Beyond #CSP: A Dichotomy for Counting Weighted Eulerian Orientations with ARS

TL;DR

This paper classifies the computational complexity of counting weighted Eulerian orientations with arrow reversal symmetry, establishing a dichotomy that they are either polynomial-time solvable or #P-hard, and reveals a deep connection with weighted #CSP problems.

Contribution

It introduces a new prime factorization framework for constraint functions and proves a complexity dichotomy for these orientation problems, linking them to weighted #CSPs through structural analysis.

Findings

All such problems are either polynomial-time computable or #P-hard.

Weighted Eulerian orientation problems encompass all weighted Boolean #CSPs.

A novel structural connection between #CSP and Eulerian orientations is established.

Abstract

We define and explore a notion of unique prime factorization for constraint functions, and use this as a new tool to prove a complexity classification for counting weighted Eulerian orientation problems with arrow reversal symmetry (ARS). We prove that all such problems are either polynomial-time computable or #P-hard. We show that the class of weighted Eulerian orientation problems subsumes all weighted counting constraint satisfaction problems (#CSP) on Boolean variables. More significantly, we establish a novel connection between #CSP and counting weighted Eulerian orientation problems that is global in nature. This connection is based on a structural determination of all half-weighted affine linear subspaces over $\mathbb{Z}_2$, which is proved using M\"obius inversion.

Equations

$${\#EO}_{\Omega }=\sum _{\sigma \in \mathrm{EO}(G)}\prod _{v\in V}{f}_v(\sigma {|}_{E_{(v)}}).$$

$${\mathrm{Holant}}_{\Omega }=\sum _{\sigma :E(G)\to \{0,1\}}\prod _{v\in V(G)}{f}_v(\sigma {|}_{E_{(v)}}).$$

$$N=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$$

$$M_{(ij)}(f)=\left[\begin{array}{cccc}{f}^{00,00\dots 0}& f^{00,00\dots 1}& \dots & f^{00,11\dots 1}\\ f^{01,00\dots 0}& f^{01,00\dots 1}& \dots & f^{01,11\dots 1}\\ f^{10,00\dots 0}& f^{10,00\dots 1}& \dots & f^{10,11\dots 1}\\ f^{11,00\dots 0}& f^{11,00\dots 1}& \dots & f^{11,11\dots 1}\end{array}\right]=\left[\begin{array}{c}{\mathbf{f}}_{ij}^{00}\\ {\mathbf{f}}_{ij}^{01}\\ {\mathbf{f}}_{ij}^{10}\\ {\mathbf{f}}_{ij}^{11}\end{array}\right].$$

$$\begin{gathered} {\mathbf{f}}_{ij}^{ab}{N}^{\otimes (n-2)}=(f^{ab,11\dots 1},f^{ab,11\dots 0},\dots ,f^{ab,00\dots 0}) \\ =(\overline{f^{\bar{a}\bar{b},00\dots 0}},\overline{f^{\bar{a}\bar{b},00\dots 1}},\dots ,\overline{f^{\bar{a}\bar{b},11\dots 1}})=\overline{{\mathbf{f}}_{ij}^{\bar{a}\bar{b}}}. \end{gathered}$$

$$\#EO(\mathcal{F},f){\equiv }_T\#EO(\mathcal{F},{\ne }_{2n}).$$

$$f(y_1,\dots ,y_n)=\sum _{\sigma :E\to \{0,1\}}\prod _{v\in V}{f}_v(\hat{\sigma }{\mid }_{E(v)}),$$

$$f$$

$$f^{010\delta }+f^{100\delta }=0.$$

$$f^{001\delta }+f^{100\delta }=0,$$

$$f^{001\delta }+f^{010\delta }=0.$$

$$M({\mathfrak{m}}_{ij}f)$$

$$=\left[\begin{array}{c}{\mathbf{f}}^{00}\\ {\mathbf{f}}^{01}\\ {\mathbf{f}}^{10}\\ {\mathbf{f}}^{11}\end{array}\right]{\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]}^{\otimes (n-2)}\left[\begin{array}{cccc}{{\mathbf{f}}^{00}}^{\mathtt{T}}& {{\mathbf{f}}^{01}}^{\mathtt{T}}& {{\mathbf{f}}^{10}}^{\mathtt{T}}& {{\mathbf{f}}^{11}}^{\mathtt{T}}\end{array}\right]$$

$$=\left[\begin{array}{c}{\mathbf{f}}^{00}\\ {\mathbf{f}}^{01}\\ {\mathbf{f}}^{10}\\ {\mathbf{f}}^{11}\end{array}\right]\left[\begin{array}{cccc}{\overline{{\mathbf{f}}^{11}}}^{\mathtt{T}}& {\overline{{\mathbf{f}}^{10}}}^{\mathtt{T}}& {\overline{{\mathbf{f}}^{01}}}^{\mathtt{T}}& {\overline{{\mathbf{f}}^{00}}}^{\mathtt{T}}\end{array}\right]$$

$$=\left[\begin{array}{cccc}0& 0& 0& |{\mathbf{f}}^{00}{|}^2\\ 0& ⟨{\mathbf{f}}^{01},{\mathbf{f}}^{10}⟩& |{\mathbf{f}}^{01}{|}^2& 0\\ 0& |{\mathbf{f}}^{10}{|}^2& ⟨{\mathbf{f}}^{10},{\mathbf{f}}^{01}⟩& 0\\ |{\mathbf{f}}^{11}{|}^2& 0& 0& 0\end{array}\right],$$

$$e\otimes f=g\otimes h.$$

$${\lambda }_1\cdot f=g^{\prime }\otimes h^{\prime },$$

$$e^{\prime }\otimes f={\lambda }_2\cdot g,$$

$$f^{\alpha \circ \beta }=g^{\alpha }\cdot h^{\beta }\ne 0.$$

$$0\ne f^{\bar{\alpha }\circ \bar{\beta }}=g^{\bar{\alpha }}\cdot h^{\bar{\beta }}.$$

$$f^{\bar{\alpha }\circ \beta }=g^{\bar{\alpha }}\cdot h^{\beta }\ne 0.$$

$$\overline{g^{\alpha }}\cdot \overline{h^{\beta }}=\overline{f^{\alpha \circ \beta }}=f^{\bar{\alpha }\circ \bar{\beta }}=g^{\bar{\alpha }}\cdot h^{\bar{\beta }}\ne 0,$$

$$g^{\alpha }\cdot h^{\bar{\beta }}=f^{\alpha \circ \bar{\beta }}=\overline{f^{\bar{\alpha }\circ \beta }}=\overline{g^{\bar{\alpha }}}\cdot \overline{h^{\beta }}\ne 0.$$

$$|g^{\alpha }{|}^2=|g^{\bar{\alpha }}{|}^2.$$

$$\overline{g^{\alpha }}\cdot \overline{h^{\beta }}=\overline{f^{\alpha \circ \beta }}=f^{\bar{\alpha }\circ \bar{\beta }}=g^{\bar{\alpha }}\cdot h^{\bar{\beta }}=\overline{g^{\alpha }}\cdot h^{\bar{\beta }},$$

$$\#EO(\{g,h\}\cup \mathcal{F}){\equiv }_T\#EO(\{f\}\cup \mathcal{F})$$

$$\lambda \cdot {\chi }_{AX=0}\cdot {\mathfrak{i}}^{Q(X)},$$

$$Q(x_1,x_2,\dots ,x_n)=a_0+\sum _{k=1}^n{a}_k{x}_k+\sum _{1⩽i<j⩽n}2b_{ij}{x}_i{x}_j,$$

$${\partial }_{(ij)}{f}^{\prime }={\partial }_{(ij)}(\bar{a}{f}_{uv}^{01}-a{f}_{uv}^{10})=\bar{a}{\partial }_{(ij)}(f_{uv}^{01})-a{\partial }_{(ij)}(f_{uv}^{10})=\bar{a}({\partial }_{(ij)}f{)}_{uv}^{01}-a({\partial }_{(ij)}f{)}_{uv}^{10}.$$

$${\partial }_{(ij)}{f}^{\prime }=\bar{a}({\partial }_{(ij)}f{)}_{uv}^{01}-a({\partial }_{(ij)}f{)}_{uv}^{10}=\bar{a}(ag)-a(\bar{a}g)\equiv 0.$$

$${\partial }_{(uv)}({\partial }_{(ij)}f)={\partial }_{(ij)}({\partial }_{(uv)}f)\equiv 0.$$

$${\partial }_{(ij)}f=b^{\prime }(x_u,x_v)\otimes g,$$

Full text

††journal: Information and Computation

1 Introduction

A most basic fact in mathematics is that (rational) integers have a unique prime factorization (UPF). When we consider algebraic extensions, certain primes remain so, while others can be further factored. For example, for algebraic integers in the extension $\mathbb{Q}[\sqrt{-2}]\supset\mathbb{Q}$, the rational prime $3$ is no longer a prime, as $3=(1+{\mathfrak{i}}\sqrt{2})(1-{\mathfrak{i}}\sqrt{2})$, but $5$ remains a prime. Also for algebraic integers, sometimes UPF still holds, but in general it fails. For example, for algebraic integers in $\mathbb{Q}[\sqrt{-5}]$, $6=2\cdot 3=(1+{\mathfrak{i}}\sqrt{5})(1-{\mathfrak{i}}\sqrt{5})$ are two distinct factorizations into irreducible elements, and thus UPF fails. Unique prime factorization is only restored in Kummer’s theory of prime ideals. Perhaps more important than what is in the details, in many mathematical investigations the idea of UPF has been a powerful guiding principle. Is there some analogous notion that is useful for the complexity classification program of counting problems?

In this paper we show that for a class (which is in fact broader than all weighted counting CSP) called counting weighted Eulerian orientation problems, we can develop an analogous notion for the purpose of complexity classification of counting problems. We prove a UPF for constraint functions, (and for the relevant constraint functions there is also a need to extend the scope where the factorization takes place). Next we develop a merging operation on constraint functions. The main technical challenge turns out to be the interplay of these merging operations and the divisibility relation in the unique prime factorization.

Let us define the counting problems that we wish to classify. Let $G$ be an undirected Eulerian graph, i.e., every vertex has even degree. An Eulerian orientation of $G$ is an orientation of its edges such that at each vertex the number of incoming edges is equal to the number of outgoing edges. Mihail and Winkler showed that counting the number of Eulerian orientations of an undirected Eulerian graph is #P-complete [28]. In this paper, we consider counting weighted Eulerian orientations (#EO problems), formulated as a partition function defined by constraint functions placed at each vertex that represent weightings of various local Eulerian configurations. There are a host of problems in statistical physics and combinatorics that can be formulated as computing this partition function.

This partition function is defined as follows. Suppose $G$ is given, and each vertex $v$ is associated with a weight function $f_{v}$ taking values in $\mathbb{C}$. The incident edges to $v$ are totally ordered and correspond to input variables to $f_{v}$. Each edge has two ends, and an orientation of the edge is denoted by assigning 0 to the head and 1 to the tail. Thus, locally at every vertex $v$, a [math] represents an incoming edge and a $1$ represents an outgoing edge. (A loop has both ends at the same vertex.) An Eulerian orientation corresponds to an assignment to each edge ($01$ or $10$) where the numbers of 0’s and 1’s at each $v$ are equal. Then a vertex $v$ contributes a weight by the local constraint function $f_{v}$ evaluated according to the local assignment. The weight of an Eulerian orientation is the product of weights over all vertices. If every constraint function $f_{v}$ is only nonzero when the numbers of input 0’s and 1’s are equal, i.e., the support of $f_{v}$ is on half weighted inputs, then only Eulerian orientations contribute nonzero values. The partition function of #EO is the sum of weights over all Eulerian orientations. This is a sum-of-product computation.

The significance of this partition function is evidenced by its appearance in several different fields. In statistical physics, the partition function of the so-called ice-type model [30, 33, 31] is the partition function of Eulerian orientations of some underlying Eulerian graph. When it is restricted on the square lattice, it is the classical six-vertex model [30]. Literally thousands of papers have been written in the literature dealing with the six-vertex model, mainly on the square lattice, but also on other graphs [3]. It is perhaps one of the three most intensely studied models in statistical physics, along with that of ferromagnetic Ising and monomer-dimer. The “exact solution” of the (unweighted) six-vertex model with periodic boundary conditions by Lieb is an important milestone in statistical physics [25, 26, 24]. In combinatorics, the resolution of the Alternating Sign Matrix conjecture is linked to the classical six-vertex model with the domain wall boundary condition [21, 29, 35, 22, 4]. Las Vergnas also observed that the evaluation of the Tutte polynomial at the point $(3,3)$ is related to the partition function of Eulerian orientations with a specific weight assignment [23]. Cai, Fu and Xia proved a complexity classification of the partition function of the six-vertex model on general 4-regular graphs [13].

The #EO problems have an intrinsic significance in the classification program for counting problems. At first glance, the #EO framework may appear to be specialized as it requires all constraint functions to be supported on half weighted inputs. However, surprisingly, we show that it encompasses all counting constraint satisfaction problems (#CSP) on Boolean variables.

Theorem 1.1**.**

Every complex-weighted #CSP problem on Boolean variables can be expressed as an #EO problem.

A more detailed statement for this theorem is given as Theorem 6.1. By this theorem, a complexity dichotomy for #EO problems would generalize the complexity dichotomy for $\operatorname{\#CSP}$ problems on the Boolean domain, which is already known444In fact, the full complexity dichotomy for $\operatorname{\#CSP}$ problems on any finite domain is also known [7, 20, 5, 10, 9]. [18, 19, 6, 16].

In physics, a symmetry called the arrow reversal symmetry (ars)555On a square lattice, when there is no external electric field, physical considerations imply that the model is unchanged by reversing all arrows [3]. This ‘zero field’ model includes the ice [30], KDP [33] and F [31] models as special cases. is usually assumed. For complex-valued local constraint functions $f$, ars requires that $f(\overline{\alpha})=\overline{f(\alpha)}$ for all $\alpha$, where $\overline{f(\alpha)}$ denotes the complex conjugation of $f(\alpha)$, and $\overline{\alpha}$ denotes the bit-wise complement of $\alpha$. For real-valued functions, this is $f(\overline{\alpha})=f(\alpha)$. This means, if we flip the orientations of all edges, the (real) function value $f$ is unchanged (or for complex $f$ it is changed to its complex conjugation). We will see that it is not only natural but also necessary for the proof that we consider complex values.

In complexity theory, there is a more intrinsic reason for considering the arrow reversal symmetry. Under the holographic transformation $Z=\frac{1}{\sqrt{2}}\left[\begin{smallmatrix}1&1\\ {\mathfrak{i}}&-{\mathfrak{i}}\end{smallmatrix}\right]$, any #EO problem is transformed to a standard Holant problem, and complex-valued constraint functions with the ars restriction translate precisely to real-valued constraint functions.

Theorem 1.2**.**

A complex-valued constraint function $f$ satisfies ars if and only if after the holographic transformation $Z$, $f$ is transformed to a real-valued function.

We will describe holographic transformation and prove this theorem in Section 7. By this theorem, the classification of #EO problems can serve as building blocks towards a classification of general Holant problems. In Section 8, we will give a partial map (Figure 2) for the complexity classification program for Holant problems on the Boolean domain, where we indicate how the framework of #EO problems fits in this program.

In this paper, we prove a complexity dichotomy theorem for #EO problems with ars.

Theorem 1.3**.**

For any set $\mathcal{F}$ of complex-valued constraint functions satisfying the ars condition, the weighted counting problem for Eulerian orientations specified by $\mathcal{F}$ is either computable in polynomial time, or #P-hard. The polynomial time tractability criterion on $\mathcal{F}$ is explicit.

The explicit tractability criterion will be stated in terms of certain constraint function sets to be defined in Section 2, where we give some preliminaries. The formal statement of the dichotomy theorem (Theorem 3.1) and a proof outline will be given in Section 3. The most technical part of the proof is an induction that guarantees a suitable interplay between the merging operation and unique factorization. However this inductive proof only works when the arity (i.e., the number of input variables) of these constraint functions is sufficiently high (arity $\geqslant 10$). For lower arity cases, we prove it separately. We will give the inductive proof and handle lower arity cases in Section 4. In addition, we discover a novel connection between #CSP and #EO problems, which will be introduced in Section 5. We establish this connection by a simulation of #CSP by #EO that is global in nature. Owing to the restriction that the support of every constraint function in #EO problems is on half weighted inputs, (provably) no local replacement reduction can work. One crucial ingredient of the proof in Section 5 is the determination of all half weighted affine linear subspaces over $\mathbb{Z}_{2}$. This result should be of independent interest. This determination uses techniques including Möbius inversion.

2 Preliminaries

2.1 Definitions and notations

A constraint function $f$, or a signature, of arity $m$ is a map $\mathbb{Z}_{2}^{m}\rightarrow\mathbb{C}$. In this paper we assume all signatures have positive arities, and since we study weighted Eulerian orientation problems, we may assume each signature $f$ has an even arity, say $2n$, for some $n=n(f)>0$. We use $f^{\alpha}$ to denote $f(\alpha)$. The support of a signature $f$ of arity $2n$ is $\mathscr{S}(f)=\{\alpha\in\mathbb{Z}_{2}^{2n}\mid f^{\alpha}\neq 0\}$. If $\mathscr{S}(f)=\emptyset$, i.e., $f$ is identically [math], we say $f$ is a zero signature and denote it by $f\equiv 0$. Otherwise, $f$ is a nonzero signature. We use ${\rm wt}({\alpha})$ to denote the Hamming weight of $\alpha\in\mathbb{Z}_{2}^{2n}$. Let $\mathscr{H}_{2n}=\{\alpha\in\mathbb{Z}_{2}^{2n}\mid{\rm wt}(\alpha)=n\}$. Note that if $\alpha$ belongs to $\mathscr{H}_{2n}$ then its complement string $\overline{\alpha}$ also belongs to $\mathscr{H}_{2n}$. A signature $f$ of arity $2n$ is an Eulerian orientation (EO) signature if $\mathscr{S}(f)\subseteq\mathscr{H}_{2n}$. Let $\mathcal{F}$ be any fixed set of EO signatures. An (EO-)signature grid $\Omega=(G,\pi)$ over $\mathcal{F}$ is a tuple, where $G=(V,E)$ is an Eulerian graph without isolated vertex (i.e., every vertex has positive even degree), $\pi$ labels each $v\in V$ with a signature $f_{v}\in\mathcal{F}$ of arity ${\operatorname{deg}(v)}$, and labels the incident edges $E(v)$ at $v$ with input variables of $f_{v}$. For any Eulerian graph $G$, let ${\rm EO}(G)$ be the set of all Eulerian orientations of $G$. We view each edge as having two ends, and an orientation of the edge is denoted by assigning 0 to the head and 1 to the tail. An Eulerian orientation corresponds to an assignment to the ends of each edge where the numbers of 0’s and 1’s at each $v$ are equal. Then a vertex $v$ contributes a weight by the local constraint function $f_{v}$ evaluated according to the local assignment. Each $\sigma\in{\rm EO}(G)$ gives an evaluation $\prod_{v\in V}f_{v}(\sigma|_{E(v)})$, where $\sigma|_{E(v)}$ assigns [math] to an incoming edge and $1$ to an outgoing edge.

