Parameterizing the Permanent: Genus, Apices, Minors, Evaluation mod 2^k
PParameterizing the Permanent:Genus, Apices, Minors, Evaluation mod 2 k Radu Curticapean ∗ , Mingji Xia † October 4, 2018
Abstract
We identify and study relevant structural parameters for the problem PerfMatch of counting perfectmatchings in a given input graph G . These generalize the well-known tractable planar case, and theyinclude the genus of G , its apex number (the minimum number of vertices whose removal renders G planar), and its Hadwiger number (the size of a largest clique minor).To study these parameters, we first introduce the notion of combined matchgates , a general techniquethat bridges parameterized counting problems and the theory of so-called Holants and matchgates: Usingcombined matchgates, we can simulate certain non-existing gadgets F as linear combinations of t = O (1)existing gadgets. If a graph G features k occurrences of F , we can then reduce G to t k graphs thatfeature only existing gadgets, thus enabling parameterized reductions.As applications of this technique, we simplify known 4 g n O (1) time algorithms for PerfMatch on graphsof genus g . Orthogonally to this, we show W [ ]-hardness of the permanent on k -apex graphs, implyingits W [ ]-hardness under the Hadwiger number. Additionally, we rule out n o ( k/ log k ) time algorithmsunder the counting exponential-time hypothesis ETH .Finally, we use combined matchgates to prove ⊕ W [ ]-hardness of evaluating the permanent modulo 2 k ,complementing an O ( n k − ) time algorithm by Valiant and answering an open question of Bj¨orklund.We also obtain a lower bound of n Ω( k/ log k ) under the parity version ⊕ ETH of the exponential-timehypothesis. ∗ Simons Institute for the Theory of Computing, Berkeley, USA, and Institute for Computer Science and Control, HungarianAcademy of Sciences (MTA SZTAKI), Budapest, Hungary. Supported by ERC Starting Grant PARAMTIGHT, No. 280152. † Institute of Software, Chinese Academy of Sciences, Beijing, China. Supported by China National 973 program2014CB340301, China Basic Research Program (973) Grant 2014CB340302, NSFC 61003030 and NSFC 61170073. a r X i v : . [ c s . CC ] N ov ontents k . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 Proof technique: Linear combinations of signatures . . . . . . . . . . . . . . . . . . . . . . . . 6 (cid:48) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 k The study of counting problems has become a classical subfield of computational complexity since Valiant’sseminal papers [51, 52] introduced the class P and established P -hardness of counting perfect matchingsin bipartite graphs. In particular, this proves P -hardness of the following generalized problem: Given agraph G with edge-weights w : E ( G ) → Q , compute the quantityPerfMatch( G ) := (cid:88) M ⊆ E ( G )perfect matching of G (cid:89) e ∈ M w ( e ) . In statistical physics, PerfMatch is known as the partition function of the dimer model [48, 35, 36],and the first nontrivial algorithms for the evaluation of this quantity stem from this area. This includesthe celebrated
FKT method , a polynomial-time algorithm for computing PerfMatch on planar graphs [36].Roughly speaking, this algorithm proceeds as follows: Given a planar graph G , it constructs a Pfaffianorientation F of G , which we may view as a subset F ⊆ E ( G ) with the following miraculous property: If wedefine a matrix A from the adjacency matrix of G by flipping the signs of edges in F , then (PerfMatch( G )) =det( A ). Overall, this yields a reduction from planar PerfMatch to the determinant.2n algebra and combinatorics, the quantity PerfMatch( G ) for a bipartite graph G with n + n vertices isbetter known as the permanent of the biadjacency matrix A of G , defined byperm( A ) = (cid:88) σ :[ n ] → [ n ]is a permutation n (cid:89) i =1 A i,σ ( i ) . The permanent is central to algebraic complexity theory, which aims at proving the permanent to be in-herently harder than the similar-looking determinant [1, 43, 4]. This would imply an algebraic analogue of P (cid:54) = NP [50].In order to obtain a more refined view on the complexity of the permanent, and to cope with its hardnessin view of practical applications, various relaxations of this problem were studied: A celebrated randomized approximation scheme [34, 33] allows one to approximate the permanent on matrices with non-negativeentries. Furthermore, on some restricted graph classes , PerfMatch can be solved in time O ( n ): Thisincludes the above-mentioned planar graphs, and in fact, all graph classes of bounded genus [29, 49, 44]. (Wewill present more classes in the remainder of the introduction.) As another relaxation, modular evaluation of the permanent was studied in Valiant’s original paper [51]: He showed that the permanent modulo m = 2 k can be computed in time n O ( k ) for all k ∈ N , but for all m containing an odd prime factor, the evaluationmodulo m is NP -hard under randomized reductions.In this paper, we consider another such refinement (and generalize existing ones) by investigating thepermanent in the framework of parameterized complexity . This area was initiated by Downey and Fellows[24, 25] and was adapted to counting problems by Flum and Grohe [26] and McCartin [42]. In parameterizedcounting complexity, the objects in study are counting problems that come with parameterizations π : { , } ∗ → N , and a central question is whether such problems are fixed-parameter tractable (fpt) . A givenproblem is fpt if it can be solved in time f ( π ( x )) | x | O (1) on input x , for a computable function f that dependsonly on the parameter value, but not on | x | . If we fail to find an fpt-algorithm for a given parameterizedproblem, we can often give evidence that no such algorithm exists by proving its W [ ]-hardness, theparameterized analogue of P -hardness. (A more detailed exposition can be found in Section 2.)By studying natural parameterizations π of the input, we obtain a fine-grained complexity analysis thatcould not be achieved by considering the input size | x | alone. For instance, consider the problem VertexCover,which asks whether a graph G on n vertices admits a vertex-cover of size k . This problem is NP -complete,but it can be solved in time n O ( k ) for every fixed k , and it is actually even fpt in the parameter k , as wecan find [24] and even count [27] vertex-covers of size k in time 2 k n O (1) . On the other hand, we can decidein polynomial time whether G contains a matching of size k , but the problem of counting k -matchings is P -complete, and in fact even W [ ]-complete when parameterized by k [13, 16]. To investigate the parameterized complexity of the permanent, we first identify interesting parameterizationsfor this problem. For instance, the maximum degree ∆( G ) of the input graph G is not particularly interesting,since the permanent is already P -complete on 3-regular graphs [17]. That is, even an n f (∆( G )) timealgorithm for some function f (and an fpt-algorithm in particular) would imply P = P . However, itturns out that the known polynomial-time solvable graph classes for PerfMatch point us towards a naturalparameter, namely the size of a smallest excluded minor. Here, a minor H of a graph G is a graph thatcan be obtained from G by deletions of edges and/or vertices, and contraction of edges. To explain thesignificance of minors for counting perfect matchings, we first survey the known algorithms for PerfMatch,all of which can be considered as generalizations of the FKT method for planar graphs. Excluding K , or K : It was shown by Little [37] and later by Vazirani [55] (who gave a parallelizedalgorithm) that PerfMatch can be solved in time O ( n ) on graphs excluding the minor K , . A similarresult was recently shown by Straub et al. [47] for graphs excluding K . Note that the FKT methodgives an O ( n ) time algorithm on graphs excluding both K , and K , whereas the two above algorithms3how that excluding either minor entails the polynomial-time solvability of PerfMatch. For the K , -free case, this was shown by constructing a Pfaffian orientation. The K -free case was shown by adifferent technique; in particular, K -free graphs do not necessarily admit Pfaffian orientations. Excluding single-crossing minors:
Extending the above item, it was recently shown by Curticapean [14]that PerfMatch can be solved in time O ( n ) on any class excluding a fixed single-crossing minor H ,i.e., a minor that can be drawn in the plane with at most one crossing, such as K , or K . In fact,it is shown that PerfMatch is fpt in the size of the smallest excluded single-crossing minor. Thisalgorithm does not inherently rely upon Pfaffian orientations, apart from a black-box algorithm forplanar PerfMatch. Bounded-genus graphs:
Another line of extensions of the FKT method is to graphs of bounded genus :It was shown independently by Gallucio and Loebl [29], Tesler [49] and Regge and Zechina [44] thatPerfMatch can be solved in time O (4 g n ) on n -vertex graphs G of genus g . In the framework offixed-parameter tractability, this can be read as PerfMatch being fpt when parameterized by the genusof G . The algorithms for the bounded-genus case proceed by expressing PerfMatch( G ) as the linearcombination of 4 g determinants derived from Pfaffian orientations. In the present paper, we give analternative proof of this theorem that proceeds by reduction to 4 g instances of planar PerfMatch.Together with the previous item, this eliminates the need for Pfaffian orientations from all knownalgorithms for PerfMatch except for the planar case.From the above list, we can draw the conclusion that every known polynomial-time solvable graph classfor PerfMatch excludes some fixed minor. This is clear for the first two items, and furthermore, the graphsof genus g ∈ N are easily seen to exclude a complete graph of size O ( g ). Since this shows that excludedminors have been a driving force behind polynomial-time algorithms for PerfMatch, it is natural to studythis problem under the more general Hadwiger number hadw( G ) := max { k ∈ N : G contains a K k -minor } . Note that planar graphs have Hadwiger number at most 4. More generally, if the genus of G or the size ofthe smallest excluded single-crossing minor is bounded, then hadw( G ) is bounded as well, but the conversedoes not hold. However, the Graph Structure Theorem [45], a celebrated result in graph minor theory [46],yields a decomposition of the graphs with fixed Hadwiger number k into graphs that have genus c = c ( k )except for c occurrences of certain defects, namely so-called vortices and apices. Such decompositions haveproven immensely useful for fpt-algorithms on graphs excluding fixed minors, see [40, 22, 21, 20, 19, 28]. If aproblem can be solved efficiently on planar instances and we can extend this to bounded-genus instances, asin the case of PerfMatch, then with a leap of faith, the Graph Structure Theorem allows us to hope for anfpt-algorithm under the more general parameterization by Hadwiger number. Our following negative resulthowever shatters these hopes for the case of PerfMatch. Theorem 1.1.
The zero-one permanent is W [ ] -hard when parameterized by the Hadwiger number. Inother words, computing PerfMatch is W [ ] -hard when parameterized by the Hadwiger number, even onbipartite graphs without edge-weights. We show this by proving the following stronger statement: Let us define the apex numberapex( G ) := min { k ∈ N | ∃ S ⊆ V ( G ) of size k : G − S is planar } . This parameter, studied in [41], measures the distance of a graph to planarity by vertex deletions. Notethat planar graphs have apex number 0. Using the apex number as parameter, we can generalize planar This statement comes with a caveat: For instance, we can determine the number of perfect matchings in a complete graphin polynomial time by means of a closed formula. The class of complete graphs clearly excludes no fixed minor. However, wecannot solve the (weighted) problem PerfMatch on this class in polynomial time, as edge-weights would allow us to simulatearbitrary graphs, for which counting perfect matchings is P -complete. G ) ≤ O (apex( G )).This allows us to obtain Theorem 1.1 as a corollary from the following result, which we consider to be ofindependent interest. Theorem 1.2.
The permanent is W [ ] -hard when parameterized by the apex number. Assuming theexponential-time hypothesis ETH , it admits no n o ( k/ log k ) time algorithm on k -apex graphs with n vertices. This contrasts with the fpt-algorithm for PerfMatch when parameterized by genus. We observe thatPerfMatch can be computed easily in time n k + O (1) on k -apex graphs by means of brute-force, so the lowerbound under ETH is almost tight. However, it should be noted that no similar algorithm is known for theHadwiger number: At least to us, it remains an important open question whether PerfMatch can be solvedin time n f ( k ) on graphs excluding the complete graph K k as minor. k In the following, we depart from structural parameters of the input graph G and consider the evaluation ofthe permanent modulo 2 k . In the seminal paper [51], not only did Valiant prove P -completeness of thepermanent, but he also studied the complexity of evaluating the permanent modulo fixed numbers m ∈ N .Observe that perm( A ) and det( A ) are equivalent modulo 2 for any matrix A , giving a polynomial-timealgorithm for the permanent modulo 2. On the other hand, for odd primes p , Valiant’s original proof showsthat the permanent modulo p is Mod p P -complete. That is, we can reduce counting satisfying assignmentsto 3-CNF formulas modulo p to the permanent modulo p . This also shows the NP -hardness of the latterproblem under randomized reductions, and this holds more generally whenever the modulus m contains anodd prime factor, that is, when m is not a power of two.For the remaining cases m = 2 k however, Valiant [51] showed an O ( n k ) time algorithm for evaluatingthe permanent modulo 2 k on n -vertex graphs, which was recently improved to n k + O (1) time by Bj¨orklund,Husfeldt and Lyckberg [3]. Given these results, it is natural to study this problem in the framework ofparameterized complexity, thus asking whether we can compute the permanent modulo 2 k in time n o ( k ) oreven f ( k ) n O (1) . This was also posed as an open problem in [3]. Please recall that this question is indeedonly interesting for m = 2 k : As stated in the previous paragraph, on all other fixed m ∈ N , the problem is NP -hard under randomized reductions.We rule out the fixed-parameter tractability of the permanent modulo 2 k by the following strongerhardness result, which also establishes an unexpected connection to the apex parameter introduced before:Evaluating the permanent modulo 2 k on k -apex graphs is ⊕ W [ ]-hard, that is, an fpt-algorithm for thisproblem would imply one for counting k -cliques modulo 2, a problem that was shown to be W [ ]-hard underrandomized reductions by a recent result of Bj¨orkund, Dell and Husfeldt [2]. We also obtain an almost-tightlower bound under ⊕ ETH , the parity version of the exponential-time hypothesis
ETH . Theorem 1.3.
