Counting perfect matchings and the eight-vertex model

TL;DR

This paper investigates the computational complexity of approximating the partition function of the eight-vertex model on 4-regular graphs, establishing connections to the hard problem of counting perfect matchings and providing new characterizations of matchgates.

Contribution

It relates the approximability of the eight-vertex model to counting perfect matchings and characterizes nonnegative 4-ary matchgates, extending previous complexity results.

Findings

Approximation of the partition function is as hard as counting perfect matchings in certain parameter regions.

Computing the partition function can be reduced to counting perfect matchings in larger parameter regions.

Identifies a parameter region where planar graph approximation is feasible but general graph approximation is as hard as counting perfect matchings.

Abstract

We study the approximation complexity of the partition function of the eight-vertex model on general 4-regular graphs. For the first time, we relate the approximability of the eight-vertex model to the complexity of approximately counting perfect matchings, a central open problem in this field. Our results extend those in arXiv:1811.03126 [cs.CC]. In a region of the parameter space where no previous approximation complexity was known, we show that approximating the partition function is at least as hard as approximately counting perfect matchings via approximation-preserving reductions. In another region of the parameter space which is larger than the previously known FPRASable region, we show that computing the partition function can be reduced to (with or without approximation) counting perfect matchings. Moreover, we give a complete characterization of nonnegatively weighted (not necessarily planar) 4-ary matchgates, which has been open for several years. The key ingredient of our proof is a geometric lemma. We also identify a region of the parameter space where approximating the partition function on planar 4-regular graphs is feasible but on general 4-regular graphs is equivalent to approximately counting perfect matchings. To our best knowledge, these are the first problems of this kind.

Figures

Tables

Table 1. Table 1 : Approximation complexity of the eight-vertex model with ( a , b , c , d ) ∉ d -SUM 𝑎 𝑏 𝑐 𝑑 d -SUM (a,b,c,d)\not\in\texttt{$d$-SUM} § § § There is a symmetry among a , b , c 𝑎 𝑏 𝑐 a,b,c for the eight-vertex model on general (not necessarily planar) 4-regular graphs, so for simplicity we assume a ≤ b ≤ c 𝑎 𝑏 𝑐 a\leq b\leq c . .

	$d=0$abc	$d>0$
$a=b=c=0$abc	P-time computable (trivial)abc	P-time computable (trivial)
$a=b=0,c>0$abc	P-time computable (trivial)abc	\pbox20cm$c=d$: P-time computable [CF17]
$c\ne d$: NP-hard [CLLY18]
$a=0,b,c>0$abc	NP-hard [CLLY18]abc	#PerfectMatchings-hard (in this paper)
$a,b,c>0$abc	NP-hard [CLL19]abc	#PerfectMatchings-hard (in this paper)

Equations

$$Z_{\text{EightVertex}}(G;a,b,c,d)=\sum _{\tau \in {\mathcal{O}}_{\mathbf{e}}(G)}{a}^{n_1+n_2}{b}^{n_3+n_4}{c}^{n_5+n_6}{d}^{n_7+n_8},$$

$${\le }_{\text{AP}}$$

$$a,b,c,d$$

$$Z^{\prime }(G^{\prime };f):=\sum _{\sigma :E^{\prime }\to \left\{0,1\right\}}\prod _{u\in U_E}({\ne }_2)\left(\sigma {|}_{E^{\prime }(u)}\right)\prod _{u\in U_V}f\left(\sigma {|}_{E^{\prime }(u)}\right).$$

$$Z_{\text{EightVertex}}(G;a,b,c,d)=\text{Holant}\left(G^{\prime };{\ne }_2|f\right)$$

$$Z_{\text{EightVertex}}(G;a,b,c,d)=\text{Holant}\left(G^{\prime };{\ne }_2|f\right),$$

$$\text{Holant}\left(G^{\prime };{\ne }_2|f\right)$$

$$=\text{Holant}\left(G^{\prime };{=}_2|Z^{\otimes 4}f\right)$$

$$=\text{Holant}\left(G;Z^{\otimes 4}f\right),$$

$$\text{Holant}\left(\left[\begin{array}{cccc}w& 0& 0& x\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ x& 0& 0& w\end{array}\right]\right)\text{mbox}and\text{Holant}\left(\left[\begin{array}{cccc}w& 0& 0& 0\\ 0& 0& x& 0\\ 0& x& 0& 0\\ 0& 0& 0& w\end{array}\right]\right)$$

$$\text{Ising}\left(\frac{w}{z}\right)\text{mbox}and\text{Holant}\left(\left[\begin{array}{cccc}w& 0& 0& 0\\ 0& 0& z& 0\\ 0& z& 0& 0\\ 0& 0& 0& w\end{array}\right]\right).$$

$$\text{Holant}\left(G;\left[\begin{array}{cccc}w& 0& 0& 0\\ 0& 0& z& 0\\ 0& z& 0& 0\\ 0& 0& 0& w\end{array}\right]\right)=z^{|V(G)|}\cdot Z_{\text{Ising}}\left(\tilde{G};\frac{w}{z}\right).$$

$$\text{Holant}\left(\left[\begin{array}{cccc}a+b+c+d& 0& 0& -a+b+c-d\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ -a+b+c-d& 0& 0& a+b+c+d\end{array}\right]\right){\le }_{\text{AP}}\text{Holant}\left(\left[\begin{array}{cccc}a+b+c+d& 0& 0& -a+b+c-d\\ 0& a-b+c-d& a+b-c-d& 0\\ 0& a+b-c-d& a-b+c-d& 0\\ -a+b+c-d& 0& 0& a+b+c+d\end{array}\right]\right).$$

$$\left[\begin{array}{cccc}a+b+c+d& 0& 0& -a+b+c-d\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ -a+b+c-d& 0& 0& a+b+c+d\end{array}\right]=\left[\begin{array}{cccc}a+b+c+d& 0& 0& -a+b+c-d\\ 0& a-b+c-d& a+b-c-d& 0\\ 0& a+b-c-d& a-b+c-d& 0\\ -a+b+c-d& 0& 0& a+b+c+d\end{array}\right]\cdot \left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 1\end{array}\right],$$

$$\text{Holant}\left(\left[\begin{array}{cccc}a+b+c+d& 0& 0& -a+b+c-d\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ -a+b+c-d& 0& 0& a+b+c+d\end{array}\right]\right){\le }_{\text{AP}}\text{Holant}\left(\left[\begin{array}{cccc}a+b+c+d& 0& 0& -a+b+c-d\\ 0& a-b+c-d& a+b-c-d& 0\\ 0& a+b-c-d& a-b+c-d& 0\\ -a+b+c-d& 0& 0& a+b+c+d\end{array}\right],\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 1\end{array}\right]\right).$$

$$\text{Holant}\left(\left[\begin{array}{cccc}a+b+c+d& 0& 0& -a+b+c-d\\ 0& a-b+c-d& a+b-c-d& 0\\ 0& a+b-c-d& a-b+c-d& 0\\ -a+b+c-d& 0& 0& a+b+c+d\end{array}\right],\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 1\end{array}\right]\right){\le }_{\text{AP}}\text{Holant}\left(\left[\begin{array}{cccc}a+b+c+d& 0& 0& -a+b+c-d\\ 0& a-b+c-d& a+b-c-d& 0\\ 0& a+b-c-d& a-b+c-d& 0\\ -a+b+c-d& 0& 0& a+b+c+d\end{array}\right]\right),$$

$$\text{Holant}\left({\ne }_2|\left[\begin{array}{cccc}d& 0& 0& a\\ 0& b& c& 0\\ 0& c& b& 0\\ a& 0& 0& d\end{array}\right],\left[\begin{array}{cccc}1& 0& 0& 1\\ 0& 1& 1& 0\\ 0& 1& 1& 0\\ 1& 0& 0& 1\end{array}\right]\right){\le }_{\text{AP}}\text{Holant}\left({\ne }_2|\left[\begin{array}{cccc}d& 0& 0& a\\ 0& b& c& 0\\ 0& c& b& 0\\ a& 0& 0& d\end{array}\right]\right),$$

$$4(Z^{-1}{)}^{\otimes 4}(({=}_4))={\left[\begin{array}{cc}1& -i\\ 1& i\end{array}\right]}^{\otimes 4}\left[{\left[\begin{array}{c}1\\ 0\end{array}\right]}^{\otimes 4}+{\left[\begin{array}{c}0\\ 1\end{array}\right]}^{\otimes 4}\right]={\left[\begin{array}{c}1\\ 1\end{array}\right]}^{\otimes 4}+{\left[\begin{array}{c}-i\\ i\end{array}\right]}^{\otimes 4}=2[1,0,1,0,1],$$

$$0<d\le a\le b\le c.$$

$$M(g_1)=\left[\begin{array}{cccc}(b+c)d& 0& 0& bc+d^2\\ 0& a(b+c)& a^2+bc& 0\\ 0& a^2+bc& a(b+c)& 0\\ bc+d^2& 0& 0& (b+c)d\end{array}\right].$$

$$(a_1,b_1,c_1,d_1)=(a(b+c),bc+d^2,a^2+bc,(b+c)d).$$

$$M(g_2)=M(f_2)\cdot N\cdot M(f_2)=\left[\begin{array}{cccc}2cd& 0& 0& c^2+d^2\\ 0& 2ab& a^2+b^2& 0\\ 0& a^2+b^2& 2ab& 0\\ c^2+d^2& 0& 0& 2cd\end{array}\right].$$

$$(a_2,b_2,c_2,d_2)=(2ab,a^2+b^2,c^2+d^2,2cd).$$

$$0\le \frac{a^{\ast }}{d^{\ast }}-1\le \frac{a}{d}-1,0\le \frac{b^{\ast }}{d^{\ast }}-1\le \frac{b}{d}-1,0\le \frac{c^{\ast }}{d^{\ast }}-1\le \frac{c}{d}-1.$$

$$0\le \frac{c^{\ast }}{d^{\ast }}-1\le {\left(\frac{c}{d}-1\right)}^2.$$

$$0\le {\alpha }_{i_1}-1\le {\alpha }_{i_0+1}-1\le {\left({\alpha }_{i_0}-1\right)}^2\le \frac{1}{2^2}.$$

$$0\le {\alpha }_{i_2}-1\le {\alpha }_{i_1+1}-1\le {\left({\alpha }_{i_1}-1\right)}^2\le \frac{1}{2^{2^2}}.$$

$$0\le \frac{\max (\check{a},\check{b},\check{c})}{\check{d}}-1\le {\alpha }_{i_k}-1\le \frac{1}{2^{2^k}}.$$

$$Z_{\text{EightVertex}}(G;a,b,c,d)=\text{Holant}\left(G^{\prime };{\ne }_2|f\right),$$

$$\{\begin{array}{ccc}{a}_1{a}_2& <& b_1{b}_2+c_1{c}_2+d_1{d}_2\\ b_1{b}_2& <& a_1{a}_2+c_1{c}_2+d_1{d}_2\\ c_1{c}_2& <& a_1{a}_2+b_1{b}_2+d_1{d}_2\\ d_1{d}_2& <& a_1{a}_2+b_1{b}_2+c_1{c}_2\end{array}$$

Full text

In a region of the parameter space where no previous approximation complexity was known, we show that approximating the partition function is at least as hard as approximately counting perfect matchings via approximation-preserving reductions. In another region of the parameter space which is larger than the previously known FPRASable region, we show that computing the partition function can be reduced to (with or without approximation) counting perfect matchings. Moreover, we give a complete characterization of nonnegatively weighted (not necessarily planar) 4-ary matchgates, which has been open for several years. The key ingredient of our proof is a geometric lemma.

