the LHS). See illustrations in Figure 1.If a gadget construction produces a constraint function g such that all of its variables are on theLHS. We claim that its arity must be a multiple of 3. This follows from the following more generalstatement that if a constraint function has n input variables on the LHS and m input variables onthe RHS, then n ≡ m mod 3. This can be easily proved by induction on the number of occurrencesof f and (= ) in a gadget construction Γ: If either f or (= ) do not occur, then the function is atensor product of (= ) or (= ), thus clearly of arity n ≡ m ≡ f or(= ) both occur and let x be an external dangling edge. It is internally connected to a vertex v ,which is labeled either f (if v ∈ U ) or (= ) (if v ∈ V ). Let v have exactly k ∈ { , , } incidentedges that are dangling edges. Now we remove v and get a gadget Γ (cid:48) with fewer occurrences of f and (= ). The induction hypothesis and a simple accounting of the arity for variables on the LHSand the RHS separately complete the induction.One idea that is instrumental in this paper is to use gadgets that produce straddled signatures.For example the gadget G in Figure 1(a) has one variable on the LHS and one variable on the3HS. Such gadgets can be iterated while respecting the bipartite structure. The signature matrix of such straddled gadgets will adopt the notation that row indices denote the input from LHS andcolumn indices denote the input from RHS. For example, the signature matrix for G in Figure 1(a)where we place [ x , x , x , x ] on the square and (= ) on the circle will be (cid:18) x x x x (cid:19) . A binarysignature is degenerate if it has determinant 0. We will use such constructions to interpolate adegenerate straddled signature so that we can “split” it to get unary signatures. To justify that wecan indeed “split” a degenerate straddled signature, whenever we connect a unary signature, wehave to “use up” another unary signature, for the pair was created by that degenerate straddledsignature. Intuitively, we can “use all other edges up” by connecting them to form known positiveglobal factors of the Holant value. This intuition is encoded into the following lemma. Effectively,we can split a degenerate binary straddled signature into unary signatures to be freely used. Lemma 2.1.
Let f and g be two nonnegative valued signatures. If a degenerate nonnegative binarystraddled signature (cid:18) xy xy (cid:19) can be interpolated in the problem Holant ( f | g ) , then Holant ( f | { g, [1 , x ] } ) ≤ T Holant ( f | g ) . (2.1) A similar statement holds for adding the unary [1 , y ] on the LHS. Remark:
The same proof applies for (cid:18) y xy x (cid:19) to get unary signatures [1 , x ] on the RHS, orunary [ y,
1] on the LHS.
Proof.
We prove (2.1). Let f and g have arity m and n respectively. We may assume that g is not a multiple of [0 , ⊗ n (including identically 0), for otherwise Holant ( f | { g, [1 , x ] } ) can becomputed in PF, since all signatures on the RHS are degenerate and can be applied directed asunary signatures on copies of f .Let k = gcd ( m, n ), and s = n/k ≥
1. Consider any bipartite signature grid Ω = (
G, π ) forHolant ( f | { g, [1 , x ] } ). Let N f , N g , N u be the numbers of occurrences of f , g , [1 , x ] respectively.Then we have mN f = nN g + N u , thus N u ≡ k . Let t = N u /k ≥
0. We may assume t ≥
1, for otherwise [1 , x ] does not occurand the reduction is trivial.We will compute (Holant ( G )) s , the s -th power of the value Holant ( G ), using an oracle forHolant ( f | g ). Since the value Holant ( G ) is nongenative, we can obtain it from (Holant ( G )) s .In G , we replace each occurrence of [1 , x ] with the binary straddled signature (cid:18) xy xy (cid:19) , withone end connected to LHS, and leaving one edge yet to be connected to RHS. This creates a totalof N u such edges. This is equivalent to connecting N u copies of the unary signature [1 , x ] to LHS,and having N u copies of the unary signature [1 , y ] yet to be connected to RHS. Now in s disjointcopies of Ω there will be sN u = stk copies of [1 , y ] to be connected, to which we create t copies of g . In other words we take g ⊗ t with total arity stk and connect all stk unary signatures [1 , y ] to it.Since y ≥ g is not a multiple of [0 , ⊗ n , we get an easily computable positive factor.4 a) Binary straddled gadget G (b) Ternary gadget G (c) Unary gadget G (d) Ternary gadget G Figure 1: Some gadgetsThe main result of this paper is the following:
Theorem 2.2.