The evaluation of the permanent modulo k is ⊕ W [ ] -hard when parameterized by k , evenwhen restricted to k -apex graphs. Assuming ⊕ ETH , there is no n o ( k/ log k ) time algorithm for this problem. Theorem 1.3 is proven by reduction from the following problem ⊕ PartitionedSub: Given vertex-coloredgraphs H and G as input, where each color in H appears exactly once, count modulo 2 the subgraphsof G that are isomorphic to H , respecting colors. It was shown that the decision version of this problem,which is W [ ]-hard, can be reduced to ⊕ PartitionedSub by means of randomized reductions [2]. Furthermore,assuming ⊕ ETH , an argument by Marx [38] implies that ⊕ PartitionedSub cannot be solved in time n o ( (cid:96)/ log (cid:96) ) for (cid:96) -edge graphs H and n -vertex graphs G .In our reduction, we transform a given instance ( H, G ) for ⊕ PartitionedSub with an (cid:96) -edge graph H to3 (cid:96) instances of the permanent modulo 2 (cid:96) +1 on 2 (cid:96) -apex graphs with O ( (cid:96) n ) vertices. Thus, if we can provebetter lower bounds for finding k -edge subgraphs, then those bounds carry over to the seemingly unrelatedproblem of evaluating permanents modulo 2 k , even on k -apex graphs. On the other hand, a randomized n o ( k ) time algorithm for the permanent modulo 2 k on k -apex graphs would imply one for PartitionedSub on k -edge graphs H , thus falsifying a hypothesis posed by Marx [38].5 .3 Proof technique: Linear combinations of signatures We phrase our proofs in the language of so-called Holant problems [8] and matchgates [8, 5, 6]. Please considerSection 3 for a more detailed introduction into these topics. In our proofs, we reformulate parameterizedcounting problems as Holant problems (specific weighted sums over assignments to the edges of graphs) andthen try to realize the occurring signatures (local constraints at vertices) by certain matchgates (gadgets).However, many required signatures cannot be realized by matchgates. The key idea in this paper is thatsuch unrealizable signatures can sometimes still be realized as linear combinations of matchgate signatures.To this end, we proceed as follows: First, we show how to simulate non-existing gadgets F as a formallinear combination of realizable gadgets F , . . . , F t , typically with t = O (1). Then, if a graph G features k occurrences of F , we can easily reduce G to t k graphs that feature only occurrences of F , . . . , F t . Eachof these t k graphs can then be handled by an algorithm (when we aim at positive results) or by an oraclecall (when proving hardness results). The generality of our technique allows it to be applied to variousparameterized problems. For instance, a recent W [ ]-hardness proof for counting k -matchings [16] can alsobe rephrased in this framework.As pointed out by Tyson Williams, a similar idea appears under the notion of vanishing signatures [30, 7]. These however apply linear combinations in a quite different setting. In particular, they consider noconnections to parameterized complexity. Organization of the paper
In Section 2, we introduce notions from parameterized complexity, exponential-time complexity, and weprove W [ ]-hardness of a modified version of the problem g · n O (1) time algorithm for PerfMatch on graphs of genus g . In Section 5, we then prove Theorem 1.2,which asserts W [ ]-hardness of PerfMatch on bipartite unweighted k -apex graphs and implies Theorem 1.1,the hardness under the Hadwiger number parameter. In Section 6, we introduce a more involved constructionand an additional technique called discrete derivatives to transform the argument from Section 5 to a proofof Theorem 1.3. For n ∈ N , we write [ n ] := { , . . . , n } . The graphs G in this paper are undirected, but they may featureparallel edges and edge-weights. All hardness results are however shown for simple graphs featuring noparallel edges and no edge-weights. We write uv ∈ E ( G ) for an edge of G , and given v ∈ V ( G ), we denotethe edges incident with v by I ( v ). Sometimes, we consider graphs to be embedded on surfaces, see [23].For numbers n ∈ N , we abbreviate ⊕ n := ( n mod 2). Given a bitstring x ∈ { , } ∗ , we write hw( x ) := (cid:80) i x i for its Hamming weight , and we define
ODD ( x ) := ⊕ hw( x ) , EVEN ( x ) := 1 − ⊕ hw( x ) . We write supp( f ) for the support of a function f . For predicates ϕ , we use the Iverson bracket notation[ ϕ ] := (cid:40) ϕ is true,0 otherwise.Let A and B be sets; we define certain abbreviations for subsets of A × B . For b ∈ B , we write( (cid:63), b ) = { ( a, b ) | a ∈ A } for the column at b . For a ∈ A , we write ( a, (cid:63) ) = { ( a, b ) | b ∈ B } for the row at6 . We use this notation only when A and B are unambiguous from the context. For k ∈ N , we say that( i, j ) ∈ [ k ] and ( i (cid:48) , j (cid:48) ) ∈ [ k ] are vertically adjacent if | i − i (cid:48) | = 1 and j = j (cid:48) . Likewise, we call such pairs horizontally adjacent if | j − j (cid:48) | = 1 and i = i (cid:48) . Parameterized counting problems are problems A /π , where A : { , } ∗ → C is a counting problem and π : { , } ∗ → N is a polynomial-time computable parameterization, see [26]. We define FPT as the classof all problems A /π such that A can be solved in time f ( π ( x )) | x | O (1) . Likewise, we define XP as the classof problems A /π that can be solved in time | x | f ( π ( x )) , where f : N → N is a computable function. In thefollowing, we define the classes W [ ], W [ ] and ⊕ W [ ] we referred to in the introduction, using the followingreduction notions. Definition 2.1 ([26]) . Let A /π and B /π (cid:48) be parameterized counting problems. • We call f : { , } ∗ → { , } ∗ a parsimonious fpt-reduction and write A /π ≤ parsfpt B /π (cid:48) if there arecomputable functions r, s such that the following holds for all x ∈ { , } ∗ :1. We have A( x ) = B( f ( x )).2. The running time of f is bounded by r ( π ( x )) · | x | O (1) .3. We have π (cid:48) ( f ( x )) ≤ s ( π ( x )).If A and B are decision problems, replace the first condition by “ x ∈ A iff f ( x ) ∈ B”, and writeA /π ≤ fpt B /π (cid:48) . • We call an algorithm T a Turing fpt-reduction and write A /π ≤ T fpt B /π (cid:48) if there are computablefunctions r and s such that the following holds for all x ∈ { , } ∗ : Firstly, the running time of T on x is bounded by r ( π ( x )) | x | O (1) . Secondly, every oracle query y issued by T on x satisfies π (cid:48) ( y ) ≤ s ( π ( x )).We introduce W [ ], ⊕ W [ ] and W [ ] as the closures of clique-related problems under fpt-reductions. Definition 2.2.
Consider the following parameterized problems and complexity classes: • Let Clique /k denote the problem of deciding , on input a graph G and k ∈ N , whether G contains a k -clique. Let W [ ] denote the set of all problems A /π with A /π ≤ fpt Clique /k . • Let /k denote the problem of determining, on input G and k , the number of k -cliques in G .Let W [ ] denote the set of all problems A /π with A /π ≤ parsfpt /k . • Let ⊕ Clique /k denote the problem of deciding , on input G and k , whether G contains an odd numberof k -cliques. Let ⊕ W [ ] denote the set of all A /π with A /π ≤ fpt ⊕ Clique /k .It is a standard assumption of parameterized complexity theory that FPT (cid:54) = W [ ] holds, implying FPT (cid:54) = W [ ]. The problem Clique /k is W [ ]-complete by definition, so this assumption can equivalently beconsidered as the statement that Clique /k is not fixed-parameter tractable. Furthermore, it has been recentlyshown in [2, Theorem 5] that ⊕ Clique /k is W [ ]-hard under randomized parameterized reductions with con-stant one-sided error. Therefore, an fpt-algorithm for ⊕ Clique /k would imply a randomized fpt-algorithmfor Clique /k , which is considered almost as unlikely as FPT = W [ ] . We also consider conditional lower bounds on the running times required to solve problems. These are basedon different exponential-time hypotheses, introduced by [31, 32] and [18].7 efinition 2.3.
The exponential-time hypothesis
ETH , introduced in [31, 32], claims that the satisfiabilityof 3-CNF formulas on n variables cannot be decided in time 2 o ( n ) n O (1) . The hypothesis ETH postulatesthe same lower bound for counting the number of satisfying assignments to 3-CNF formulas, and ⊕ ETH postulates the same for computing the parity of the number of satisfying assignments.The hypothesis
ETH implies a lower bound for Clique /k , and thus also FPT (cid:54) = W [ ]: It was shown in[11, 12] that Clique /k cannot be solved in time f ( k ) · n o ( k ) on n -vertex graphs, for any computable function f . Furthermore, if a problem A /π cannot be solved in time f ( k ) · n g ( k ) under ETH , and we can reduce A /π to B /π (cid:48) with a reduction f that satisfies π (cid:48) ( f ( x )) ∈ O ( π ( x )) for all x , then it can be seen that B /π (cid:48) cannotbe solved in time f (cid:48) ( k ) · n Ω( g ( k )) under ETH , for any computable function f (cid:48) .By an isolation argument similar to the Valiant-Vazirani theorem [54], it was shown in [10] that a 2 o ( n ) time algorithm for counting satisfying assignments to 3-CNF formulas modulo 2 implies a randomized 2 o ( n ) time algorithm for deciding the existence of a satisfying assignment. In other words, a randomized version rETH of ETH implies ⊕ ETH ; see also [18] for more details.
We will reduce from the problem GridTiling of deciding the existence of grid tilings, as well as its countingversion ⊕ GridTiling. The decision version GridTiling wasintroduced by Marx [39] in order to obtain lower bounds for planar multiway cut, but grid tilings have sinceproven to be generally useful for proving hardness of problems on planar structures [40].
Definition 2.4.
The inputs to the problem GridTiling are numbers n, k ∈ N , together with a set C ⊆ [ k ] and a function T : C → [ n ] . The task is to decide whether there exists a grid tiling of T , i.e., a function a : [ k ] → [ n ] such that:1. For horizontally adjacent κ, κ (cid:48) ∈ [ k ] , the first components of a ( κ ) and a ( κ (cid:48) ) agree.2. For vertically adjacent κ, κ (cid:48) ∈ [ k ] , the second components of a ( κ ) and a ( κ (cid:48) ) agree.3. For all κ ∈ C , we have a ( κ ) ∈ T ( κ ).On the same inputs, we also define the problem number of gridtilings, and the problem ⊕ GridTiling, which asks to determine the parity of this number. All three problemsare parameterized by k .It should be noted that our definition of GridTiling is actually a generalized version of Marx’s originalformulation [39]: In his definition, the set C of any instance is fixed to C = [ k ] . That is, the third conditionof Definition 2.4 is required to apply for all κ ∈ [ k ] , whereas in our formulation, only a subset is relevant.In particular, we may choose sparse subsets C with |C| = O ( k ), which will make the generalized grid tilingproblems very useful in proving lower bounds under the exponential-time hypotheses.By reduction from k -cliques, Marx showed that GridTiling is complete for W [ ]. A simple adaptationof this reduction shows that the same holds for its counting and parity version, where W [ ] and ⊕ W [ ]take the part of W [ ]. In the remainder of this subsection, we give a different reduction, which choosespartitioned subgraph isomorphisms rather than k -cliques as a reduction source. This allows us to transferan almost-tight conditional lower bound for subgraph isomorphisms under ETH to GridTiling.
Definition 2.5.