We also identify a region of the parameter space where approximating the partition function on planar 4-regular graphs is feasible but on general 4-regular graphs is equivalent to approximately counting perfect matchings. To our best knowledge, these are the first problems of this kind.

1 Introduction

The eight-vertex model is defined over 4-regular graphs, the states of which are the set of even orientations, i.e. those with an even number of arrows into (and out of) each vertex. There are eight permitted types of local configurations around a vertex—hence the name eight-vertex model (see Figure 1).

Classically, the eight-vertex model is defined by statistical physicists on a square lattice region where each vertex of the lattice is connected by an edge to four nearest neighbors. In general, the eight configurations 1 to 8 in Figure 1 are associated with eight possible weights $w_{1},\ldots,w_{8}$. By physical considerations, the total weight of a state remains unchanged if all arrows are flipped, assuming there is no external electric field. In this case we write $w_{1}=w_{2}=a$, $w_{3}=w_{4}=b$, $w_{5}=w_{6}=c$, and $w_{7}=w_{8}=d$. This complementary invariance is known as arrow reversal symmetry or zero field assumption.

Even in the zero-field setting, this model is already enormously expressive: its special case when $d=0$, the zero-field six-vertex model, has sub-models such as the ice ($a=b=c$), KDP, and Rys $F$ models; on the square lattice, some other important models such as the dimer and zero-field Ising models can be reduced to it [Bax72]. After the eight-vertex model was introduced in 1970 by Sutherland [Sut70], and Fan and Wu [FW70], Baxter [Bax71, Bax72] achieved a good understanding of the zero-field case in the thermodynamic limit on the square lattice (in physics it is called “exactly solved”).

In this paper, we assume the arrow reversal symmetry and further assume that $a,b,c,d\geq 0$, as is the case in classical physics. Given a 4-regular graph $G$, we label four incident edges of each vertex from 1 to 4. The partition function of the eight-vertex model with parameters $(a,b,c,d)$ on $G$ is defined as

[TABLE]

where $\mathcal{O}_{\bf e}(G)$ is the set of all even orientations of $G$, and $n_{i}$ is the number of vertices in type $i$ in $G$ ($1\leq i\leq 8$, locally depicted as in Figure 1) under an even orientation $\tau\in\mathcal{O}_{\bf e}(G)$.

In terms of the exact computational complexity, a complexity dichotomy is given for the eight-vertex model on 4-regular graphs for all eight parameters [CF17]. This is studied in the context of a classification program for the complexity of counting problems [CC17], where the eight-vertex model serves as important basic cases for Holant problems defined by not necessarily symmetric constraint functions. It is shown that every setting is either P-time computable (and some are surprising) or #P-hard. However, most cases for P-time tractability are due to nontrivial cancellations. In our setting where $a,b,c,d$ are nonnegative, the problem of computing the partition function of the eight-vertex model is #P-hard unless: (1) $a=b=c=d$ (this is equivalent to the unweighted case); (2) at least three of $a,b,c,d$ are zero; or (3) two of $a,b,c,d$ are zero and the other two are equal. In addition, on planar graphs it is also P-time computable for parameter settings $(a,b,c,d)$ with $a^{2}+b^{2}=c^{2}+d^{2}$, using the FKT algorithm.

Since exact computation is hard in most cases, one natural question is what is the approximate complexity of counting and sampling of the eight-vertex model. To our best knowledge, prior to [CLLY18], there is only one previous result in this regard due to Greenberg and Randall [GR10]. They showed that on square lattice regions a specific Markov chain (which flips the orientations of all four edges along a uniformly picked face at each step) is torpidly mixing when $d$ is large. It means that when sinks and sources have large weights, this particular chain cannot be used to approximately sample eight-vertex configurations on the square lattice according to the Gibbs measure. Recently, similar torpid mixing results have been achieved for the six-vertex model on the square lattice [Liu18].

[CLLY18] gave the first classification results for the approximate complexity of the eight-vertex model on general and planar 4-regular graphs, and they conform to phase transition in physics. This is an extension to the work on the approximability of the six-vertex model [CLL19]. In order to state the results in [CLLY18] and in this work, we adopt the following notations assuming $a,b,c,d\geq 0$.

• •

DO = $\{(a,b,c,d)\;|\;a\leq b+c+d,~{}~{}b\leq a+c+d,~{}~{}c\leq a+b+d,~{}~{}d\leq a+b+c\}$;

• •

$d$-SUM = $\{(a,b,c,d)\;|\;a+d\leq b+c,~{}~{}b+d\leq a+c,~{}~{}c+d\leq a+b\}$;

• •

SQ-SUM = $\{(a,b,c,d)\;|\;a^{2}\leq b^{2}+c^{2}+d^{2},~{}~{}b^{2}\leq a^{2}+c^{2}+d^{2},~{}~{}c^{2}\leq a^{2}+b^{2}+d^{2},~{}~{}d^{2}\leq a^{2}+b^{2}+c^{2}\}$.

Remark 1.1*.*

$d$-SUM $\subset\texttt{DO}$, SQ-SUM $\subset\texttt{DO}$.

Physicists have shown an order-disorder phase transition for the eight-vertex model on the square lattice between parameter settings outside DO and those inside DO (see Baxter’s book [Bax82] for more details). In [CLLY18], it was shown that: (1) approximating the partition function of the eight-vertex model on general 4-regular graphs outside DO is NP-hard, (2) there is an FPRAS‡‡‡Suppose $f:\Sigma^{*}\rightarrow\mathbb{R}$ is a function mapping problem instances to real numbers. A fully polynomial randomized approximation scheme (FPRAS) [KL83] for a problem is a randomized algorithm that takes as input an instance $x$ and $\varepsilon>0$, running in time polynomial in $n$ (the input length) and $\varepsilon^{-1}$, and outputs a number $Y$ (a random variable) such that $\operatorname{Pr}\left[(1-\varepsilon)f(x)\leq Y\leq(1+\varepsilon)f(x)\right]\geq\frac{3}{4}.$ for general 4-regular graphs in the region $d$-SUM $\bigcap$ SQ-SUM, and (3) there is an FPRAS for planar 4-regular graphs in the extra region $\{(a,b,c,d)\;|\;a+d\leq b+c,~{}~{}b+d\leq a+c,~{}~{}c+d\geq a+b\}$ $\bigcap$ SQ-SUM.

In this paper we make further progress in the classification program of the approximate complexity of the eight-vertex model on 4-regular graphs in terms of the parameters (see Figure 2). For the first time, the complexity of approximating the partition function of the eight-vertex model (#EightVertex($a,b,c,d$)) is related to that of approximately counting perfect matchings (#PerfectMatchings).

Theorem 1.1**.**

For any four positive numbers $a,b,c,d>0$ such that (a,b,c,d)\not\in\textup{{d-SUM}}, the problem #EightVertex($a,b,c,d$) is at least as hard to approximate as counting perfect matchings:

[TABLE]

Remark 1.2*.*

The theorem is stated for the case where all four parameters are positive. The same proof also works for the case when there is exactly one zero among the nonnegative values $\{a,b,c\}$. A complete account for four nonnegative values $\{a,b,c,d\}$ is given in the Appendix A.

The proof of Theorem 1.1 is in Section 3. Our proof for the hardness result has several ingredients:

1. (1)

We express the eight-vertex model on a 4-regular graph $G$ as an edge-2-coloring problem on $G$ using Valiant’s holographic transformation [Val08]. 2. (2)

We show that some special cases in this edge-2-coloring problem on $G$ is equivalent to the zero-field Ising model on its crossing-circuit graph $\tilde{G}$. Thus known #PerfectMatchings-equivalence result for the Ising model [GJ08, Lemma 7] directly transfers to the special cases under certain parameter settings. 3. (3)

We further show that for any parameter setting outside $d$-SUM, approximating the partition function of the eight-vertex model is at least as hard as the #PerfectMatchings-equivalent special cases of the edge-2-coloring problem via approximation-preserving reductions (introduced in [DGGJ04]).

Theorem 1.2**.**

For any $(a,b,c,d)\in\textup{{SQ-SUM}}$,

[TABLE]

The proof of Theorem 1.2 is in Section 4. To prove the easiness result, we again express the eight-vertex model in the Holant framework (see Section 2) and show that the constraint functions of the eight-vertex model in SQ-SUM can be implemented by constant-size matchgates with nonnegatively weighted edges (Definition 4.1). We note that allowing nonnegative edge-weights does not add more computational power [McQ13, Proposition 5]. The crucial ingredient of our proof is a geometric lemma (Lemma 4.3) in 3-dimensional space.

Our result is tight in the sense that no constraint functions of the eight-vertex model with parameter settings outside the region SQ-SUM can be implemented by a matchgate (Lemma 4.4). Moreover, the general version of our result also works for the eight-vertex model without the arrow reversal symmetry. It is open if computing the partition function in \texttt{DO}\setminus(\texttt{d-SUM}\bigcup\texttt{SQ-SUM}) is #PerfectMatchings-equivalent or not.

As part of this work, we give a complete characterization of the constraint functions that can be expressed by 4-ary matchgates in Theorem 4.1. This solves an important question that has been open for several years [McQ13, BGJ*+*17]. We believe it is of independent interest.

Corollary 1.3**.**

For any four positive numbers $a,b,c,d>0$ such that (a,b,c,d)\in\textup{{SQ-SUM}}\setminus\textup{{d-SUM}},

[TABLE]

Note that for the eight-vertex model in the region $\{(a,b,c,d)\;|\;a+d\leq b+c,~{}~{}b+d\leq a+c,~{}~{}c+d>a+b\}$ $\bigcap$ SQ-SUM, computing $Z_{\textup{{EightVertex}}}(a,b,c,d)$ is (1) #P-complete in exact computation [CF17], (2) #PerfectMatchings-equivalent in approximate computation on general 4-regular graphs (Corollary 1.3), and (3) admits an FPRAS in approximate computation on planar 4-regular graphs [CLLY18]. To our best knowledge, these are the first identified problems of this kind.