Holant ([ x , x , x , x ] | (= )) where x i ≥ for i = 0 , , , is -hard except inthe following cases, for which the problem is in FP .1. [ x , x , x , x ] is degenerate;2. x = x = 0 ;3. [( x = x = 0) ∧ ( x = x )] or [( x = x = 0) ∧ ( x = x )] . In case 1 the signature [ x , x , x , x ] decomposes into three unary signatures. In case 2[ x , , , x ] is a generalized equality. In case 3 the signature is in the affine class; see more de-tails about these tractable classes in [4]. Therefore the Holant problem is in FP in cases 1–3. Themain claim lies in that all other cases are Theorem 2.3.
Suppose a, b ∈ C , and let X = ab , Z = (cid:16) a + b (cid:17) . Then Holant ([ a, , b ] | (= )) is -hard except in the following cases, for which the problem is in P .1. X = 1 ;2. X = Z = 0 ;3. X = − and Z = 0 ;4. X = − and Z = − . In fact, since this paper mainly concerns with nonnegative valued functions, when establishing X = 1. When x and x are not both 0, we can normalize the signature. If x (cid:54) = 0 we divide the signatureby x , and get the form [1 , a, b, c ], with a, b, c ≥
0. If x = 0, but x (cid:54) = 0, we can flip all 0 and 1inputs, which amounts to a reversal of the signature and get the above form. This does not changethe complexity since the Holant value is only modified by a known nonzero factor.5onsider arguably the simplest possible gadget G in Figure 1. We have the signature matrix G = (cid:18) ba c (cid:19) . Let ∆ = (cid:112) (1 − c ) + 4 ab , and assume for now ∆ (cid:54) = 0, and we take the positivesquare root. Note that ∆ = 0 iff ( c = 1) ∧ ( ab = 0). The matrix G has two distinct eigenvalues λ = − ∆+(1+ c )2 and µ = ∆+(1+ c )2 . If a (cid:54) = 0, let x = ∆ − (1 − c )2 a and y = ∆+(1 − c )2 a . The matrix for G hasthe Jordan Normal Form (cid:18) ba c (cid:19) = (cid:18) − x y (cid:19) (cid:18) λ µ (cid:19) (cid:18) − x y (cid:19) − . (3.2)We now interpolate a binary degenerate straddled signature which will be used several times later. Lemma 3.1.
Given the binary straddled signature G = (cid:18) ba c (cid:19) with a (cid:54) = 0 and ∆ > , we caninterpolate unary signatures [1 , x ] on RHS or [ y, on LHS.Proof. For ∆ >
0, we have x + y = ∆ /a >
0. Consider D = 1 x + y (cid:18) y xy x (cid:19) = (cid:18) − x y (cid:19) (cid:18) (cid:19) (cid:18) − x y (cid:19) − . Given any signature grid Ω where the binary degenerate straddled signature D appears n times, weform gadgets G s where 0 ≤ s ≤ n by iterating the G gadget s times and replacing each occurrenceof D with G s . Denote the resulting signature grid as Ω s . We stratify the assignments in the Holantsum for Ω according to assignments to (cid:18) λ µ (cid:19) as:- (0 , i times;- (1 , j times;with i + j = n ; all other assignments will contribute 0 in the Holant sum. Let c i,j be the sum overall such assignments of the products of evaluations (including the contributions from (cid:18) − x y (cid:19) and its inverse). Then we have Holant Ω s = (cid:88) i + j = n (cid:0) λ i µ j (cid:1) s · c i,j and Holant Ω = c ,n . Since ∆ >
0, the coefficients form a full rank Vandermonde matrix. Thus wecan interpolate D by solving the linear system of equations in polynomial time. Ignoring a nonzerofactor, we may split D into unary signatures [ y,
1] on LHS or [1 , x ] on RHS.We have ∆ ≥ | − c | , and x, y ≥
0. Thus we can separate D and obtain unary signatures asspecified in Lemma 2.1.The following lemma lets us interpolate unary signatures on the RHS from a binary gadget witha straddled signature and a suitable unary signature s on the RHS. Mathematically, the proof isessentially the same as in [15], but technically Lemma 3.2 applies to binary straddled signatures. Lemma 3.2.