For k ∈ N , a [ k ] -colored graph is a pair ( H, c ), where H is a graph and c : V ( H ) → [ k ] is acoloring. We call ( H, c ) colorful if c is bijective. This implies of course that H has k vertices.For [ k ]-colored graphs ( H, c ) and (
G, c (cid:48) ), we say that (
H, c ) is color-preserving isomorphic to (
G, c (cid:48) ) ifthere exists an isomorphism f from H to G such that c ( v ) = c (cid:48) ( f ( v )) holds for all v ∈ V ( H ). To simplifynotation, we will often write G rather than ( G, c ) for a colored graph.The problem PartitionedSub is defined as follows: The input consists of [ k ]-colored graphs H and G ,where H is colorful. The task is to decide whether there exists a copy of H in G , which is a (not necessarilyinduced) subgraph F of G such that H and F are color-preserving isomorphic. Likewise, the problem8PartitionedSub asks to determine the number of copies of H in G , and ⊕ PartitionedSub asks to determineits parity. All problems are parameterized by k .It can be shown by a parsimonious reduction from Clique that the problem PartitionedSub is W [ ]-complete, and this implies similar statements for its other variants as well. We omit the elementary proof. Lemma 2.6.
The three variants of
PartitionedSub are complete for W [ ] , W [ ] or ⊕ W [ ] , respectively.Remark . Let H and G be [ k ]-colored such that H is colorful; we assume V ( H ) = [ k ] without limitationof generality. If F is a H -copy in G and uv ∈ E ( F ) is an edge with endpoints of colors i and j for some i, j ∈ [ k ], then the edge ij must be present in H .We may therefore assume the following: Whenever an instance ( H, G ) to PartitionedSub is given, thenfor all i, j ∈ [ k ] with ij / ∈ E ( H ), the graph G contains no edges between i -colored and j -colored vertices.Otherwise, we can delete these edges without affecting the set of color-preserving H -copies in G .In the following, we consider instances ( H, G ) for PartitionedSub with n = | V ( G ) | and k = | V ( H ) | . Wecan solve each such instance in time n O ( k ) by brute force, and by reduction from Clique, it was shown thatalgorithms with running time f ( k ) · n o ( k ) would refute ETH , see [11, 12].This lower bound alone would however not suffice for our purposes of proving tight lower bounds: In thereductions from PartitionedSub to the permanent we construct later, each edge of H will incur some constantparameter blowup. As an example, on instances ( H, G ), our reduction images for the permanent will have O ( | E ( H ) | ) apices, which amounts to O ( k ) apices if H is a k -clique. Thus, if we reduced from Clique for ourlower bound, then ETH could only rule out algorithms with running time n o ( √ t ) for the permanent on t -apexgraphs. This is however obviously far from the upper bound of O ( n t +3 ) time obtained by the brute-forcealgorithm, and we would not consider such a result to be satisfactory.To avoid this problem, we use a refined lower bound for PartitionedSub, shown also by Marx, whichallows to assume that H has constant degree, and thus, only O ( k ) edges, see [38, Corollary 6.3]. Theorem 2.8 ([38]) . Assuming
ETH , there is a universal constant C ∗ such that PartitionedSub cannotbe solved in time f ( k ) · n o ( k/ log k ) , for any computable function f , even on instances ( H, G ) where H hasmaximum degree at most C ∗ . The same applies to the variants and ⊕ PartitionedSub ,assuming
ETH and ⊕ ETH respectively.
Using Lemma 2.6 and Theorem 2.8, we can then prove similar lower bounds for grid tilings.
Theorem 2.9.
The three variants of
GridTiling are complete for W [ ] , W [ ] and ⊕ W [ ] , respectively.Furthermore, the problems admit no n o ( k/ log k ) time algorithms, even on instances with |C| = O ( k ) , unless ETH , ETH or ⊕ ETH fails, respectively.Proof.
Let G and H be [ k ]-vertex-colored, where we assume V ( G ) = [ n ] and V ( H ) = [ k ]. Replace each edge uv in G by the directed edges uv and vu , then add all self-loops to G to obtain a colored directed graph G (cid:48) .Define the colorful directed graph H (cid:48) by applying the same operations on H . Then we can observe that thecolor-preserving H -copies in G stand in bijection with the color-preserving H (cid:48) -copies in G (cid:48) .For i, j ∈ [ k ], write E i,j = E i,j ( G (cid:48) ) for the set of directed edges in G (cid:48) from i -colored vertices to j -coloredvertices. By Remark 2.7, we may assume that E i,j = ∅ if ij / ∈ E ( H (cid:48) ). Note that E i,j ⊆ [ n ] ; we use this todefine an instance ( n, k, C , T ) for GridTiling by declaring C := E ( H (cid:48) ) and T ( i, j ) := E i,j for all ij ∈ E ( H (cid:48) ).We then claim that the grid tilings of this instance correspond bijectively to the H (cid:48) -copies in G (cid:48) . Thisgives a parsimonious reduction from a : [ k ] → [ n ] encodes an edge-subset S a ⊆ E ( G (cid:48) ) with | S a | = | E ( H ) | that picks exactly one element from E i,j for each ij ∈ E ( H (cid:48) ). If the edges in S a are incident with exactly k distinct vertices, then S a inducesa H (cid:48) -copy in G (cid:48) . By the first two properties of Definition 2.4, the edge set S a contains exactly k distinctendpoints and k distinct starting points. Since E i,i for i ∈ [ k ] contains only self-loops, the sets of endpointsand starting points of edges in S a are identical, which implies that S a is a H (cid:48) -copy in G (cid:48) . Conversely, every H (cid:48) -copy in G (cid:48) can be mapped to such a grid tiling by reversing this operation.9n the following, we add a small technical extension to Theorem 2.9 that allows us to assume each inputinstance to be balanced along rows or columns in a certain way. While it is almost trivial to ensure thisbalance property by adding dummy elements, it turns out to be very useful in our reductions from GridTiling. Lemma 2.10.
Let A = ( n, k, C , T ) be an instance for GridTiling and let W be either of the words “hori-zontal” or “vertical”. In polynomial time, we can then compute a number T ∈ N and a grid tiling instance A (cid:48) = ( n (cid:48) , k, C , T (cid:48) ) with n (cid:48) = O ( k n ) such that:1. The instances A and A (cid:48) have precisely the same grid tilings.2. For u ∈ [ n ] , write ( u, (cid:63) ) := { ( u, v ) | v ∈ [ n ] } . For v ∈ [ n ] , write ( (cid:63), v ) := { ( u, v ) | u ∈ [ n ] } .(a) If W is “horizontal”, then for all κ ∈ C and u ∈ [ n (cid:48) ] , we have |T (cid:48) ( κ ) ∩ ( u, (cid:63) ) | = T .(b) If W is “vertical”, then for all κ ∈ C and v ∈ [ n (cid:48) ] , we have |T (cid:48) ( κ ) ∩ ( (cid:63), v ) | = T .Proof. We show the statement if W is “vertical”; the horizontal case is shown in exactly the same manner.Let us first define T κ,v := |T ( κ ) ∩ ( (cid:63), v ) | for κ ∈ [ k ] and v ∈ [ n ] , that is, the number of elements in the v -th column of T ( κ ). Then we define T := max κ ∈ [ k ] , v ∈ [ n ] T κ,v and let n (cid:48) := n + k T . Consider [ n (cid:48) ] to be partitioned into [ n ] and k consecutive “dummy” blocks B κ for κ ∈ [ k ] , with | B κ | = T . We keep C unchanged and modify T to a function T (cid:48) that maps from C into thepower-set of [ n (cid:48) ] : For κ ∈ [ k ] and v ∈ [ n ], we simply add T − T κ,v arbitrary distinct dummy elements from { ( f, v ) | f ∈ B κ } to T ( κ ) in order to obtain T (cid:48) ( κ ).This ensures the vertical balance property defined in the statement of the lemma, and we observe that T (cid:48) has the same grid tilings as T : Every grid tiling of T is also one of T (cid:48) . Furthermore, dummy elementscannot be chosen in any grid tiling of T (cid:48) since, for all κ and κ (cid:48) , the dummy elements in T (cid:48) ( κ ) and T (cid:48) ( κ (cid:48) )have disjoint first coordinates, which are also distinct from [ n ]. Thus, in particular, any assignment usingdummy elements cannot satisfy the first condition of a grid tiling required in Definition 2.4. In the following, we give a introduction to what we call the
Holant framework , a toolbox introduced by[53, 8, 9]. Some of this material is abridged from [15]. We use Holant problems as an intermediate step forreducing problems, such as counting grid tilings, to the permanent.
The input to a Holant problem is a so-called signature graph, that is, a graph with certain functions associatedwith its vertices.
Definition 3.1. A signature graph is an edge-weighted graph Ω which may feature parallel edges, and whichhas a vertex function f v : { , } I ( v ) → C associated with each v ∈ V (Ω). We also call f v the signature of v .If v has degree d and an edge-ordering I ( v ) = { e , . . . , e d } is specified, we also consider f v : { , } d → C .The Holant of Ω is a particular sum over edge assignments x ∈ { , } E (Ω) . For x ∈ { , } E (Ω) , we saythat e ∈ E (Ω) is active in x if x ( e ) = 1 holds, and we tacitly identify x with the set of active edges in x . Given a subset S ⊆ E (Ω), we write x | S for the restriction of x to S , which is the unique assignment in { , } S that agrees with x on S . 10 efinition 3.2 (adapted from [53]) . Let Ω be a signature graph with edge weights w : E (Ω) → C and avertex function f v : { , } I ( v ) → C for each v ∈ V (Ω). For x ∈ { , } E (Ω) , we defineval Ω ( x ) := (cid:89) v ∈ V (Ω) f v ( x | I ( v ) ) , (1) w Ω ( x ) := (cid:89) e ∈ x w ( e ) . (2)We say that x satisfies Ω if val Ω ( x ) (cid:54) = 0 holds. Furthermore, we defineHolant(Ω) := (cid:88) x ∈{ , } E (Ω) w Ω ( x ) · val Ω ( x ) . (3)A particularly useful type of vertex functions is that of Boolean functions , whose ranges are restricted to { , } rather than C . If all signatures appearing in a signature graph Ω (cid:48) are Boolean, then Holant(Ω (cid:48) ) simplysums over those assignments x ∈ { , } E (Ω (cid:48) ) that satisfy all constraints imposed by the vertex functions,and each x is weighted by w Ω (cid:48) ( x ). As an example, we use Boolean functions to reformulate PerfMatch as aHolant problem. Example 3.3.
Given an edge-weighted graph G , let f v : { , } I ( v ) → { , } for v ∈ V ( G ) be the vertexfunction defined by f v ( x ) = (cid:40) if hw( x ) = 1 , otherwise . (4) Let Ω denote the signature graph obtained from G by associating f v with v , for all v ∈ V ( G ) . Then Holant(Ω) ranges over those assignments x ∈ { , } E (Ω) in which each vertex is incident with exactly one active edge.Each such x is weighted by w Ω ( x ) = (cid:81) e ∈ x w ( e ) . This is precisely the expression of PerfMatch( G ) . In some occasions, we can simulate signatures f appearing in a signature graph Ω by gadgets, i.e., signaturegraphs on “basic” signatures that realize f . We call such gadgets gates , similar to the F -gates in [9], and wewill be particularly interested in matchgates . These are gates Γ that feature, at each vertex v ∈ V (Γ), theperfect matching signature from Example 3.3 that maps x ∈ { , } I ( v ) to HW = ( x ) := [hw( x ) = 1] . The formal definition of gates and matchgates follows.
Definition 3.4. A gate is a signature graph Γ containing a set D ⊆ E (Γ) of dangling edges , all of whichhave edge-weight 1. A dangling edge is an “edge” that is incident with only one vertex. We consider thedangling edges of Γ to be labeled as 1 , . . . , | D | .Given a signature graph Ω, a vertex v ∈ V (Ω) of degree | D | , and an ordering of I ( v ) as I ( v ) = { e , . . . , e | D | } , we can insert Γ at v by deleting v , placing a copy of Γ into G , and identifying e i withthe i -labeled dangling edge of Γ, for all i .For disjoint sets A , B , and for x ∈ { , } A and y ∈ { , } B , write xy ∈ { , } A ∪ B for the assignmentthat agrees with x on A , and with y on B . We say that xy extends x . The signature of Γ is the functionSig(Γ) : { , } D → C that maps x ∈ { , } D toSig(Γ , x ) = (cid:88) y ∈{ , } E (Γ) \ D w Γ ( xy ) · val Γ ( xy ) . (5)We also say that Γ realizes Sig(Γ). If all v ∈ V (Γ) feature the function HW = defined above, then Γ is a matchgate . Finally, we call Γ planar if it can be drawn in the plane with all dangling edges on the outerface, such that they appear in the order 1 , . . . , | D | in a clockwise traversal of this face.11y the following lemma, if Γ realizes a signature f , and v is a vertex with signature f in a signaturegraph Ω, then we can insert Γ at v in a way that preserves Holants. In other words, we can treat Γ as if itwere a single vertex of signature Sig(Γ). This will be used to reduce Holant(Ω) to PerfMatch if all signaturesin Ω can be realized by matchgates. For a proof, see Chapter 2 of [15]. Lemma 3.5.