2 Preliminaries

Given a 4-regular graph $G=(V,E)$, the edge-vertex incidence graph $G^{\prime}=(U_{E},U_{V},E^{\prime})$ is a bipartite graph where $(u_{e},u_{v})\in U_{E}\times U_{V}$ is an edge in $E^{\prime}$ iff $e\in E$ in $G$ is incident to $v\in V$. We model an orientation ($w\rightarrow v$) on an edge $e=\left\{w,v\right\}\in E$ from $w$ into $v$ in $G$ by assigning $1$ to $(u_{e},u_{w})\in E^{\prime}$ and [math] to $(u_{e},u_{v})\in E^{\prime}$ in $G^{\prime}$. A configuration of the eight-vertex model on $G$ is an edge 2-coloring on $G^{\prime}$, namely $\sigma:E^{\prime}\rightarrow\{0,1\}$, where for each $u_{e}\in U_{E}$ its two incident edges are assigned 01 or 10, and for each $u_{v}\in U_{V}$ the sum of values $\sum_{i=1}^{4}\sigma(e_{i})\equiv 0\pmod{2}$, over the four incident edges of $u_{v}$. Thus we model the even orientation rule of $G$ on all $v\in V$ by requiring “two-0-two-1/four-0/four-1” locally at each vertex $u_{v}\in U_{V}$.

The “one-0-one-1” requirement on the two edges incident to a vertex in $U_{E}$ is a binary Disequality constraint, denoted by $(\neq_{2})$. The values of a 4-ary constraint function $f$ can be listed in a matrix $M(f)=\left[\begin{smallmatrix}f_{0000}&f_{0010}&f_{0001}&f_{0011}\\ f_{0100}&f_{0110}&f_{0101}&f_{0111}\\ f_{1000}&f_{1010}&f_{1001}&f_{1011}\\ f_{1100}&f_{1110}&f_{1101}&f_{1111}\end{smallmatrix}\right]$, called the constraint matrix of $f$. For the eight-vertex model satisfying the even orientation rule and arrow reversal symmetry, the constraint function $f$ at every vertex $v\in U_{V}$ in $G^{\prime}$ has the form $M(f)=\left[\begin{smallmatrix}d&0&0&a\\ 0&b&c&0\\ 0&c&b&0\\ a&0&0&d\end{smallmatrix}\right]$, if we locally index the left, down, right, and up edges incident to $v$ by 1, 2, 3, and 4, respectively according to Figure 1. Thus computing the partition function $Z_{\textup{{EightVertex}}}(G;a,b,c,d)$ is equivalent to evaluating

[TABLE]

where $E^{\prime}(u)$ denotes the incident edges of $u\in U_{E}\cup U_{V}$. In fact, in this way we express the partition function of the eight-vertex model as the Holant sum in the framework for Holant problems:

[TABLE]

where we use $\textup{Holant}(H;g\ |\ f)$ to denote the Holant sum $\sum_{\sigma:E\rightarrow\left\{0,1\right\}}\prod_{u\in U}g\left(\sigma|_{E(u)}\right)\prod_{u\in V}f\left(\sigma|_{E(u)}\right)$ on a bipartite graph $H=(U,V,E)$ for the Holant problem $\textup{Holant}(g\ |\ f)$. Each vertex in $U$ (or $V$) is assigned the constraint function $g$ (or $f$, respectively). The constraint function $g$ is considered as a row vector (or covariant tensor), whereas the signature $f$ is considered as a column vector (or contravariant tensor). (See [CC17] for more on Holant problems.) The following proposition says that an invertible holographic transformation does not change the complexity of the Holant problem in the bipartite setting.

Proposition 2.1** ([Val08]).**

Suppose $T\in\mathbb{C}^{2}$ is an invertible matrix. Let $d_{1}=\operatorname{arity}(g)$ and $d_{2}=\operatorname{arity}(f)$. Define $g^{\prime}=g\left(T^{-1}\right)^{\otimes d_{1}}$ and $f^{\prime}=T^{\otimes d_{2}}f$. Then for any bipartite graph $H$, $\textup{Holant}(H;g\ |\ f)=\textup{Holant}(H;g^{\prime}\ |\ f^{\prime})$.

3 #PerfectMatchings-hardness

Our proof strategy for Theorem 1.1 is as follows. In Lemma 3.1, we express the eight-vertex model on a 4-regular graph $G$ as a Holant problem; this is an equivalent form of the orientation problem expressed as an edge-2-coloring problem on $G$, and is achieved using a holographic transformation. In Lemma 3.3, we establish the equivalence between some special cases of this edge-2-coloring problem and the zero-field Ising model. Thus a known result for the Ising model (Proposition 3.2) indicates the #PerfectMatchings-equivalence of the special cases under certain parameter settings (Corollary 3.4). It follows from Lemma 3.5 that for any $(a,b,c,d)$ with $a+d>b+c$ (and symmetrically $b+d>a+c$ or $c+d>a+b$), approximately computing the partition function is at least as hard as the #PerfectMatchings-equivalent special cases under approximation-preserving reductions.

Lemma 3.1**.**

$2^{|V(G)|}\cdot Z_{\textup{{EightVertex}}}(G;a,b,c,d)=\textup{Holant}\left(G;\left[\begin{smallmatrix}a+b+c+d&0&0&-a+b+c-d\\ 0&a-b+c-d&a+b-c-d&0\\ 0&a+b-c-d&a-b+c-d&0\\ -a+b+c-d&0&0&a+b+c+d\end{smallmatrix}\right]\right)$.

Proof.

Using the binary disequality function $(\not=_{2})$ for the orientation of any edge, we can express the partition function of the eight-vertex model $G$ as a Holant problem on its edge-vertex incidence graph $G^{\prime}$,

[TABLE]

where $f$ is the 4-ary constraint function with $M(f)=\left[\begin{smallmatrix}d&0&0&a\\ 0&b&c&0\\ 0&c&b&0\\ a&0&0&d\end{smallmatrix}\right]$. According to Proposition 2.1, under a $Z$-transformation where $Z=\frac{1}{\sqrt{2}}\left[\begin{smallmatrix}1&1\\ i&-i\end{smallmatrix}\right]$, we have

[TABLE]

and a direct calculation shows that $M(Z^{\otimes 4}f)=\frac{1}{2}\left[\begin{smallmatrix}a+b+c+d&0&0&-a+b+c-d\\ 0&a-b+c-d&a+b-c-d&0\\ 0&a+b-c-d&a-b+c-d&0\\ -a+b+c-d&0&0&a+b+c+d\end{smallmatrix}\right]$. ∎

Proposition 3.2** ([GJ08, Lemma 7]).**

Suppose $\beta<-1$. Then $\textup{{\#PerfectMatchings}}\equiv_{\textup{AP}}\textup{{Ising}}(\beta)$.

Lemma 3.3**.**

The Ising problem $\textup{{Ising}}\left(\frac{w}{x}\right)$ is equivalent to the Holant problem $\textup{Holant}\left(\left[\begin{smallmatrix}w&0&0&x\\ 0&0&0&0\\ 0&0&0&0\\ x&0&0&w\end{smallmatrix}\right]\right)$. In particular, $\textup{{Ising}}\left(\frac{w}{x}\right)\equiv_{\textup{AP}}\textup{Holant}\left(\left[\begin{smallmatrix}w&0&0&x\\ 0&0&0&0\\ 0&0&0&0\\ x&0&0&w\end{smallmatrix}\right]\right)$.

Remark 3.1*.*

A non-homogenized form of the Ising model is $\tilde{Z}_{\textup{{Ising}}}(G;x,w):=\sum\limits_{\sigma:V\rightarrow\{0,1\}}w^{\textup{mono}(\sigma)}x^{|E|-\textup{mono}(\sigma)}$. If $x\not=0$ then $\tilde{Z}_{\textup{{Ising}}}(G;x,w)=x^{|E|}Z_{\textup{{Ising}}}(G;\frac{w}{x})$. If $x=0$ then in $\tilde{Z}_{\textup{{Ising}}}$ all vertices in each component must take the same assignment (all [math] or all $1$). In this case both $\tilde{Z}_{\textup{{Ising}}}(G;x,w)$ and the Holant problem in Lemma 3.3 are trivially solvable in polynomial time.

Proof.

For the problem $\textup{Holant}\left(\left[\begin{smallmatrix}w&0&0&x\\ 0&y&z&0\\ 0&z&y&0\\ x&0&0&w\end{smallmatrix}\right]\right)$, the roles of $x,y,z$ are interchangeable by relabeling the edges. For example, if the constraint function $f(x_{1},x_{2},x_{3},x_{4})$ has the constraint matrix $\left[\begin{smallmatrix}w&0&0&x\\ 0&y&z&0\\ 0&z&y&0\\ x&0&0&w\end{smallmatrix}\right]$, then the constraint function $f(x_{1},x_{3},x_{2},x_{4})$ has the constraint matrix $\left[\begin{smallmatrix}w&0&0&z\\ 0&y&x&0\\ 0&x&y&0\\ z&0&0&w\end{smallmatrix}\right]$. It follows that

[TABLE]

are exactly the same problem. So to prove the lemma it suffices to prove the equivalence of

[TABLE]

First we show that $\textup{Holant}\left(\left[\begin{smallmatrix}w&0&0&0\\ 0&0&z&0\\ 0&z&0&0\\ 0&0&0&w\end{smallmatrix}\right]\right)$ can be expressed as $\textup{{Ising}}\left(\frac{w}{z}\right).$

Given a 4-regular graph $G=(V,E)$ as an instance of $\textup{Holant}\left(\left[\begin{smallmatrix}w&0&0&0\\ 0&0&z&0\\ 0&z&0&0\\ 0&0&0&w\end{smallmatrix}\right]\right)$, we can partition $E$ into a set $\mathcal{C}$ of circuits (in which vertices may repeat but edges cannot) in the following way: at every vertex $v\in V$, denote the four edges incident to $v$ by $e_{1},e_{2},e_{3},e_{4}$ in a cyclic order according to the local labeling of the signature function; we make $e_{1}$ and $e_{3}$ into adjacent edges in a single circuit, and similarly we make $e_{2}$ and $e_{4}$ into adjacent edges in a single circuit (note that these may be the same circuit). We say each circuit in $\mathcal{C}$ is a crossing circuit of $G$. For the graph $G$, we define its crossing-circuit graph $\tilde{G}=(\mathcal{C},\tilde{E})$, with possible multiloops and multiedges, as follows: its vertex set $\mathcal{C}$ consists of the crossing circuits; for every $v\in V$, if circuits $C_{1}$ and $C_{2}$ intersect at $v$, then there is an edge $\tilde{e}_{v}\in\tilde{E}$ labeled by $v$. Note that it is possible that $C_{1}=C_{2}$, and for such a self-intersectison point the edge $\tilde{e}_{v}$ is a loop. Each $C\in\mathcal{C}$ may have multiple loops, and for distinct circuits $C_{1}$ and $C_{2}$ there may be multiple edges between them. The edge set $\tilde{E}$ of $\tilde{G}$ is in 1-1 correspondence with $V$ of $G$.