Let M ∈ R × be the signature matrix for a binary straddled gadget which is diago-nalizable with distinct eigenvalues, and let s = [ a, b ] be a unary signature on RHS that is not a roweigenvector of M . Then { s · M j } j ≥ can be used to interpolate any unary signature on RHS. A Basic Lemma
In this section, we prove Lemma 4.2 which will be invoked in the remaining cases.When ¬ ( x = x = 0), we normalize the signature [ x , x , x , x ] to be [1 , a, b, c ], where a, b, c ≥
0. We first deal with a special case when a = b and c = 1, which will be used in the proof ofLemma 4.2. By a slight abuse of notation, we say [1 , a, b, c ] is , a, b, c ] | (= )) is Lemma 4.1. [1 , a, a, is -hard unless a = 0 or a = 1 in which case the problem is in FP .Proof. When a = 0 or 1, the signature [1 , a, a,
1] is clearly in FP. Suppose a (cid:54) = 0 ,
1, we have∆ = 2 a >
0. By Lemma 3.1, we can interpolate the unary signature [1 , x ] = [1 ,
1] on RHS. Connect[1 ,
1] on RHS to [1 , a, a,
1] on LHS, we get the binary signature [1 + a, a, a ] on LHS. InvokeTheorem 2.3, as a (cid:54) = 1, the problem [1 + a, a, a ] is , a, a,
1] is
Lemma 4.2.
When ab (cid:54) = 0 , the problem [1 , a, b, c ] is -hard unless it is degenerate in which casethe problem is in FP .Proof. As ab (cid:54) = 0, we have a (cid:54) = 0, and ∆ >
0. By Lemma 3.1, we can interpolate the unarysignature [1 , x ] on RHS or [ y,
1] on LHS where x = ∆ − (1 − c )2 a and y = ∆+(1 − c )2 a . Consider the unarygadget G in Figure 1 where we place the circles to be (= ), the triangles to be [ y, , a, b, c ]. It has the unary signature [ y + yb, ya + c ] on the RHS. By Lemma 3.2, wecan interpolate any unary gadget, in particular ∆ := [1 ,
0] and ∆ := [0 , G is a row eigenvector of G . By Equation (3.2), the row eigenvectors of G are therows of (cid:18) − x y (cid:19) − , namely proportional to [1 , − y ] and [1 , x ]. Since y > ab (cid:54) = 0, the onlyexception is ya + cy + yb = x .Observe that xy = ba . Solve for y we get (cid:0) b − a (cid:1) y = (cid:0) ac − b (cid:1) . If b = a , then ac = b and thus [1 , a, b, c ] is degenerate. Otherwise, y = ac − b b − a . Now plug into y = ∆+(1 − c )2 a , we have∆ + (1 − c ) = a ( ac − b ) b − a which implies (cid:0) a − b − ab (1 − c ) (cid:1) ( ab − c ) = 0 (the high order multi-variable polynomial magically factors out).We now divide our discussion into three cases: (1) (cid:2)(cid:0) a − b − ab (1 − c ) (cid:54) = 0 (cid:1) ∧ ( ab − c (cid:54) = 0) (cid:3) ,(2) a − b − ab (1 − c ) = 0, and (3) ab − c = 0. Case 1: (cid:0) a − b − ab (1 − c ) (cid:54) = 0 (cid:1) ∧ ( ab − c (cid:54) = 0)We can interpolate ∆ and ∆ on RHS. By connecting ∆ and ∆ on RHS to [1 , a, b, c ] on LHS,we get the binary signatures [1 , a, b ] and [ a, b, c ] on LHS. That is we haveHolant ([1 , a, b ] | (= )) ≤ T Holant ([1 , a, b, c ] | (= ))and Holant ([ a, b, c ] | (= )) ≤ T Holant ([1 , a, b, c ] | (= )) . By Theorem 2.3, the problem Holant ([1 , a, b, c ] | (= )) is a · ba = 1 and ab · cb = 1,in which case the signature [1 , a, b, c ] is degenerate and thus the problem is in FP. Case 2: a − b − ab (1 − c ) = 0 7e have 1 − c = a − b ab and thus ∆ = (cid:112) (1 − c ) + 4 ab = a + b ab . Thus the unary signatureinterpolated by Lemma 3.1 on RHS is [1 , x ] = [1 , ∆ − (1 − c )2 a ] = [1 , b a ]. Connect [1 , x ] on RHS to[1 , a, b, c ] on LHS, we get the binary signature [1 + b a , a + b a , b + b ca ] on LHS. By Theorem 2.3,the problem [1 + b a , a + b a , b + b ca ] is , a, b, c ] is b a )( b + b ca ) = ( a + b a ) , which, after substituting c , simplifies to (cid:0) a − b (cid:1) (cid:0) a + ab + 2 b (cid:1) = 0.Since a, b >
0, we have a + ab +2 b (cid:54) = 0 and thus a − b = 0. This combined with a − b = ab (1 − c )gives c = a , and thus [1 , a, b, c ] is degenerate. Case 3: ab − c = 0Observe that under this case we have ∆ = 1 + c . The unary signature interpolated on RHS is[1 , x ] = [1 , ∆ − (1 − c )2 a ] = [1 , ca ] = [1 , b ]. We connect [1 , x ] on RHS to [1 , a, b, c ] on LHS to get binarysignature [1 + ab, a + b , b + bc ] on LHS. That is we haveHolant (cid:0) [1 + ab, a + b , b + bc ] | (= ) (cid:1) ≤ T Holant ([1 , a, b, c ] | (= ))By Theorem 2.3, this problem is ab )( b + bc ) = ( a + b ) which implies (cid:0) a − b (cid:1) (cid:0) b − (cid:1) = 0. If a − b = 0, since we are under the case ab − c = 0, the signature[1 , a, b, c ] is degenerate. If b − b is a positive real number, we have b = 1. Thus thesignature [1 , a, b, c ] is simply [1 , a, , a ].Now consider the ternary gadget G = [2 + 2 a , a + 2 a , a + 2 a , a ] on LHS where weplace the circles to be (= ) and the squares to be [1 , a, , a ]. This signature by G has the form inLemma 4.1, and we see that it is a = 2 a + 2 a . So the problem [1 , a, , a ] is , a, b, c ] is a − a − a + 1 = ( a − ( a + 1) = 0,which implies a = 1. When a = 1, the signature [1 , a, , a ] is degenerate.The proof is now complete.We are now ready to prove a series of lemmas which lead to our main result, Theorem 2.2. Thebasic idea is to reduce cases to Lemma 4.2, and handle exceptional cases separately. ¬ ( x = x = 0) With the help of Lemma 4.2 we can quickly finish this case. We may normalize [ x , x , x , x ] to[1 , a, b, c ] by dividing a nonzero factor, and flipping all input 0’s and 1’s if necessary (i.e., takingthe reversal of the signature). The exceptional cases in Lemma 4.2 for [1 , a, b, c ] are a = 0 or b = 0.We first consider the case a (cid:54) = 0 but b = 0. Lemma 5.1. If a (cid:54) = 0 and b = 0 , then [1 , a, b, c ] is -hard.Proof. By gadget G where we place the squares to be [1 , a, , c ] and the circles to be (= ), we havea ternary signature [1 + 3 a , a + a , a , a + c ] on LHS. Thus Lemma 4.2 applies (after normalizing)and [1 + 3 a , a + a , a , a + c ] is a + a ) = (1 + 3 a )( a ) and( a ) = ( a + a )( a + c ). The first equality forces a = 1 for positive a . Then we get = 1 + c ≥ a = 0. We have the following subcases (1) c = 1, (2) c = 0, and (3)( c (cid:54) = 1) ∧ ( c (cid:54) = 0). These are handled by the following three lemmas.8 emma 5.2. If a = 0 and c = 1 , then [1 , , b, is -hard unless b = 0 , in which case the problemis in FP .Proof. By flipping 0 and 1 in the input, we equivalently consider the signature [1 , b, , b = 0 in which case it is in FP. Lemma 5.3. If a = 0 and c = 0 , then [1 , , b, is -hard unless b = 0 or b = 1 , in which casethe problem is in FP .Proof. By gadget G where we place [1 , , b,