Let Ω be a signature graph, let v ∈ V (Ω) be arbitrary, and let f v denote the vertex function of v in Ω . Furthermore, let Γ be a (match-)gate with Sig(Γ) = f v , and let Ω (cid:48) be obtained from Ω by inserting Γ at v . Then we have Holant(Ω) = Holant(Ω (cid:48) ) . If Ω and Γ are planar and Ω is given together with a plane embedding, then the following holds: If we order I ( v ) according to its clockwise ordering in the embedding and insert Γ under this order, then Ω (cid:48) is planar. In the remainder of this subsection, we consider specific matchgates that will be relevant later. To simplifyour presentation, we abbreviate the following 4-bitstrings. Each corresponds to a specific assignment to theedges incident with a vertex of degree 4.:= 0000 , := 0101 , := 1010 , := 1111 , := 1000 , := 0010 , := 1101 , := 0111 . In Figure 1, we define a signature
PASS of arity 4 and two signatures
PRE and
ACT of arity 6. Note that
PASS essentially acts as a “crossing” signature: It enforces equality on its western and eastern dangling edges(numbered 4 and 2), as well as on its northern and southern dangling edges (numbered 1 and 3). However,if all dangling edges are active, then the output of
PASS is − PASS to admit a planar matchgate Γ
PASS , shown in Figure 1. We verified that Sig(Γ
PASS ) =
PASS holds by means ofa computer program: For all x ∈ { , } , we showed mechanically that Sig(Γ PASS , x ) =
PASS ( x ) holds. Notethat this verification can also be carried out by hand. For more details, consider Appendix C of [15]. Itshould also be noted that planar matchgates for PASS were already studied in [53, 6].Next, we consider the signatures
PRE and
ACT , each of arity 6. We consider their last two inputs (thedangling edges with numbers 5 and 6) as “switches”, which will later be connected to apices. It is crucial toobserve that
PRE ( x
00) =
ACT ( x
00) =
PASS ( x ) ∀ x ∈ { , } . That is, if the two switch edges are not active, then
PRE and
ACT behave exactly like
PASS on their non-switch inputs. If both switches are active, then some differences occur, namely, the restriction to non-switchedges must be in state or for
PRE or ACT to yield a nonzero value. Furthermore, if only one of the twoswitches is active, then
ACT yields value zero, while
PRE still allows such assignments (such as 01). Weverified with a computer program that
PRE = Sig(Γ
PRE ) holds for the matchgate Γ
PRE from Figure 1. In thefollowing, we prove manually that
ACT = Sig(Γ
ACT ) holds.
Lemma 3.6.
We have
ACT = Sig(Γ
ACT ) with the matchgate Γ ACT from Figure 1.Proof.
Note that Γ
ACT has a green vertex of signature
PRE , and some additional part (a ring of
PASS signatures,and an edge of weight ) which we call the even filter . Observe also that, for all x ∈ { , } and y ∈ { , } ,we have the identity PRE ( xy ) = (cid:40) ACT ( xy ) if hw( x ) even , arbitrary otherwise . (6)The even filter now ensures the following, for all x ∈ { , } and y ∈ { , } : • If hw( x ) is not even, then Sig(Γ ACT , xy ) = 0, regardless of the value of
PRE on xy . • If hw( x ) is even, then Sig(Γ ACT , xy ) =
PRE ( xy ). By (6), this implies Sig(Γ ACT , xy ) =
ACT ( xy ).12 ASS ( x ) := − x =1 x ∈ { , , } PRE ( x ) := PASS ( y ) x = y x ∈ { , } x ∈ { , } x ∈ { , } ACT ( x ) := PASS ( y ) x = y x ∈ { , } Figure 1: The matchgates Γ
PASS , Γ
PRE and Γ
ACT and the signatures
PASS , PRE and
ACT . Note that Γ
PASS hasfour dangling edges, numbered 1 to 4, whereas Γ
PRE and Γ
ACT each have six dangling edges, numbered 1 to6. The signature
PASS is defined on assignments x ∈ { , } , while PRE and
ACT are defined on assignments x ∈ { , } . These strings correspond canonically to assignments at the dangling edges of Γ PASS , Γ
PRE andΓ
ACT . All black vertices are assigned HW = . In the gate Γ ACT , all red vertices are assigned
PASS , and the greenmiddle vertex is assigned
PRE . Note that we can also view Γ
ACT as a matchgate by realizing its signatureswith the matchgates Γ
PASS and Γ
ACT . All matchgates are planar after removal of the dangling edges 5 and 6,which will later connect to apex vertices.Since
ACT ( xy ) (cid:54) = 0 implies x ∈ { , , , } , which in turn implies that hw( x ) is even, this will prove thelemma. To compute Sig(Γ ACT , xy ) for x ∈ { , } and y ∈ { , } , we consider the satisfying assignments w to E (Γ ACT ) that extend xy . The dummy edge of weight / is present in any assignment w and contributes afactor / to val( w ). (In this proof, we write val( w ) instead of val Γ ACT ( w ) to avoid double indexing.) At eachred vertex, the signature PASS ensures that opposing edges have the same assignment under w . This fixesthe value of all black edges and ensures that val( w ) contains the factor PRE ( xy ), contributed from the greenvertex with signature PRE .It remains to assign values to the red edges: Due to the signature
PASS at red vertices, this is possiblewith at most two satisfying assignments w , w ∈ { , } E (Γ ACT ) : w : All red edges are active. Then every red vertex in state yields a factor PASS ( ) = −
1, while allother red vertices are in one of the states or and yield value 1. The number of red vertices instate is hw( x ), so the value of Γ ACT on w isval( w ) = 12 · ( − hw( x ) · PRE ( xy ) .w : No red edges are active. Then every red vertex is in one of the states or and hence yields value1. Thus, the value of Γ ACT on w is val( w ) = 12 · PRE ( xy ) .
13t follows that for all x ∈ { , } and y ∈ { , } , we haveSig(Γ ACT , xy ) = val( w ) + val( w )= 12 · (cid:16) ( − hw( x ) · PRE ( xy ) + PRE ( xy ) (cid:17) = (cid:40) PRE ( xy ) if hw( x ) even,0 otherwise . = ACT ( xy )This proves the lemma. We introduce our main tool for the later sections, a technique that allows us to simulate signatures by linearcombinations of other signatures, in particular, of matchgate signatures.
Definition 3.7.
Let f = c · f + . . . + c t · f t be a signature, where c , . . . , c t ∈ C are coefficients and f , . . . , f t are signatures, and the linear combination is point-wise. Then we say that f is t -combined from constituents f , . . . , f t .We apply such linear combinations as follows: Assume we are given a signature graph that features k occurrences of some interesting signature f which cannot be realized by matchgates. If we can express f as a linear combination of t constituents that do admit matchgates, then the following lemma allows us tocompute Holant(Ω) from the Holants of t k derived signature graphs whose signatures all admit matchgates. Lemma 3.8.
Let Ω be a signature graph, let k, t ∈ N and let w , . . . , w k be distinct vertices of Ω such that thefollowing holds: For all κ ∈ [ k ] , the signature f κ at w κ admits coefficients c κ, , . . . , c κ,t ∈ C and signatures g κ, , . . . , g κ,t such that f κ = (cid:80) ti =1 c κ,i · g κ,i . Given a tuple θ ∈ [ t ] k , let Ω θ be defined by replacing, for each κ ∈ [ k ] , the vertex function f κ at w κ with g κ,θ ( κ ) . Then we have Holant(Ω) = (cid:88) θ ∈ [ t ] k (cid:32) k (cid:89) κ =1 c κ,θ ( κ ) (cid:33) · Holant(Ω θ ) . (7) Proof.
Choose any fixed single κ ∈ [ k ]. For i ∈ [ t ], let Ω i denote the signature graph obtained from Ω byreplacing f κ with g κ,i . By elementary manipulations, we haveHolant(Ω) = (cid:88) x ∈{ , } E (Ω) f κ ( x ) · (cid:89) v ∈ V (Ω) \{ w } f v ( x )= (cid:88) x ∈{ , } E (Ω) (cid:32) t (cid:88) i =1 c κ,i · g κ,i ( x ) (cid:33) · (cid:89) v ∈ V (Ω) \{ w } f v ( x )= t (cid:88) i =1 c κ,i · (cid:88) x ∈{ , } E (Ω) g κ,i ( x ) (cid:89) v ∈ V (Ω) \{ w } f v ( x )= t (cid:88) i =1 c κ,i · Holant(Ω i ) . Then apply this identity inductively for κ = 1 , . . . , k . Each step reduces the number of combined signa-tures by one, and elementary algebraic manipulations imply (7).14hen using Lemma 3.8 for positive results, as in Section 4, then the right-hand side of (7) is “easy”,in the sense that the values Holant(Ω θ ) for all θ can be obtained efficiently, e.g., by reduction to planarPerfMatch. In the same way, Lemma 3.8 also allows us to prove hardness results under Turing reductions,as we do in Sections 5 and 6: In this case, the left-hand side is “hard” and could be computed from oracleaccess to the values Holant(Ω θ ) for all θ . In this section, we present a first application of the framework of combined signatures: We show that,for graphs of genus k , the quantity PerfMatch( G ) can be expressed as a linear combination of 4 k valuesPerfMatch( G i ), where G i is a planar graph for all i ∈ [4 k ]. The linear combinations resemble those usedin [29, 49, 44], but unlike these papers, we can state our linear combinations without any necessity forPfaffian orientations. That is, we obtain a parameterized reduction with black-box access to counting perfectmatchings in planar graphs. Following [49], we assume that the graph G in question is given to us together with a plane model: Allvertices of G are drawn in a polygon P with 2 k sides. If there is a set of d i parallel edges x i = x i x i · · · x id i leaving P from one side and going into P through another side, we denote the two sides by a i and a − i respectively. Since the edges are parallel, when we walk along the sides of P counterclockwise, we meet theexits of edges in the order x i x i · · · x id i on side a i , then the entrances of edges in the order x id i x i ( d i − · · · x i on side a − i . If G can be embedded on an orientable compact boundaryless surface S of genus k , then it canbe drawn such that there are no edges crossing inside P , and the sides of P are a a a − a − a a a − a − · · · a k − a k a − k − a − k . The side pair a i , a − i represents boundaries to be glued together. When G is drawn on the surface S , theedge bunches x and x overpass each other without any edges crossing; see the left picture of Figure 2 forsuch a situation, which we call a grid cap .We use linear combinations of matchgates to simulate the grid cap by a planar graph. Write x − i to denote x id i x i ( d i − · · · x i . Then the grid cap realizes a function that is defined on assignments ( x , x , y , y ) to itsdangling edges as follows: O ( x , x , y , y ) = [ y = x − ] · [ y = x − ] . The straightforward idea is to place a
PASS matchgate at each crossing of overpassing edges, as shown in themiddle of Figure 2. Let us denote by C ( x , x , y , y ) the signature of the resulting gate. In any satisfyingassignment ( x , x , y , y ) to its dangling edges, there are hw( x ) · hw( x ) instances of PASS in state , eachof which gives a factor −
1, while all other instances of
PASS (in states , , ) give a factor 1, so C ( x , x , y , y ) = ( − ODD ( x ) · ODD ( x ) · [ y = x − ] · [ y = x − ] . We can thereforce conclude that O can be expressed as a linear combination of signatures of type C , eachof which is the signature of a planar matchgate. Lemma 4.1.
Every grid cap gate is a linear combination of matchgates, given by O ( x , x , y , y ) = 12 (1 + ( − ODD ( x ) + ( − ODD ( x ) + ( − ODD ( x )+ ODD ( x )+1 ) · C ( x , x , y , y ) . Proof.
Observe first that O ( x , x , y , y ) = 12 (1 + ( − ODD ( x ) + ( − ODD ( x ) + ( − ODD ( x )+ ODD ( x )+1 ) · ( − ODD ( x ) · ODD ( x ) . X X Y Y Y X X X X Y Y Y Y X X X Y Y Y X X X X Y Y Y Y X X X X Y Y Y Y Figure 2: The first two subfigures show a grid cap and the matchgate realizing one of the constituents usedto realize the grid cap. The third subfigure shows the matchgate used to simulate a cross cap. In thesematchgates, all vertices are assigned the signature
PASS .From this, we can conclude that O ( x , x , y , y ) = 12 C ( x , x , y , y ) + 12 ( − ODD ( x ) C ( x , x , y , y ) ++ 12 ( − ODD ( x ) C ( x , x , y , y ) −
12 ( − ODD ( x ) ( − ODD ( x ) C ( x , x , y , y ) . The extra factor ( − ODD ( x ) can be realized by giving weight − x i in the matchgate C . Hence, all the four functions can be realized by some matchgates similar to C after introduction ofadditional − G can be embedded on a non-orientable surface S , which is the connected sum of a surface of orientable genus k with either a projectiveplane or a Klein bottle, then it can be drawn without crossings inside P , such that the sides of P are a a a − a − a a a − a − · · · a k − a k a − k − a − k a k +1 a k +2 , and a a a − a − a a a − a − · · · a k − a k a − k − a − k a k +1 a k +2 a k +3 a k +4 , respectively. Here, the side pair a i a i means that, when a bunch of edges x i = x i x i · · · x id i leaves the interiorof P through the first side a i and then enters back into P through the second side a i , then we meet the exitsand entrances in the order x i x i . Such a bunch of edges is called a cross cap , and it realizes a function O ( x, y ) = [ y = x ] . If we draw it on the plane and replace each crossing by a
PASS matchgate, as shown in the right part ofFigure 2, we get a matchgate realizing C ( x, y ) = ( − hw( x )2 ) · [ y = x ] . From this, we obtain a linear combination for cross cap gates from planar matchgates:
Lemma 4.2.