Observe that the problem $\textup{Holant}\left(\left[\begin{smallmatrix}w&0&0&0\\ 0&0&z&0\\ 0&z&0&0\\ 0&0&0&w\end{smallmatrix}\right]\right)$ requires that every valid configuration $\sigma$ (that contributes a non-zero term) obeys the following rule at each vertex $v$:

• •

Assuming $e_{1},e_{2},e_{3},e_{4}$ are the four edges incident to $v$ in cyclic order, then $\sigma(e_{1})=\sigma(e_{3})$ (denoted by $b_{1}$) and $\sigma(e_{2})=\sigma(e_{4})$ (denoted by $b_{2}$). That is to say, all edges in a crossing circuit must have the same assignment (either all [math] or all $1$). Therefore, the valid configurations $\sigma$ on the edges of $G$ are in 1-1 correspondence with $0,1$-assignments $\sigma^{\prime}$ on the vertices of $\tilde{G}$.

• •

Under $\sigma$, the local weight on $v$ is $w$ if $b_{1}=b_{2}$ and is $z$ otherwise. Suppose crossing circuits $C_{1}$ and $C_{2}$ intersect at $v$ (they could be identical). Then in $\tilde{G}$, $\sigma^{\prime}$ has local weight $w$ on the edge $\tilde{e}_{v}$ if $\sigma^{\prime}(C_{1})=\sigma^{\prime}(C_{2})$ and has local weight $z$ otherwise.

This means

[TABLE]

Next we show that $\textup{{Ising}}\left(\frac{w}{z}\right)$ can be expressed as $\textup{Holant}\left(\left[\begin{smallmatrix}w&0&0&0\\ 0&0&z&0\\ 0&z&0&0\\ 0&0&0&w\end{smallmatrix}\right]\right).$ Note that every graph $G=(V,E)$ (without isolated vertices) is the crossing-circuit graph of some 4-regular graph $\bar{G}$. To define $\bar{G}$ from $G$, one only needs to do the following: (1) transform each vertex $v\in V$ into a closed cycle $C_{v}$; (2) for each loop at $v\in V$, make a self-intersection on $C_{v}$; and (3) for each non-loop edge $\{u,v\}\in E$ ($u$ and $v$ are two distinct vertices), make $C_{u}$ and $C_{v}$ intersect in a “crossing” way at a vertex in $\bar{G}$ (by first creating a vertex $p$ on $C_{u}$ and another vertex $p^{\prime}$ on $C_{v}$, then merging $p$ and $p^{\prime}$ with local labeling $1,3$ on $C_{u}$ and $2,4$ on $C_{v}$). Then the above proof holds for the reverse direction. ∎

Corollary 3.4**.**

Suppose $x\neq 0$ and $\frac{w}{x}<-1$. Then $\textup{{\#PerfectMatchings}}\equiv_{\textup{AP}}\textup{Holant}\left(\left[\begin{smallmatrix}w&0&0&x\\ 0&0&0&0\\ 0&0&0&0\\ x&0&0&w\end{smallmatrix}\right]\right).$

Lemma 3.5**.**

Suppose $d>0$ and at most one of $a,b,c$ is zero. Then $\textup{Holant}\left(\left[\begin{smallmatrix}a+b+c+d&0&0&-a+b+c-d\\ 0&0&0&0\\ 0&0&0&0\\ -a+b+c-d&0&0&a+b+c+d\end{smallmatrix}\right]\right)\leq_{\textup{AP}}\textup{{\#EightVertex}}(a,b,c,d)$.

Proof.

According to Lemma 3.1, we only need to prove

[TABLE]

Figure 3 describes a simple gadget construction in the Holant framework. In Figure 3 if we place the constraint function $f_{1}$ with $M_{x_{i}x_{j},x_{l}x_{k}}(f_{1})$ and $f_{2}$ with $M_{x_{p}x_{q},x_{s}x_{r}}(f_{2})$ at the two degree 4 vertices, then the constraint function $f_{3}$ of the 4-ary construction is $M(f_{3})=M_{x_{i}x_{j},x_{s}x_{r}}(f_{3})=M_{x_{i}x_{j},x_{l}x_{k}}(f_{1})\cdot M_{x_{p}x_{q},x_{s}x_{r}}(f_{2})$. Since

[TABLE]

we know that the Holant problem below on the left can be expressed by the Holant problem on the right, and therefore

[TABLE]

Notice that $\left[\begin{smallmatrix}1&0&0&0\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&0&1\end{smallmatrix}\right]$ is the signature matrix of the arity 4 equality function $(=_{4})$. Therefore, (3.1) is true if we can show

[TABLE]

which is equivalent to the following in the orientation view according to the proof of Lemma 3.1

[TABLE]

by the holographic transformation

[TABLE]

a constant multiple of the even parity function, which has the signature matrix $\left[\begin{smallmatrix}1&0&0&1\\ 0&1&1&0\\ 0&1&1&0\\ 1&0&0&1\end{smallmatrix}\right]$.

Next we show how to get (3.2). Given the constraint function $f$ with matrix $\left[\begin{smallmatrix}d&0&0&a\\ 0&b&c&0\\ 0&c&b&0\\ a&0&0&d\end{smallmatrix}\right]$ in #EightVertex($a,b,c,d$), we construct a 4-ary signature $\check{f}$ with constraint matrix $\left[\begin{smallmatrix}\check{d}&0&0&\check{a}\\ 0&\check{b}&\check{c}&0\\ 0&\check{c}&\check{b}&0\\ \check{a}&0&0&\check{d}\end{smallmatrix}\right]$ using a polynomial number of vertices and edges such that $\check{a}$, $\check{b}$, $\check{c}$, and $\check{d}$ are all exponentially close to $1$ after normalization, i.e., to be $2^{-n^{C}}$ close to 1, for any $C>0$, with a construction of $n^{O(1)}$ size in polynomial time.

We assume we start with the following condition:

[TABLE]

If this is not the case, we can obtain a 4-ary construction that realizes this condition using constantly many vertices. With some preliminary construction we can further assume $1\leq d\leq a\leq b\leq c\leq\frac{3}{2}d$ initially. (See Appendix B for details.) Note that starting with the constraint function $f$ with matrix $M(f)=\left[\begin{smallmatrix}d&0&0&a\\ 0&b&c&0\\ 0&c&b&0\\ a&0&0&d\end{smallmatrix}\right]$, we can arbitrarily permute $a,b,c$ by relabeling the edges, and so we get constraint functions $f_{1}$ with $M(f_{1})=\left[\begin{smallmatrix}d&0&0&b\\ 0&a&c&0\\ 0&c&a&0\\ b&0&0&d\end{smallmatrix}\right]$ and $f_{2}$ with $M(f_{2})=\left[\begin{smallmatrix}d&0&0&c\\ 0&a&b&0\\ 0&b&a&0\\ c&0&0&d\end{smallmatrix}\right]$. There are two constructions $G_{1}$ and $G_{2}$ which we use as basic steps; both constructions start with a constraint function $f$ with parameters satisfying (3.3).

$\boldsymbol{G_{1}}$: connect two vertices with constraint functions $f_{1}$ and $f_{2}$ respectively as in Figure 3. Since we are in the orientation view, we place the constraint function $(\neq_{2})$ on the two degree 2 vertices connecting the two degree 4 vertices. Then the constraint function $g_{1}$ of the construction $G_{1}$ is obtained by matrix multiplication $M(g_{1})=M_{x_{i}x_{j},x_{s}x_{r}}(g_{1})=M(f_{1})\cdot N\cdot M(f_{2})$, where $N=\left[\begin{smallmatrix}&&&1\\ &&1&\\ &1&&\\ 1&&&\end{smallmatrix}\right]$. Thus

[TABLE]

The constraint function $g_{1}$ has four new parameters, denoted by

[TABLE]

We make the following observations; all of them can be easily verified using (3.3):

• •

$d_{1}$ is the weight on sink and source and $0<d_{1}\leq a_{1},b_{1},c_{1}$.

• •

$c_{1}=\max(a_{1},b_{1},c_{1},d_{1})$.

• •

$\frac{a_{1}}{d_{1}}=\frac{a}{d},\frac{b_{1}}{d_{1}}\leq\frac{b}{d},\frac{c_{1}}{d_{1}}\leq\frac{c}{d}$.

• •

$c_{1}d_{1}\leq a_{1}b_{1}$ because $c_{1}d_{1}-a_{1}b_{1}=-(b+c)(a-d)(bc-ad)\leq 0$. 2. 2.

$\boldsymbol{G_{2}}$: connect two vertices with constraint functions $f_{2}$ as in Figure 3. Denote the constraint function of $G_{2}$ by $g_{2}$. We have

[TABLE]

The constraint function $g_{2}$ has four new parameters, denoted by

[TABLE]

The following observations can also be easily verified using (3.3):

• •

$d_{2}$ is the weight on sink and source and if $cd\leq ab$, then $0<d_{2}\leq a_{2},b_{2},c_{2}$.

• •

$\frac{a_{2}}{d_{2}}\leq\frac{a}{d},\frac{b_{2}}{d_{2}}\leq\frac{b}{d},\frac{c_{2}}{d_{2}}\leq\frac{c}{d}$.

• •

$\frac{c_{2}-d_{2}}{d_{2}}=\frac{(c-d)^{2}}{2cd}\leq\frac{1}{2}\left(\frac{c-d}{d}\right)^{2}\leq\left(\frac{c-d}{d}\right)^{2}$.

Based on the two basic constructions above, we construct the constraint function $\check{f}$ in logarithmically many rounds recursively, each of the $O(\log n)$ rounds uses the constraint function constructed in the previous round. We now describe a single round in this construction, which consists of two steps. In step 1 we use a signature with some parameter setting $(a,b,c,d)$ satisfying (3.3) and apply $G_{1}$ to two copies of the signature. If the resulting parameter $b_{1}<a_{1}$ we switch the roles of $a_{1}$ and $b_{1}$, and obtain $(a^{\prime}_{1},b^{\prime}_{1},c^{\prime}_{1},d^{\prime}_{1})=(b_{1},a_{1},c_{1},d_{1})$, again satisfying (3.3), as well as $c_{1}d_{1}\leq a_{1}b_{1}$. In step 2, we apply $G_{2}$ to two copies of the signature constructed in step 1 (with the switching of the roles of $a_{1}$ and $b_{1}$ if it is needed). Denote the parameters of the resulting signature by $(a^{*},b^{*},c^{*},d^{*})$. Altogether each round uses four copies of the signature from the previous round, starting with the initial given signature. Therefore in polynomial time we can afford to carry out $C\log n$ rounds for any constant $C$. Note that, if we consider the normalized quantities $(\frac{a}{d},\frac{b}{d},\frac{c}{d},\frac{d}{d})$, then the respective quantities in each step $G_{1}$ and $G_{2}$ do not increase their distances to 1, i.e.,

[TABLE]

This is true even if the $G_{2}$ construction in step 2 is applied in the case when the roles of $a_{1}$ and $b_{1}$ are switched for the signature from step 1, when that switch is required ($b_{1}<a_{1}$) as described. More importantly, based on the properties of $G_{1}$ and $G_{2}$, we know that the (normalized) gap between $d$ and the previous largest entry $c$ shrinks quadratically fast, as measured by the new $c^{*}$ normalized with $d^{*}$. More precisely,

[TABLE]

Note that $c^{*}$ may no longer be the largest among $a^{*},b^{*},c^{*}$; however we will permute them to get $\widetilde{a},\widetilde{b},\widetilde{c}$ so that (3.3) is still satisfied before proceeding to the next round. This completes the description of our construction in one round which obtains $(\widetilde{a},\widetilde{b},\widetilde{c},\widetilde{d})$ from $(a,b,c,d)$.