Definition 2.1** (#EO problems).**

*A #EO problem #EO$(\mathcal{\mathcal{F}})$ specified by a set $\mathcal{F}$ of EO signatures is the following: The input is an *(EO-)signature grid $\Omega=(G,\pi)$ over $\mathcal{F}$; the output is the partition function of $\Omega$,

[TABLE]

Example 2.2** (Unweighted #EO problem).**

Let $\mathcal{F}=\{f_{2},f_{4},\ldots f_{2n},\ldots\},$ where $f_{2n}^{\alpha}=1$ when ${\rm wt}(\alpha)=n$ and $f_{2n}^{\alpha}=0$ otherwise. Then #EO$(\mathcal{\mathcal{F}})$ counts the number of Eulerian orientations.

Example 2.3** (Six-vertex models).**

Let $f_{\rm six}$ be an EO signature of arity $4$, where $f_{\rm six}^{0011}=f_{\rm six}^{1100}=a$, $f_{\rm six}^{0101}=f_{\rm six}^{1010}=b$, $f_{\rm six}^{0110}=f_{\rm six}^{1001}=c$, (where $a,b,c\in\mathbb{R}^{+}$). Then #EO$(f_{\rm six})$ is the classical six-vertex model satisfying ars with real parameters $(a,b,c)$.

#EO problems can be viewed as special cases of Holant problems. A (general) signature grid $\Omega=(G,\pi)$ over a set of arbitrary (not necessarily EO) signatures $\mathcal{F}$ is a tuple as before, where $G$ is a graph, each vertex $v$ is assigned some $f_{v}\in\mathcal{F}$ of arity $\deg(v)$, with incident edges $E(v)$ labeled as input variables. We consider all 0-1 edge assignments $\sigma:E(G)\rightarrow\{0,1\}$. Each $\sigma$ gives an evaluation $\prod_{v\in V(G)}f_{v}(\sigma|_{E(v)})$, where $\sigma|_{E(v)}$ denotes the restriction of $\sigma$ to $E(v)$.

Definition 2.4** (Holant problems).**

The input to the problem Holant$(\mathcal{F})$ is a signature grid $\Omega=(G,\pi)$ over $\mathcal{F}$. The output is the partition function,

[TABLE]

The bipartite Holant problems $\operatorname{Holant}\left(\mathcal{F}\mid\mathcal{G}\right)$ are Holant problems over bipartite graphs $H=(U,V,E)$, where each vertex in $U$ or $V$ is labeled by a signature in $\mathcal{F}$ or $\mathcal{G}$ respectively.

We use $\leqslant_{T}$ (and $\equiv_{T}$) to denote polynomial-time Turing reductions (and equivalences, respectively). In this paper without loss of generality we assume all signatures have positive arity, i.e., any signature grid $G$ has no isolated vertices. We use $\neq_{2}$ to denote the binary Disequality signature with truth table $(f^{00},f^{01},f^{10},f^{11})=(0,1,1,0)$. It can also be expressed by the matrix

[TABLE]

with one variable indexing rows and the other indexing columns respectively.

Lemma 2.5**.**

#EO$({\mathcal{F}})\equiv_{T}\operatorname{Holant}(\neq_{2}\mid\mathcal{F}).$**

Proof.

If $\Omega=(G,\pi)$ is an instance of #EO$({\mathcal{F}})$, we add a middle vertex on each edge of $G$ and label it by $\neq_{2}$. This defines an instance $\Omega^{\prime}$ of $\operatorname{Holant}(\neq_{2}\mid\mathcal{F})$ with a bipartite graph $H$ (which is the edge-vertex incidence graph of $G$), where every edge of $G$ is broken into two. There is a 1-1 correspondence of the terms in the partition functions ${\rm\#EO}_{\Omega}$ and $\operatorname{Holant}_{\Omega^{\prime}}$. The process is obviously reversible. ∎

Counting constraint satisfaction problems (#CSP) can also be expressed as Holant problems (Lemma 1.2 in [8]). We use $=_{n}$ to denote the Equality signature of arity $n$, which takes value $1$ on the all-0 or all-1 inputs, and [math] elsewhere. Let $\mathcal{EQ}=\{=_{1},=_{2},\ldots,=_{n},\ldots\}$ denote the set of all Equality signatures.

Lemma 2.6**.**

$\operatorname{\#CSP}(\mathcal{F})\equiv_{T}\operatorname{Holant}\left(\mathcal{EQ}\mid\mathcal{F}\right)$.

A signature $f$ of arity $n\geqslant 2$ can be expressed as a $4\times 2^{n-2}$ matrix $M_{(ij)}(f)$, which lists the $2^{n}$ values of $f$ with variables $(x_{i},x_{j})\in\{0,1\}^{2}$ as row index and the assignments of the other $n-2$ variables as column index, both in lexicographic order. That is,

[TABLE]

We use ${\bf{f}}^{ab}_{ij}$ to denote the row vector indexed by $(x_{i},x_{j})=(a,b)$ in $M_{(ij)}(f)$, $\overline{{\bf{f}}^{ab}_{ij}}$ denotes its conjugate, and ${{\bf{f}}^{ab}_{ij}}^{\tt T}$ denotes its transpose. Consider the reversal vector ${\bf{f}}^{ab}_{ij}N^{\otimes(n-2)}$ of ${\bf{f}}^{ab}_{ij}$. We have

[TABLE]

The second equality holds by ars. By taking transpose, we have $N^{\otimes(n-2)}{{\bf{f}}^{ab}_{ij}}^{\tt T}={\overline{{\bf{f}}^{\bar{a}\bar{b}}_{ij}}}^{\tt T}$. When $(ij)$ is clear from the context, we omit the subscript $(ij)$.

We generalize the notion of binary Disequality to signatures of higher arities. A signature $f$ of arity $2n$ is called a Disequality signature of arity $2n$, denoted by $\neq_{2n}$, if $f=1$ when $x_{1}=x_{2}=\ldots=x_{n}\neq x_{n+1}=x_{n+2}=\ldots=x_{2n},$ and 0 otherwise. By permuting its variables the Disequality signature of arity $2n$ also defines ${2n\choose n}/2$ functions which we also call Disequality signatures. Alternatively we can define a Disequality signature as follows. A signature $f$ of arity $2n$ is a Disequality signature if $f^{\alpha}=f^{\bar{\alpha}}=1$ for some $\alpha$ with ${\rm wt}(\alpha)=n$, and $f^{\beta}=0$ elsewhere. Since there are ${2n\choose n}/2$ many possible ways to choose a pair $\{\alpha,\bar{\alpha}\}$, there are ${2n\choose n}/2$ many Disequality signatures of arity $2n$. They are equivalent for the complexity of Holant problems; once we have one we have them all, i.e., if $f$ is obtained from $\neq_{2n}$ by relabeling its variables, then for any set $\mathcal{F}$ of EO signatures,

[TABLE]

Let $\mathcal{DEQ}=\{\neq_{2},\neq_{4},\ldots,\neq_{2n},\ldots\}$ denote the set of all Disequality signatures.

2.2 Gadget construction

One basic tool used throughout this paper is gadget construction. An $\mathcal{F}$-gate (or simply a gadget) is similar to a signature grid $(G,\pi)$ for $\operatorname{Holant}(\mathcal{F})$ except that $G=(V,E,D)$ is a graph with internal edges $E$ and dangling edges $D$. The dangling edges $D$ define input variables for the $\mathcal{F}$-gate. We order the regular edges in $E$ from $1$ to $m$, and order the dangling edges in $D$ from $1$ to $n$. Then the $\mathcal{F}$-gate defines a function $f$

[TABLE]

where $(y_{1},\dotsc,y_{n})\in\{0,1\}^{n}$ is an assignment on the dangling edges, $\hat{\sigma}$ is the extension of $\sigma$ on $E$ by the assignment $(y_{1},\ldots,y_{n})$, and $f_{v}$ is the signature assigned at each vertex $v\in V$. This function $f$ is called the signature of the $\mathcal{F}$-gate. There may be no internal edges in an $\mathcal{F}$-gate. In this case, $f$ is simply a tensor product of these signatures $f_{v}$, i.e., $f={\bigotimes}_{v\in V}f_{v}$ (with possibly a permutation of variables). We say a signature $f$ is realizable from a signature set $\mathcal{F}$ by gadget construction if $f$ is the signature of an $\mathcal{F}$-gate. In the following, when we describe a gadget construction we identify a vertex in the graph $G=(V,E,D)$ for the gadget with the signature labeling that vertex.

If $f$ is realizable from a set $\mathcal{F}$, then we can freely include $f$ in the set of signatures while preserving the complexity (Lemma 1.3 in [8]). Note that by Lemma 2.5, in the setting of #EO($\mathcal{F}$) problems (i.e., for $\operatorname{Holant}\left(\neq\mid\mathcal{F}\right)$), every edge in a gadget construction is effectively labeled by $\neq_{2}$.

A basic gadget construction is merging. Given a signature $f$ of arity $n$, we can connect two variables $x_{i}$ and $x_{j}$ of $f$ using $\neq_{2}$, and it gives a signature of arity $n-2$. We use $\partial_{(ij)}f$ to denote this signature and $\partial_{(ij)}f=f^{01}_{ij}+f^{10}_{ij}$, where $f^{ab}_{ij}$ denotes the signature obtained by setting $(x_{i},x_{j})=(a,b)\in\{0,1\}^{2}$. We use $f^{ab}_{ij}$ to denote a function, and ${\bf f}^{ab}_{ij}$ to denote a vector that lists the truth table of $f^{ab}_{ij}$ in a given order. We may further merge variables $x_{u}$ and $x_{v}$ of $\partial_{(ij)}f$ for any $\{u,v\}$ disjoint with $\{i,j\}$, and we use $\partial_{(uv)(ij)}f=\partial_{(uv)}(\partial_{(ij)}f)$ to denote the realized signature. Note that these two merging operations commute, $\partial_{(uv)(ij)}f=\partial_{(ij)(uv)}f$. This is illustrated in the following commutative diagram.

[TABLE]

(We adopt the notation $\partial$ for the similarity of the merging operation with taking partial derivatives. They both reduce the number of variables, they both are linear, and under mild smoothness conditions we know $\frac{\partial^{2}f}{\partial x\partial y}=\frac{\partial^{2}f}{\partial y\partial x}$. The commutativity of $\partial_{(ij)}f$ and $\partial_{(uv)}f$ is a key property in our proof.) If by merging any two variables of $f$, we can only realize the zero signature, then we show that $f$ itself is “almost” a zero signature.

Lemma 2.7**.**

Let $f$ be a signature of arity $n\geqslant 3$. If for every pair of distinct indices $\{i,j\}$, by merging variables $x_{i}$ and $x_{j}$ of $f$ we have $\partial_{(ij)}f\equiv 0$, then $f^{\alpha}=0$ for any $\alpha$ with $0<{\rm wt}(\alpha)<n$. In particular, if $f$ is an EO signature, then $f\equiv 0$.

Proof.

Suppose there exists some $\alpha$, where $0<{\rm wt}(\alpha)<n$, such that $f^{\alpha}\neq 0$. Since $\alpha$ is not all-0 nor all-1 and $\alpha$ has length at least $3$, we can find three bits in some order such that on these three bits, $\alpha$ takes value $001$ or $110$. Without loss of generality, we assume they are the first three bits of $\alpha$ and we denote $\alpha$ by $001\delta$ or $110\delta$ ($\delta$ maybe empty). We first consider the case that $\alpha=001\delta$. Consider another two strings $\beta=010\delta$ and $\gamma=100\delta$. Note that if we merge variables $x_{1}$ and $x_{2}$ of $f$, we get $\partial_{(12)}f$, its entry $(\partial_{(12)}f)^{0\delta}$ on the input $0\delta$ (for bit positions 3 to $n$) is the sum of $f^{010\delta}$ and $f^{100\delta}$. Since $\partial_{(12)}f\equiv 0$, we have

[TABLE]

Similarly, by merging variables $x_{1}$ and $x_{3}$, we have

[TABLE]

and by merging variables $x_{2}$ and $x_{3}$, we have

[TABLE]

These three equations have only a trivial solution, $f^{001\delta}=f^{010\delta}=f^{100\delta}=0$. A contradiction.

If $\alpha=110\delta$, the proof is symmetric. ∎

The following lemma makes sure that a realized signature is suitable for #EO problems.

Lemma 2.8**.**

Any signature realizable from a set $\mathcal{F}$ of EO signatures satisfying ars is also an EO signature satisfying ars.

Proof.

By definition $\partial_{(ij)}f=f^{01}_{ij}+f^{10}_{ij}$. Hence for any EO signature satisfying ars, after merging any two variables, the realized signature is still an EO signature satisfying ars. Then, suppose $f$ is realized by a graph $G$ with dangling edges and $n$ vertices labeled by signatures $f_{1},f_{2},\ldots,f_{n}\in\mathcal{F}$. We first cut all internal edges in $G$ and get the signature $f^{\prime}=f_{1}\otimes f_{2}\otimes\cdots\otimes f_{n}$. Clearly $f^{\prime}$ is an EO signature satisfying ars since all $f_{i}$ are. Then, $f$ can be realized by merging (with $\neq_{2}$) all cut edges of $f^{\prime}$ in a sequence. After each merging operation, the realized signature is an EO signature satisfying ars, and hence $f$ is an EO signature satisfying ars. ∎

Having established Lemma 2.8, we have the following reduction.

Lemma 2.9**.**

If $f$ is realizable from a set $\mathcal{F}$ of EO signatures, then ${\rm\#EO}(\{f\}\cup\mathcal{F})\equiv_{T}\rm{\#EO}(\mathcal{F})$.

A particular gadget construction often used in this paper is mating. Given an EO signature $f$ of arity $n\geqslant 3$, we connect two copies of $f$ in the following manner: Fix a set $S$ of $n-2$ variables among all $n$ variables of $f$. For each $x_{k}\in S$, connect $x_{k}$ of one copy of $f$ with $x_{k}$ of the other copy using $\neq_{2}$. The two variables $x_{i},x_{j}$ that are not in $S$ are called dangling variables. Then, this mating construction realizes a signature of arity $4$, denoted by $\mathfrak{m}_{ij}f$. It can be represented by matrix multiplication. We have

[TABLE]

where $\langle\cdot,\cdot\rangle$ denotes the (complex) inner product and $|\cdot|$ denotes the norm defined by this inner product. Note that $|\langle{\bf f}^{01},{\bf f}^{10}\rangle|^{2}\leqslant|{\bf f}^{01}|^{2}|{\bf f}^{10}|^{2}$ by Cauchy-Schwarz inequality.

2.3 Signature factorization

Recall that in this paper all signatures have positive arity. A nonzero signature $g$ divides $f$, denoted by $g\mid f$, if there is a signature $h$ such that $f=g\otimes h$ (with possibly a permutation of variables) or there is a constant $\lambda$ such that $f=\lambda\cdot g$. In the latter case, if $\lambda\neq 0$, then we also have $f\mid g$ since $g=\frac{1}{\lambda}\cdot f$. For nonzero signatures, if both $g\mid f$ and $f\mid g$, then they are nonzero constant multiples of each other, and we say $g$ is an associate of $f$, denoted by $g\sim f$. In terms of this division relation, we can define irreducible signatures and prime signatures. We will show that they are equivalent, and this gives us the unique prime factorization of signatures666The factorization of signatures is synonymous with the decomposition of multipartite quantum states in quantum information theory. There, the uniqueness of decomposition is usually assumed as a common knowledge. To our best knowledge, we are not aware of any formal proof..

Definition 2.10** (Irreducible signatures).**

A nonzero signature $f$ is irreducible if $g\mid f$ implies that $g\sim f$. We say a signature is reducible if it is not irreducible or it is a zero signature. By definition, if a signature $f$ of arity greater than 1 is reducible, then there is a factorization $f=g\otimes h$, for some signatures $g$ and $h$ (of positive arities).

Definition 2.11** (Prime signatures).**

A nonzero signature $f$ is a prime signature, if for any nonzero signatures $g$ and $h$, $f\mid g\otimes h$ implies that $f\mid g$ or $f\mid h$.

Lemma 2.12**.**

The notions of irreducible signatures and prime signatures are equivalent.

Proof.

Suppose $f$ is a prime signature. If $f$ is not irreducible, then there is a nonzero signature $g$ such that $g\mid f$ but not $g\sim f$. So there is a signature $h$ (of arity $\geqslant 1$) such that $f=g\otimes h$, up to a permutation of variables ($h\not\equiv 0$ due to $f\not\equiv 0$). Then $f\mid g\otimes h$ and by being a prime, either $f\mid g$ or $f\mid h$. This is impossible because both $g$ and $h$ have lower arity than $f$.

Now, suppose $f$ is irreducible and let $f\mid g\otimes h$, where $g$ and $h$ are nonzero signatures (of arity $\geqslant 1$). If $f\sim g\otimes h$, then $f=(\lambda g)\otimes h$ for some constant $\lambda\not=0$. This contradicts $f$ being irreducible. Thus, there is a nonzero signature $e$ (of arity $\geqslant 1$) such that, up to a permutation of variables,

[TABLE]

Consider the scope of $f$, i.e., its set of variables. Suppose it intersects with the scopes of both $g$ and $h$. Since $e\not\equiv 0$, we can pick an input $\beta$ of $e$ such that $e^{\beta}=\lambda_{1}\neq 0$. By setting the variables in the scope of $e$ to $\beta$ on both sides of (2.2), we have

[TABLE]

where $g^{\prime}$ and $h^{\prime}$ denote the resulting signatures from $g$ and $h$ respectively, both of which have a non-empty scope, i.e., having arity $\geqslant 1$. This is a contradiction to $f$ being irreducible.

Hence the scope of $f$ is a subset of the scope of either $g$ or $h$. Suppose it is $g$, then the scope of $h$ is a subset of the scope of $e$. Since $h\not\equiv 0$, we can pick an input $\alpha$ of $h$ such that $h^{\alpha}=\lambda_{2}\neq 0$. By setting the variables in the scope of $h$ to $\alpha$ on both sides of (2.2), we have

[TABLE]

where $e^{\prime}$ denotes the resulting signature by setting $\alpha$ in $e$. Thus, we have $f\mid g$. Similarly, if the scope of $f$ is a subset of the scope of $h$, then we have $f\mid h$. ∎

A prime factorization of a signature $f$ is $f=g_{1}\otimes\ldots\otimes g_{k}$ up to a permutation of variables, where each $g_{i}$ is a prime signature (irreducible). Start with any nonzero signature, if we keep factoring reducible signatures and induct on arity, any nonzero $f$ has a factorization into irreducible (prime) signatures. The following important lemma says that the prime factorization of a nonzero signature is unique up to the order of the tensor factors and constant scaling factors. It can be proved using Lemma 2.12 and a standard argument, which we omit.