Every cross cap gate is a linear combination of matchgates, given by O ( x, y ) = 1 − i · i hw( x ) · C ( x, y ) + 1 + i · ( − i ) hw( x ) · C ( x, y ) . roof. The sequence ( − hw( x )2 ) indexed by hw( x ) is1 , , − , − , , , − , − , . . . It must be a linear combination of 4 sequences w hw( x ) , for w ∈ { , i, − , − i } , all of which have the sameperiod 4, since the length 4 initial segments of the 4 sequences form a full rank Vandermonde matrix. Infact, it can be expressed as a linear combination of two such sequences, as we can observe that( − hw( x )2 ) = 1 − i i hw( x ) + 1 + i − i ) hw( x ) . The extra factor i hw( x ) can be realized by giving weight i instead of 1 to each input edge in C .Using the fact that G is embedded as a plane model, and using the combined signatures for grid capsand cross caps from the last two lemmas, we then obtain the following known theorem. Theorem 4.3. [49] Let G be a graph that is embedded on a surface. Then PerfMatch( G ) is a summationof PerfMatch of k , k +1 or k +2 planar graphs, respectively, if the surface is the connected sum of anorientable surface of genus k with the plane, the projective plane, or the Klein bottle, respectively.Proof. By Lemma 4.1 and 4.2, use Lemma 3.8 on the k grid caps and 0, 1 or 2 cross caps. For a matrix A , let A ⊗ k denote the matrix obtained from the k -fold Kronecker product A ⊗ . . . ⊗ A . Theessence of Lemma 4.1 is that we can use the four matchgates to realize all four columns of the basis (cid:18) − (cid:19) ⊗ , so that we can then obtain any other function by linear combinations. The same observation also holds fora larger base (cid:18) − (cid:19) ⊗ m . We give an example: In a cross cap of m edges, we may replace each edge by a bunch of parallel edges,and call the result a grated cross cap . All the (cid:0) m (cid:1) latent crossings of the cross cap become grid caps in thegrated cross cap. Fact 4.4.
Every grated cross cap gate over m bunches of edges, as defined above, can be expressed as a linearcombination of m planar matchgates. In fact, these 2 m basis matchgates are powerful enough to express (as a linear combination) any functionthat depends only upon the parities p , . . . , p m of active edges in the m edge bunches. However, amongthese functions, we currently only know one interesting function, i.e., the grid cap. Even the grated crosscap seems too artificial to be related with a natural tractability result. A similar generalization applies toLemma 4.2, where the functions to be expressed may also depend upon residuals of the numbers of activeedges in the m edge bunches, in this case however modulo 4 rather than 2. In this section, we prove Theorem 1.1 by an application of our framework of combined signatures. We use A to G ( A ). Border vertices c κ for κ ∈ { N , W , S , E }× [ k ] and their incident edges are colored gray. Cell vertices c κ for κ ∈ C are colored red, while vertices c κ for κ ∈ [ k ] \ C are colored black. Horizontally or vertically adjacent vertices are connected by an edge bundleof n parallel edges. The right part of the figure shows the gates Φ and Φ (cid:48) ( A ). Each white vertex is assigned PASS , each black vertex is assigned
ACT , and each gray vertex is assigned HW = . Edges from apices in Φ (cid:48) aredrawn dashed. Note that, due to the balance property of T , we may assume that every column has the samenumber T of occurrences of ACT .1. We express the solution to the instance as Holant( G ) for a signature graph G in Section 5.1.2. We realize the signatures of G in Section 5.2. At this point however, we require combined signatures,and this is where we depart from the usual reductions from GridTiling.Large parts of this section will be reused in Section 6 with an added layer of technicalities. In the following, let A = ( n, k, C , T ) be a fixed instance to T ≤ n such thatfor all κ ∈ C and all v ∈ [ n ], there are exactly T elements of type ( (cid:63), v ) in T ( κ ). This will become relevantin Section 5.2.First, we reformulate A as the Holant of a signature graph G = G ( A ). This graph G consists of a k × k square grid of cells, and 4 k additional border vertices adjacent to the borders of the grid, as seen in the leftpart of Figure 3. Note that G is planar. We denote its vertices by c κ for κ ∈ Ξ, whereΞ := [ k ] ∪ { N , W , S , E } × [ k ] . For i ∈ [ k ], we declare ( N , i ) to be vertically adjacent to (1 , i ), and ( S , i ) to ( k, i ). Likewise, we declare( W , i ) to be horizontally adjacent to ( i, E , i ) to ( i, k ). We refer to the neighbors of any index κ ∈ Ξ orvertex c κ ∈ V ( G ) using cardinal directions in the obvious way, e.g., we may speak of the northern neighborof a vertex. Between any pair of vertices c κ and c κ (cid:48) with adjacent indices κ and κ (cid:48) , we place a set E κ,κ (cid:48) of n parallel edges, which we call an edge bundle .We proceed to define the signatures of G . In the assignments a ∈ { , } E ( G ) we are interested in, eachedge bundle features exactly one active edge, which is used to encode a number from [ n ]. At border vertices,we place the signature HW = to ensure this. The signatures of cells c κ with κ ∈ [ k ] are then defined so thateach cell propagates the number x W ∈ [ n ] encoded by its western incident edge bundle to the east, and itsnorthern number x N ∈ [ n ] to the south, while checking along the way whether ( x W , x N ) ∈ T ( κ ) holds. Remark . We adhere to the following notational conventions in this section: • For v ∈ [ n ], we often identify the string 0 v − n − v ∈ { , } n with the number v when it is clear fromthe context which of these two objects we currently refer to.18 For κ ∈ [ k ] , the 4 n incident edges of each vertex c κ are ordered such that all northern edges appearfirst, in a block of length n , followed by the n eastern, the n southern, and finally the n western edges. • We implicitly consider strings x ∈ { , } n to be decomposed into x = x N x E x S x W with four bistrings x N , x E , x S , x W ∈ { , } n corresponding to the four cardinal directions.Using these conventions, we then define the following predicates for strings x ∈ { , } n : ϕ one ( x ) ≡ hw( x N ) = 1 ∧ hw( x W ) = 1 ,ϕ prop ( x ) ≡ x N = x S ∧ x W = x E . If a function f satisfies ϕ prop ( x ) for each x ∈ supp( f ), then we call f propagating . For each κ ∈ [ k ] , weplace a specific propagating signature f κ at the vertex c κ in order to complete G to a signature graph whosesatisfying assignments correspond bijectively to the grid tilings of A = ( n, k, C , T ). Definition 5.2.
Let let A = ( n, k, C , T ) an instance to the grid tiling problem, as described above. For all κ ∈ [ k ] \ C , we define the vertex function f κ : { , } n → { , } of c κ such that, for all x ∈ { , } n satisfyingthe predicate ϕ one ( x ), we have f κ ( x ) := [ ϕ prop ( x )] . Note that no requirement is imposed upon f κ ( x ) on those x ∈ { , } n that fail to satisfy ϕ one ( x ). For allremaining κ , namely all κ ∈ C , we define the vertex function g κ of c κ on such x ∈ { , } n by declaring g κ ( x ) := [ ϕ prop ( x ) ∧ ( x W , x N ) ∈ T ( κ )]This finishes the definition of the signature graph G = G ( A ). In the following, we verify by a simpleargument that G indeed encodes A properly. Lemma 5.3.
The grid tilings of A correspond bijectively to the satisfying assignments x ∈ { , } E ( G ) of G ,and each satisfying assignment x additionally has val G ( x ) = 1 .Proof. Every grid tiling a : [ k ] → [ n ] can be transformed into an assignment x ( a ) ∈ { , } E ( G ) as follows:For each κ ∈ [ k ] , with a ( κ ) = ( u, v ), declare the u -th edge in the western edge bundle of c κ and the v -thedge in the northern edge bundle of c κ to be active. At vertices c ( k,(cid:63) ) , copy the assignment from northernedges to southern edges, and at c ( (cid:63),k ) , copy the assignment from western edges to eastern edges. Declareall other edges to be inactive. It follows from the definition of f κ at κ ∈ C and g κ at κ ∈ [ k ] \ C thatval G ( x ( a )) = 1 holds.For the converse direction, we show that every satisfying assignment x ∈ { , } E ( G ) can be written as x = x ( a ) for some grid tiling a , where x ( a ) is defined as in the previous paragraph. Note that this alsoimplies val G ( x ) = 1. By the signature HW = , every border vertex is incident with exactly one active edge in x . Hence, the restriction of x to I ( c , ) satisfies ϕ one ; call this restricted assignment y . • If (1 , ∈ [ k ] \C , then the vertex function of c , is f , . Since f , ( y ) = 1, and since f , is propagatingon inputs satisfying ϕ one , we also have ϕ prop ( y ). • If (1 , ∈ C , then we additionally have ( y W , y N ) ∈ T (1 ,
1) by definition of g , .By induction along rows and columns, we obtain, for every κ ∈ [ k ] , that the partial assignment y at I ( c κ )satisfies ϕ prop ( y ) and ( y W , y N ) ∈ T ( κ ) if κ ∈ C . Hence x = x ( a ) holds for a unique grid tiling a .In the next subsection, we realize each signature f κ for κ ∈ C as a planar matchgate, and each g κ for κ ∈ [ k ] \ C as a linear combination of two matchgate signatures that have maximum apex number 2. Notethat the remaining signatures HW = occurring in G are planar. Since G itself is planar and features at most O ( k ) signatures g κ , the graphs realizing G will feature at most O ( k ) apices, and we will use this to obtainthe desired parameterized reduction and lower bound under ETH .19 .2 Realizing cell signatures
It can be shown (under no additional assumptions) that some of the signatures g κ for κ ∈ [ k ] are non-planar.From a complexity viewpoint, if all such signatures were planar and we knew explicit planar matchgates,then we could reduce FP = P by the FKT method.Rather than trying to use planar matchgates, we show that each signature g κ can be realized as a specific linear combination of the signatures of one planar and one 2-apex matchgate. Note again that at least onenon-planar constituent is necessary, as we could otherwise show FPT = W [ ].In the remainder of this section, we consider κ ∈ [ k ] to be fixed, we write A = T ( κ ) and we recall that A ⊆ [ n ] . The constituents for g κ will be the signatures of two gates Φ and Φ (cid:48) ( A ), which use as buildingblocks the signatures PASS and
ACT from Section 3.
Definition 5.4.
Let n ∈ N and let A ⊆ [ n ] . We define gates Φ and Φ (cid:48) = Φ (cid:48) ( A ) with 4 n dangling edges (thatis, with n dangling edges for each cardinal direction) as follows. Consider also the right part of Figure 3. • To obtain the gate Φ, arrange vertices b τ for τ ∈ [ n ] in a n × n grid and assign the signature PASS toeach such vertex. Add a single edge of weight − HW = . • A similar construction yields the gate Φ (cid:48) : Starting from Φ, remove the extra edge of weight −
1, addapex vertices a and a with signatures HW = , and for all τ ∈ A , do the following:1. Replace the signature PASS at b τ with ACT .2. Add the edges a b τ and a b τ . Declare these to be the last two edges in the edge ordering of I ( v τ ).Recall that PASS is realized by the planar matchgate Γ
PASS , so we can also view the gate Φ as a planarmatchgate after realizing all signatures by matchgates. We will later switch between these views dependingon the application. Note also that the 2-coloring of Γ
PASS can be extended to one of Φ. Likewise,
ACT isrealized by the matchgate Γ
ACT , which is planar when ignoring its dangling edges 5 and 6. That is, afterrealizing each occurrence of
ACT by Γ
ACT , the resulting matchgate obtained from Φ (cid:48) is planar after removalof a and a .Our goal for this subsection is to realize the signatures f κ and g κ from Definition 5.2. In the following,we prove that f κ = Sig(Φ) and that g κ can be realized by a linear combination of Sig(Φ) and Sig(Φ (cid:48) ). It willbe crucial for our calculations to assume our instance A for GridTiling to be balanced: By Lemma 2.10, weassume there is some T ∈ N such that | A ∩ ( (cid:63), v ) | = T for all v ∈ [ n ]. That is, in the right part of Figure 3,we may assume that every column of Φ (cid:48) ( A ) features the same number T of vertices with signature ACT . Lemma 5.5.