We will construct the final signature $\check{f}$ by $O(\log n)$ rounds of this construction. Also we will follow each value $a,b,c$ individually as they get transformed through each round. To state it formally, starting with the normalized triple $(\frac{a}{d},\frac{b}{d},\frac{c}{d})$, we define a successor triple $(\frac{a^{*}}{d^{*}},\frac{b^{*}}{d^{*}},\frac{c^{*}}{d^{*}})$, so that each entry has the respective successor (e.g., the entry $\frac{a}{d}$ has successor $\frac{a^{*}}{d^{*}}$). This is well-defined because $(a_{1},b_{1},c_{1},d_{1})$ and $(a_{2},b_{2},c_{2},d_{2})$ are homogeneous functions of $(a,b,c,d)$. Note that even though from one round to the next, we may have to rename $a^{*},b^{*},c^{*}$ so that the permutated triple $\widetilde{a},\widetilde{b},\widetilde{c}$ satisfies (3.3), the successor sequence as the rounds progress stays with an individual value. E.g., starting from $(a,b,c,d)$, if after one round $a^{*}=\max(a^{*},b^{*},c^{*},d^{*})=\widetilde{c}$, then the successor of $\frac{a}{d}$ after two rounds is $\frac{(\widetilde{c})^{*}}{(d^{*})^{*}}$. Now define $(\alpha_{k},\beta_{k},\gamma_{k})$ to be the (ordered) triple $(\frac{a}{d},\frac{b}{d},\frac{c}{d})$, or its successor triple, at the beginning of the $k$-th round, for $k\geq 1$.

Let $\check{f}$ be the 4-ary signature constructed after $3(k+1)$ rounds. By the Pigeonhole Principle, after $3(k+1)$ rounds, at least one of $a,b,c$ has the property that in at least $k+1$ many rounds the corresponding $\frac{a}{d},\frac{b}{d},\frac{c}{d}$ or its successors are the maximum (normalized) value in that round, and thus its next successor gets shrunken quadratically in that round. Suppose this is $a$; the same proof works if it is $b$ or $c$. Let $\alpha_{i}$ be the maximum (normalized) value at the beginning of round $i$ in $k+1$ rounds, where $i\in\{i_{0},\ldots,i_{k}\}$, and $1\leq i_{0}<i_{1}<\ldots<i_{k}\leq 3(k+1)$. Since initially we have $1\leq d\leq a\leq b\leq c\leq\frac{3}{2}d$,

[TABLE]

Then

[TABLE]

By induction $0\leq\alpha_{i_{k}}-1\leq\frac{1}{2^{2^{k}}}$. At the end of $3(k+1)$ rounds, if $\check{f}$ has parameters $(\check{a},\check{b},\check{c},\check{d})$, then

[TABLE]

Therefore, after logarithmically many rounds, using polynomially many vertices, we can get a 4-ary construction with parameters $\check{a}$, $\check{b}$, $\check{c}$, and $\check{d}$ that are exponentially close to 1 after normalizing by $\check{d}$. Thus (3.2) is proved. ∎

4 #PerfectMatchings-easiness

In this section, we address two problems:

What are the constraint functions that can be realized by 4-ary matchgates (Definition 4.1)?

Although the set of constraint functions that can be realized by planar matchgates with complex edge weights have been completely characterized [CC17], the set of constraint functions that can be realized by general (not necessarily planar) matchgates with nonnegative real edge weights is not fully understood, even for matchgates of arity 4. This type of matchgates plays a crucial role in the study of the approximate complexity of counting problems, as we will see in this paper.

In Theorem 4.1, we give a complete characterization of constraint functions of arity 4 that can be realized by matchgates with nonnegative real edges. Our method is primarily geometric. 2. 2.

Theorem 1.1 shows that for positive parameters (a,b,c,d)\not\in\texttt{d-SUM} the problem #EightVertex($a,b,c,d$) is at least as hard as counting perfect matchings approximately. Here we ask the reverse question: For what parameter settings $(a,b,c,d)$ does #EightVertex($a,b,c,d$) $\leq_{\textup{AP}}$ #PerfectMatchings?

We know that

[TABLE]

where $f$ is the 4-ary constraint function with $M(f)=\left[\begin{smallmatrix}d&0&0&a\\ 0&b&c&0\\ 0&c&b&0\\ a&0&0&d\end{smallmatrix}\right]$. Considering the fact that $(\neq_{2})$ can be easily realized by a matchgate (a vertex with two dangling edges), Theorem 1.2 is a direct consequence of Lemma 4.2 which says that any constraint function in SQ-SUM is realizable by some 4-ary matchgate of constant size (with nonnegative edge weights, but not necessarily planar) (see Definition 4.1). Our theorem works for the eight-vertex model with parameter settings $\mathcal{S}^{\textup{E}}_{\leq^{2}}$ (defined below) not necessarily satisfying the arrow reversal symmetry.

Moreover, Lemma 4.4 indicates that our result is tight in the sense that $\mathcal{S}^{\textup{E}}_{\leq^{2}}$ captures precisely the set of all constraint functions that can be realized by 4-ary matchgates (with even support, i.e., nonzero only on inputs of even Hamming weight). A similar statement holds for $\mathcal{S}^{\textup{O}}_{\leq^{2}}$. the corresponding set with odd support.

Definition 4.1**.**

We use the term a $k$-ary matchgate to denote a graph $\Gamma$ having $k$ “dangling” edges, labelled $i_{1},\dots,i_{k}$. Each dangling edge has weight $1$ and each non-dangling edge $e$ is equipped with a nonnegative weight $w_{e}$. A configuration is a $0,1$-assignment to the edges. A configuration is a perfect matching if every vertex has exactly one incident edge assigned $1$. The matchgate implements the constraint function $f$, where $f(b_{1},\dots,b_{k})$ for $(b_{1},\dots,b_{k})\in\{0,1\}^{k}$ is the sum, over perfect matchings, of the product of the weight of edges with assignment $1$, where the dangling edge $i_{j}$ is assigned $b_{j}$, and the empty product has weight $1$.

Remark 4.1*.*

Contrary to Definition 4.1 which does not require planarity, planar matchgates with complex edge weights has been completely characterized [Val02, CC17]. As computing the weighted sum of perfect matchings is in polynomial time over planar graphs by the FKT algorithm [TF61, Kas61, Kas67], problems that can be locally expressed by planar matchgates are tractable over planar graphs.

Notation*.*

$\mathcal{S}^{\textup{E}}_{\leq^{2}}=\{f\ |\ M(f)=\left[\begin{smallmatrix}d_{1}&0&0&a_{1}\\ 0&b_{1}&c_{1}&0\\ 0&c_{2}&b_{2}&0\\ a_{2}&0&0&d_{2}\end{smallmatrix}\right]\text{ satisfying }\left\{\begin{smallmatrix}a_{1}a_{2}\ \leq\ b_{1}b_{2}+c_{1}c_{2}+d_{1}d_{2}\\ b_{1}b_{2}\ \leq\ a_{1}a_{2}+c_{1}c_{2}+d_{1}d_{2}\\ c_{1}c_{2}\ \leq\ a_{1}a_{2}+b_{1}b_{2}+d_{1}d_{2}\\ d_{1}d_{2}\ \leq\ a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}\end{smallmatrix}\right.,a_{1},a_{2},b_{1},b_{2},c_{1},c_{2},d_{1},d_{2}\geq 0.\}$, $\mathcal{S}^{\textup{O}}_{\leq^{2}}=\{f\ |\ M(f)=\left[\begin{smallmatrix}0&d_{1}&a_{1}&0\\ b_{1}&0&0&c_{1}\\ c_{2}&0&0&b_{2}\\ 0&a_{2}&d_{2}&0\end{smallmatrix}\right]\text{ satisfying }\left\{\begin{smallmatrix}a_{1}a_{2}\ \leq\ b_{1}b_{2}+c_{1}c_{2}+d_{1}d_{2}\\ b_{1}b_{2}\ \leq\ a_{1}a_{2}+c_{1}c_{2}+d_{1}d_{2}\\ c_{1}c_{2}\ \leq\ a_{1}a_{2}+b_{1}b_{2}+d_{1}d_{2}\\ d_{1}d_{2}\ \leq\ a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}\end{smallmatrix}\right.,a_{1},a_{2},b_{1},b_{2},c_{1},c_{2},d_{1},d_{2}\geq 0.\}$.

Theorem 4.1**.**

Denote by $\mathcal{M}$ the set of constraint functions that can be realized by 4-ary matchgates. Then $\mathcal{M}=\mathcal{S}^{\textup{E}}_{\leq^{2}}\bigcup\mathcal{S}^{\textup{O}}_{\leq^{2}}$.

Remark 4.2*.*

Note that any constraint function in $\mathcal{M}$ must satisfy either even parity (nonzero only on inputs of even Hamming weight) or odd parity (nonzero only on inputs of odd Hamming weight). Theorem 4.1 for the even parity part ($\mathcal{S}^{\textup{E}}_{\leq^{2}}$) is a combination of Lemma 4.2 and Lemma 4.4. The odd parity part can be proved similarly.

Lemma 4.2**.**

Suppose $f\in\mathcal{S}^{\textup{E}}_{\leq^{2}}$. Then there is a 4-ary matchgate of constant size whose constraint function is $f$.

Proof.

We first note that if any of the four inequalities in the definition of $\mathcal{S}_{\leq^{2}}$ is an equality, then the remaining three inequalities automatically hold, since the 8 values $a_{1},\ldots,d_{2}$ are all nonnegative.

Given a constraint function $\left[\begin{smallmatrix}d_{1}&0&0&a_{1}\\ 0&b_{1}&c_{1}&0\\ 0&c_{2}&b_{2}&0\\ a_{2}&0&0&d_{2}\end{smallmatrix}\right]$, first we construct a matchgate for $d_{1}d_{2}=a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}$. If $d_{1}d_{2}=0$ then all four products $a_{1}a_{2}=b_{2}b_{2}=c_{1}c_{2}=d_{1}d_{2}=0$, and one can easily adapt from the following proof to show that the signature is realizable as a matchgate signature. So it suffices to implement the normalized version $\left[\begin{smallmatrix}a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}&0&0&a_{1}\\ 0&b_{1}&c_{1}&0\\ 0&c_{2}&b_{2}&0\\ a_{2}&0&0&1\end{smallmatrix}\right]$. Our construction is a weighted $K_{4}$ depicted in Figure 4a. Let $e_{1},e_{2},e_{3},e_{4}$ be the dangling edges incident to vertices $1,2,3,4$, respectively. Denote by $w_{ij}$ the weight on the edge between vertex $i$ and vertex $j$. One can check that the following weight assignment meets our need: $w_{12}=a_{1},w_{34}=a_{2},w_{14}=b_{1},w_{23}=b_{2},w_{13}=c_{1},w_{24}=c_{2}$.