Lemma 2.13** (Unique prime factorization).**

Every nonzero signature $f$ has a prime factorization. If $f$ has prime factorizations $f=g_{1}\otimes\ldots\otimes g_{k}$ and $f=h_{1}\otimes\ldots\otimes h_{\ell}$, both up to a permutation of variables, then $k=\ell$ and after reordering the factors we have $g_{i}\sim h_{i}$ for all $i$.

If a vertex $v$ in a signature grid is labeled by a reducible signature $f=g\otimes h$, we can replace the vertex $v$ by two vertices $v_{1}$ and $v_{2}$ and label $v_{1}$ with $g$ and $v_{2}$ with $h$, respectively. The incident edges of $v$ become incident edges of $v_{1}$ and $v_{2}$ respectively according to the partition of variables of $f$ in the tensor product of $g$ and $h$. This does not change the Holant value. Clearly, $f=g\otimes h$ is realizable from $\{g,h\}$. On the other hand, Lin and Wang proved in [27] (Corollary 3.3) that, from a reducible signature $f=g\otimes h$ we can freely replace $f$ by $g$ and $h$ while preserving the complexity of a Holant problem. To apply their theorem to #EO problems we need to take care of one subtlety, namely from the factorization $f=g\otimes h$ we have to get EO signatures $g$ and $h$, so that we can discuss #EO problems involving signatures $g$ and $h$. This is true for EO signatures $f$ satisfying ars.

Lemma 2.14**.**

Let $f$ be a nonzero reducible EO signature satisfying ars. Then,

for any factorization $f=g\otimes h$, $g$ and $h$ are both EO signatures; 2. 2.

there exists a factorization $f=g\otimes h$ such that $g$ and $h$ are both EO signatures satisfying ars.

Proof.

Since $f\not\equiv 0$, we know $g\not\equiv 0$ and $h\not\equiv 0$ for any factorization $f=g\otimes h$. For a contradiction, suppose there is a factorization $f=g\otimes h$ such that $g$ is not an EO signature. Then, there is an input $\alpha$ of $g$ such that $g^{\alpha}\neq 0$, and ${\rm wt}(\alpha)\neq{\rm wt}(\bar{\alpha})$. (This is true no matter whether $g$ has even or odd arity.) Since $h\not\equiv 0$, there is an input $\beta$ of $h$ such that $h^{\beta}\neq 0$. Note that $\alpha\circ\beta$ is an input of $f$ and we have

[TABLE]

Moreover, since $f$ satisfies ars, we have

[TABLE]

Then, we know $g^{\bar{\alpha}}\neq 0$, and hence

[TABLE]

However, notice that ${\rm wt}(\alpha\circ\beta)={\rm wt}(\alpha)+{\rm wt}(\beta)\neq{\rm wt}(\bar{\alpha})+{\rm wt}(\beta)={\rm wt}(\bar{\alpha}\circ\beta)$. This implies that $\mathscr{S}(f)\not\subseteq\mathscr{H}_{{\rm arity}(f)}$, contradicting $f$ being an EO signature. Thus, for any $f=g\otimes h$, $f$ and $g$ are both EO signatures. 2. 2.

By the first part of the proof, it suffices to show that there exists a factorization $f=g\otimes h$ such that $g$ and $h$ both satisfy ars. Suppose $f=g\otimes h$. Since $f\not\equiv 0$, there is $\alpha\circ\beta$ such that $f^{\alpha\circ\beta}=g^{\alpha}\cdot h^{\beta}\neq 0.$ Since $f$ satisfy ars, we have

[TABLE]

and also

[TABLE]

Multiply these two equalities, and cancel a nonzero common factor, we have

[TABLE]

Since $g^{\alpha}$ and $g^{\bar{\alpha}}$ have the same norm, we can pick a scalar $\lambda=1/(g^{\alpha}g^{\bar{\alpha}})^{1/2}$ such that $\overline{\lambda g^{\alpha}}=\lambda g^{\bar{\alpha}}.$ We have $f=(\lambda g)\otimes(\frac{1}{\lambda}h)$ and we will show $\lambda g$ and $\frac{1}{\lambda}h$ satisfy the ars condition. We rename $\lambda g$ and $\frac{1}{\lambda}h$ by $g$ and $h$, and now we can assume there is an $\alpha$ such that $\overline{g^{\alpha}}=g^{\bar{\alpha}}\neq 0$. For any input $\beta$ of $h$, we have

[TABLE]

and hence, $\overline{h^{\beta}}=h^{\bar{\beta}}$. Hence $h\not\equiv 0$ satisfies the ars condition. We can pick a particular $\beta$ such that $\overline{h^{\beta}}=h^{\bar{\beta}}\neq 0$. Then, for any input $\alpha^{\prime}$ of $g$, since $f$ satisfies the ars condition, we have $\overline{g^{\alpha^{\prime}}}\cdot\overline{h^{\beta}}=g^{\overline{\alpha^{\prime}}}\cdot h^{\overline{\beta}}=g^{\overline{\alpha^{\prime}}}\cdot\overline{h^{\beta}}$, and hence $\overline{g^{\alpha^{\prime}}}=g^{\overline{\alpha^{\prime}}}$. That is, $g$ also satisfies ars. ∎

Remark**.**

The first part does not hold without assuming ars. For example, $f=(0,0,1,0)=(0,1)\otimes(1,0)$, where $(0,0,1,0)$ is an EO signature but $(0,1)$ and $(1,0)$ are not. 2. 2.

The second part does not hold when we restrict signatures to real valued signatures. For example, $f=(0,1,-1,0)^{\otimes 2}=\left[\begin{smallmatrix}0&0&0&0&\\ 0&1&-1&0&\\ 0&-1&1&0&\\ 0&0&0&0&\\ \end{smallmatrix}\right]$ is a real valued reducible signature satisfying ars. (Note that while $f$ satisfies ars, the binary factor $(0,1,-1,0)$ does not satisfy ars.) In fact it cannot be factorized into two real valued signatures that also satisfy ars. To see that, first $f$ cannot have a unary tensor factor by Lemma 2.14. So any factorization of $f$ must be with two binary signatures. A real valued binary EO signature satisfying ars has the form $(0,a,a,0)$ where $a\in\mathbb{R}$. Thus, if $f$ is a tensor product of such two signatures, we have $f=(0,a,a,0)\otimes(0,b,b,0)$ up to a permutation of variables, which implies all nonzero entries of $f$ are the same, a contradiction. However, $f=(0,\mathfrak{i},-\mathfrak{i},0)\otimes(0,-\mathfrak{i},\mathfrak{i},0)$, which is a tensor product of two complex valued signatures satisfying ars. Thus, by going to the complex field we restored a closure property for prime factorizations of signatures satisfying ars.

In the following, when we say that a nonzero EO signature $f$ satisfying ars has a factorization $g\otimes h$, we always assume $g$ and $h$ are EO signatures satisfying ars. The following lemma follows from Corollary 3.3 of Lin and Wang [27].

Lemma 2.15**.**

If a nonzero EO signature $f$ satisfying ars has a factorization $g\otimes h$, then

[TABLE]

for any EO signature set $\mathcal{F}$. In this case, we also say $g$ and $h$ are realizable from $f$.

Definition 2.16**.**

We use $\mathcal{B}$ to denote the set of signatures that are tensor products of (one or more) binary EO signatures satisfying ars.

Note that a binary EO signature $b(x_{i},x_{j})$ over variables $x_{i},x_{j}$ satisfying ars has a particular form $(0,a,\bar{a},0)$ for some $a\in\mathbb{C}$, and it is irreducible when $a\neq 0$. Signatures in $\mathcal{B}$ satisfy a closure property: If $f\in\mathcal{B}$ and $b(y,z)\in\mathcal{B}$ where $y$ and $z$ are distinct from variables of $f$, and we connect one variable $x$ of $f$ via $\neq_{2}$ with $y$, the new signature also belongs to $\mathcal{B}$. This is easily verified by $\left[\begin{smallmatrix}0&a\\ \bar{a}&0\end{smallmatrix}\right]\left[\begin{smallmatrix}0&1\\ 1&0\end{smallmatrix}\right]\left[\begin{smallmatrix}0&b\\ \bar{b}&0\end{smallmatrix}\right]=\left[\begin{smallmatrix}0&ab\\ \overline{ab}&0\end{smallmatrix}\right]$. We use the notation $b^{\mathfrak{i}}(\cdot,\cdot)$ to denote the binary signature $(0,\mathfrak{i},-\mathfrak{i},0)$ in $\mathcal{B}$. For signatures in $\mathcal{B}$, we have the following result.

Lemma 2.17**.**

Suppose $f\in\mathcal{B}$. Then $\partial_{(ij)}f\equiv 0$ iff the signature $b^{\mathfrak{i}}(x_{i},x_{j})=(0,\mathfrak{i},-\mathfrak{i},0)$ divides $f$.

Proof.

If $b^{\mathfrak{i}}(x_{i},x_{j})|f$, then $f=b^{\mathfrak{i}}(x_{i},x_{j})\otimes g$, where $g$ is a constant or a signature on variables other than $x_{i},x_{j}$. We have $\partial_{(ij)}f=(\mathfrak{i}-\mathfrak{i})\cdot g\equiv 0$.

Now, suppose $\partial_{(ij)}f\equiv 0$. If $f\equiv 0$, then it is trivial. Otherwise, $f\not\equiv 0$. Consider the unique prime factorization of $f$. If $x_{i}$ and $x_{j}$ appear in one binary signature $b(x_{i},x_{j})=(0,a,\bar{a},0)$, then $a\neq 0$, and $f=b(x_{i},x_{j})\otimes g$, where $g$ is a constant or a signature on variables other than $x_{i},x_{j}$ and $g\not\equiv 0$ due to $f\not\equiv 0$. Then, we have $\partial_{(ij)}f=(a+\bar{a})g\equiv 0$, which means $a+\bar{a}=0$. That is, $a=\lambda\mathfrak{i}$ for some $\lambda\in\mathbb{R}$. So, we have $b^{\mathfrak{i}}(x_{i},x_{j})|f$.

Otherwise, $x_{i}$ and $x_{j}$ appear in separate binary signatures $b_{1}(x_{i},x_{i^{\prime}})=(0,a,\bar{a},0)$ and $b_{2}(x_{j},x_{j^{\prime}})=(0,b,\bar{b},0)$ in the unique prime factorization of $f$. That is, $f=b_{1}(x_{i},x_{i^{\prime}})\otimes b_{2}(x_{j},x_{j^{\prime}})\otimes g$, where $g$ is a constant or a signature on variables other than $x_{i},x_{i^{\prime}},x_{j},x_{j^{\prime}}$ and $g\not\equiv 0$. Then $\partial_{(ij)}f=b^{\prime}(x_{i^{\prime}},x_{j^{\prime}})\otimes g$ where $b^{\prime}(x_{i^{\prime}},x_{j^{\prime}})=(0,\bar{a}b,a\bar{b},0)\not\equiv 0$. Hence, $\partial_{(ij)}f\not\equiv 0$. A contradiction. ∎

2.4 Tractable signatures and #P-hardness results

We give some known signature sets that define polynomial time computable (tractable) counting problems. In this paper, we need two families: affine signatures and product-type signatures.

Definition 2.18**.**

A signature $f(x_{1},\ldots,x_{n})$ of arity $n$ is affine if it has the form

[TABLE]

where $\lambda\in\mathbb{C}$, $X=(x_{1},x_{2},\dotsc,x_{n},1)$, $A$ is a matrix over $\mathbb{Z}_{2}$, $Q(x_{1},x_{2},\ldots,x_{n})\in\mathbb{Z}_{4}[x_{1},x_{2},\ldots,x_{n}]$ is a quadratic (total degree at most 2) multilinear polynomial with the additional requirement that the coefficients of all cross terms are even, i.e., $Q$ has the form

[TABLE]

and $\chi$ is a 0-1 indicator function such that $\chi_{AX=0}$ is $1$ iff $AX=0$. We use $\mathscr{A}$ to denote the set of all affine signatures.

If the support set $\mathscr{S}(f)$ is an affine linear subspace, then we say $f$ has affine support. Clearly, any affine signature has affine support.

Definition 2.19**.**

A signature on a set of variables $X$ is of product type if it can be expressed as a product of unary functions, binary equality functions $(1,0,0,1)$, and binary disequality functions $(0,1,1,0)$, each on one or two variables of $X$. We use $\mathscr{P}$ to denote the set of product-type functions.

Note that the products of unary, binary equality, and binary disequality functions in the definition of $\mathscr{P}$ can be on overlapping variables. A simple and important observation is that $\mathcal{B}\subseteq\mathscr{P}$. By definition, it is easy to verify the following lemmas.

Lemma 2.20**.**

Any signature of product type has affine support.

Please see Definition 2.22 in Section 2.5 of [12] for a proof of Lemma 2.20. More information on $\mathscr{A}$ and $\mathscr{P}$ can be found in [8].

Lemma 2.21**.**

Let $f$ be an EO signature of arity $4$ satisfying ars with support size $4$, say $f^{\alpha}$, $f^{\beta}$, $f^{\overline{\alpha}}$ and $f^{\overline{\beta}}\neq 0$ where $\alpha\neq\beta,\overline{\beta}$. Then $f\in\mathscr{P}$ if and only if $|f^{\alpha}|=|f^{\beta}|$.

Proof.

Suppose $f\in\mathscr{P}$. Then by Lemma 2.20 it has affine support. Being an EO signature with support size 4, we can show that, after renaming its 4 variables we may assume the support is defined by $(x_{1}\neq x_{2})\wedge(x_{3}\neq x_{4})$. No binary equality is used in its definition for being in $\mathscr{P}$, and exactly these two binary disequalities are used. Then $f$ takes values $ac,ad,bc,bd$ on $0101,0110,1001,1010$ for some $a,b,c,d\neq 0$. By ars, we have $bd=\overline{ac}$ and $ad=\overline{bc}$. It follows that $|a|=|b|$. Similarly $|c|=|d|$. Therefore all nonzero values of $f$ have the same norm. Hence $|f^{\alpha}|=|f^{\beta}|$.

Conversely, suppose $f^{\alpha}=re^{i\varphi}$ and $f^{\beta}=re^{i\psi}$, for some $r>0$ and $\varphi$, $\psi$. By renaming variables we may assume $\alpha=0101$, $\beta=0110$. Let $a=re^{i\frac{\varphi+\psi}{2}}$ and $c=e^{i\frac{\varphi-\psi}{2}}$. Then the unary functions $(a,\bar{a})$ on $x_{1}$ and $(c,\bar{c})$ on $x_{3}$, times $(x_{1}\neq x_{2})\wedge(x_{3}\neq x_{4})$ defines $f\in\mathscr{P}$. ∎

The following tractable result is known [16], since $(\neq_{2})\in\mathscr{A}\cap\mathscr{P}$.

Theorem 2.22**.**

Let $\mathcal{F}$ be any set of complex-valued signatures on Boolean variables. If $\mathcal{F}\subseteq\mathscr{A}$ or $\mathcal{F}\subseteq\mathscr{P}$, then $\operatorname{Holant}(\neq_{2}\mid\mathcal{F})$ is tractable.

Problems defined by $\mathscr{A}$ and by $\mathscr{P}$ are tractable [8]. For an EO signature $f$ of arity $4$, the complexity classification of #EO$(f)$ is known [13]. (This is the six-vertex model for general 4-regular graphs.) For the special case when the $4$-ary EO signature $f$ further satisfies ars, the tractability criterion in [13] simplifies because in this case $f\in\mathscr{A}$ implies that $f\in\mathscr{P}$. We restate this complexity classification of #EO$(f)$ for our setting.

Theorem 2.23**.**

Let $f$ be an EO signature of arity $4$ satisfying ars. Then #EO$(f)$ is #P-hard unless $f\in\mathscr{P}$.

The complexity classification of #CSP is also known [16].

Theorem 2.24**.**

Let $\mathcal{F}$ be any set of complex-valued signatures on Boolean variables. Then $\operatorname{\#CSP}(\mathcal{F})$ is ${\#\rm{P}}$-hard unless $\mathcal{F}\subseteq\mathscr{A}$ or $\mathcal{F}\subseteq\mathscr{P}$, in which cases the problem is tractable.

3 Main Result and Proof Outline

Starting from this section, a signature means a complex-valued EO signature satisfying the ars condition and $\mathcal{F}$ denotes a set of such signatures. The main result of this paper is

Theorem 3.1**.**

Let $\mathcal{F}$ be a set of EO signatures satisfying ars. Then, $\rm{\#EO}(\mathcal{F})$ is #P-hard unless $\mathcal{F}\subseteq\mathscr{A}$ or $\mathcal{F}\subseteq\mathscr{P}$, in which cases it is tractable.

Since a complexity dichotomy of ${\rm\#EO}(f)$ for a single signature $f$ of arity $4$ is known (Theorem 2.23), we wish to leverage this knowledge and realize arity 4 signatures from a given set of signatures, to which we can apply the known tractability criteria. We will use the mating construction to realize signatures of arity $4$, then apply the Cauchy-Schwarz inequality. Consider a nonzero signature $f\in\mathcal{F}$. We may assume that $f$ is irreducible. Otherwise we can replace $f$ by its irreducible factors without changing the complexity due to Lemma 2.15. We have the following lemma.

Lemma 3.2**.**

Let $f\in\mathcal{F}$ be a nonzero irreducible signature of arity $n\geqslant 4$. Then one of the following alternatives holds:

#EO$(\mathcal{F})$* is *#P-hard,

• 2.

$\rm{\#EO}(\{\neq_{4}\}\cup\mathcal{F})\leqslant_{T}\rm{\#EO}(\mathcal{F})$, or

• 3.

there exists a nonzero constant $\lambda$, such that for all pairs of distinct indices $\{i,j\}$, $M(\mathfrak{m}_{ij}f)=\lambda N^{\otimes 2}$, where $N^{\otimes 2}=\left[\begin{smallmatrix}0&0&0&1\\ 0&0&1&0\\ 0&1&0&0\\ 1&0&0&0\\ \end{smallmatrix}\right]$.

We call the third statement the property of (mutual) orthogonality.

Proof.