Recall the definition of the predicates ϕ one and ϕ prop on the preceding page. Let x ∈ { , } n be an assignment that satisfies the predicate ϕ one . Then Sig(Φ , x ) = (cid:40) if ¬ ϕ prop ( x ) , if ϕ prop ( x ) . (8)Sig(Φ (cid:48) ( A ) , x ) = if ¬ ϕ prop ( x ) (cid:40) − T if ( x W , x N ) / ∈ A − T + 2 if ( x W , x N ) ∈ A if ϕ prop ( x ) . (9) Note that f κ = Sig(Φ) for κ ∈ [ k ] \ C . For κ ∈ C and for x ∈ { , } n satisfying ϕ one , we have g κ ( x ) = T · Sig(Φ , x ) + 12 · Sig(Φ (cid:48) ( T ( κ )) , x ) . (10)In Section 5.3, we prove Lemma 5.5 by inspecting the possible satisfying assignments to Φ and Φ (cid:48) . Beforedoing this, let us first show how Lemma 5.5 implies Theorem 1.2. We will require parts of this argumentagain in Section 6. 20 roof of Theorem 1.2. Using Lemma 5.3, we know that Holant( G ) counts precisely the grid tilings of A . ByTheorem 2.9, this problem is W [ ]-hard and cannot be solved in time f ( k ) · n o ( k/ log k ) , even on instances A = ( n, k, C , T ) with |C| = O ( k ).Using the linear combination (10) and Lemma 3.8 about the linear combinations of signatures, as wellas Lemma 3.5 about inserting matchgates into signature graphs, we obtainHolant( G ) = 12 |C| (cid:88) ω : C→ [2] T d ( ω ) · PerfMatch( H ω ) . (11)For ω : C → [2], the number d ( ω ) is the number of 1-entries in ω , and the graph H ω is obtained as follows: • For κ ∈ [ k ] \ C , insert the matchgate Φ at the cell vertex c κ . • For κ ∈ C with ω ( κ ) = 1, insert the matchgate Φ at c κ as well. • For κ ∈ C with ω ( κ ) = 2, insert the matchgate Φ (cid:48) ( T ( κ )) at c κ .Since G is planar, and since Φ is planar and Φ (cid:48) ( T ( κ )) for κ ∈ C has at most 2 apices, it follows thatapex( H ω ) ≤ |C| for all ω : C → [2], and this proves the required parameter bound. By 2-coloring thematchgates Φ and Φ (cid:48) , it can furthermore be verified that each graph H ω is bipartite.Additionally, by construction of the matchgates Γ PASS and Γ
ACT , every graph H ω features only edge-weightsfrom the set {− , / , } . Non-unit edge-weights in H ω appear only at edges uv ∈ E ( H ω ) not incident withapices. We can hence use standard weight simulation techniques to remove the edge-weights − / , as in[51] or Chapter 1 of [15], while maintaining the apex number. We consequently obtain W [ ]-completenessof the permanent under the apex parameter and the claimed lower bound under ETH . Remark . The following might prove useful for later applications: By construction, the apices in theconstructed graphs H ω form an independent set, for any ω : [ k ] → [2], and each non-apex vertex in H ω isincident with at most one apex. This last condition holds because the matchgate Γ ACT has no vertex withtwo incident dangling edges. Φ and Φ (cid:48) In the remainder of this section, we provide the deferred proof of Lemma 5.5. To this end, we calculatethe signatures of Φ and Φ (cid:48) by analyzing, for any given assignment x ∈ { , } n to their dangling edges, thepossible satisfying assignments xy extending x . ΦLet x ∈ { , } n be an assignment to the dangling edges of Φ that satisfies ϕ one ( x ), and let xy ∈ { , } E (Φ) be a satisfying assignment to Φ that extends x . We show that, whenever ϕ prop ( x ) holds, then y is unique and xy has value 1, so Sig(Φ , x ) = val Φ ( xy ) = 1. Furthermore, we show that, if x does not satisfy the predicate ϕ prop , then no such y exists, and hence Sig(Φ , x ) = 0 . Recall from Remark 5.1 that we implicitly decompose the string x into x N , x E , x S , x W . Write W ∈ [ n ]and N ∈ [ n ] for the unique non-zero index in x W ∈ { , } n and x N ∈ { , } n , respectively. These numbersare well-defined because x satisfies ϕ one ( x ) by assumption. Then all western and eastern edges of verticesin row ( W, (cid:63) ) are active in xy , see Figure 4: The western edge of the vertex b W, is active by definition, andsince xy satisfies Φ and PASS at b W, , this vertex must be in state or , so its eastern edge is also active.The same follows inductively for all vertices in the row ( W, (cid:63) ). By the same argument, rotated about 90degrees, all northern and southern edges of vertices in row ( (cid:63), N ) are active in xy .By a similar argument, no other edges are active, and we conclude that y is uniquely determined by x .Furthermore, if E and S denote the active indices in x E and x S , then we observe that W = E and N = S ,since otherwise xy could not satisfy b W,n and b n,N . Hence, xy satisfies Φ only if ϕ prop ( x ) holds. We obtainSig(Φ , x ) = 0 if ¬ ϕ prop ( x ) . y to E (Φ) that extends x . Active edges are drawn with thicker lines thannon-active edges. Note that the edge of weight − HW = at its endpoints must be active in any satisfyingassignment.If ϕ prop ( x ) holds, then b W,N is in state under xy , while the n − W, (cid:63) ) are instate , the n − (cid:63), N ) are in state , and the remaining n − n + 1 vertices arein state . Furthermore, we have the additional active edge of weight −
1. Hence, in conclusion, ϕ prop ( x )implies Sig(Φ , x ) = val(Φ , xy )= ( − · PASS ( ) · PASS ( ) n − · PASS ( ) n − · PASS ( ) n − n +1 = 1 . This proves (8). Φ (cid:48) ( A )Let Φ (cid:48) = Φ (cid:48) ( A ) for some fixed A ⊆ [ n ] , let D ⊆ E (Φ (cid:48) ) denote the dangling edges of Φ (cid:48) and let F = I ( a ) ∪ I ( a ) denote the set of edges incident with either of the apices a or a in Φ (cid:48) . Let x ∈ { , } n be an assignment to D that satisfies the predicate ϕ one ( x ), and let xyz ∈ { , } E (Φ (cid:48) ) be a satisfying assign-ment to the edges of Φ (cid:48) that extends x , with y ∈ { , } E (Φ (cid:48) ) \ ( F ∪ D ) ,z ∈ { , } F . We consider the restriction of xyz to xy , that is, to edges not incident with any apex. By definition of PASS and
ACT , we have, for every vertex b ∈ V (Φ (cid:48) ) \ { a , a } , that( xy ) | I ( b ) ∈ { , , , } . (12)Recall from Remark 5.1 that we decompose x into x N , x E , x S , x W , and write W ∈ [ n ] and N ∈ [ n ] forthe unique non-zero index in x W ∈ { , } n and x N ∈ { , } n , respectively. Since ( xy ) | I ( b ) ∈ supp( PASS )holds by (12) and the definition of
PASS , the same argument as in the previous subsection for Φ shows thatthe western and eastern edges of all vertices in row (
W, (cid:63) ) are active under xy , as well as the northern andsouthern edges of all vertices in the column ( (cid:63), N ). Likewise, as seen in the previous subsection, it showsthat no other edges in E (Φ (cid:48) ) \ F are active, that y is unique if ϕ prop ( x ) holds, and that y does not existotherwise. This last statement implies thatSig(Φ (cid:48) , x ) = 0 if ¬ ϕ prop ( x ) .
22n the following, let x ∈ { , } D be an assignment to the dangling edges of Φ (cid:48) that satisfies ϕ prop ( x ), andlet xy ∈ { , } E (Φ (cid:48) ) \ F be its unique extension to edges not incident with apices, as seen for Φ. We considerthe possible assignments z ∈ { , } F to the apex edges such that xyz satisfies Φ (cid:48) . Here, while the choice of y was unique, the choice of z is not unique.By virtue of HW = at the apex vertices a and a , there are unique indices τ, τ (cid:48) ∈ A such that the edges a b τ and a b τ (cid:48) are active in xy . By definition of ACT , we actually have τ = τ (cid:48) , since all elements in supp( ACT )end on 00 or 11. We write τ ∗ := τ = τ (cid:48) for the unique “apex-matched” index, and b ∗ := b τ ∗ for the unique“apex-matched” vertex. By definition of ACT , we have( xyz ) | I ( b ∗ ) ∈ { , } . It follows that the second component of τ ∗ must be equal to N , since only vertices in ( (cid:63), N ) have stateor under xy . There are T vertices with signature ACT in row ( (cid:63), N ), by the balance property of ourinstance T to GridTiling, and we can choose any of these vertices to be apex-matched. To determine theset of such possible choices, we distinguish two cases, depending on whether ( W, N ) ∈ A or not.( W, N ) / ∈ A : The apex-matched vertex must be in state 11 under xyz . It cannot be in state 11,since only b W,N can have state among its first four edges, but b W,N has
PASS assigned, since(
W, N ) / ∈ A . This gives T assignments z such that xyz satisfies Φ (cid:48) . Each of the T assignments xyz satisfies val Φ (cid:48) ( xyz ) = −
1, because there is (i) one vertex in state 00, which contributes a factor of − Φ (cid:48) ( xyz ), and (ii) some number of vertices in states 00, 00 and 00, which however allcontribute a unit factor to val Φ (cid:48) ( xyz ). This implies that Sig(Φ (cid:48) , x ) = − T if both ( W, N ) / ∈ A and ϕ prop ( x ) hold.( W, N ) ∈ A : The apex-matched vertex may be in state 11 or 11. We make a distinction into these twoindividual sub-cases:11 : We proceed as in the case of (
W, N ) / ∈ A , but we have only T − b W,N must have state among its first four edges and can thus not be instate 11. This gives T − z with val Φ (cid:48) ( xyz ) = PASS ( ) = − z . (In theexpression of val Φ (cid:48) ( xyz ), we ignored the vertices in states 00, 00 and 00 that contribute aunit factor.)11 : Since only b W,N can have state among its first four edges, the apex-matched vertex mustbe b W,N . This gives one assignment z , and val Φ (cid:48) ( xyz ) = ACT ( ) = 1. Again, we ignored unitfactors.In total, if both (
W, N ) ∈ A and ϕ prop ( x ) hold, then we obtainSig(Φ (cid:48) , xyz ) = ( T − · ( −
1) + 1 = − T + 2This proves (9), and thus Lemma 5.5. The proof of Theorem 1.2 is completed. k We prove Theorem 1.3, which asserts ⊕ W [ ]-hardness of evaluating the permanent mod 2 k . We reduce fromthe problem ⊕ GridTiling, the parity version of GridTiling from Definition 2.4. From a high level, the proofresembles that of Theorem 1.2, but the setting of modular evaluation requires us to apply linearly combinedsignatures in a more intricate way. 23 .1 The main idea
Our reduction is based upon the following observation: Let A = ( n, k, C , T ) be an instance for ⊕ GridTiling.For ω : C → [2], recall the graphs H ω and the numbers d ( ω ) from the last section. We can rewrite (11) as2 |C| · T ) = (cid:88) ω : C→ [2] T d ( ω ) · perm( H ω ) . (13)Theorem 2.9 asserts that computing ⊕ GridTiling( T ) is ⊕ W [ ]-hard. Let M := 2 |C| and assume we couldevaluate perm( H ω ) modulo 2 M for all ω . Using arithmetic in Z / M Z , we could then evaluate the entireright-hand-side of (13), and this allows us to compute M · T ) ≡ M (cid:40) M if T ) is odd , T ) is even.Hence, it seems that we could solve ⊕ GridTiling( T ) with an oracle for the permanent modulo 2 M = 2 |C| +1 ,and we might be tempted to believe that we just proved Theorem 1.3.However, the above argument suffers from a fatal gap: The graphs H ω from the previous section featureedges of weight , a number that does not exist in the rings Z / k Z for k ∈ N . In other words, the proof failsfor the surprisingly philosophical reason that the instances H ω constructed in the previous section do noteven exist modulo 2 k . More precisely, it is the matchgate Γ ACT used to realize the signature
ACT that featuresthis offending weight, and it is incurred by the part that we called the even filter . To obtain graphs H ω thatavoid edge-weights with even denominators, we therefore construct cell gates using the signature PRE ratherthan its more benign version
ACT . This adds several complications to our arguments, which we can howeverhandle with a suitable linear combination.