0] at the squares and (= ) at the circles, we have thebinary degenerate straddled signature (cid:18) b (cid:19) . We can thus split and get unary signature [1 , b ]on RHS. Connect [1 , b ] on RHS and [1 , , b,
0] on LHS, we get the binary signature [1 , b , b ] on LHS.By Theorem 2.3, [1 , b , b ] is b = 0 or b = 1. When b = 0, the original signature[1 , , b,
0] becomes [1 , , ,
0] which is degenerate and thus the problem is tractable. When b = 1,the signature [1 , , b,
0] becomes [1 , , ,
0] which is an affine signature and thus the problem is alsotractable ([4] p.70).
Lemma 5.4. If a = 0 , c (cid:54) = 0 , and c (cid:54) = 1 , then [1 , , b, c ] is -hard unless b = 0 , in which case theproblem is in FP .Proof. By flipping 0 and 1 in the input, we equivalently consider the signature [ c, b, , b = 0 in which case [ c, , ,
1] is in FP.The discussion for the case ¬ ( x = x = 0) is now complete. x = x = 0 If both x = x = 0, the signature is identically zero and the problem is trivially in FP.Suppose exactly one of x and x is 0. In this case, by normalizing and possibly flipping 0 and1 in the input, it suffices to consider the ternary signature [0 , , , Lemma 6.1.
The problem
Holant ([0 , , , | (= )) is -hard.Proof. We begin with a reduction from Restricted Exact Cover by 3-Set (RX3C) [11]. This is arestricted version of the Set Exact Cover problem where every set has exactly 3 elements and everyelement is in exactly 3 sets. Given any instance for RX3C, construct the 3-regular bipartite graphwhose vertices on LHS are elements of the set X and vertices on RHS are 3-element subsets of X . Connect an edge between one vertex v on LHS to one vertex C on RHS if and only if v ∈ C .Then every nonzero term in the Holant sum for Holant ([0 , , , | (= )) exactly corresponds toone solution of RX3C. We observe that the reduction given in [11] (from the Exact Cover by 3 Setproblem (X3C)) is parsimonious. And it is well known that SAT reduces to X3C parsimoniouslyvia 3-Dimensional Matching (3DM). Thus we have the reduction chain: ≤ T ≤ T ≤ T ≤ T Holant ([0 , , , | (= ))We conclude that the problem [0 , , ,
0] is x · x (cid:54) = 0. We normalize the signature to be [0 , , b, emma 6.2. If b (cid:54) = 0 , then the problem Holant ([0 , , b, | (= )) is -hard.Proof. By gadget G where we place the squares to be [0 , , b,
0] and the circles to be (= ), wehave the ternary signature [3 b , b , b + b , b ] on LHS. Lemma 4.2 applies (after normalizing)and the problem [3 b , b , b + b , b ] is b , b , b + b , b ] being degenerate is (cid:40) b b = b b + b b b + b = b + b b The second equation simplifies to ( b − = 0. Thus when b (cid:54) = 1, we conclude that [0 , , b,
0] is , , , G , where we place the squares to be [0 , , ,
0] andthe circles to be [1 , , , , , ,
3] on LHS. ByLemma 4.2 this problem is , , , | (= )) when restricted to planar graphs is in FP. Problem : Pl-