We consider the signature $\mathfrak{m}_{ij}f$ realized by mating two copies of $f$ for all pairs of distinct indices $\{i,j\}\subseteq[n]$. If #EO$(\mathfrak{m}_{ij}f)$ is already #P-hard, then #EO$(\mathcal{F})$ is also #P-hard since ${\rm\#EO}(\mathfrak{m}_{ij}f)\leqslant_{T}\rm{\#EO}(\mathcal{F})$. Since we already have a complexity dichotomy for arity 4 signatures, we may assume that $\mathfrak{m}_{ij}f$ satisfies the tractability condition and that #EO$(\mathfrak{m}_{ij}f)$ is computable in polynomial time for every pair $\{i,j\}$. Recall the form (2.1). If there exists some $\{i,j\}$, such that $\mathfrak{m}_{ij}f\equiv 0$, then ${\bf f}_{ij}^{00}={\bf f}_{ij}^{01}={\bf f}_{ij}^{10}={\bf f}_{ij}^{11}\equiv 0$, which implies $f\equiv 0$. A contradiction. So we have $\mathfrak{m}_{ij}f\not\equiv 0$, for all pairs $\{i,j\}$. Then by Theorem 2.23, #EO$(\mathfrak{m}_{ij}f)$ is tractable if and only if $\mathfrak{m}_{ij}f\in\mathscr{P}$. By Lemma 2.20, we know $\mathfrak{m}_{ij}f$ has affine support, and being nonzero it has support size either $2$ or $4$ (by the form in (2.1), the support size is not 1). There are two cases depending on the support size of $\mathfrak{m}_{ij}f$ for all pairs $\{i,j\}$.

There exists some pair $\{i,j\}$ such that $\mathfrak{m}_{ij}f$ has support size $2$. Then,

• (a)

Either $M(\mathfrak{m}_{ij}f)$ has the form $\lambda_{ij}\left[\begin{smallmatrix}0&0&0&1\\ 0&0&0&0\\ 0&0&0&0\\ 1&0&0&0\\ \end{smallmatrix}\right]$ where $\lambda_{ij}=|{\bf f}_{ij}^{00}|^{2}=|{\bf f}_{ij}^{11}|^{2}\neq 0$,

• (b)

or $M(\mathfrak{m}_{ij}f)$ has the form $\lambda_{ij}\left[\begin{smallmatrix}0&0&0&0\\ 0&0&1&0\\ 0&1&0&0\\ 0&0&0&0\\ \end{smallmatrix}\right]$ where $\lambda_{ij}=|{\bf f}_{ij}^{01}|^{2}=|{\bf f}_{ij}^{10}|^{2}\neq 0$.

The form that $\langle{\bf f}_{ij}^{01},{\bf f}_{ij}^{10}\rangle\neq 0$ while $|{\bf f}_{ij}^{01}|^{2}=0$ cannot occur since $|\langle{\bf f}_{ij}^{01},{\bf f}_{ij}^{10}\rangle|\leqslant|{\bf f}_{ij}^{01}||{\bf f}_{ij}^{10}|$. In both forms, $\neq_{4}$ is realizable since $\lambda_{ij}\not=0$. Thus, #EO$(\{\neq_{4}\}\cup\mathcal{F})\leqslant_{T}{\rm\#EO}(\mathcal{F})$. 2. 2.

For all pairs $\{i,j\}$, $\mathfrak{m}_{ij}f$ has support size $4$. By Lemma 2.21,

• (a)

Either $M(\mathfrak{m}_{ij}f)$ has the form $\lambda_{ij}\left[\begin{smallmatrix}0&0&0&1\\ 0&0&1&0\\ 0&1&0&0\\ 1&0&0&0\\ \end{smallmatrix}\right]$ where $\lambda_{ij}=|{\bf f}_{ij}^{00}|^{2}=|{\bf f}_{ij}^{11}|^{2}=|{\bf f}_{ij}^{01}|^{2}=|{\bf f}_{ij}^{10}|^{2}\neq 0$,

• (b)

or $M(\mathfrak{m}_{ij}f)$ has the form $\left[\begin{smallmatrix}0&0&0&0\\ 0&\langle{\bf f}_{ij}^{01},{\bf f}_{ij}^{10}\rangle&|{\bf f}_{ij}^{01}|^{2}&0\\ 0&|{\bf f}_{ij}^{10}|^{2}&\langle{\bf f}_{ij}^{10},{\bf f}_{ij}^{01}\rangle&0\\ 0&0&0&0\\ \end{smallmatrix}\right]$, where $|\langle{\bf f}_{ij}^{01},{\bf f}_{ij}^{10}\rangle|^{2}=|{\bf f}_{ij}^{01}|^{2}|{\bf f}_{ij}^{10}|^{2}\not=0$.

Again, the form that $\langle{\bf f}_{ij}^{01},{\bf f}_{ij}^{10}\rangle\neq 0$ while $|{\bf f}_{ij}^{01}|^{2}=0$ cannot occur. In the first form, four vectors form a set of mutually orthogonal vectors of nonzero equal norm. In the second form, by Cauchy-Schwarz, it means that ${\bf f}_{ij}^{01}=c{\bf f}_{ij}^{10}$ for some $c\in\mathbb{C}$. In addition, we know $|c|=1$ due to $|{\bf f}_{ij}^{01}|=|{\bf f}_{ij}^{10}|$ by ars. Since $|{\bf f}_{ij}^{00}|^{2}=|{\bf f}_{ij}^{11}|^{2}=0$, we have ${\bf f}_{ij}^{00}={\bf f}_{ij}^{11}=\bf{0}$, the all-zero vector. Thus, $f$ is factorizable as a tensor product $f=b(x_{i},x_{j})\otimes g$, for some $g$ and some binary signature $b(x_{i},x_{j})=(0,a,\overline{a},0)$, a contradiction because $f$ is irreducible.

Thus, in this case, $M(\mathfrak{m}_{ij}f)=\lambda_{ij}N^{\otimes 2}$ for all pairs $\{i,j\}$. Then, we show that all $\lambda_{ij}$ have the same value. If we mate further the four dangling variables of $\mathfrak{m}_{ij}f$, i.e., we connect the remaining $x_{i}$ in one copy of $f$ with another copy using $(\neq_{2})$, and similarly for $x_{j}$, which totally mates two copies of $f$, we get a value $4\lambda_{ij}$. This value clearly does not depend on the particular indices $\{i,j\}$. We denote the common value $\lambda_{ij}$ by $\lambda$. Thus, there exists $\lambda\neq 0$ such that for all pairs $\{i,j\}$, $M(\mathfrak{m}_{ij}f)=\lambda N^{\otimes 2}.$ ∎

Now we return to the proof of Theorem 3.1. By Lemma 3.2, we have two main cases depending on whether $\neq_{4}$ can be realized by $\mathfrak{m}_{ij}f$ from $\mathcal{F}$. We give a proof outline to show how they will be handled.

The signature $\neq_{4}$ cannot be realized by $\mathfrak{m}_{ij}f$ from $\mathcal{F}$. That is, every irreducible signature (or factor of signatures) in $\mathcal{F}$ satisfies the orthogonality property stated in Lemma 3.2.

We show that this case happens only if $\mathcal{F}\subseteq\mathcal{B}$ (Theorem 4.8). We want to prove this by induction. The general strategy is to start with any signature $f\in\mathcal{F}$ of arity $2n$ that is not in $\mathcal{B}$, we realize a signature $g$ of arity $2n-2$ that is also not in $\mathcal{B}$, i.e. #EO$(\{g\}\cup\mathcal{F})\leqslant_{T}{\rm\#EO}(\mathcal{F})$ (Lemma 4.6). If we can reduce the arity down to 4 (this is by a sequence of reductions that is constant in length independent of the problem instance size of the graph), then we can show it is impossible for such a signature to satisfy the orthogonality. Thus, we can use it to realize $\neq_{4}$ or a #P-hard signature by Lemma 3.2. However, our induction proof only works when the arity $2n\geqslant 10$ (there is an intrinsic reason for this). Therefore we must establish the base cases at arity $4,6$ and $8$. Fortunately, using the orthogonality of $f$, we can prove our theorem for signatures of arity $4,6$ and $8$ separately (Lemma 4.7).

For the induction proof, we use merging to realize signatures of lower arity. It naturally reduces the arity by two. Given a signature $f$ $\notin\mathcal{B}$ of arity $2n\geqslant 10$, if $\partial_{(ij)}f\notin\mathcal{B}$ for some $\{i,j\}$, then we are done. So we may assume for every $\{i,j\}$, $\partial_{(ij)}f\in\mathcal{B}$. We further inquire whether for every $\{i,j\}$, $\partial_{(ij)}f\not\equiv 0$. If for some $\{i,j\}$, $\partial_{(ij)}f\equiv 0$, then it turns out to be relatively easy to handle (Lemma 2.17). So we may assume for every $\{i,j\}$, $\partial_{(ij)}f\not\equiv 0$. We aim to show that there is a binary signature $b(x_{u},x_{v})$ such that $b(x_{u},x_{v})\mid f$. If so, the “quotient” gives us a signature not in $\mathcal{B}$, but of arity $2n-2$, by Lemma 2.15. In some cases we have to replace $f$ by another $f^{\prime}$ to accomplish that.

Assuming $\partial_{(ij)}f\in\mathcal{B}$ for all $\{i,j\}$, we prove there is a $b(x_{u},x_{v})$ such that $b(x_{u},x_{v})\mid f$ or $b(x_{u},x_{v})\mid f^{\prime}$ in the following steps:

1. (a)

If there is a binary signature $b(x_{u},x_{v})$ such that $b(x_{u},x_{v})\mid\partial_{(ij)}f$ for every $\{i,j\}$ disjoint with $\{u,v\}$, then $b(x_{u},x_{v})\mid f$ (Lemma 4.1). 2. (b)

We have assumed $\partial_{(ij)}f\in\mathcal{B}$ for all $\{i,j\}$. Suppose there is one $\partial_{(uv)}f\equiv 0$. We show that the binary signature $b^{\mathfrak{i}}(x_{u},x_{v})=(0,\mathfrak{i},-\mathfrak{i},0)$ divides $\partial_{(ij)}f$ for every $\{i,j\}$ disjoint with $\{u,v\}$ (Lemma 4.2). 3. (c)

Now, we further assume $\partial_{(ij)}f\not\equiv 0$ for all $\{i,j\}$. We want to show that if a binary signature $b(x_{u},x_{v})$ divides a “triangle”, i.e. $b(x_{u},x_{v})\mid\partial_{(rs)}f,\partial_{(st)}f,\partial_{(rt)}f$ (we say $f$ satisfies the $\Delta$-property), it divides $\partial_{(ij)}f$ for every $\{i,j\}$ disjoint with $\{u,v\}$ (Lemma 4.4). To prove this, we need the following delicate lemma. 4. (d)

If a binary signature $b(x_{u},x_{v})$ divides “two pairs”, i.e. $b(x_{u},x_{v})\mid\partial_{(st)}f,\partial_{(s^{\prime}t^{\prime})}f$, where $\{s,t\}$ and $\{s^{\prime},t^{\prime}\}$ are distinct but not necessarily disjoint, then it divides $\partial_{(ij)}f$ for any $\{i,j\}$ which is disjoint with $\{u,v\}\cup\{s,t\}\cup\{s^{\prime},t^{\prime}\}$ that satisfies $\partial_{(st)(ij)}f\not\equiv 0$ and $\partial_{(s^{\prime}t^{\prime})(ij)}f\not\equiv 0$ (Lemma 4.3). 5. (e)

Finally, we show that either (i) $f$ satisfies the $\Delta$-property, or (ii) we can realize a signature $f^{\prime}$, where $f^{\prime}\not\in\mathcal{B}$ has the same arity as $f$, such that either $\partial_{(ij)}f\not\in\mathcal{B}$ for some $\{i,j\}$, or $f^{\prime}$ satisfies the $\Delta$-property (Lemma 4.5).

These steps will accomplish the arity reduction inductive step.

This case is handled in Section 4. We will see that the unique prime factorization plays an important role in the proof. 2. 2.

Otherwise, we have $\rm{\#EO}(\{\neq_{4}\}\cup\mathcal{F})\leqslant_{T}\rm{\#EO}(\mathcal{F})$.

The signature $\neq_{4}$ can be used to realize any $(\neq_{2k})\in\mathcal{DEQ}$ (Lemma 5.1), and then the problem $\rm{\#EO}(\mathcal{DEQ}\cup\mathcal{F})$ can be expressed as $\operatorname{Holant}\left(\mathcal{DEQ}\mid\mathcal{F}\right)$ (Lemma 5.2). The next idea is to simulate #CSP$(\mathcal{G})\equiv_{T}\operatorname{Holant}\left(\mathcal{EQ}\mid\mathcal{G}\right)$ using $\operatorname{Holant}\left(\mathcal{DEQ}\mid\mathcal{F}\right)$ for some $\mathcal{G}$ closely related to $\mathcal{F}$, and we can apply the dichotomy of #CSP (Theorem 2.24) to get hardness results. The challenge is to simulate $\mathcal{EQ}$ using $\mathcal{DEQ}$ and $\mathcal{F}$. After some reflection one can observe that it is impossible to realize $\mathcal{EQ}$ by direct gadget constructions. Since signatures in $\mathcal{DEQ}$ and $\mathcal{F}$ are EO signatures satisfying ars, by Lemma 2.8 any gadget realizable from them is also an EO signature. But clearly, any $(=_{k})\in\mathcal{EQ}$ is not an EO signature. However we found an alternative way to simulate $\mathcal{EQ}$ globally, and this is achieved depending crucially on some special properties of $\mathcal{F}$, as follows:

1. (a)

First, using ars we show that #CSP$(|\mathcal{F}|^{2})\leqslant_{T}\operatorname{Holant}\left(\mathcal{DEQ}\mid\mathcal{F}\right)$ (Lemma 5.3), where $|\mathcal{F}|^{2}$ denotes the set of signatures by taking norm squares of signatures in $\mathcal{F}$, namely $|\mathcal{F}|^{2}=\{|f|^{2}\mid f\in\mathcal{F}\}$. This directly implies that $\operatorname{Holant}\left(\mathcal{DEQ}\mid\mathcal{F}\right)$ is #P-hard unless every signature in $\mathcal{F}$ has affine support (Corollary 5.4). 2. (b)

Then, consider an EO signature with affine support. We show its support has a special structure called pairwise opposite (Definition 5.5 and Lemma 5.7). 3. (c)

Finally, given the support of every signature $f\in\mathcal{F}$ is pairwise opposite, we show #CSP$(\mathcal{F})\leqslant_{T}\operatorname{Holant}\left(\mathcal{DEQ}\mid\mathcal{F}\right)$ (Lemma 5.8) by a global simulation, and hence the problem $\operatorname{Holant}\left(\mathcal{DEQ}\mid\mathcal{F}\right)$ is #P-hard unless $\mathcal{F}\subseteq\mathscr{A}$ or $\mathcal{F}\subseteq\mathscr{P}$ (Corollary 5.9).

It follows that, in this case, we have $\rm{\#EO}(\{\neq_{4}\}\cup\mathcal{F})$ is #P-hard unless $\mathcal{F}\subseteq\mathscr{A}$ or $\mathcal{F}\subseteq\mathscr{P}$ (Theorem 5.10). This case is handled in Section 5. We will introduce the pairwise opposite structure and show the global reductions from #CSP to #EO problems.

As observed earlier $\mathcal{B}\subseteq\mathscr{P}$. Thus #EO$(\mathcal{F})$ is tractable if $\mathcal{F}\subseteq\mathcal{B}$, by Lemma 2.5 and Theorem 2.22. In Section 4, we show that if $\mathcal{F}\not\subseteq\mathcal{B}$ then either #EO$(\mathcal{F})$ is #P-hard, or we have #EO$(\{\neq_{4}\}\cup\mathcal{F})\leqslant_{T}$ #EO$(\mathcal{F})$. In Section 5, we show that $\rm{\#EO}(\{\neq_{4}\}\cup\mathcal{F})$ is #P-hard unless $\mathcal{F}\subseteq\mathscr{A}$ or $\mathcal{F}\subseteq\mathscr{P}$. This completes the proof of Theorem 3.1.

4 Interplay of Unique Prime Factorization and $\partial_{(ij)}$

Operations

In this section, we show that if $\mathcal{F}\not\subseteq\mathcal{B}$ then either #EO$(\mathcal{F})$ is #P-hard or we can realize $\neq_{4}$, i.e., #EO$(\{\neq_{4}\}\cup\mathcal{F})\leqslant_{T}$ #EO$(\mathcal{F})$, and then the results from Section 5 take over. Suppose $\mathcal{F}\not\subseteq\mathcal{B}$, then it contains some signature $f\notin\mathcal{B}$, and we prove the statement by induction on the arity of $f$. The general strategy is that we start with any signature $f$ of arity $2n\geqslant 10$ that is not in $\mathcal{B}$, and realize a signature $g$ of arity $2n-2$ that is also not in $\mathcal{B}$. As stated in Section 1 this induction only works for arity $2n\geqslant 10$. We prove the base cases of the induction separately, when $f$ has arity 4, 6 or 8.

For the inductive step, we consider $\partial_{(ij)}f$ for all $\{i,j\}$. If there exists $\{i,j\}$ such that $\partial_{(ij)}f\notin\mathcal{B}$, then we can realize $g=\partial_{(ij)}f$ which has arity $2n-2$, and we are done. Thus, we assume $\partial_{(ij)}f\in\mathcal{B}$ for all $\{i,j\}$. We denote this property by $f\in\int\mathcal{B}$. Under the assumption that $f\in\int\mathcal{B}$, our goal is to show that there is a binary signature $b(x_{u},x_{v})$ such that either $b(x_{u},x_{v})\mid f$ or there exists another $f^{\prime}\not\in\mathcal{B}$ realizable from $f$, such that $f^{\prime}$ has the same arity as $f$, and $b(x_{u},x_{v})\mid f^{\prime}$. In the second case we may again assume $f^{\prime}\in\int\mathcal{B}$, for otherwise we may take $\partial_{(ij)}f^{\prime}$ for some $\{i,j\}$. Now we may replace $f$ by $f^{\prime}$ in the second case. From the factorization $f=b(x_{u},x_{v})\otimes g$, it follows from the definition of $\mathcal{B}$ that $g\notin\mathcal{B}$ since $f\notin\mathcal{B}$. From the factorization of $f$, we can realize $g$ from $f$ by Lemma 2.15, and we are done. We carry out our induction proof in the next five lemmas.

For convenience, we use the following notations.

$\mathcal{B}=\{\mbox{tensor products of one or more binary EO signatures satisfying {\sc ars}}\}$.

• 2.

$f\in\int\mathcal{B}$ denotes the property that $\partial_{(ij)}f\in\mathcal{B}$ for all $\{i,j\}$.

• 3.

$f\in\int\mathcal{B}_{\not\equiv 0}$ denotes the property that $\partial_{(ij)}f\in\mathcal{B}$ and $\partial_{(ij)}f\not\equiv 0$ for all $\{i,j\}$.

• 4.

We say $f$ satisfies the $\Delta$-property, if there exist three distinct indices $\{r,s,t\}$ and a binary signature $b(x_{u},x_{v})$ such that $\{u,v\}\cap\{r,s,t\}=\emptyset$, and $b(x_{u},x_{v})\mid\partial_{(rs)}f,\partial_{(st)}f,\partial_{(rt)}f$.