For $a_{1}a_{2}=b_{1}b_{2}+c_{1}c_{2}+d_{1}d_{2}$, without loss of generality we assume $a_{1}a_{2}\not=0$ and we normalize $a_{1}=1$. Then our construction is shown in Figure 4b where we set $w_{11^{\prime}}=1,w_{22^{\prime}}=1,w_{1^{\prime}2^{\prime}}=d_{2},w_{34}=d_{1},w_{1^{\prime}4}=c_{2},w_{2^{\prime}3}=c_{1},w_{1^{\prime}3}=b_{2},w_{2^{\prime}4}=b_{1}$. One can verify that it realizes the normalized constraint function $\left[\begin{smallmatrix}d_{1}&0&0&1\\ 0&b_{1}&c_{1}&0\\ 0&c_{2}&b_{2}&0\\ b_{1}b_{2}+c_{1}c_{2}+d_{1}d_{2}&0&0&d_{2}\end{smallmatrix}\right]$. The construction for $b_{1}b_{2}=a_{1}a_{2}+c_{1}c_{2}+d_{1}d_{2}$ and $c_{1}c_{2}=a_{1}a_{2}+b_{1}b_{2}+d_{1}d_{2}$ are symmetric to the above case.

It remains to show that the interior

[TABLE]

can all be reached. We first deal with the case when all eight parameters are strictly positive and leave the other cases to the end of this proof. We use a weighted $K_{6}$ to be our matchgate depicted in Figure 4c, and set $w_{12}=r_{1},w_{34}=r_{2},w_{14}=s_{1},w_{23}=s_{2},w_{13}=t_{1},w_{24}=t_{2},w_{15}=p_{1},w_{25}=p_{2},w_{35}=p_{3},w_{45}=p_{4},w_{16}=q_{1},w_{26}=q_{2},w_{36}=q_{3},w_{46}=q_{4},w_{56}=1$. Then the matchgate has a singature with the following parameters

[TABLE]

Note that all the edge weights have to be nonnegative. By properly setting the edge weights in the matchgate, we show that we can achieve any relative ratios among the eight given positive values $a_{1},a_{2},b_{1},b_{2},c_{1},c_{2},d_{1},d_{2}$ that satisfy (4.1). Our first step is to achieve any relative ratios among the four product values $a_{1}a_{2}$, $b_{1}b_{2}$, $c_{1}c_{2}$, $d_{1}d_{2}$ satisfying (4.1); and the second step is to adjust the relative ratio within the pairs $\{a_{1},a_{2}\}$, $\{b_{1},b_{2}\}$, $\{d_{1},d_{2}\}$ and $\{c_{1},c_{2}\}$ without affecting the product values. This can be justified by the observation that, by a scaling a global positive constant can be easily achieved, and all appearances of $a_{1}$ and $a_{2}$ in (4.1) are as a product $a_{1}a_{2}$, and similarly for $b_{1},b_{2},c_{1},c_{2},d_{1},d_{2}$.

For the fourteen edge weights $r_{1},\ldots,q_{4}$ to be determined, let

[TABLE]

and define

[TABLE]

Note that $A^{\prime},B^{\prime},C^{\prime},R,S,T$ are all nonnegative and so are $A,B,C,D$.

Our goal is to choose the fourteen edge weights $r_{1},\ldots,q_{4}$ so that $A,B,C,D$ are all positive and satisfy

[TABLE]

Note that, by definition, the left-side of (4.4) is precisely the right-side of (4.4) when $a_{1},\ldots,d_{2}$ are replaced by $a^{\prime}_{1},\ldots,d^{\prime}_{2}$ respectively. Denote the products $a_{1}a_{2},b_{1}b_{2},c_{1}c_{2},d_{1}d_{2}$ by $a^{**},b^{**},c^{**},d^{**}$ respectively. Then (4.4) is a set of four linear equations $M\cdot\left[\begin{smallmatrix}a^{**}\\ b^{**}\\ c^{**}\\ d^{**}\end{smallmatrix}\right]=\left[\begin{smallmatrix}A\\ B\\ C\\ D\end{smallmatrix}\right]$, where $M=\frac{1}{2}\left[\begin{smallmatrix}-1&1&1&1\\ 1&-1&1&1\\ 1&1&-1&1\\ 1&1&1&-1\end{smallmatrix}\right]$. Note that $M$ is invertible and $M^{-1}=M$, so (4.4) is equivalent to $M\cdot\left[\begin{smallmatrix}A\\ B\\ C\\ D\end{smallmatrix}\right]=\left[\begin{smallmatrix}a^{**}\\ b^{**}\\ c^{**}\\ d^{**}\end{smallmatrix}\right]$, having an identical form. Since the requirement (4.1) in terms of $a^{**},b^{**},c^{**},d^{**}$ translates into the requirement $A,B,C,D$ being strictly positive via $M$, it is not surprising that the requirement $a^{**},b^{**},c^{**},d^{**}$ being strictly positive translates into the requirement

[TABLE]

and that $A,B,C,D$ are positive is the same as (4.1).

Furthermore, let $\left\{\begin{smallmatrix}X=S+T\\ Y=R+T\\ Z=R+S\end{smallmatrix}\right.$, then the requirement $R,S,T$ being positive is equivalent to the requirement $\left\{\begin{smallmatrix}Y+Z>X\\ X+Z>Y\\ X+Y>Z\end{smallmatrix}\right.$. This is because $\left[\begin{smallmatrix}0&1&1\\ 1&0&1\\ 1&1&0\end{smallmatrix}\right]\cdot\left[\begin{smallmatrix}R\\ S\\ T\end{smallmatrix}\right]=\left[\begin{smallmatrix}X\\ Y\\ Z\end{smallmatrix}\right]$ is the same as $\frac{1}{2}\left[\begin{smallmatrix}-1&1&1\\ 1&-1&1\\ 1&1&-1\end{smallmatrix}\right]\cdot\left[\begin{smallmatrix}X\\ Y\\ Z\end{smallmatrix}\right]=\left[\begin{smallmatrix}R\\ S\\ T\end{smallmatrix}\right]$.

The crucial ingredient of our proof is a geometric lemma in 3-dimensional space. Suppose $a^{**},b^{**},c^{**},d^{**}$ are positive and they satisfy (4.1). This defines $\left[\begin{smallmatrix}\tilde{A}\\ \tilde{B}\\ \tilde{C}\\ \tilde{D}\end{smallmatrix}\right]=M\cdot\left[\begin{smallmatrix}a^{**}\\ b^{**}\\ c^{**}\\ d^{**}\end{smallmatrix}\right]$. By a scaling we may assume $\tilde{D}=1$. Hence $(\tilde{A},\tilde{B},\tilde{C},\tilde{D})$ are positive and satisfy (4.5). Thus $(\tilde{A},\tilde{B},\tilde{C})$ belongs to the set $U$ in the statement of Lemma 4.3.

By Lemma 4.3, there exist (strictly) positive tuples $(\tilde{A}^{\prime},\tilde{B}^{\prime},\tilde{C}^{\prime})$ and $(\tilde{X},\tilde{Y},\tilde{Z})$ such that

[TABLE]

satisfying $\tilde{A}^{\prime}+\tilde{B}^{\prime}+\tilde{C}^{\prime}=1$ and $\left\{\begin{smallmatrix}\tilde{Y}+\tilde{Z}>\tilde{X}\\ \tilde{X}+\tilde{Z}>\tilde{Y}\\ \tilde{X}+\tilde{Y}>\tilde{Z}\end{smallmatrix}\right.$. By the previous observation this indicates that there exist (strictly) positive $\tilde{A}^{\prime},\tilde{B}^{\prime},\tilde{C}^{\prime},\tilde{R},\tilde{S},\tilde{T}$ such that $\left\{\begin{smallmatrix}\tilde{A}=\tilde{A}^{\prime}+\tilde{S}+\tilde{T}\\ \tilde{B}=\tilde{B}^{\prime}+\tilde{R}+\tilde{T}\\ \tilde{C}=\tilde{C}^{\prime}+\tilde{R}+\tilde{S}\\ \tilde{D}=\tilde{A}^{\prime}+\tilde{B}^{\prime}+\tilde{C}^{\prime}\end{smallmatrix}\right.$.

We first set $p_{i},q_{i}\ (1\leq i\leq 4)$ so that $(A^{\prime},B^{\prime},C^{\prime})=c\cdot(\tilde{A}^{\prime},\tilde{B}^{\prime},\tilde{C}^{\prime})$ for some constant $c$. To achieve this, set $q_{1}=q_{2}=q_{3}=q_{4}=1$, and let $o_{1},o_{2},o_{3}$ be positive, and then set $p_{1}=\sqrt{\frac{o_{2}o_{3}}{o_{1}}}$, $p_{2}=\sqrt{\frac{o_{3}o_{1}}{o_{2}}}$, $p_{3}=\sqrt{\frac{o_{1}o_{2}}{o_{3}}}$, and $p_{4}=\frac{1}{p_{1}p_{2}p_{3}}$. We have $p_{1}p_{2}p_{3}p_{4}=1$, and $p_{2}p_{3}=o_{1},p_{3}p_{1}=o_{2},p_{1}p_{2}=o_{3}$. Then set $\left\{\begin{smallmatrix}A^{\prime}=p_{1}p_{2}+\frac{1}{p_{1}p_{2}}=o_{3}+\frac{1}{o_{3}}\\ B^{\prime}=p_{2}p_{3}+\frac{1}{p_{2}p_{3}}=o_{1}+\frac{1}{o_{1}}\\ C^{\prime}=p_{3}p_{1}+\frac{1}{p_{3}p_{1}}=o_{2}+\frac{1}{o_{2}}\end{smallmatrix}\right.$, which can be independently any positive numbers at least 2, by choosing $o_{1},o_{2},o_{3}$ to be suitable positive numbers. This allows us to get $A^{\prime},B^{\prime},C^{\prime}$ such that $(A^{\prime},B^{\prime},C^{\prime})=c\cdot(\tilde{A}^{\prime},\tilde{B}^{\prime},\tilde{C}^{\prime})$ for some constant $c$. Then it is obvious that $r_{1},r_{2},s_{1},s_{2},t_{1},t_{2}$ can be set so that $(R,S,T)=c\cdot(\tilde{R},\tilde{S},\tilde{T})$. Compute $A,B,C,D$ according to (4.3). As a consequence, $(A,B,C,D)=c\cdot(\tilde{A},\tilde{B},\tilde{C},\tilde{D})$ is a valid solution.