Input : A planar 3-uniform 3-regular hypergraph G . Output : The number of subsets of hyperedges that cover every vertex with no vertex coveredthree times.This is exactly the problem Pl-Holant([0 , , , | (= )), the restriction of Holant ([0 , , , | (= ))to planar graphs. Its P-time tractability is seen by the following holographic reduction using theHadamard matrix H = (cid:18) − (cid:19) to counting weighted perfect matchings. Under this transfor-mation both signatures [0 , , ,
0] and (= ) are transformed to matchgate signatures. H ⊗ (= ) = H ⊗ (cid:34)(cid:18) (cid:19) ⊗ + (cid:18) (cid:19) ⊗ (cid:35) = (cid:18) (cid:19) ⊗ + (cid:18) − (cid:19) ⊗ = [2 , , , , and [0 , , , H − ) ⊗ = 18 (cid:2) (1 , ⊗ − (1 , ⊗ − (0 , ⊗ (cid:3) H ⊗ = 14 [3 , , − , . Since both these are matchgate signatures, the problem is reduced to counting weighted perfectmatchings on planar graphs. Thus the planar problem Pl- Holant ([0 , , , | (= )) can be com-puted in polynomial time using Kasteleyn’s algorithm (see [16] and [4]). For readers unfamiliarwith holographic algorithms, we will describe this algorithm for Pl- in more detail in an Appendix. Acknowledgement
We sincerely thank Shuai Shao for many helpful discussions.10 ppendix
The holographic algorithm for Pl- is achieved by a many-to-many reduction from this problem to the problem of counting (weighted) planar perfect matchings(Pl- PM ). Given planar weighted graph G = ( V, E, w ) where w : E → R is a weight functionfor the edge set, any perfect matching M ⊆ E has weight w ( M ) = (cid:81) e ∈ M w ( e ), and the partitionfunction of the problem Pl- PM is G ) = (cid:88) perfect matchings M w ( M ) . Notice that when all weights w ( e ) = 1, i.e., in the unweighted case, G ) simply counts thenumber of perfect matchings in G . Kasteleyn’s algorithm can compute this partition function G ) in polynomial time on planar graphs. It is important to note that Kasteleyn’s algorithmcan handle the case when w ( e ) are both positive and negative real (even complex) numbers, whichis important in this case for us.The values of Holant ([0 , , , | (= )) and Holant (cid:0) [3 , , − , | [2 , , , (cid:1) on the same in-stance graph G are exactly the same by Valiant’s Holant Theorem [16], as[0 , , , H − ) ⊗ = 14 [3 , , − ,
0] and H ⊗ (= ) = [2 , , , . Notice that this is a many-to-many transformation among solutions of one problem Holant ([0 , , , | (= ))(subsets of hyperedges that cover every vertex with no vertex covered three times) and anotherHolant (cid:0) [3 , , − , | [2 , , , (cid:1) .Next we will use the following gadgets, called matchgates, to “implement” the constraint func-tions [3 , , − ,
0] and [2 , , ,
0] by perfect matchings. Consider the following two matchgates. (a) Matchgate for [3 , , − ,
0] (b) Matchgate for [2 , , , Figure 2: MatchgatesFor the matchgate in Figure 2 (a), we consider its perfect matchings, when any subset ∅ ⊆ S ⊆ { , , } of the external nodes labeled 1 , , S = ∅ there is a unique perfectmatching M with weight w ( M ) = 3 /
4. If | S | = 2, we get w ( S ) = − /
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