Lemma 4.1**.**

Let $f\in\int\mathcal{B}$ be a signature of arity $2n\geqslant 6$. If there exists a binary signature $b(x_{u},x_{v})$ such that $b(x_{u},x_{v})\mid\partial_{(ij)}f$ for all $\{i,j\}$ disjoint with $\{u,v\}$, then $b(x_{u},x_{v})\mid f$.

Proof.

Recall that $f_{uv}^{bc}$ denotes the signature obtained by setting variables $(x_{u},x_{v})$ of $f$ to $(b,c)\in\{0,1\}^{2}$. These are called the pinning operations on $\{u,v\}$. Clearly, for any $\{i,j\}$ disjoint with $\{u,v\}$, the pinning operations on $\{u,v\}$ commute with the merging operation $\partial_{(ij)}$, and so we have $(\partial_{(ij)}f)_{uv}^{bc}=\partial_{(ij)}(f_{uv}^{bc})$.

We may assume the binary signature has the form $b(x_{u},x_{v})=(0,a,\bar{a},0)$, where $a\neq 0$. Consider the signature $f^{\prime}\vcentcolon=\bar{a}f_{uv}^{01}-af_{uv}^{10}$. It is a signature on variables of $f$ other than $x_{u}$ and $x_{v}$. For any $\{i,j\}$ disjoint with $\{u,v\}$, by merging variables $x_{i}$ and $x_{j}$ of $f^{\prime}$, and recalling that $\partial_{(ij)}$ is a linear operator, we have

[TABLE]

By assumption, $\partial_{(ij)}f=b(x_{u},x_{v})\otimes g$, where $g$ is a signature on variables other than $x_{u},x_{v},x_{i},x_{j}$. (Since $\partial_{(ij)}f$ has arity at least 4, $g$ is not a constant.) Then we have

[TABLE]

Note that $f^{\prime}$ is also an EO signature. By Lemma 2.7, we have $f^{\prime}\equiv 0$, and hence $\bar{a}f_{uv}^{01}\equiv af_{uv}^{10}$. Moreover, by the factorization of $\partial_{(ij)}f$, we have $\partial_{(ij)}(f_{uv}^{00})=(\partial_{(ij)}f)_{uv}^{00}\equiv 0$ and $\partial_{(ij)}(f_{uv}^{11})=(\partial_{(ij)}f)_{uv}^{11}\equiv 0$ for any $\{i,j\}$ disjoint with $\{u,v\}$. Also, since $2n\geqslant 6$, $f_{uv}^{00}(\alpha)=f_{uv}^{11}(\alpha)=0$ when ${\rm wt}(\alpha)=0$ or $2n-2$. By Lemma 2.7 again, we have $f_{uv}^{00}=f_{uv}^{11}\equiv 0$. Hence, $f=(f_{uv}^{00},f_{uv}^{01},f_{uv}^{10},f_{uv}^{11})=(0,a,\bar{a},0)\otimes(\frac{1}{a}f_{uv}^{01})$, and we have $b(x_{u},x_{v})\mid f$. ∎

Notice that for arity $2n\geqslant 6$, if $b(x_{u},x_{v})\mid f$ and thus $f=b(x_{u},x_{v})\otimes g$, then by the definition of $\mathcal{B}$, from $f\notin\mathcal{B}$ we obtain $g\notin\mathcal{B}$, which has arity $2n-2$, completing the induction step using Lemma 2.15. Therefore, to apply Lemma 4.1 we want to show that there is a binary signature $b(x_{u},x_{v})$ such that $b(x_{u},x_{v})\mid\partial_{(ij)}f$ for every $\{i,j\}$ disjoint with $\{u,v\}$. We first consider the case that $\partial_{(uv)}f\equiv 0$ for some $\{u,v\}$.

Lemma 4.2**.**

Suppose $f$ has arity $\geqslant 4$ and $f\in\int\mathcal{B}$. If $\partial_{(uv)}f\equiv 0$ for some $\{u,v\}$, then the binary signature $b^{\mathfrak{i}}(x_{u},x_{v})=(0,\mathfrak{i},-\mathfrak{i},0)$ satisfies $b^{\mathfrak{i}}(x_{u},x_{v})\mid\partial_{(ij)}f$ for all $\{i,j\}$ disjoint with $\{u,v\}$.

Proof.

For any $\{i,j\}$ disjoint with $\{u,v\}$, the operations $\partial_{(ij)}$ and $\partial_{(uv)}$ commute. Since $\partial_{(uv)}f\equiv 0$, we have

[TABLE]

Since $\partial_{(ij)}f\in\mathcal{B}$, by Lemma 2.17, we have $b^{\mathfrak{i}}(x_{u},x_{v})\mid\partial_{(ij)}f$. ∎

In the following, for convenience we denote $\partial_{(ij)}(\partial_{(uv)}f)$ by $\partial_{(ij)(uv)}f$.

Now, we assume $\partial_{(ij)}f\in\mathcal{B}$ and $\partial_{(ij)}f\not\equiv 0$ for all $\{i,j\}$. We denote this property by $f\in\int\mathcal{B}_{\not\equiv 0}$. Each $\partial_{(ij)}f$ has a unique prime factorization. We will show that once we can find some binary signature $b(x_{u},x_{v})$ that divides a “triangle”, i.e. $b(x_{u},x_{v})|\partial_{(rs)}f,\partial_{(st)}f,\partial_{(rt)}f$ for three distinct $\{r,s,t\}$ disjoint with $\{u,v\}$, then it divides $\partial_{(ij)}f$ for all $\{i,j\}$ disjoint with $\{u,v\}$. We first consider the case that $b(x_{u},x_{v})$ divides “two pairs”. The statement of the following lemma is delicate.

Lemma 4.3**.**

Let $f$ be a signature of arity $2n\geqslant 8$ and $f\in\int\mathcal{B}_{\not\equiv 0}$. Suppose there exist two pairs of indices $\{s,t\}$ and $\{s^{\prime},t^{\prime}\}$ that are distinct but not necessarily disjoint, and a binary signature $b(x_{u},x_{v})$, where $\{u,v\}\cap(\{s,t\}\cup\{s^{\prime},t^{\prime}\})=\emptyset$, such that $b(x_{u},x_{v})\mid\partial_{(st)}f,\partial_{(s^{\prime}t^{\prime})}f$. Then for any $\{i,j\}$ disjoint with $\{u,v\}\cup\{s,t\}\cup\{s^{\prime},t^{\prime}\}$, if $\partial_{(st)(ij)}f\not\equiv 0$ and $\partial_{(s^{\prime}t^{\prime})(ij)}f\not\equiv 0$, then $b(x_{u},x_{v})\mid\partial_{(ij)}f$.

Proof.

By hypothesis $f\in\int\mathcal{B}_{\not\equiv 0}$, so for any $\{i,j\}$, we have $\partial_{(ij)}f\in\mathcal{B}$ and is nonzero, and thus it has a unique factorization with binary prime factors. Let $\{i,j\}$ be disjoint with $\{u,v\}\cup\{s,t\}\cup\{s^{\prime},t^{\prime}\}$. Suppose it satisfies the condition $\partial_{(st)(ij)}f\not\equiv 0$ and $\partial_{(s^{\prime}t^{\prime})(ij)}f\not\equiv 0$. We first prove that $x_{u}$ and $x_{v}$ must appear in one single binary prime factor $b^{\prime}(x_{u},x_{v})$ in the factorization of $\partial_{(ij)}f$. That is,

[TABLE]

where $g\not\equiv 0$ is a signature on variables other than $x_{u},x_{v},x_{i},x_{j}$. For a contradiction, suppose variables $x_{u}$ and $x_{v}$ appear in two distinct binary prime factors $b_{1}(x_{u},x_{u^{\prime}})$ and $b_{2}(x_{v},x_{v^{\prime}})$ in the prime factorization of $\partial_{(ij)}f$. Then,

[TABLE]

where $g^{\prime}\not\equiv 0$ is a signature on variables other than $x_{u},x_{u^{\prime}},x_{v},x_{v^{\prime}},x_{i},x_{j}$. By hypothesis, $b(x_{u},x_{v})\mid\partial_{(st)}f$, thus $\partial_{(st)}f=b(x_{u},x_{v})\otimes h$ for some $h$ on variables other than $x_{u},x_{v},x_{s},x_{t}$, which certainly include $x_{i},x_{j}$. Thus $\partial_{(ij)(st)}f=b(x_{u},x_{v})\otimes\partial_{(ij)}h$, and we have $b(x_{u},x_{v})|\partial_{(ij)(st)}f=\partial_{(st)(ij)}f$. By hypothesis for this $\{i,j\}$ we have $\partial_{(st)(ij)}f\not\equiv 0$. This implies that after merging variables $x_{s}$ and $x_{t}$ of $\partial_{(ij)}f$, $x_{u}$ and $x_{v}$ form a nonzero binary signature. By the form (4.2) of $\partial_{(ij)}f$, the only way $x_{u}$ and $x_{v}$ can form a nonzero binary signature in $\partial_{(st)(ij)}f$ is that the merge operation is actually merging $x_{u^{\prime}}$ and $x_{v^{\prime}}$. We conclude that $\{s,t\}=\{u^{\prime},v^{\prime}\}$. We can repeat the same proof replacing $\{s^{\prime},t^{\prime}\}$ for $\{s,t\}$, and since $b(x_{u},x_{v})\mid\partial_{(s^{\prime}t^{\prime})(ij)}f$ and $\partial_{(s^{\prime}t^{\prime})(ij)}f\not\equiv 0$, we have $\{s^{\prime},t^{\prime}\}=\{u^{\prime},v^{\prime}\}$. Hence, we have $\{s,t\}=\{s^{\prime},t^{\prime}\}$. This is a contradiction, and (4.2) does not hold.

Thus (4.1) holds. Since $\{s,t\}$ is disjoint with $\{u,v,i,j\}$, by the form (4.1) of $\partial_{(ij)}f$, when merging variables $x_{s}$ and $x_{t}$ of $\partial_{(ij)}f$, we actually merge variables $x_{s}$ and $x_{t}$ of $g$ and the binary signature $b^{\prime}(x_{u},x_{v})$ is not affected. Thus,

[TABLE]

That is, $b^{\prime}(x_{u},x_{v})\mid\partial_{(st)(ij)}f.$ By hypothesis we also have $b(x_{u},x_{v})\mid\partial_{(st)}f$. By the fact that $\{i,j\}$ is disjoint with $\{u,v,s,t\}$, we have $b(x_{u},x_{v})\mid\partial_{(ij)(st)}f=\partial_{(st)(ij)}f.$ Thus $b(x_{u},x_{v})$ and $b^{\prime}(x_{u},x_{v})$ both divide $\partial_{(st)(ij)}f\not\equiv 0$. By the unique factorization lemma (Lemma 2.13), we have $b(x_{u},x_{v})=\lambda b^{\prime}(x_{u},x_{v})$ for some $\lambda\neq 0$. In particular, by (4.1), $b(x_{u},x_{v})\mid\partial_{(ij)}f$. ∎

Now we come to the pivotal “triangle” lemma. Recall that the $\Delta$-property was defined just before Lemma 4.1. Suppose $f$ satisfies the $\Delta$-property, i.e., there is a binary $b(x_{u},x_{v})$ that divides a “triangle”, $b(x_{u},x_{v})\mid\partial_{(rs)}f,\partial_{(st)}f,\partial_{(rt)}f$. A key step in the proof of Lemma 4.4 is to show that for any $\{i,j\}$ disjoint with $\{u,v,r,s,t\}$, among the three iterated “derivatives” $\partial_{(rs)(ij)}f,\partial_{(st)(ij)}f$ and $\partial_{(rt)(ij)}f$, at most one of them can be identically zero. Then Lemma 4.3 applies.

Lemma 4.4**.**

Let $f\in\int\mathcal{B}_{\not\equiv 0}$ have arity $2n\geqslant 10$. Suppose $f$ satisfies the $\Delta$-property. Then there is a binary signature $b(x_{u},x_{v})$ such that for any $\{i,j\}$ disjoint with $\{u,v\}$, we have $b(x_{u},x_{v})\mid\partial_{(ij)}f$.

Proof.

By the $\Delta$-property, there is a binary signature $b(x_{u},x_{v})$ and $\{r,s,t\}$ disjoint with $\{u,v\}$ such that $b(x_{u},x_{v})\mid\partial_{(rs)}f,\partial_{(st)}f,\partial_{(rt)}f$. For any $\{i,j\}$ disjoint with $\{u,v\}$, we first consider the case that $\{i,j\}$ is also disjoint with $\{r,s,t\}$. Our idea is to show that among $\partial_{(rs)(ij)}f,\partial_{(st)(ij)}f$ and $\partial_{(rt)(ij)}f$, at most one of them can be a zero signature. This implies that there are two among these that are not identically zero. Then by Lemma 4.3, we have $b(x_{u},x_{v})\mid\partial_{(ij)}f$.

By Lemma 2.17, $\partial_{(rs)(ij)}f\equiv 0$ iff the binary signature $b^{\mathfrak{i}}(x_{r},x_{s})=(0,\mathfrak{i},-\mathfrak{i},0)$ divides $\partial_{(ij)}f$. Similarly, $\partial_{(st)(ij)}f\equiv 0$ iff $b^{\mathfrak{i}}(x_{s},x_{t})\mid\partial_{(ij)}f$, and $\partial_{(rt)(ij)}f\equiv 0$ iff $b^{\mathfrak{i}}(x_{r},x_{t})\mid\partial_{(ij)}f$. By hypothesis, $f\in\int\mathcal{B}_{\not\equiv 0}$, so $\partial_{(ij)}f\not\equiv 0$. The signature $\partial_{(ij)}f\in\mathcal{B}$ has a unique prime factorization. By Lemma 2.13, since the three signatures $b^{\mathfrak{i}}(x_{r},x_{s}),b^{\mathfrak{i}}(x_{s},x_{t})$ and $b^{\mathfrak{i}}(x_{r},x_{t})$ are on pairwise overlapping sets of variables, at most one of them can be a tensor factor of $\partial_{(ij)}f$. Thus, among $\partial_{(rs)(ij)}f,\partial_{(st)(ij)}f$ and $\partial_{(rt)(ij)}f$, at most one of them can be a zero signature, which implies $b(x_{u},x_{v})\mid\partial_{(ij)}f$, by Lemma 4.3, for all $\{i,j\}$ disjoint with $\{u,v,r,s,t\}$.

Now suppose $\{i,j\}$ is disjoint with $\{u,v\}$, but not disjoint with $\{r,s,t\}$. In the union $\{i,j\}\cup\{r,s,t\}\cup\{u,v\}$, there are at most $6$ distinct indices. Since the arity of $f$ is at least $10$, there are three indices $\{r^{\prime},s^{\prime},t^{\prime}\}$ such that $\{r^{\prime},s^{\prime},t^{\prime}\}$ is disjoint with $\{i,j\}\cup\{r,s,t\}\cup\{u,v\}$. Since $\{r^{\prime},s^{\prime}\}$ is disjoint with $\{u,v,r,s,t\}$, we can replace $\{i,j\}$ by $\{r^{\prime},s^{\prime}\}$ in the proof above for the case when $\{i,j\}$ is disjoint with $\{u,v,r,s,t\}$, and derive $b(x_{u},x_{v})\mid\partial_{(r^{\prime}s^{\prime})}f$. By the same reason, we also have $b(x_{u},x_{v})\mid\partial_{(s^{\prime}t^{\prime})}f$, and $b(x_{u},x_{v})\mid\partial_{(r^{\prime}t^{\prime})}f$. In other words we found a new “triangle”, that is, $f$ satisfies the $\Delta$-property with the binary signature $b(x_{u},x_{v})$ and the triple $\{r^{\prime},s^{\prime},t^{\prime}\}$ replacing $\{r,s,t\}$. Note that now $\{i,j\}$ is disjoint with $\{r^{\prime},s^{\prime},t^{\prime}\}$. So, we can apply the proof above with $\{r,s,t\}$ now replaced by $\{r^{\prime},s^{\prime},t^{\prime}\}$, and we conclude that $b(x_{u},x_{v})\mid\partial_{(ij)}f$. ∎

Remark**.**

This is the first place we require the arity of $f$ to be at least $10$.

We go for the kill in the next lemma.

Lemma 4.5**.**

Let $f\in\mathcal{F}$ be a signature of arity $2n\geqslant 10$, $f\notin\mathcal{B}$ and $f\in\int\mathcal{B}_{\not\equiv 0}$. Then

either $f$ satisfies the $\Delta$-property;

• 2.

*or there is a signature $f^{\prime}\not\in\mathcal{B}$ that has the same arity as $f$, such that ${\rm\#EO}(f^{\prime}\cup\mathcal{F})\leqslant_{T}{\rm\#EO}(\mathcal{F})$, and the following hold: either (1) $f^{\prime}\not\in\int\mathcal{B}$ or (2) $f^{\prime}$ satisfies the $\Delta$-property. *

Proof.

Consider $\partial_{(12)}f$. Since $\partial_{(12)}f\in\mathcal{B}$ and $\partial_{(12)}f\not\equiv 0$, without loss of generality, we may assume in the unique prime factorization of $\partial_{(12)}f$, variables $x_{3}$ and $x_{4}$ appear in one binary prime factor, $x_{5}$ and $x_{6}$ appear in one binary prime factor, and so on. That is,

[TABLE]

Case 1. For all $1\leqslant k\leqslant n-1$, $b_{k}(x_{2k+1},x_{2k+2})\neq$ a scalar multiple of $(0,\mathfrak{i},-\mathfrak{i},0)$.

Then by Lemma 2.17, $\partial_{(34)(12)}f\not\equiv 0$, and clearly, $b_{k}(x_{2k+1},x_{2k+2})|\partial_{(34)(12)}f$ for $k\geqslant 2$. In particular, we have

[TABLE]

since $f$ has arity at least $10$.

Now consider $\partial_{(34)}f$. We have $\partial_{(34)}f\in\mathcal{B}$, $\partial_{(34)}f\not\equiv 0$, and $\partial_{(12)(34)}f=\partial_{(34)(12)}f\not\equiv 0$.

• 1.

If $x_{1}$ and $x_{2}$ appear in one binary prime factor $b^{\prime}_{1}(x_{1},x_{2})$ in the unique prime factorization of $\partial_{(34)}f$, then after merging variables $x_{1}$ and $x_{2}$, the binary signature $b^{\prime}_{1}(x_{1},x_{2})$ becomes a nonzero constant, but all other binary prime factors of $\partial_{(34)}f$ are unchanged and appear in the prime factorization of $\partial_{(12)(34)}f$. By commutativity $\partial_{(12)(34)}f$ = $\partial_{(34)(12)}f$, and by (4.3) the prime factors of $\partial_{(12)(34)}f$ are precisely $b_{k}(x_{2k+1},x_{2k+2})$, for $2\leqslant k\leqslant n-1$, we conclude that the unique prime factorization of $\partial_{(34)}f$ has the following form (up to a nonzero constant)

[TABLE]

• 2.