Let A ⊆ [ n ] be fixed in the following, and recall the gates Φ and Φ (cid:48) from Definition 5.4. Note that Φfeatures only occurrences of PASS , which is realized by the matchgate Γ
PASS on edge-weights − k . This does not apply to the gate Φ (cid:48) ( A ), as the matchgate Γ ACT realizing
ACT features the weight . We modify Φ (cid:48) ( A ) to a new gate Γ( A ) by replacing all occurrences of ACT with
PRE . Definition 6.1.
For A ⊆ [ n ] , let the gate Γ( A ) on 4 n dangling edges be defined exactly as the gate Φ (cid:48) ( A )from Definition 5.4, but replace every occurrence of ACT by PRE .For all u, v ∈ [ n ], let α u,v denote the number of occurrences of PRE among vertices b τ with τ ∈{ (1 , v ) , . . . , ( u − , v ) } . Likewise, let β u,v denote the number of occurrences of PRE among vertices b τ with τ ∈ { ( u + 1 , v ) , . . . , ( n, v ) } .Figuratively speaking, α u,v is the number of occurrences of PRE in the column above ( u, v ), and β u,v isthe number of occurrences below it. In Section 5.2, we used the vertical balance property to ensure that α u,v + β u,v is equal to T − u, v ) ∈ A , and equal to T when ( u, v ) / ∈ A . In this section, this verticalbalance will not be required, but horizontal balance will prove useful instead, for different reasons. For theremainder of our proofs, we define the following auxiliary polynomials, for all u, v, w ∈ [ n ]: q u := (cid:88) z ∈ [ n ] α u,z · β u,z − (cid:18) α u,z (cid:19) − (cid:18) β u,z (cid:19) , (14) p u,v,w := ( α u,v − β u,v ) · ( β u,w − α u,w ) , (15) r u,v := (cid:88) z ∈ [ n ] \{ v } ( u,z ) ∈ A β u,z , (16) s u,v := (cid:88) z ∈ [ n ] \{ v } ( u,z ) ∈ A α u,z . (17)24sing these polynomials, we can express the signature of Γ. Lemma 6.2.
Let A ⊆ [ n ] , let Γ = Γ( A ) and let x ∈ { , } n satisfy ϕ one . Recall the conventions fromRemark 5.1, including that we implicitly decompose the string x into x N , x E , x S , x W . • If x W (cid:54) = x E or hw( x S ) (cid:54) = 1 , then Sig(Γ , x ) = 0 . • If ϕ prop ( x ) is true (i.e., we have x W = x E and additionally x N = x S ), write u := x W and v := x N ,with u, v ∈ [ n ] . Note that these numbers are well-defined. We call such assignments x wanted , and wehave Sig(Γ , x ) = (cid:40) q u − r u,v − s u,v − α u,v − β u,v if ( u, v ) / ∈ Aq u − r u,v − s u,v + 1 if ( u, v ) ∈ A • If ϕ prop ( x ) is false (i.e, we have x W = x E , but x N (cid:54) = x S ) , then write u := x W , v := x N , and w := x S .We call such assignments x unwanted, and we have Sig(Γ , x ) = p u,v,w if ( u, v ) / ∈ A, ( u, w ) / ∈ Ap u,v,w + α u,v − β u,v if ( u, v ) / ∈ A, ( u, w ) ∈ Ap u,v,w + β u,w − α u,w if ( u, v ) ∈ A, ( u, w ) / ∈ Ap u,v,w + β u,w − α u,w + α u,v − β u,v + 1 if ( u, v ) ∈ A, ( u, w ) ∈ A The full proof of this lemma requires a somewhat tedious calculation, which is deferred to Section 6.4.Note that the entries of Sig(Γ) are polynomials in the indeterminates α u,v and β u,v for u, v ∈ [ n ]Taking Lemma 6.2 for granted at the moment, we note that the gate Γ essentially discriminates betweensix different assignment types, depending on whether x is wanted (giving 2 types) or unwanted (giving 4types, depending on whether ( x W , x N ) and ( x W , x S ) are each contained in A ). However, the actual value ofSig(Γ , x ) is not constant for each of the six types, as it depends on u, v, w and the concrete values for α u (cid:48) ,v (cid:48) and β u (cid:48) ,v (cid:48) for all u (cid:48) , v (cid:48) ∈ [ n ]. Compare this to the gate Φ (cid:48) from Section 5.3.2, which attains one of the threefixed values { , − T, − T + 2 } due to vertical balance. It turns out that the simple balance argument used inthe last section does not work in this setting; our technical efforts in the remainder of the proof thereforeaim at the following two goals: Goal 1:
Ensure that the four unwanted cases (as defined above) cancel out.
Goal 2:
Ensure that the two wanted cases (as defined above) do not depend upon the actual value of( x W , x N ), but only on the information whether ( x W , x N ) ∈ A or ( x W , x N ) / ∈ A .In the following, we show how to attain these goals by considering a particular linear combination ofmatchgate signatures that could be considered as the “derivative” of a matchgate. Recall the construction of Γ from Definition 6.1. In the following, we construct a gate Γ ↑ from Γ by addingseveral “dummy” vertices. Then we consider the differenceSig(Γ ↑ ) − Sig(Γ) . The gate Γ ↑ is obtained from Γ by adding dummy rows of vertices with signature PRE , and this allows us toobtain Sig(Γ ↑ ) by a simple substitution on the indeterminates of Sig(Γ). Definition 6.3.
We define a dummy gate as in Figure 5: Starting from a vertex with signature
PRE , addseveral vertices of signature HW = to its western and eastern dangling edges to force these edges to be inactive,as shown in the left part of the figure. We then define a dummy row by arranging n dummy gates horizontallyas shown in the right part of the figure. 25igure 5: A dummy gate is shown on the left. On the right, we see Γ ↑ , which is obtained from Γ by addingrows of dummy gates, shown red. Each gray vertex is assigned HW = , and the apices connect to all blackvertices (assigned PRE ) and all red vertices (whose signature is realized by the dummy gate). White verticesare assigned
PASS , and they are not adjacent to apices.Starting from Γ, define a gate Γ ↑ by adding a dummy row above the row (1 , (cid:63) ), and a dummy row belowthe row ( n, (cid:63) ), as shown in Figure 5. We connect apex a to the dangling edge 5 of each dummy gate, and a to the dangling edge 6.Furthermore, we define algebraic manipulations on multivariate polynomials that capture the effect ofintroducing dummy rows into Γ as described above. Definition 6.4.
Let p be any multivariate polynomial on the indeterminates α u,v and β u,v for u, v ∈ [ n ].Write x ← y for the operation of substituting x with y in p . Then we define p ↑ to be the polynomial obtainedfrom p after performing the substitutions α u,v ← α u,v + 1 and β u,v ← β u,v + 1 for all u, v ∈ [ n ].We also define the following discrete derivative operator D on such polynomials p : D ( p ) := p ↑ − p. The following is then easily observed:
Lemma 6.5.
We have
Sig(Γ ↑ ) = (Sig(Γ)) ↑ , and in particular, we have D (Sig(Γ)) = Sig(Γ ↑ ) − Sig(Γ) . Note that the operator D indeed resembles a derivative: We have linearity by D ( p + q ) = D ( p ) + D ( q ),and applying D to a polynomial p of degree d gives one of degree d −
1. We will use these properties of D to effect two useful modifications on the polynomials in (14)-(16), and thus ultimately on Sig(Γ). Thesecorrespond to the two goals described at the end of Section 6.2.1. Concerning the first goal, our choice of D ensures that “unwanted” polynomials vanish under D . Forinstance, for all u, v, w ∈ [ n ], the polynomial p u,v,w from (15) maps to D ( p u,v,w ) = (( α u,v + 1) − ( β u,v + 1)) · (( β u,w + 1) − ( α u,w + 1)) − ( α u,v − β u,v ) · ( β u,w − α u,w )= 0 . (18)By our calculation of Sig(Γ), this implies that D (Sig(Γ)) vanishes on assignments x with x N (cid:54) = x S and( x W , x N ) / ∈ A and ( x W , x S ) / ∈ A . The other unwanted cases will be handled by similar arguments.26. Under the operator D , linear terms, such as α u,v or β u,v for u, v ∈ [ n ], are mapped to D ( α u,v ) = ( α u,v + 1) − α u,v = 1 , (19) D ( β u,v ) = ( β u,v + 1) − β u,v = 1 . (20)This helps us to attain the second goal, since the original terms depend on the concrete values of α u,v or β u,v in A , whereas the resulting constants do not. It will also turn out that only linear terms needto be considered.In the following, we show that D (Sig(Γ)) essentially realizes the function g κ , up to some additive term onassignments x with ϕ prop . This allows us to write g κ as a linear combination of the matchgate signaturesSig(Γ ↑ ) and Sig(Γ). As a technical requirement, we use Lemma 2.10 to ensure that the set A in the definitionof Γ = Γ( A ) is horizontally balanced. Lemma 6.6.
Assume the existence of a number T ∈ N such that A features exactly T elements of type ( u, (cid:63) ) , for all u ∈ [ n ] . Let Γ = Γ( A ) and write D := D (Sig(Γ)) = Sig(Γ ↑ ) − Sig(Γ) . We then have D = if ¬ ϕ prop ( x ) (cid:40) n − T − x W , x N ) / ∈ An − T + 2 ( x W , x N ) ∈ A if ϕ prop ( x ) Proof.
We prove the identity using linearity of D . For all u, v, w ∈ [ n ], consider the effect of D on thepolynomials from (14)-(17). For instance, we have seen in (18) and (19)-(20) that D ( p u,v,w ) = 0 ,D ( α u,v ) = D ( β u,v ) = 1 . Likewise, we can show that D ( q u ) = (cid:88) v ∈ [ n ] n,D ( r u,v ) = D ( s u,v ) = (cid:88) z ∈ [ n ] \{ v } ( u,z ) ∈ A (cid:40) T ( u, v ) / ∈ A,T − u, v ) ∈ A. Together with linearity of D and the expression of Sig(Γ) from 6.2, this proves the claim by a simplecalculation for each of the six assignment types. Corollary 6.7.