To adjust the relative ratio between $\{d_{1},d_{2}\}$, say increasing $\frac{d_{2}}{d_{1}}$ by $\delta$, while keeping all product values and the relative ratios within the other three pairs, just increase $r_{1},r_{2},s_{1},s_{2},t_{1},t_{2}$ by $\delta^{1/2}$ and increase $p_{i},q_{i}\ (1\leq i\leq 4)$ by $\delta^{1/4}$. Similarly, to increase $\frac{a_{2}}{a_{1}}$ by $\delta$ alone without affecting the other products and ratios, just increase $r_{2}$ by $\delta^{1/2}$ and $p_{3},p_{4},q_{3},q_{4}$ by $\delta^{1/4}$, and decrease $r_{1}$ by $\delta^{1/2}$ and $p_{1},p_{2},q_{1},q_{2}$ by $\delta^{1/4}$. The other cases are symmetric.

Finally we deal with the cases when there are zeros among the eight parameters. Note that at most one of $a_{1}a_{2},b_{1}b_{2},c_{1}c_{2},d_{1}d_{2}$ is zero, because if at least two products are zero, say $a_{1}a_{2}=b_{1}b_{2}=0$, then (4.1) forces a contradiction that $c_{1}c_{2}<d_{1}d_{2}$ and $d_{1}d_{2}<c_{1}c_{2}$. In the case $d_{1}d_{2}=0$:

• •

$d_{1}=0,d_{2}\neq 0$: We make the modification that $w_{12}=w_{34}=w_{14}=w_{23}=w_{13}=w_{24}=0$, i.e. $r_{1},r_{2},s_{1},s_{2},t_{1},t_{2}=0$.

• •

$d_{1}=d_{2}=0$: We make the further modification that $w_{56}=0$.

• •

$d_{1}\neq 0,d_{2}=0$: We connect the four dangling edges in Figure 4c to four degree 2 vertices, respectively. This switches the role of $d_{1}$ and $d_{2}$ in the previous proof.

One can check our proof is still valid in the above three cases. If $a_{1}a_{2}=0$, then we connect the dangling edges on vertices 1, 2 to two degree 2 vertices (similar to the operation from Figure 4a to Figure 4b). This switches the role of $d_{1},d_{2}$ with $a_{2},a_{1}$ and the proof folllows. The proofs for $b_{1}b_{2}=0$ and $c_{1}c_{2}=0$ are symmetric. ∎

Now we give the crucial geometric lemma.

Lemma 4.3**.**

Let $U=\{(x,y,z)\in\mathbb{R}^{3}_{>0}\ |\ x<y+z+1,y<x+z+1,z<x+y+1,1<x+y+z\}$, $V=\{(x,y,z)\in\mathbb{R}^{3}_{>0}\ |\ x+y+z=1\}$, and $W=\{(x,y,z)\in\mathbb{R}^{3}_{>0}\ |\ y+z>x,x+z>y,x+y>z\}$. Then $U$ is the Minkowski sum of $V$ and $W$, namely, $U$ consists of precisely those points $\boldsymbol{u}\in{\mathbb{R}}^{3}$, such that $\boldsymbol{u}=\boldsymbol{v}+\boldsymbol{w}$ for some $\boldsymbol{v}\in V$ and $\boldsymbol{w}\in W$. The same statement is true for the closures of $U$, $V$ and $W$.

Proof.

Observe that $U$, $V$, and $W$ are the interiors of a polyhedron with 7 facets, a regular triangle, and a polyhedron with 3 facets, respectively.

The polyhedron for $W$ is the intersection of three half spaces bounded by three planes, $\left\{\begin{smallmatrix}(\pi_{1}):y+z\geq x\\ (\pi_{2}):x+z\geq y\\ (\pi_{3}):x+y\geq z\end{smallmatrix}\right.$, where the planes are defined by equalities. Note that these inequalities imply that $x,y,z\geq 0$, thus this polyhedron has only three facets. We can find the intersection of each pair of the three planes for $W$ as $\left\{\begin{smallmatrix}\pi_{1}\cap\pi_{2}:x=y\geq 0,z=0\\ \pi_{1}\cap\pi_{3}:x=z\geq 0,y=0\\ \pi_{2}\cap\pi_{3}:y=z\geq 0,x=0\end{smallmatrix}\right.$. Note that these intersections lie on the planes $z=0$, $y=0$, and $x=0$, respectively.

Let $\boldsymbol{e}_{1}=(1,0,0),\boldsymbol{e}_{2}=(0,1,0),\boldsymbol{e}_{3}=(0,0,1)$. Then the triangle for $V$ is just the convex hull of $\boldsymbol{e}_{1},\boldsymbol{e}_{2},\boldsymbol{e}_{3}$. Suppose we shift the origin of $W$ from $\boldsymbol{0}$ to $\boldsymbol{e}_{1}$ and denote the resulting (interior of a) polyhedron by $W^{\boldsymbol{e}_{1}}$, then we have the defining inequalities $\left\{\begin{smallmatrix}(\pi_{1}^{\boldsymbol{e}_{1}}):y+z\geq(x-1)\\ (\pi_{2}^{\boldsymbol{e}_{1}}):(x-1)+z\geq y\\ (\pi_{3}^{\boldsymbol{e}_{1}}):(x-1)+y\geq z\end{smallmatrix}\right.$, where the shifted planes are defined by the corresponding equalities. By symmetry, if we shift the origin of $W$ to $\boldsymbol{e}_{2}$ and to $\boldsymbol{e}_{3}$, we have respectively $W^{\boldsymbol{e}_{2}}$ with $\left\{\begin{smallmatrix}(\pi_{1}^{\boldsymbol{e}_{2}}):(y-1)+z\geq x\\ (\pi_{2}^{\boldsymbol{e}_{2}}):x+z\geq(y-1)\\ (\pi_{3}^{\boldsymbol{e}_{2}}):x+(y-1)\geq z\end{smallmatrix}\right.$, and $W^{\boldsymbol{e}_{3}}$ with $\left\{\begin{smallmatrix}(\pi_{1}^{\boldsymbol{e}_{3}}):y+(z-1)\geq x\\ (\pi_{2}^{\boldsymbol{e}_{3}}):x+(z-1)\geq y\\ (\pi_{3}^{\boldsymbol{e}_{3}}):x+y\geq(z-1)\end{smallmatrix}\right.$. Note that the shifted planes $\pi_{1}^{\boldsymbol{e}_{1}}$, $\pi_{2}^{\boldsymbol{e}_{2}}$, and $\pi_{3}^{\boldsymbol{e}_{3}}$ contain three distinct facets of $U$, and they coincide exactly with a facet of $W^{\boldsymbol{e}_{1}}$, $W^{\boldsymbol{e}_{2}}$, and $W^{\boldsymbol{e}_{3}}$, respectively.

By sliding $W$ with its origin along the line $x+y=1,z=0$ from $\boldsymbol{e}_{1}$ to $\boldsymbol{e}_{2}$, we have a partial coverage of $U$ by the shifting copies of $W$ from $W^{\boldsymbol{e}_{1}}$ and $W^{\boldsymbol{e}_{2}}$:

• •

The shifted ray of $\pi_{1}\cap\pi_{2}:x=y\geq 0,z=0$ moves from $\pi_{1}^{\boldsymbol{e}_{1}}\cap\pi_{2}^{\boldsymbol{e}_{1}}:(x-1)=y\geq 0,z=0$ to $\pi_{1}^{\boldsymbol{e}_{2}}\cap\pi_{2}^{\boldsymbol{e}_{2}}:x=(y-1)\geq 0,z=0$. Notice that this is a parallel transport, and stays on the plane $z=0$, and thus it swipes another facet of $U$ on $z=0$ bounded by the two lines $x-y=-1$ and $x-y=1$.

• •

The shifted ray of $\pi_{1}\cap\pi_{3}:x=z\geq 0,y=0$ moves from $\pi_{1}^{\boldsymbol{e}_{1}}\cap\pi_{3}^{\boldsymbol{e}_{1}}:(x-1)=z\geq 0,y=0$ to $\pi_{1}^{\boldsymbol{e}_{2}}\cap\pi_{3}^{\boldsymbol{e}_{2}}:x=z\geq 0,(y-1)=0$; the shifted ray of $\pi_{2}\cap\pi_{3}:y=z\geq 0,x=0$ moves from $\pi_{2}^{\boldsymbol{e}_{1}}\cap\pi_{3}^{\boldsymbol{e}_{1}}:y=z\geq 0,(x-1)=0$ to $\pi_{2}^{\boldsymbol{e}_{2}}\cap\pi_{3}^{\boldsymbol{e}_{2}}:(y-1)=z\geq 0,x=0$. Notice that both stay on the plane $x+y-z=1$ which is $\pi_{3}^{\boldsymbol{e}_{1}}=\pi_{3}^{\boldsymbol{e}_{2}}$.

It follows that the part of $U$ satisfying $x+y-z>1$ is covered by the Minkowski sum of $W$ and the line segment on $x+y=1,z=0$ from $\boldsymbol{e}_{1}$ to $\boldsymbol{e}_{2}$ (which is a side of the triangle $V$).

Symmetrically, after sliding $W$ with its origin from $\boldsymbol{e}_{2}$ to $\boldsymbol{e}_{3}$ along the line $y+z=1,x=0$ we get the parallel tranport from $W^{\boldsymbol{e}_{2}}$ to $W^{\boldsymbol{e}_{3}}$. Also after sliding $W$ with its origin from $\boldsymbol{e}_{3}$ back to $\boldsymbol{e}_{1}$ along the line segment $x+z=1,y=0$ we get the parallel tranport from $W^{\boldsymbol{e}_{3}}$ to $W^{\boldsymbol{e}_{1}}$. After these, the only subset in $U$ that is left uncovered by shifting copies of $W$ is $U\cap\{(x,y,z)\ |\left\{\begin{smallmatrix}-x+y+z\leq 1\\ x-y+z\leq 1\\ x+y-z\leq 1\end{smallmatrix}\right.\}=\{(x,y,z)\in\mathbb{R}^{3}_{>0}\ |\left\{\begin{smallmatrix}x+y+z\geq 1\\ -x+y+z\leq 1\\ x-y+z\leq 1\\ x+y-z\leq 1\end{smallmatrix}\right.\}$ — a tetrahedron (Figure 5b). However this subset can be covered by the rays $\{\boldsymbol{v}+\lambda(1,1,1)\ |\ \boldsymbol{v}\in V,\lambda>0\}$. Note that $\lambda(1,1,1)\in W$ for all $\lambda>0$. This completes the proof. ∎

Lemma 4.4**.**

Suppose $f$ is the constraint function of a 4-ary matchgate with $M(f)=\left[\begin{smallmatrix}d_{1}&0&0&a_{1}\\ 0&b_{1}&c_{1}&0\\ 0&c_{2}&b_{2}&0\\ a_{2}&0&0&d_{2}\end{smallmatrix}\right]$. Then $f\in\mathcal{S}^{\textup{E}}_{\leq^{2}}$. In particular, if $f$ satisfies arrow reversal symmetry, $f\in\textup{{SQ-SUM}}$.