If $x_{1}$ and $x_{2}$ appear in two distinct binary prime factors $b^{\prime\prime}_{1}(x_{1},x_{i})$ and $b^{\prime\prime}_{2}(x_{2},x_{j})$ in the unique prime factorization of $\partial_{(34)}f$, then after merging variables $x_{1}$ and $x_{2}$, from (4.3) we have

[TABLE]

for some nonzero constant $c$. On the other hand, from the form of $\partial_{(34)}f$, the two variables $x_{i}$ and $x_{j}$ form a new nonzero binary $b^{\prime\prime}(x_{i},x_{j})$. Thus the pair $\{i,j\}$ is either $\{5,6\}$, or $\{7,8\}$, etc. and we may assume $(i,j)=(5,6)$ by renaming the variables. Thus, we have

[TABLE]

(In the following proof we can use any $b_{j}$, for $4\leqslant j\leqslant n-1$; for definiteness we set $j=4$, and since $n\geqslant 5$ this choice $b_{4}$ is permissible.) In both cases above, we have $b_{4}(x_{9},x_{10})|\partial_{(34)}f$, and $\partial_{(78)(34)}f\not\equiv 0$ since $b_{3}(x_{7},x_{8})\neq(0,\mathfrak{i},-\mathfrak{i},0)$ by assumption. Moreover, note that in both cases, $x_{6}$ and $x_{7}$ do not appear as the two variables of a single binary signature tensor factor of $\partial_{(34)}f$. The same is true for $x_{6}$ and $x_{8}$. This implies that $\partial_{(67)(34)}f\not\equiv 0$ and $\partial_{(68)(34)}f\not\equiv 0$. So we have derived

[TABLE]

Clearly, by (4.3), we also have

[TABLE]

Apply Lemma 4.3 three times (with $\{u,v\}=\{9,10\},\{s,t\}=\{1,2\},\{s^{\prime},t^{\prime}\}=\{3,4\},$ and taking $\{i,j\}=\{6,7\},\{7,8\},\{6,8\}$ separately), we have

[TABLE]

Thus $f$ satisfies the $\Delta$-property ($\{u,v\}=\{9,10\}$ and $\{r,s,t\}=\{6,7,8\}$) and we are done.

Case 2. There is a binary signature $b_{k-1}(x_{2k-1},x_{2k})$ in the factorization of $\partial_{(12)}f$ such that $b_{k-1}(x_{2k-1},x_{2k})=$ a scalar multiple of $(0,\mathfrak{i},-\mathfrak{i},0)$. Then by Lemma 2.15, we have the reduction #EO$((0,\mathfrak{i},-\mathfrak{i},0),f)\leqslant_{T}$#EO$(f)$. Connecting the variable $x_{2k-1}$ of $f$ with $(0,\mathfrak{i},-\mathfrak{i},0)$, we can realize a signature $f^{\prime}$. Consider $\partial_{(12)}f^{\prime}$. Again the operations commute: it is the same as connecting the variable $x_{2k-1}$ of $\partial_{(12)}f$ with $(0,\mathfrak{i},-\mathfrak{i},0)$. Since $\partial_{(12)}f$ is a tensor product of binary signatures, connecting the variable $x_{2k-1}$ of $\partial_{(12)}f$ with $(0,\mathfrak{i},-\mathfrak{i},0)$ is just connecting the variable $x_{2k-1}$ of the binary $b_{k-1}(x_{2k-1},x_{2k})$ with $(0,\mathfrak{i},-\mathfrak{i},0)$, which gives a binary $(0,1,1,0)$. That is, $\partial_{(12)}f^{\prime}$ is still a tensor product of the same binary signatures as in $\partial_{(12)}f$ except that $b_{k-1}(x_{2k-1},x_{2k})=(0,\mathfrak{i},-\mathfrak{i},0)$ is replaced by $b^{\prime}_{k-1}(x_{2k-1},x_{2k})=(0,1,1,0)$. Similarly, for any binary signature $b_{\ell-1}(x_{2\ell-1},x_{2\ell})=(0,\mathfrak{i},-\mathfrak{i},0)$ in $\partial_{(12)}f$, we modify it in this way (together all at once). Thus, we can realize a signature $f^{\prime}$ by connecting some variables with $(0,\mathfrak{i},-\mathfrak{i},0)$ such that

[TABLE]

where $b^{\prime}_{k}(x_{2k+1},x_{2k+2})\neq$ a scalar multiple of $(0,\mathfrak{i},-\mathfrak{i},0)$ for any $1\leqslant k\leqslant n-1$. Moreover, we know $f^{\prime}\notin\mathcal{B}$ since $f\notin\mathcal{B}$; this follows from the closure property of $\mathcal{B}$ under the operation of connecting a variable by $(0,\mathfrak{i},-\mathfrak{i},0)$ via $\neq_{2}$, and the fact that if we connect three times $(0,\mathfrak{i},-\mathfrak{i},0)$ via $\neq_{2}$ in a chain from $f^{\prime}$, we get $f$ back: $\left(N\left[\begin{smallmatrix}0&\mathfrak{i}\\ -\mathfrak{i}&0\end{smallmatrix}\right]\right)^{4}=I$.

If $f^{\prime}\notin\int\mathcal{B}$, we are done. Otherwise, $f^{\prime}\in\int\mathcal{B}$. If there is $\{u,v\}$ such that $\partial_{(uv)}f^{\prime}\equiv 0$, then by Lemma 4.2, we have $b^{\mathfrak{i}}(x_{u},x_{v})\mid\partial_{(ij)}f^{\prime}$ for any $\{i,j\}$ disjoint with $\{u,v\}$ where $b^{\mathfrak{i}}(x_{u},x_{v})=(0,\mathfrak{i},-\mathfrak{i},0)$. Then clearly $f^{\prime}$ satisfies the $\Delta$-property. Otherwise, $f^{\prime}\in\int\mathcal{B}_{\not\equiv 0}$. As we just proved in Case 1, now replacing $f$ by $f^{\prime}$, we have $b^{\prime}_{4}(x_{9},x_{10})\mid\partial_{(67)}f^{\prime},\partial_{(78)}f^{\prime},\partial_{(68)}f^{\prime}.$ This completes the proof.

∎

Remark**.**

This proof also requires the arity of $f$ to be at least $10$.

Lemma 4.6** (Induction).**

If $\mathcal{F}$ contains a signature $f\notin\mathcal{B}$ of arity $2n\geqslant 10$, then there is a signature $g\notin\mathcal{B}$ of arity $2n-2$ such that #EO$(\{g\}\cup\mathcal{F})\leqslant_{T}{\rm\#EO}(\mathcal{F})$.

Proof.

If $f\not\in\int\mathcal{B}$, then there exists $\{i,j\}$ such that $\partial_{(ij)}f\notin\mathcal{B}$, and we are done by choosing $g=\partial_{(ij)}f$. Thus, we assume $f\in\int\mathcal{B}$. If $\partial_{(uv)}f\equiv 0$ for some indices $\{u,v\}$, then by Lemmas 4.2 and 4.1, the binary signature $b^{\mathfrak{i}}(x_{u},x_{v})=(0,\mathfrak{i},-\mathfrak{i},0)$ divides $f$. That is, $f=b^{\mathfrak{i}}(x_{u},x_{v})\otimes g$ where $g$ is a signature of arity $2n-2$, and $g\notin\mathcal{B}$ since $f\notin\mathcal{B}$. By Lemma 2.15, we have #EO$(\{g\}\cup\mathcal{F})\leqslant_{T}{\rm\#EO}(\mathcal{F})$. So we may assume $f\in\int\mathcal{B}_{\not\equiv 0}$. Now we apply Lemma 4.5. If the first alternative of Lemma 4.5 holds, then $f$ satisfies the $\Delta$-property. Then by Lemmas 4.4 and 4.1, there is a binary signature $b(x_{u},x_{v})$ such that $b(x_{u},x_{v})\mid f$. This divisibility of $f$ produces a signature not in $\mathcal{B}$ of arity $2n-2$ similar to what we have just proved, and we are done. If the second alternative of Lemma 4.5 holds, then we have a signature $f^{\prime}\not\in\mathcal{B}$ having the same arity as $f$. We have ${\rm\#EO}(\{f^{\prime}\}\cup\mathcal{F})\leqslant_{T}{\rm\#EO}(\mathcal{F})$. If $f^{\prime}\not\in\int\mathcal{B}$, then there exists $\{i,j\}$ such that $\partial_{(ij)}f^{\prime}\notin\mathcal{B}$, and we can take $\partial_{(ij)}f^{\prime}$ as $g$, and so we are done. Otherwise, by the conclusion of Lemma 4.5, $f^{\prime}$ satisfies the $\Delta$-property. Similar to the proof above for $f$, there is a binary signature $b(x_{u},x_{v})$ such that $b(x_{u},x_{v})\mid f^{\prime}$. This divisibility of $f^{\prime}$ produces a signature not in $\mathcal{B}$ of arity $2n-2$. This completes the inductive step. ∎

Now, we use the orthogonality property to prove the base cases.

Lemma 4.7** (Base cases).**

*If $\mathcal{F}$ contains a signature $f\notin\mathcal{B}$ of arity $4$, $6$ or $8$, then either *#EO$(\mathcal{F})$ is #P-hard or #EO$(\{\neq_{4}\}\cup\mathcal{F})\leqslant_{T}$ #EO$(\mathcal{F})$.

Proof.

Again by Lemma 2.15, we may assume $f$ is irreducible. Otherwise, we just need to analyze each irreducible factor of $f$. More specifically, if $f\notin\mathcal{B}$ and $f$ is reducible, then there exists an irreducible factor $g$ of $f$ such that $g\notin\mathcal{B}$, and $g$ has arity $4$ or $6$. If we can use $g$ to realize a #P-hard signature or $\neq_{4}$, we can also use $f$ to do so.

By Lemma 3.2, we may assume that $f$ satisfies the orthogonality. Otherwise, we are done.

Therefore, we have

[TABLE]

for any $(a,b)\in\{0,1\}^{2}$, and any pair $\{i,j\}$. This readily leads to a contradiction for signatures of arity $4$ as follows. Suppose $f$ is an irreducible signature on four variables $x_{1},x_{2},x_{3},x_{4}$. Let $(i,j,k,\ell)$ be an arbitrary permutation of $\{1,2,3,4\}$. Consider the vector ${\bf f}_{ij}^{00}$. It has only one possible nonzero entry ${f}_{ijk\ell}^{0011}$ since the support of $f$ is on half weight. Thus,

[TABLE]

for any $(x_{i},x_{j},x_{k},x_{\ell})=(0,0,1,1)$. Since $(i,j,k,\ell)$ is an arbitrary permutation of $\{1,2,3,4\}$, $f_{ijk\ell}^{0011}$ is an arbitrary entry of $f$ at half weight, and since $f$ is nonzero, every weight two entry of $f$ has the same nonzero norm $\sqrt{\lambda}$. However, consider the vector ${\bf f}_{ij}^{01}$, it has two nonzero entries $f_{ijk\ell}^{0101}$ and $f_{ijk\ell}^{0110}$. Hence,

[TABLE]

which means $\lambda=0$. This is a contradiction.

Before we go into the technical details of the proof for signatures of arity $6$ and $8$, we first give some intuitions. By considering the norm-squares of entries in $f$ as unknowns, the orthogonality property of $f$ actually gives a linear system. Our proof is to show that when $f$ has small arity $4,6,8$, the solution region of such a system only has the trivial zero point. We illustrate this perspective by the arity $4$ case. Suppose $f$ has arity $4$. It has ${4\choose 2}=6$ possible nonzero entries. These entries satisfy the orthogonality condition. We have

[TABLE]

for any $\{i,j\}\subseteq\{1,2,3,4\}$. There are ${4\choose 2}\times 4=24$ many equations in total. If we view these norm-squares of entries $|f^{0011}|^{2}$, $|f^{0101}|^{2}$, $|f^{0110}|^{2}$, $|f^{1001}|^{2}$, $|f^{1010}|^{2}$, $|f^{1100}|^{2}$ (we omit subscripts here) and the value $\lambda$ as variables, those equations are linear equations on these variables. By ars, we have $|f^{0011}|^{2}=|f^{1100}|^{2}$, $|f^{0101}|^{2}=|f^{1010}|^{2}$, and $|f^{0110}|^{2}=|f^{1001}|^{2}$. So there are only four variables. Our idea is to show that the matrix of this linear system which has $24$ many rows but only $4$ columns has full rank. We only need $4$ rows to prove this. In our proof for arity 4, we picked the following 4 rows and showed that the induced linear system has full rank:

[TABLE]

For the arity $6$ case, we will basically show the same thing (i.e., the linear system has only the trivial zero solution) with some carefully chosen rows. For arity $8$ case, we will use the fact that the variables take nonnegative values and we show the linear system has no nonnegative solution except the zero solution.

An intuitive reason why this proof could succeed for signatures of small arity is that in these cases, we have more equations than variables, which leads to an over-determined linear system. For the general case of arity $n$, there are $4{n\choose 2}$ many equations but ${n\choose n/2}/2+1$ many variables. Since $4{n\choose 2}\ll{n\choose n/2}/2+1$ when $n$ is large, this method will not work for large $n$. This is why we cannot hope to apply this proof to signatures of large arity.

Now, we give the formal proof for signatures of arity 6 and 8. In what follows we assume $f$ has arity $\geqslant 6$. Given a vector ${\bf f}_{ij}^{ab}$, we can pick a third variable $x_{k}$ and separate ${\bf f}_{ij}^{ab}$ into two vectors ${\bf f}_{ijk}^{ab0}$ and ${\bf f}_{ijk}^{ab1}$ according to $x_{k}=0$ or $1$. By setting $(a,b)=(0,0)$, we have

[TABLE]

Similarly, we consider the vector ${\bf f}_{ik}^{00}$ and separate it according to $x_{j}=0$ or $1$. We have

[TABLE]

Comparing equations (4.4) and (4.5), we have $|{\bf f}_{ijk}^{001}|^{2}=|{\bf f}_{ijk}^{010}|^{2}$. Moreover, by ars, we have $|{\bf f}_{ijk}^{010}|^{2}=|{\bf f}_{ijk}^{101}|^{2}.$ Thus, we have $|{\bf f}_{ijk}^{001}|^{2}=|{\bf f}_{ijk}^{101}|^{2}$. Note that the vector ${\bf f}_{jk}^{01}$ can be separated into two vectors ${\bf f}_{ijk}^{001}$ and ${\bf f}_{ijk}^{101}$ according to $x_{i}=0$ or $1$. Therefore,

[TABLE]

Thus, we have $|{\bf f}_{ijk}^{001}|^{2}=|{\bf f}_{ijk}^{101}|^{2}=\lambda/2$. Then, by equation (4.4), we have $|{\bf f}_{ijk}^{000}|^{2}=\lambda/2$, and again by ars, we also have $|{\bf f}_{ijk}^{111}|^{2}=|{\bf f}_{ijk}^{000}|^{2}=\lambda/2$. Note that the indices $i,j,k$ can be arbitrary three distinct indices, by symmetry we have

[TABLE]

for $f$ of arity $\geqslant 6$, and for all $(x_{i},x_{j},x_{k})=(a,b,c)\in\{0,1\}^{3}.$

This leads to a contradiction for signatures of arity $6$. Suppose $f$ is an irreducible signature on $6$ variables $x_{1},x_{2},\ldots,x_{6}$. Let $(i,j,k,i^{\prime},j^{\prime},k^{\prime})$ be an arbitrary permutation of $\{1,2,\ldots,6\}$. Note that the vector ${\bf f}_{ijk}^{000}$ has only one possible nonzero entry $f_{ijki^{\prime}j^{\prime}k^{\prime}}^{000111}$. Thus, by (4.6) we have

[TABLE]

for any $(x_{i},x_{j},x_{k},x_{i^{\prime}},x_{j^{\prime}},x_{k^{\prime}})=(0,0,0,1,1,1)$. That is, any entry of $f$ at half weight has the same nonzero norm $\sqrt{\lambda/2}$. However, the vector ${\bf f}_{ijk}^{001}$ has ${3\choose 2}=3$ nonzero entries. But,

[TABLE]

which means $\lambda=0$. This is a contradiction.

For signatures of arity $8$, we need to go further and use the fact that the norm-square is nonnegative. Given a vector ${\bf f}_{ijk}^{abc}$, we can continue to pick a fourth variable $x_{\ell}$ and separate ${\bf f}_{ijk}^{abc}$ into two vectors ${\bf f}_{ijk\ell}^{abc0}$ and ${\bf f}_{ijk\ell}^{abc1}$ according to $x_{\ell}=0$ or $1$. By setting $(a,b,c)=(0,0,0)$, we have from (4.6)

[TABLE]

Similarly, we consider the vector ${\bf f}_{ij\ell}^{001}$ and separate it according to $x_{k}=0$ or $1$. We have

[TABLE]

Comparing equations (4.7) and (4.8), we have $|{\bf f}_{ijk\ell}^{0000}|^{2}=|{\bf f}_{ijk\ell}^{0011}|^{2}$. This leads to a contradiction for signatures of arity $8$.