Write S := n − T − and recall the matchgate Φ from Section 5.3.1 with Sig(Φ , x ) = (cid:40) if ϕ prop ( x ) , otherwise . Then the following linear combination realizes the signature g κ : D − S · Sig(Φ)4 = Sig(Γ ↑ ) − Sig(Γ) − S · Sig(Φ)4 . Note that each of the constituent gates Γ ↑ , Γ and Φ has at most two apices and features only edge-weightsfrom the set {− , } . Furthermore, each of these gates admits a -coloring. Using Corollary 6.7, we can complete the proof of Theorem 1.3. Recall that we aim at a reduction from ⊕ GridTiling to the permanent modulo 2 k . 27 roof of Theorem 1.3. Let A = ( n, k, C , T ) be an instance for the ⊕ W [ ]-complete problem ⊕ GridTiling. Toprove the lower bound under ⊕ ETH , we may assume |C| = O ( k ) by Theorem 2.9. Furthermore, by horizontalbalance via Lemma 2.10, we may assume that we are given a number T ∈ N such that |T ( κ ) ∩ ( u, (cid:63) ) | = T for all κ ∈ C and u ∈ [ n ].Recall Definition 5.2 and Lemma 5.3 of Section 5.1: These allow us to compute a signature graph G withsignatures f κ at κ ∈ [ k ] \ C and signatures g κ at κ ∈ C such that A ) = Holant( G ) . As shown in Lemma 5.5, we can realize f κ by the planar matchgate Φ on edge-weights {− , } . Further-more, as shown in Lemma 6.6, we can realize g κ for each κ ∈ C as the linear combination of three 2-apexmatchgates on edge-weights {− , } : Let Γ κ := Γ( T ( κ )) be as in Definition 6.1, and let Γ κ, ↑ be obtainedfrom Γ κ as in Definition 6.3. Then, similarly to the proof of Theorem 1.2, we obtain with Lemma 6.6 andLemma 3.8 about the linear combinations of signatures that4 |C| · Holant( G ) = (cid:88) ω : C→ [3] ( − d ( ω ) · ( − S ) e ( ω ) · PerfMatch( H ω ) . (21)Here, for each ω : C → [3], the number d ( ω ) is defined to be the number of 2-entries in ω , and e ( ω ) is thenumber of 3-entries. The graph H ω is obtained as follows: For κ ∈ [ k ] \ C , insert the matchgate Φ at thecell vertex c κ . For all κ ∈ C , insert Γ κ, ↑ or Γ κ or Φ at c κ if ω ( κ ) is 1 or 2 or 3, respectively.Let M := 2 |C| . With an oracle for computing PerfMatch( H ω ) modulo 2 M for all ω , we can compute theright-hand side of (21) modulo 2 M via arithmetic in Z / M Z . We then obtain the value (modulo 2 M ) of M · Holant( G ) = M · A ) ≡ M (cid:40) M if A ) odd , A ) even.Each graph H ω is bipartite, has at most 2 |C| = O ( k ) apices, and the computation is modulo 2 M =2 O ( k ) . We have thus shown a parameterized Turing reduction from ⊕ GridTiling to the evaluation of thepermanent on O ( k )-apex graphs modulo 2 O ( k ) . Since Theorem 2.9 asserts the ⊕ W [ ]-completeness of theformer problem, the theorem follows. Γ In the remainder of this section, we prove Lemma 6.2. Let x ∈ { , } n be an assignment to the danglingedges of Γ. The statement of the lemma is shown by inspecting the possible satisfying extensions of x ,as we did when calculating Sig(Φ (cid:48) ). To understand the following proof, we therefore recommend recallingSection 5.3.2, since that section contains a similar, yet substantially simpler argument.Let F ⊆ E (Γ) denote the edges of Γ that are incident with apices. Given x , let xyz ∈ { , } E (Γ) bean assignment extending x such that Sig(Γ , xyz ) (cid:54) = 0, with y ∈ { , } E (Γ) \ F and z ∈ { , } F . Due to HW = at the apex vertices a and a of Γ, there are apex-matched indices τ , τ ∈ A and apex-matched vertices b := b τ and b := b τ such that a b and a b are active in xyz . However, opposing Section 5.3.2, it maywell be that τ (cid:54) = τ , and this makes our calculations somewhat more difficult. In particular, the assignment y is no longer uniquely determined by x .For each assignment x , we partition the satisfying extending assignments xyz to Γ into six partitionclasses {P i ( x ) } i ∈ [6] , corresponding to the states of the (at most two distinct) apex-matched vertices. Moreprecisely, for i ∈ [6], we let P i ( x ) := { xyz ∈ { , } E (Γ) | xyz | I ( b ) and xyz | I ( b ) are as in row i of Table 1 } . Note that b and b depend upon the assignment xyz . To give an example, in row 1, and thus in class P ,we consider extending assignments xyz that have only one vertex with active edges leading to an apex, andthe local assignment at this vertex reads 11. More formally, we have b = b ∧ xyz | I ( b ) = 11 . u, v ) / ∈ A ( u, v ) ∈ A ( u,v ) / ∈ A ( u,w ) / ∈ A ( u,v ) / ∈ A ( u,w ) ∈ A ( u,v ) ∈ A ( u,w ) / ∈ A ( u,v ) ∈ A ( u,w ) ∈ A
11 0 1 0 0 0 011 − α u,v − β u,v − α u,v − β u,v , q u q u p u,v,w p u,v,w p u,v,w p u,v,w , − r u,v − r u,v + α u,v α u,v − β u,v α u,v − β u,v , − s u,v − s u,v + β u,v β u,w − α u,w β u,w − α u,w ,
01 0 0 0 0 0 1Table 1: The six assignment types of the cell are listed as columns, and the possible states of the (at mosttwo) apex-matched vertices are listed as rows. The signature of Γ on each of the six assignment types is givenas the sum of the elements in the corresponding column. Note that the table is divided into four quadrants.We have essentially already calculated the top left quadrant in Section 5.3 when we calculated Sig(Φ (cid:48) ).As another example, in row 3, we have b (cid:54) = b ∧ xyz | I ( b ) = 10 ∧ xyz | I ( b ) = 01 . It is evident that, given x ∈ { , } n , we haveSig(Γ , x ) = (cid:88) i ∈ [6] (cid:88) xyz ∈P i ( x ) val Γ ( xyz ) (cid:124) (cid:123)(cid:122) (cid:125) =: P i ( x ) . (22)In Table 1, we calculate P i ( x ) for all i ∈ [6] and all six types of assignments x to dangling edgesdistinguished by the signature: The entry in this table at row i ∈ [6] and column j ∈ [6] denotes the number P i ( x ) on assignments x of the j -th type. Note that the table is divided into four quadrants, as indicated bythe double lines in Table 1. In Section 5.3.2, we have essentially already calculated the values in the top leftquadrant. In the following, we calculate the remaining quadrants.Before doing so, we first need to make some general observations: In each satisfying assignment xyz extending x , all western and eastern edges of vertices in the row ( x W , (cid:63) ) are active, and no other westernand eastern edges are active. This is because for any vertex b ∈ V (Γ) \ { a , a } , the signatures PASS and
PRE imply that the assignment xy | I ( b ) has one of the states , , , (cid:124) (cid:123)(cid:122) (cid:125) “tame” , , , , (cid:124) (cid:123)(cid:122) (cid:125) “wild” . (23)In each such state, be it tame or wild, the western incident edge is active iff the eastern edge is active aswell. By an argument as in Section 5.3.1, this implies x W = x E for the assignment x . Note that a similarstatement from north to south is not necessarily true, as witnessed by vertices in a “wild” state.If b (cid:54) = b , this implies xyz | I ( b ) ∈ { , } and xyz | I ( b ) = { , } . Because all other verticesare in tame states and thus enforce equality on their northern and southern dangling edges, the vertex b “shoots” a ray of active vertical edges to the north (transmitted by vertices in state , , 00, 00).This ray may either leave the cell, or it hits b . We conclude that, for any column j ∈ [ n ], • x N ( j ) = x S ( j ) iff column ( (cid:63), j ) contains neither b nor b , or it contains both, • x N ( j ) = 1 ∧ x S ( j ) = 0 iff column ( (cid:63), j ) contains b but not b , • x N ( j ) = 0 ∧ x S ( j ) = 1 iff column ( (cid:63), j ) contains b but not b .We are now ready to calculate the remaining quadrants of Table 1. Recall that we use the abbreviations u := x W , v := x N and w := x S . 29igure 6: Relevant states in the bottom right quadrant. The vertices b and b are shown as black dots,crossings with the horizontal path are shown as turquoise dots. Top right quadrant: If x N (cid:54) = x S , then P ( x ) = P ( x ) = ∅ . This is because all vertices in assignments xyz ∈ P ( x ) ∪ P ( x ) are in tame states, which would imply x N = x S . This explains all zeros in the top rightquadrant of Table 1. Bottom right quadrant (0/1 entries): If x N (cid:54) = x S and ( x W , x N ) / ∈ A , then no satisfying assignmenthas a vertex in state : By our general observation, the index of this vertex would be ( x W , x N ), but thisvertex has no adjacent apex, since ( x W , x N ) / ∈ A , and it can thus only be in a tame state. Likewise, if( x W , x S ) / ∈ A , then no satisfying assignment has a vertex in state . This explains all zeros in the bottomright quadrant of Table 1, and it also explains the bottom right entry of 1. Bottom right quadrant (other entries):
By our general observation, the vertex b must be located inthe column ( (cid:63), v ) and b must be located in the column ( (cid:63), w ) .Consider the third row in the right quadrant and Figure 6. Because of the states of b and b , neither ofthem is on the horizontal path u . This gives α u,v + β u,v choices for b . When b is above ( u, v ), there are α u,v possibilities, and the northbound ray emitted by b does not cross the horizontal path in ( u, (cid:63) ) described inthe general observations. When b is below ( u, v ), there are β u,v possibilities, and the northbound ray crossesthe horizontal path in ( u, (cid:63) ), so the vertex at ( u, v ) contributes a factor − PASS ( ) or
PRE ( 00). Bya similar analysis for b as for b , we obtain four cases, shown in Figure 6 and we see that, for inputs x ofthe third type in Table 1, we have P ( x ) = α u,v · β u,w − α u,v · α u,w − β u,v · β u,w + β u,v · α u,w = ( α u,v − β u,v ) · ( β u,w − α u,w )= p u,v,w The calculation of the remaining rows of Table 1 is similar, except that b or b may appear on thehorizontal path ( u, (cid:63) ) by the or state, so only one or fewer factors of p u,v,w remain. Bottom left quadrant (zero entries):
The argument for the zero entries in the bottom right quadrantapplies here as well.
Bottom left quadrant (other entries):
It can be verified that b and b must be located in the samecolumn, as otherwise it would be impossible to have x N = x S . In particular, either they are in some column( (cid:63), j ) with j (cid:54) = v , or they are in the column ( (cid:63), v ). We calculate the weighted sum over the relevant extensionsin Table 2, and then use it to get the bottom left quadrant of Table 1. To verify the completeness of ourreasoning, we advise to tick the corresponding cells of the table while reading.Let us assume first that b and b appear in a column ( (cid:63), j ) of Γ with j (cid:54) = v . These situations arecovered in columns 1, 2, 4, and 5 of Table 2. Then, after fixing the positions of b and b , the uniquepossible assignment realizing this choice contains the horizontal path ( u, (cid:63) ), a vertical path ( (cid:63), v ) and a path30tates of b , b u,v ) / ∈ Aj (cid:54) = v ( u,j ) / ∈ A ( u,v ) / ∈ Aj (cid:54) = v ( u,j ) ∈ A ( u,v ) / ∈ Aj = v ( u,v ) ∈ Aj (cid:54) = v ( u,j ) / ∈ A ( u,v ) ∈ Aj (cid:54) = v ( u,j ) ∈ A ( u,v ) ∈ Aj = v , t j t j t j t j t j t j ,
01 0 − β u,j − β u,j α u,v ,
01 0 − α u,j − α u,j β u,v ,
01 0 0 0 0 0 0Table 2: A detailed table of the bottom left quadrant of Table 1.connecting b and b . The vertex at ( u, v ) yields the value −
1, since it is in state or 00. Whether thevertex at ( u, j ) also yields − b b crosses the horizontal path ( u, (cid:63) ).Consider the first row of Table 2 for columns with j (cid:54) = v . When b and b are in states 10 and 01respectively, there are α u,j β u,j choices for b and b such that the line segment b b crosses the horizontalpath (and in this case, we have two crossings, each of which yields a factor − (cid:0) α u,j (cid:1) + (cid:0) β u,j (cid:1) choices of b and b such that the crossing does not occur (yielding one crossing in total and a factor − x with b and b in states 10 and 01 is equal to t j = α u,j β u,j − (cid:18) α u,j (cid:19) − (cid:18) β u,j (cid:19) . We observe that no extension to x can have the vertices b and b in states 10 and 01, as these stateswould force the vertices b and b to appear in different columns of Γ. Hence, the number of extensions inrow 4 are all zero. Note also that, in columns 1, 3, and 4, no states other than 10 and 01 can appear:Every other state would require ( u, j ) ∈ A , since only such vertices can possibly be in wild states.The calculations so far have settled columns 1 and 4; we now consider column 2. If and only if b islocated on ( u, j ), then the vertices b and b are in states 10 ,
01. Then the vertex b at ( u, j ) gives PRE ( 01) = 1, and b gives PRE ( 10) = 1. The vertex at ( u, v ) is in state or 00 and consequentlyyields −
1. We observe that there are β u,j choices for b . This settles row 2 of column 2. A symmetricargument applies in row 3 of column 2, when the vertices b and b are in states 10 , b and b do not appear in the v -th column of Γ.This settles all columns with j (cid:54) = v ; we will henceforth consider the case j = v as in columns 3 and 6. Inthese columns, the vertices b and b must be situated in column ( (cid:63), v ) of Γ. Furthermore, we again havethe horizontal path passing through row ( u, (cid:63) ).Consider row 1, corresponding to states 10 ,
01. Here, it is irrelevant whether ( u, v ) ∈ A or not,since none of b or b can be located at ( u, v ), as the horizontal path could otherwise not pass through thesevertices. There are α u,j β u,j possible positions for b and b such that b lies above the horizontal path ( u, (cid:63) )and b lies below it. In both situations, no crossing occurs. Furthermore, there are (cid:0) α u,j (cid:1) + (cid:0) β u,j (cid:1) possiblepositions for b and b such that both lie above or both lie below the horizontal path, introducing preciselyone crossing with the path. Hence, the weighted sum over extensions is again given by t j , with j = v .This settles column 3; it remains to consider column 6. Consider its second row. Because b is in state01, it is located at ( u, v ), and shoots a ray to the south. There are α u,v positions left for b to shoot aray to the north. Similarly, the third entry is β u,v . It is important to note here that no crossing occurs, asopposed to, say, column 5.We have now calculated all entries of the table. If we sum the first 3 columns and the last 3 columns,respectively, we get the bottom left quadrant of Table 1. (Note that each block of 3 columns actuallycorresponds to n choices for j , so each sum involves n terms.) Conclusion of the calculation.
This explains all entries of Table 1. Given an assignment x having oneof the types indicated in the columns of Table 1, the value Sig(Γ , x ) is then obtained by summing along thecorresponding column as in (22). 31 cknowledgement We wish to thank an anonymous reviewer for extensive and helpful comments on the submitted version.
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