Remark 4.3*.*

The last part $d_{1}d_{2}\ \leq\ a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}$ was proved in [BGJ*+*17, Lemma 56]. The proofs for other three parts are symmetric and similar to the proof for the last part. For completeness, here we give the proof for the first part $a_{1}a_{2}\ \leq\ b_{1}b_{2}+c_{1}c_{2}+d_{1}d_{2}$.

Proof.

Consider a 4-ary matchgate $\Gamma$ with constraint function $f$. Given that $M(f)=\left[\begin{smallmatrix}d_{1}&0&0&a_{1}\\ 0&b_{1}&c_{1}&0\\ 0&c_{2}&b_{2}&0\\ a_{2}&0&0&d_{2}\end{smallmatrix}\right]$, $a_{1}a_{2}\ \leq\ b_{1}b_{2}+c_{1}c_{2}+d_{1}d_{2}$ is equivalent as

[TABLE]

Let $I=\{i_{1},i_{2},i_{3},i_{4}\}$ be the set of dangling edges of $\Gamma$. For $X\subseteq I$, let $M_{X}$ denote the set of perfect matchings that include dangling edges in $X$ (by assigning them $1$) and exclude dangling edges in $I\setminus X$ (by assigning them [math]). We exhibit an injective map

[TABLE]

which is weight-preserving in the sense that for matchings $m_{1},m_{2},m_{3},m_{4}$ with $\mu(m_{1},m_{2})=(m_{3},m_{4})$, we have $w(m_{1})w(m_{2})=w(m_{3})w(m_{4})$. The existence of $\mu$ implies (4.6).

Given $(m_{1},m_{2})\in M_{\{i_{1},i_{2}\}}\times M_{\{i_{3},i_{4}\}}$, consider $m_{1}\oplus m_{2}$ and note that this is a collection of cycles together with two paths. Let $\pi$ be the path connecting the dangling edge $i_{1}$ to some other dangling edge; let $\pi^{\prime}$ be the path connecting the remaining two dangling edges. Let $m_{3}:=m_{1}\oplus\pi$ and $m_{4}:=m_{2}\oplus\pi$. Then we have the following

• •

If $\pi$ connects $i_{1}$ to $i_{2}$, then $m_{3}\in M_{\emptyset}$ and $m_{4}\in M_{I}$;

• •

If $\pi$ connects $i_{1}$ to $i_{3}$, then $m_{3}\in M_{\{i_{2},i_{3}\}}$ and $m_{4}\in M_{\{i_{1},i_{4}\}}$;

• •

If $\pi$ connects $i_{1}$ to $i_{4}$, then $m_{3}\in M_{\{i_{2},i_{4}\}}$ and $m_{4}\in M_{\{i_{1},i_{3}\}}$.

The construction is invertible, since if $(m_{3},m_{4})$ is in the image of the above mapping, then $m_{3}\oplus m_{4}=m_{1}\oplus m_{2}$. From $m_{1}\oplus m_{2}$, we can recover $\pi$ (as the unique path that connects $i_{1}$ to one of the other dangling edges in $\{i_{2},i_{3},i_{4}\}$). Then we can recover $m_{1}$ and $m_{2}$ as $m_{3}\oplus\pi$ and $m_{4}\oplus\pi$ respectively. Therefore, $\mu:(m_{1},m_{2})\rightarrow(m_{3},m_{4})$ is an injection.

To see that $\mu$ is weight-preserving, observe that the each of the edges in $\pi$ appears in exactly one of $m_{1}$ and $m_{2}$ and in exactly one of $m_{3}$ and $m_{4}$ and that $m_{i}\setminus\pi=m_{i+2}\setminus\pi$ for $i\in\{1,2\}$. Hence,

[TABLE]

∎

Appendix

Appendix A

Appendix B

Given a constraint function $f$ with $M(f)=\left[\begin{smallmatrix}d&0&0&a\\ 0&b&c&0\\ 0&c&b&0\\ a&0&0&d\end{smallmatrix}\right]$ and $a,b,c\geq 0,d>0$, we show how to obtain a 4-ary construction with constraint function $\hat{f}$ such that $M(\hat{f})=\left[\begin{smallmatrix}\hat{d}&0&0&\hat{a}\\ 0&\hat{b}&\hat{c}&0\\ 0&\hat{c}&\hat{b}&0\\ \hat{a}&0&0&\hat{d}\end{smallmatrix}\right]$ and $\hat{a},\hat{b},\hat{c},\hat{d}$ satisfy the condition $0<\hat{d}\leq\hat{a}\leq\hat{b}\leq\hat{c}\leq\frac{3}{2}\hat{d}$, using copies of the constraint function $f$. (When used in the context of Lemma 3.5, the number of copies of $f$ used is a constant.) By a scaling we get

[TABLE]

We first deal with the case where one of $a,b,c$ is equal to 0. By relabeling we may assume $b=0$. Then we use the construction $G_{1}$ in the proof of Lemma 3.5 to get a signature with parameters $(a_{1},b_{1},c_{1},d_{1})=(ac,d^{2},a^{2},cd)$ with no zero entries.

In the following we may assume all entries are positive $a,b,c,d>0$.

Our first step is to satisfy (3.3). If all four entries are equal $a=b=c=d$ then we are done (the signature is also in the affine family and thus the problem is also P-time tractable.) So we may assume at least one of $a,b,c\neq d$. By relabeling we may assume $a\neq d$. We may also assume at least one of the following three displayed inequalities does not hold, and by relabeling we may assume $a+d<b+c$. Indeed, if all three inequalities

[TABLE]

hold, then by connecting two copies of $f$ using the 4-ary construction in Figure 3 and relabeling edges, we get $(a^{\prime},b^{\prime},c^{\prime},d^{\prime})=(2bc,a^{2}+d^{2},b^{2}+c^{2},2ad)$. This signature has $a^{\prime}+d^{\prime}<b^{\prime}+c^{\prime}$.

Under a holographic transformation (in the proof of Lemma 3.1), in the edge-2-coloring view $\left[\begin{smallmatrix}d&0&0&a\\ 0&b&c&0\\ 0&c&b&0\\ a&0&0&d\end{smallmatrix}\right]$ becomes $\frac{1}{2}\left[\begin{smallmatrix}a+b+c+d&0&0&-a+b+c-d\\ 0&a-b+c-d&a+b-c-d&0\\ 0&a+b-c-d&a-b+c-d&0\\ -a+b+c-d&0&0&a+b+c+d\end{smallmatrix}\right]=P\cdot\left[\begin{smallmatrix}a+d&0&0&0\\ 0&c-b&0&0\\ 0&0&a-d&0\\ 0&0&0&c+b\end{smallmatrix}\right]\cdot P^{-1}$, where $P=\left[\begin{smallmatrix}1&0&0&1\\ 0&1&1&0\\ 0&-1&1&0\\ -1&0&0&1\end{smallmatrix}\right]$. Connect $k$ copies of $f$ using the 4-ary construction in Figure 3, we get the constraint function $f^{\prime}$ with $M(f^{\prime})=\left[\begin{smallmatrix}d^{\prime}&0&0&a^{\prime}\\ 0&b^{\prime}&c^{\prime}&0\\ 0&c^{\prime}&b^{\prime}&0\\ a^{\prime}&0&0&d^{\prime}\end{smallmatrix}\right]$ which in the edge-2-coloring view becomes $P\left[\begin{smallmatrix}a^{\prime}+d^{\prime}&0&0&0\\ 0&c^{\prime}-b^{\prime}&0&0\\ 0&0&a^{\prime}-d^{\prime}&0\\ 0&0&0&c^{\prime}+b^{\prime}\end{smallmatrix}\right]P^{-1}$. By construction, it is also equal to

[TABLE]

It follows that

[TABLE]

Since $b+c$ is strictly larger than $a+d$, and also $b+c=|b+c|>|c-b|$ and $|a-d|$, one can see that for a large $k$, we can get $f^{\prime}$ such that $b^{\prime}\approx c^{\prime}\gg a^{\prime},d^{\prime}$, and all entries are positive. If furthermore $d^{\prime}\leq a^{\prime}$, then we have arrived at a signature $f^{\prime}$ satisfying (3.3) with a possible relabeling. If $d^{\prime}>a^{\prime}$, using the construction in Figure 3 with two copies of $f^{\prime}$ (and two $(\neq_{2})$ in the middle as we are in the orientation view), we can get $f^{\prime\prime}$ with $M(f^{\prime\prime})=M(f^{\prime})\cdot N\cdot M(f^{\prime})=\left[\begin{smallmatrix}2a^{\prime}d^{\prime}&0&0&{a^{\prime}}^{2}+{d^{\prime}}^{2}\\ 0&2b^{\prime}c^{\prime}&{b^{\prime}}^{2}+{c^{\prime}}^{2}&0\\ 0&{b^{\prime}}^{2}+{c^{\prime}}^{2}&2b^{\prime}c^{\prime}&0\\ {a^{\prime}}^{2}+{d^{\prime}}^{2}&0&0&2a^{\prime}d^{\prime}\end{smallmatrix}\right]$. The four parameters for $f^{\prime\prime}$ is $(a^{\prime\prime},b^{\prime\prime},c^{\prime\prime},d^{\prime\prime})=({a^{\prime}}^{2}+{d^{\prime}}^{2},2b^{\prime}c^{\prime},{b^{\prime}}^{2}+{c^{\prime}}^{2},2a^{\prime}d^{\prime})$ which satisfies (3.3) (with a possible relabeling) when $b^{\prime},c^{\prime}\gg a^{\prime},d^{\prime}$.

Now that we have a constraint function $f^{\prime}$ (or $f^{\prime\prime}$ if the last step is needed) satisfying (3.3), we obtain $\hat{f}$ using 4-ary constructions with $f^{\prime}$ (or $f^{\prime\prime}$ respectively). This is done by repeatedly applying $G_{2}$ (defined in Lemma 3.5) in which each round starts with the function constructed from the previous round, and possibily relabeling the input values $(a,b,c,d)$, so that the construction $G_{2}$ starts with some $(a,b,c,d)$ satisfying (3.3). Note that for any $(a,b,c,d)$ satisfying (3.3) and $c>\frac{3}{2}d$, $G_{2}$ shrinks the distance between $c$ and $d$ by at least a constant (after normalization). This can be checked by $\frac{c}{d}-\frac{c_{2}}{d_{2}}=\frac{\left(\frac{c}{d}+1\right)\left(\frac{c}{d}-1\right)}{2\cdot\frac{c}{d}}>\frac{5}{12}$ when $c>\frac{3}{2}d$. Therefore, starting with $f^{\prime}$ and iterative applying $G_{2}$ (constantly many times), we can obtain $\hat{f}$ so that $1\leq\hat{d}\leq\hat{a}\leq\hat{b}\leq\hat{c}\leq\frac{3}{2}\hat{d}$.