Suppose $f$ is an irreducible signature on $8$ variables $x_{1},x_{2},\ldots,x_{8}$. Let $(i,j,k,\ell,i^{\prime},j^{\prime},k^{\prime},\ell^{\prime})$ be an arbitrary permutation of $\{1,2,\ldots,8\}$. The vector ${\bf f}_{ijk\ell}^{0000}$ has only one possible nonzero entry $f_{ijk\ell i^{\prime}j^{\prime}k^{\prime}\ell^{\prime}}^{00001111}$. Thus,

[TABLE]

The vector ${\bf f}_{ijk\ell}^{0011}$ has ${4\choose 2}=6$ possible nonzero entries including $f_{ijk\ell i^{\prime}j^{\prime}k^{\prime}\ell^{\prime}}^{00110011}$. Thus,

[TABLE]

where $\Delta$ denotes the sum of norm-squares of the other $5$ entries in ${\bf f}_{ijk\ell}^{0011}$ and we know $\Delta\geqslant 0$. Since the left-hand sides of equations (4.9) and (4.10) are equal, we have

[TABLE]

Similarly, consider vectors ${\bf f}_{iji^{\prime}j^{\prime}}^{0000}$ and ${\bf f}_{iji^{\prime}j^{\prime}}^{0011}$. We have $|{\bf f}_{iji^{\prime}j^{\prime}}^{0000}|^{2}=|{\bf f}_{iji^{\prime}j^{\prime}}^{0011}|^{2}$. The vector ${\bf f}_{iji^{\prime}j^{\prime}}^{0000}$ has only one possible nonzero entry. Thus,

[TABLE]

The vector ${\bf f}_{iji^{\prime}j^{\prime}}^{0011}$ has $6$ possible nonzero entries. Thus,

[TABLE]

where $\Delta^{\prime}$ denotes the sum of norm-squares of the other $5$ entries in ${\bf f}_{iji^{\prime}j^{\prime}}^{0011}$ and we know $\Delta^{\prime}\geqslant 0$. Thus, we have

[TABLE]

Comparing equations (4.11) and (4.12), we have $\Delta=-\Delta^{\prime}$, which means $\Delta=\Delta^{\prime}=0$ due to $\Delta\geqslant 0$ and $\Delta^{\prime}\geqslant 0$. Since $\Delta$ is the sum of $5$ norm-squares, each of which is nonnegative, $\Delta=0$ means each norm-square in the sum $\Delta$ is [math]. In particular, $|f_{ijk\ell i^{\prime}j^{\prime}k^{\prime}\ell^{\prime}}^{00111100}|^{2}$ is a term in the sum $\Delta$. We have $|f_{ijk\ell i^{\prime}j^{\prime}k^{\prime}\ell^{\prime}}^{00111100}|^{2}=0$. Since the order of indices is picked arbitrarily, all entries of $f$ are zero. Thus, $f$ is a zero signature. A contradiction. ∎

Theorem 4.8**.**

*If $\mathcal{F}\not\subseteq\mathcal{B}$, then either *#EO$(\mathcal{F})$ is #P-hard or #EO$(\{\neq_{4}\}\cup\mathcal{F})\leqslant_{T}$ #EO$(\mathcal{F})$.

Proof.

The base case is Lemma 4.7 and the inductive step is Lemma 4.6. Done by induction. ∎

5 Reduction from #CSP to #EO Problems

In this section, we will show $\rm{\#EO}(\{\neq_{4}\}\cup\mathcal{F})$ is #P-hard unless $\mathcal{F}\subseteq\mathscr{A}$ or $\mathcal{F}\subseteq\mathscr{P}$. The first steps are simple; the availability of $\neq_{4}$ allows us to realize any $(\neq_{2k})$ and therefore all of $\mathcal{DEQ}$.

Lemma 5.1**.**

$\rm{\#EO}(\mathcal{DEQ}\cup\mathcal{F})\leqslant_{T}\rm{\#EO}(\{\neq_{4}\}\cup\mathcal{F})$.

Proof.

Connecting $\neq_{2k}$ ($k\geqslant 2$) and $\neq_{4}$ using $\neq_{2}$ we get $\neq_{2k+2}$. Every occurrence of signatures in $\mathcal{DEQ}$ can be realized by a linear size gadget. Then we have $\rm{\#EO}(\mathcal{DEQ}\cup\mathcal{F})\leqslant_{T}\rm{\#EO}(\{\neq_{4}\}\cup\mathcal{F})$. ∎

Recall that $\rm{\#EO}(\mathcal{DEQ}\cup\mathcal{F})$ is just $\operatorname{Holant}\left(\neq_{2}\mid\mathcal{DEQ}\cup\mathcal{F}\right)$ expressed in the Holant framework. We show that after we get $\mathcal{DEQ}$ on the right hand side (RHS) in the above Holant problem, we can also realize $\mathcal{DEQ}$ on the left-hand side (LHS).

Lemma 5.2**.**

$\operatorname{Holant}\left(\mathcal{DEQ}\mid\mathcal{F}\right)\leqslant_{T}\operatorname{Holant}\left(\neq_{2}\mid\mathcal{DEQ}\cup\mathcal{F}\right)$, which is equivalent to $\rm{\#EO}(\mathcal{DEQ}\cup\mathcal{F})$.

Proof.

In $\operatorname{Holant}\left(\neq_{2}\mid\mathcal{DEQ}\cup\mathcal{F}\right)$ we take $2k$ copies of $\neq_{2}$ on the LHS and connect one variable of each copy of $\neq_{2}$ to all $2k$ variables of one copy of $\neq_{2k}$ on the RHS. This gives us the constraint function $\neq_{2k}$ on the LHS. ∎

Now, consider an arbitrary instance of $\operatorname{Holant}\left(\mathcal{DEQ}\mid\mathcal{F}\right)$; it is given by a bipartite graph. Similar to how we express #CSP$(\mathcal{F})$ using $\operatorname{Holant}\left(\mathcal{EQ}\mid\mathcal{F}\right)$, in $\operatorname{Holant}\left(\mathcal{DEQ}\mid\mathcal{F}\right)$ we can view vertices on the LHS (labeled by $(\neq_{2k})\in\mathcal{DEQ}$) as variables, and vertices on the RHS (labeled by $f\in\mathcal{F}$) as constraints. However, the difference here is that in this setting, both a variable itself and its negation appear as input variables of constraints, and they always appear the same number of times. More specifically, for a variable vertex $x$ labeled by $\neq_{2k}$, the entire set of $2k$ edges incident to $x$ can be divided into two subsets, each of which consisting of $k$ edges. In each subset, every edge takes the same value, while two edges in different sets always take opposite values. Then, we can view the $k$ edges in one subset as the variable $x$ appearing $k$ times, while another $k$ edges in the other subset as its negation $\overline{x}$ appearing $k$ times.

Recall that signatures $f\in\mathcal{F}$ satisfy ars. Suppose $f\in\mathcal{F}$ has arity $2n$. Then, consider the function $f(\overline{x_{1}},\overline{x_{2}},\ldots,\overline{x_{2n}})$. That is, we replace the input variables by their negations. Then we have $f(\overline{x_{1}},\overline{x_{2}},\ldots,\overline{x_{2n}})=\overline{f(x_{1},x_{2},\ldots,x_{2n})}$ by ars. Define the norm square function $|f|^{2}$, which takes value $|f(x_{1},\ldots,x_{2n})|^{2}$ on input $(x_{1},\ldots,x_{2n})$. Then, we have

[TABLE]

and this gives the following reduction.

Lemma 5.3**.**

Let $|\mathcal{F}|^{2}=\{|f|^{2}\mid f\in\mathcal{F}\}.$ Then #CSP$(|\mathcal{F}|^{2})\leqslant_{T}\operatorname{Holant}\left(\mathcal{DEQ}\mid\mathcal{F}\right).$

Proof.

Given an instance $I$ of #CSP$(|\mathcal{F}|^{2})$ over $m$ variables. Suppose it contains $\ell$ occurrences of constraints $|f_{i}|^{2}\in|\mathcal{F}|^{2}$ $(i\in[\ell])$ of arity $2n_{i}$, and $f_{i}$ is applied to the variables $x_{i_{1}},\ldots,x_{i_{2n_{i}}}$. Then

[TABLE]

Notice that in the final form of (5.1), for each variable $x\in\{x_{1},\ldots,x_{m}\}$, both itself and its negation appear as input variables to various constraints $f_{i}\in\mathcal{F}$. Moreover, there is a one-to-one correspondence between each occurrence of $x$ and that of $\bar{x}$. Thus, $x$ and $\bar{x}$ appear the same number of times. Thus the partition function ${\rm\#CSP}(I)$ for the #CSP$(|\mathcal{F}|^{2})$ problem can be expressed as the partition function of an instance of $\operatorname{Holant}\left(\mathcal{DEQ}\mid\mathcal{F}\right)$ of polynomially bounded size. ∎

Directly by this reduction, we have the following hardness result. Corollary 5.4 follows from Theorem 2.24.

Corollary 5.4**.**

$\operatorname{Holant}\left(\mathcal{DEQ}\mid\mathcal{F}\right)$* is #P-hard if there is some $f\in\mathcal{F}$ such that $\mathscr{S}(f)$ is not affine.*

Proof.

By the definition of $|f|^{2}$, we know $\mathscr{S}(|f|^{2})=\mathscr{S}(f)$. Thus, there is some $|f|^{2}\in|\mathcal{F}|^{2}$ such that $\mathscr{S}(|f|^{2})$ is not affine. This implies that $|\mathcal{F}|^{2}\not\subseteq\mathscr{A}$. Moreover, by Lemma 2.20, we also have $|\mathcal{F}|^{2}\not\subseteq\mathscr{P}$. By Theorem 2.24, #CSP$(|\mathcal{F}|^{2})$ is #P-hard and hence, by Lemma 5.3, $\operatorname{Holant}\left(\mathcal{DEQ}\mid\mathcal{F}\right)$ is #P-hard. ∎

Now, we may assume every signature $f\in\mathcal{F}$ has affine support. Quite amazingly, if an EO signature has affine support, then its support must have a special structure, called pairwise opposite.

Definition 5.5** (Pairwise opposite).**

Let $\mathscr{S}\subseteq\mathbb{Z}_{2}^{2n}$ be an affine linear subspace. We say $\mathscr{S}$ is pairwise opposite if we can partition the $2n$ variables into $n$ pairs such that on $\mathscr{S}$, two variables of each pair always take opposite values. If $\mathscr{S}$ is pairwise opposite, we fix a pairing. Then each pair under this paring is called an opposite pair.

Example 5.6**.**

Let $\mathscr{S}=\{(x_{1},x_{2},\ldots,x_{2n})\mid x_{1},\ldots,x_{2n}\in\mathbb{Z}_{2},\overline{x_{i}}=x_{n+i}\ (i\in[n])\}$. Then $\mathscr{S}$ is pairwise opposite. Moreover, any affine linear subspace of $\mathscr{S}$ is pairwise opposite.

For instance, let $C$ be the Hamming $(7,4)$-code. We consider its dual Hamming code $C^{\perp}$. $C^{\perp}$ is a linear subspace of $\mathbb{Z}_{2}^{7}$ of dimension 3. Let

[TABLE]

*Then $\mathscr{S}_{C}$ is pairwise opposite. This $\mathscr{S}_{C}$ is introduced in [17] related to a certain tractable family of signatures for a class of Holant problems. *

Note that if an affine linear subspace $\mathscr{S}\subseteq\mathbb{Z}_{2}^{2n}$ is pairwise opposite, then $\mathscr{S}\subseteq\mathscr{H}_{2n}$. Now, we show the other direction is also true. This result should be of independent interest.

Lemma 5.7**.**

Let $\mathscr{S}\subseteq\mathbb{Z}_{2}^{2n}$ be an affine linear subspace. If $\mathscr{S}\subseteq\mathscr{H}_{2n}$, then $\mathscr{S}$ is pairwise opposite.

Proof.

The lemma is trivially true if $|\mathscr{S}|=0,1$. Suppose $\dim(\mathscr{S})=k\geqslant 1$. We can pick a set of free variables $F=\{x_{1},\ldots,x_{k}\}$, then on $\mathscr{S}$, every variable $x$ is expressible as a unique affine linear combination over $\mathbb{Z}_{2}$ of these free variables, $x=\lambda_{1}x_{1}+\ldots+\lambda_{k}x_{k}+\lambda_{k+1}$, where $\lambda_{1},\ldots,\lambda_{k+1}\in\mathbb{Z}_{2}$. (If $x$ takes a constant value on $\mathscr{S}$, it is still an affine linear combination of these free variables.)

We separate out all $2n$ variables into two types, those with $\lambda_{k+1}=0$ (linear form) and those with $\lambda_{k+1}=1$ (affine, but not linear form). If we set all free variables $x_{1},\ldots,x_{k}$ to 0, we get a vector $\alpha\in\mathscr{S}$ with ${\rm wt}(\alpha)=n$. Each $x$ of the first type contributes zero and each $x$ of the second type contributes one. Hence among all $2n$ variables, there are exactly $n$ variables of each type, and the chosen free variables are among the first type. Without loss of generality, we may assume variables of the first and second type are $U=\{x_{1},\ldots,x_{n}\}$ and $V=\{x_{n+1},\ldots,x_{2n}\}$.

For any variable $x=\lambda_{1}x_{1}+\ldots+\lambda_{k}x_{k}+\lambda_{k+1}$, with respect to this unique affine linear expression, let $\Lambda(x)=\{i\in[k]\mid\lambda_{i}=1\}$, the set of free variables that do appear in the expression of $x$. We have,

[TABLE]

Clearly, for $i\in[k]$, $\Lambda(x_{i})=\{i\}$. For any subset $I\subseteq[k]$, we let

[TABLE]

Define $V^{\subseteq}(I)$ and $V^{=}(I)$ analogously, with $V$ in place of $U$. For any subset $I\subseteq[k]$, let $\alpha^{I}\in\mathscr{S}$ be the vector determined by setting free variables $x_{i}=1$ for $i\in I$ and $x_{i}=0$ for $i\in[k]-I$. Within the $2n$ bit positions in the vector $\alpha^{I}$, for any variable $x\in U$,

[TABLE]

Symmetrically for any variable $x\in V$, we have

[TABLE]

Let $U^{\rm odd}(I)=\{x\in U\mid|I\cap\Lambda(x)|\text{ is odd}\}$ and $V^{\rm odd}(I)=\{x\in V\mid|I\cap\Lambda(x)|\text{ is odd}\}$. Since

[TABLE]

we have $|U^{\rm odd}(I)|=|V^{\rm odd}(I)|$, for all $I\subseteq[k]$.

Claim 1. For all $I\subseteq[k]$,

[TABLE]

To prove this Claim, we count the contributions of every $x\in U$ to both sides of the equation. For $x\in U$, let $m(x)=|I\cap\Lambda(x)|$. This $x$ contributes one or zero to the LHS, according to whether $m(x)$ is odd or even respectively. On the RHS, its contribution is

[TABLE]

which is also precisely one or zero according to whether $m(x)$ is odd or even respectively.

The same statement is true for $V^{\rm odd}(I)$ replacing $U$ by $V$, with the same proof.

Claim 2. For all $I\subseteq[k]$,

[TABLE]

We show next that $|U^{\rm\subseteq}(I)|=|V^{\rm\subseteq}(I)|$ for all $I\subseteq[k]$. If $I=\emptyset$, then $U^{\subseteq}(I)=U$ and $V^{\subseteq}(I)=V$, and so they have the same cardinality, both being $n$. Inductively, for any $I\subseteq[k]$, suppose we already know that $|U^{\subseteq}(J)|=|V^{\subseteq}(J)|$, for all proper subsets $J\subsetneq I$, then since $|U^{\rm odd}(I)|=|V^{\rm odd}(I)|$, by the two Claims we have $|U^{\subseteq}(I)|=|V^{\subseteq}(I)|$ as well, since the coefficient $(-2)^{|I|-1}\not=0$.

Then, by definition

[TABLE]

By the Möbius inversion formula, we have

[TABLE]

Indeed,

[TABLE]

and for a proper containment $I\subsetneq J^{\prime}$ the coefficient of $|U^{=}(J^{\prime})|$ is $(1-1)^{|J^{\prime}|-|I|}=0$, and it is 1 for $I=J^{\prime}$.

The same statement is true for $V$. Thus, we have $|U^{=}(I)|=|V^{=}(I)|$ for all $I\subseteq[k]$.

This allows us to set up a pairing between $U$ and $V$ such that for each pair of paired variables $(x,y)\in U\times V$, we have $\Lambda(x)=\Lambda(y)$. For any $I\subseteq[k]$, we arbitrarily pick a pairing between $U^{=}(I)$ and $V^{=}(I)$. This is achievable because they have the same cardinality. Since the following decompositions for both $U$ and $V$ are disjoint unions

[TABLE]

we get a global pairing between $U$ and $V$, such that for each pair of paired variables $(x,y)\in U\times V$, we have $\Lambda(x)=\Lambda(y)$. Recall that on $\mathscr{S}$, since $x\in U$, we have $x=\sum_{i\in\Lambda(x)}x_{i}$; meanwhile since $y\in V$ we have $y=1+\sum_{i\in\Lambda(y)}x_{i}$. It follows that $\overline{x}=y$ on $\mathscr{S}$. ∎

Now, we are going to simulate #CSP$(\mathcal{F})$ using $\operatorname{Holant}\left(\mathcal{DEQ}\mid\mathcal{F}\right)$ when $\mathcal{F}$ consists of signatures with affine support. Suppose $f(x_{1},\ldots,x_{2n})\in\mathcal{F}$ has affine support, by Lemma 5.7, we know $\mathscr{S}(f)$ is pairwise opposite. By permuting variables, we may assume for $i\in[n]$, $(x_{i},x_{n+i})$ is paired as an opposite pair. Then, we have the following reduction.

Lemma 5.8**.**

Suppose $\mathcal{F}$ is a set of EO signatures. If every signature $f\in\mathcal{F}$ has affine support, then #CSP$(\mathcal{F})\leqslant_{T}\operatorname{Holant}\left(\mathcal{DEQ}\mid\mathcal{F}\right).$

Proof.

Given an instance $I$ of #CSP$(\mathcal{F})$ over $m$ variables $V=\{x_{1},\ldots,x_{m}\}$. Suppose it contains $\ell$ constraints $f_{i}$ $(i\in[\ell])$ of arity $2n_{i}$, and $f_{i}$ is applied to the variables $x_{i_{1}},\ldots,x_{i_{2n_{i}}}$. We define a graph $G=(V,E)$, where $V$ is the variable set and $(x,y)\in E$ if variables $x,y$ appear as an opposite pair in some $\mathscr{S}(f_{i})$. Consider all connected components of $G$. We get a partition of $V$. Pick a representative variable in each connected component and define $V^{\tt r}$ to be the set of representative variables. Without loss of generality, we assume $V^{\tt r}=\{x_{1},\ldots,x_{m^{\tt r}}\}$. For each variable $x\in V$, we use $x^{\tt r}\in V^{\tt r}$ to denote its representative variable. By the definition of opposite pairs, for any assignment with a nonzero contribution, we have $x=\overline{x^{\tt r}}$ if there is a path of odd length from $x$ to $x^{\tt r}$ and $x=x^{\tt r}$ if there is a path of even length from $x$ to $x^{\tt r}$ (if $x^{\tt r}$ is $x$ itself, we say there is a path of length [math] from $x^{\tt r}$ to $x$). If for some $x$, we have both $x=\overline{x^{\tt r}}$ and $x=x^{\tt r}$, (that is, the connected component containing $x$ is not a bipartite graph), then we know #CSP$(I)\equiv 0$ since $x=\bar{x}$ is impossible. Otherwise, for each variable $x\in V$ we have either $x=\overline{x^{\tt r}}$ or $x=x^{\tt r}$, but not both.

Then, for any nonzero term in the sum

[TABLE]

the assignment of all variables in $V$ can be uniquely extended from its restriction on representative variables $V^{\tt r}$. Moreover, since $\mathscr{S}(f_{i})$ is pairwise opposite, for each opposite pair $(x_{i_{s}},x_{i_{n+s}})$, we know exactly one variable is equal to $x^{\tt r}_{i_{s}}$ while the other one is equal to $\overline{x^{\tt r}_{i_{s}}}$. Thus each pair $(x_{i_{s}},x_{i_{n+s}})$ is either $(x^{\tt r}_{i_{s}},\overline{x^{\tt r}_{i_{s}}})$ or $(\overline{x^{\tt r}_{i_{s}}},x^{\tt r}_{i_{s}})$. We will write this as $(\widehat{x^{\tt r}_{i_{s}}},\overline{\widehat{x^{\tt r}_{i_{s}}}})$. Then, we have

[TABLE]

The final form of (5.2) is an instance of $\operatorname{Holant}\left(\mathcal{DEQ}\mid\mathcal{F}\right)$. ∎

By this reduction, we have the following hardness result.

Corollary 5.9**.**

If every signature $f\in\mathcal{F}$ has affine support, then $\operatorname{Holant}\left(\mathcal{DEQ}\mid\mathcal{F}\right)$ is #P-hard unless $\mathcal{F}\subseteq\mathscr{A}$ or $\mathcal{F}\subseteq\mathscr{P}$.

Theorem 5.10**.**

$\rm{\#EO}(\{\neq_{4}\}\cup\mathcal{F})$* is #P-hard unless $\mathcal{F}\subseteq\mathscr{A}$ or $\mathcal{F}\subseteq\mathscr{P}$.*

Proof.

It follows from Lemmas 5.1, 5.2, Corollaries 5.4 and 5.9. ∎

Combining Theorems 2.22, 4.8 and 5.10, we can finish the proof of the main Theorem 3.1.

Proof.

(of Theorem 3.1) If $\mathcal{F}\subseteq\mathscr{A}$ or $\mathcal{F}\subseteq\mathscr{P}$, then by Theorems 2.22, $\rm{\#EO}(\mathcal{F})$ is tractable. Suppose $\mathcal{F}\not\subseteq\mathscr{A}$ and $\mathcal{F}\not\subseteq\mathscr{P}$, then certainly $\mathcal{F}\not\subseteq\mathcal{B}$ as $\mathcal{B}\subseteq\mathscr{P}$. Then Theorems 4.8 and 5.10 complete the proof. ∎

6 #EO Encompasses #CSP

In this section, we will show that the #EO framework encompasses #CSP.

Let $g$ be an arbitrary signature of arity $n>0$ (with no assumption on satisfying ars or to be EO). We associate $g$ with an EO signature $\widetilde{g}$ of arity $2n$ in the following way. We define

[TABLE]

Clearly, $\widetilde{g}$ is an EO signature. Moreover, its support is pairwise opposite, i.e., $x_{i}$ and $x_{n+i}$ form an opposite pair. We say $x_{i}$ is in the first half of the inputs of $\widetilde{g}$, while $x_{n+i}$ is in the second half. We define $\widetilde{\mathcal{G}}=\{\widetilde{g}\mid g\in\mathcal{G}\}$ for an arbitrary signature set $\mathcal{G}$. We show that #CSP is expressible in the #EO framework by the following theorem.

Theorem 6.1**.**

For every signature set $\mathcal{G}$ and the EO signature set $\widetilde{\mathcal{G}}$ defined above, we have

[TABLE]

Remark**.**

*Before we give the proof, we remark that this theorem is not merely stating that for an arbitrary $\operatorname{\#CSP}(\mathcal{G})$ problem, one can reduce every instance of $\operatorname{\#CSP}(\mathcal{G})$ to an instance of a suitable ${\rm\#EO}(\widetilde{\mathcal{G}})$ problem. Theorem 6.1 is stronger and categorical: For every signature set $\mathcal{G}$ in the $\operatorname{\#CSP}$ framework, there is a (uniformly constructible) EO signature set $\widetilde{\mathcal{G}}$ such that $\operatorname{\#CSP}(\mathcal{G})$ is the same as the #EO problem ${\rm\#EO}(\widetilde{\mathcal{G}})$. In particular, a complexity dichotomy for #EO problems would generalize the complexity dichotomy for $\operatorname{\#CSP}$ problems on Boolean variables (which is already known). This is similar to the fact that the $\operatorname{\#CSP}$ framework is encompassed by the Holant framework (but not vice versa, thus the latter is the more expressive one). In contrast, an instance dependent reduction in one direction would not have the consequence that a complexity dichotomy for the latter generalizes a complexity dichotomy for the former. *

Proof.

We first show that every instance of $\operatorname{\#CSP}(\mathcal{G})$ is expressible canonically as an instance of ${\rm\#EO}(\widetilde{\mathcal{G}}),$ thus, $\operatorname{\#CSP}(\mathcal{G})\leqslant_{T}{\rm\#EO}(\widetilde{\mathcal{G}}).$ Let $G=(U,V,E)$ be a bipartite graph representing an instance $I$ of $\operatorname{\#CSP}(\mathcal{G})$, where each $u\in U$ is a variable and each $v\in V$ is labeled by a constraint function $g\in\mathcal{G}$. We will modify the instance $I$ to an instance $\widetilde{I}$ of ${\rm\#EO}(\widetilde{\mathcal{G}})$ that evaluates to the same value, as follows.

For every $u\in U$, we create $k=\deg_{G}(u)$ vertices denoted by $u^{i}$ ($1\leqslant i\leqslant k$). (For example, in Figure 1, vertices $u_{1}$, $u_{2}$ and $u_{3}$ are decomposed into $3$, $2$ and $1$ vertices respectively.) Then we connect the $k$ edges originally incident to $u$ to these $k$ new vertices, so that each new vertex is incident to exactly one edge. (To be specific we assume the edges at $u$ in $I$ are ordered from 1 to $k$, and we connect the $i$-th edge to $u^{i}$. These are edges drawn by solid lines in Figure 1(b).) We denote this graph by $G^{\prime}$. Each $u^{i}$ in $G^{\prime}$ has degree $1$ and the degree of each $v\in V$ does not change. 2. 2.

For each edge $e^{i}=(u^{i},v)$ in the graph $G^{\prime}$, we add an edge $\bar{e}^{i}=(u^{i+1},v)$ to $G^{\prime}$ and we call them a pair. (Here if $\deg_{G}(u)=k$ then we use $u^{k+1}$ to denote $u^{1}$; we will add a multiple edge if $e^{i+1}=(u^{i+1},v)$ is already in $G^{\prime}$. These edges $\bar{e}^{i}$ are drawn by dashed lines in Figure 1(b).) This defines a graph $\widetilde{G}$. Each $u^{i}$ in $\widetilde{G}$ has degree $2$ and we label it by $\neq_{2}$. If $\deg_{G}(v)=n$ and is labeled by the constraint function $g\in\mathcal{G}$, then $v$ in $\widetilde{G}$ has degree $2n$ and we label it by the corresponding $\widetilde{g}\in\widetilde{\mathcal{G}}$. We place the signature $\widetilde{g}$ in a way such that every pair of edges $e^{i}=(u^{i},v)$ and $\overline{e}^{i}=(u^{i+1},v)$ incident to the same $v$ appears as an opposite pair in the inputs of the function $\widetilde{g}$, and $e^{i}$ appears in the first half of the inputs of $\widetilde{g}$ while $\overline{e}^{i}$ appears in the second half. Recall that $\widetilde{g}$ is defined to be pairwise opposite such that its $j$-th variable in the first half is paired with its $(n+j)$-th variable in the second half. This defines an instance $\widetilde{I}$ of ${\rm\#EO}(\widetilde{\mathcal{G}})$.

We show that ${\rm\#EO}_{\widetilde{I}}$ has the same value as the instance $I$ for $\operatorname{\#CSP}(\mathcal{G})$. Consider each variable $u\in U$. Suppose it has $\deg_{G}(u)=k$ in the instance $I$. It corresponds to $k$ vertices $u^{1},\ldots,u^{k}$ and $2k$ edges $e^{1},\overline{e}^{1},\ldots,e^{i},\overline{e}^{i},\ldots,e^{k}$ and $\overline{e}^{k}$. These $2k$ edges form a circuit $C_{u}$. For example, in Figure1, $u_{1}^{1},v_{1},u_{1}^{2},v_{1},u_{1}^{3},v_{2}$ back to $u_{1}^{1}$ is such a circuit where the edges are successively $e_{1}^{1},\overline{e}_{1}^{1},e_{1}^{2},\overline{e}_{1}^{2},e_{1}^{3},\overline{e}_{1}^{3}$ (edges drawn by solid lines and dashed lines alternate). Note that, for every pair of edges $e^{i}$ and $\overline{e}^{i}$, we placed the signature $\widetilde{g}$ such that $e^{i}$ and $\overline{e}^{i}$ appear as an opposite pair. Thus, we may assume $e^{i}$ and $\overline{e}^{i}$ take opposite values in the evaluation of ${\rm\#EO}_{\widetilde{I}}$. Also, since each $u^{i}$ is labeled by $\neq_{2}$, we may also assume $\overline{e}^{i}$ and $e^{i+1}$ take opposite values in the evaluation. (This is really a consequence of the definition of #EO problems.) Thus, for any (possible) nonzero term in the sum ${\rm\#EO}_{\widetilde{I}}$, as a consequence of the support of signatures in $\widetilde{G}$ and $\neq_{2}$, we know on each circuit $C_{u}$ all edges must take values $(0,1,0,1,\cdots,0,1)$ or $(1,0,1,0,\cdots,1,0)$, i.e., the values of $0,1$ alternate. Therefore, on the circuit $C_{u}$, we have $e^{1},e^{2},\ldots,e^{k}$ all take the same 0-1 value, which corresponds to the 0-1 assignment on the variable $u$ in the #CSP instance $I$. Recall in the definition of $\widetilde{g}$, its value can be determined by the first half of its inputs. By the placement of $\widetilde{g}$, the first half of its inputs are edges in the graph $G^{\prime}$ (drawn by solid lines). Therefore, the contribution of $\widetilde{g}$ to ${\rm\#EO}_{\widetilde{I}}$ is exactly the same as the contribution of $g$ in the #CSP instance $I$. Thus, these two instances have the same value.

For the other direction, we first note that $\operatorname{\#CSP}(\mathcal{G}\cup\{\neq_{2}\})\leqslant_{T}\operatorname{\#CSP}(\mathcal{G})$. If $\operatorname{\#CSP}(\mathcal{G})$ is #P-hard, the reduction holds trivially since every #CSP problem can be reduced in P-time to a #P-hard problem. Otherwise, by Theorem 2.24, $\operatorname{\#CSP}(\mathcal{G})$ is tractable and $\mathcal{G}\subseteq\mathscr{A}$ or $\mathscr{P}$. Since $(\neq_{2})\in\mathscr{A}\cap\mathscr{P}$, we have $\mathcal{G}\cup\{\neq_{2}\}\subseteq\mathscr{A}$ or $\mathscr{P}$. Thus, $\operatorname{\#CSP}(\mathcal{G}\cup\{\neq_{2}\})$ is tractable. Then, again the reduction holds trivially. Then, we will show that ${\rm\#EO}(\widetilde{\mathcal{G}})\leqslant_{T}\operatorname{\#CSP}(\mathcal{G}\cup\{\neq_{2}\})$.

Consider an arbitrary instance $I^{\prime}$ of ${\rm\#EO}(\widetilde{\mathcal{G}})$. Because every signature in $\widetilde{\mathcal{G}}$ has the pairing structure among its variables, we can decompose the graph of $I^{\prime}$ into edge disjoint circuits, by always following the paired variables at each constraint vertex. For each edge disjoint circuit, we choose an arbitrary default orientation. The circuit visits constraint vertices in some order according to the default orientation. The visit follows successive pairs of edges. Recall that as a consequence of the support of constraint functions, on each circuit, all these pairs of edges in the successive order must take the same ordered pair of values $(x,\bar{x})$, where $x\in\{0,1\}$. Thus, we can define a Boolean variable $x$ from the edges on each such circuit. From this a corresponding instance $I$ for $\operatorname{\#CSP}(\mathcal{G}\cup\{\neq_{2}\})$ can be obtained that has the same value as $I^{\prime}$ in ${\rm\#EO}(\widetilde{\mathcal{G}})$.

More specifically, suppose $g(x_{1},\ldots,x_{n})\in\mathcal{G}$ and let $g^{\prime}(x_{1},\ldots,x_{n})=g(x^{\epsilon_{1}}_{1},\ldots,x^{\epsilon_{n}}_{n})$, where each $x^{\epsilon_{i}}_{i}$ is either $x_{i}$ or $\overline{x_{i}}$. To discuss the complexity of $\operatorname{\#CSP}(\mathcal{G}\cup\{\neq_{2}\})$, using $(\neq_{2})$ we may assume every function obtained by flipping any number of variables in a function $g\in\mathcal{G}$ is also in $\mathcal{G}$.

Now, consider the default orientation of each circuit. At constraint vertices, the default orientation visits successive pairs of edges corresponding to paired inputs of constraint functions, say, $\{x_{j},x_{n+j}\}$. If the default orientation always visits in the order $x_{j}$ followed by $x_{n+j}$, then this is exactly how the canonical construction given above and we can recover an instance $I$ for $\operatorname{\#CSP}(\mathcal{G})$ with the same value. If at some constraint $\widetilde{g}$ of arity $2n$ the default orientation happens to visit in the order $x_{n+j}$ followed by $x_{j}$, we can use one copy of $\neq_{2}$ to modify the original function $g$ to get another constraint $g^{\prime}$, so that the corresponding $\widetilde{g^{\prime}}$ is just $\widetilde{g}$ with a flip between its variables $x_{n+j}$ and $x_{j}$. Then according to the default orientation the visit is in the order $x_{j}$ followed by $x_{n+j}$. ∎

7 Holographic Transformation to Real-Valued Holant Problems

We show that the ars condition corresponds to real-valued signatures under a holographic transformation without restricting on EO signatures. First we introduce the idea of holographic transformations. It is convenient to consider bipartite graphs. For a general graph, we can always transform it into a bipartite graph while preserving the Holant value, as follows. For each edge in the graph, we replace it by a path of length two. (This operation is called a 2-stretch of the graph and yields the edge-vertex incidence graph.) Each new vertex is assigned the binary Equality signature $(=_{2})=(f^{00},f^{01},f^{10},f^{11})=(1,0,0,1)$. For an invertible $2$-by-$2$ matrix $T\in{\rm GL}_{2}({\mathbb{C}})$ and a signature $f$ of arity $n$, written as a column vector (contravariant tensor) $f\in\mathbb{C}^{2^{n}}$, we denote by $T^{-1}f=(T^{-1})^{\otimes n}f$ the transformed signature. For a signature set $\mathcal{F}$, define $T^{-1}\mathcal{F}=\{T^{-1}f\mid f\in\mathcal{F}\}$ the set of transformed signatures. For signatures written as row vectors (covariant tensors) we define $fT$ and $\mathcal{F}T$ similarly. In the special case of the matrix $Z=\frac{1}{\sqrt{2}}\left[\begin{smallmatrix}1&1\\ \mathfrak{i}&-\mathfrak{i}\end{smallmatrix}\right]$, we also define $\widehat{\mathcal{F}}=Z\mathcal{F}$ and $\widehat{f}=Zf$. Note that $Z^{-1}=\frac{1}{\sqrt{2}}\left[\begin{smallmatrix}1&-\mathfrak{i}\\ 1&\mathfrak{i}\end{smallmatrix}\right]$, and $(=_{2})Z^{\otimes 2}=(\neq_{2})$, i.e., $Z$ transforms binary Equality to binary Disequality.

Let $T\in{\rm GL}_{2}({\mathbb{C}})$. The holographic transformation defined by $T$ is the following operation: given a signature grid $\Omega=(H,\pi)$ of $\operatorname{Holant}\left(\mathcal{F}\mid\mathcal{G}\right)$, for the same bipartite graph $H$, we get a new signature grid $\Omega^{\prime}=(H,\pi^{\prime})$ of $\operatorname{Holant}\left(\mathcal{F}T\mid T^{-1}\mathcal{G}\right)$ by replacing each signature in $\mathcal{F}$ or $\mathcal{G}$ with the corresponding signature in $\mathcal{F}T$ or $T^{-1}\mathcal{G}$.

Theorem 7.1** (Valiant’s Holant Theorem [34]).**

For any $T\in{\rm GL}_{2}({\mathbb{C}})$,

[TABLE]

Therefore, a holographic transformation does not change the value and thus the complexity of a Holant problem in the bipartite setting. Note that #EO$(\mathcal{F})$ can be expressed as $\operatorname{Holant}\left(\neq_{2}\mid\mathcal{F}\right)$ and $(\neq_{2})(Z^{-1})^{\otimes 2}=(=_{2})$. Thus, we have

[TABLE]

We can determine $\widehat{\mathcal{F}}$, for $\mathcal{F}$ consisting of (not necessarily EO) signatures satisfying ars, as follows.

Theorem 7.2**.**

A complex-valued signature $f$ satisfies ars if and only if $\widehat{f}=Zf$ is real-valued.

Proof.

We first prove that if $f$ satisfies ars then $\widehat{f}$ is real.

We have $2^{n/2}\widehat{f}=\left[\begin{smallmatrix}1&1\\ \mathfrak{i}&-\mathfrak{i}\end{smallmatrix}\right]^{\otimes n}f$, and thus for all $(a_{1},\ldots,a_{n})\in\{0,1\}^{n}$,

[TABLE]

Then,

[TABLE]

Hence, $\widehat{f}$ is real.

Now in the opposite direction, suppose $\widehat{f}$ is real. We have $2^{n/2}f=\left[\begin{smallmatrix}1&-\mathfrak{i}\\ 1&\mathfrak{i}\end{smallmatrix}\right]^{\otimes n}\widehat{f}$, and thus for all $(a_{1},\ldots,a_{n})\in\{0,1\}^{n}$,

[TABLE]

So

[TABLE]

Then,

[TABLE]

Hence, $f$ satisfies ars. ∎

Due to this holographic transformation, the classification of #EO problems with ars is not only interesting in its own right, but also it serves as a basic building block in the classification program for real-valued Holant problems over signature sets that are not necessarily symmetric. The signatures with ars are precisely those signatures that can be transformed into real-valued signatures in the Holant setting.

8 Conclusion and Outlook

The main technical contribution of this paper is to introduce unique prime factorization for signatures as a method to prove complexity dichotomies. Combined with merging operations, it is a powerful technique to analyze the complexity of signatures, and build inductive proofs. This method should be more widely applicable in the study of Holant problems. The results of this paper can serve as building blocks towards a classification of real-valued Holant problems (with no symmetry assumptions on the signatures). Under a suitable holographic transformation, these #EO problems with ars correspond to precisely a class of real valued Holant problems. It also seems that the EO signatures, which have support on half Hamming weight, are where some of the most intricate cases for the final classification lie.

We give the following Figure 2 as a partial map for the complexity classification program for Holant problems on the Boolean domain. We indicate how the framework of #EO problems considered in this paper fits in this program.

The complexity classification for Holant problems is built on the dichotomy for #CSP on the Boolean domain [16]. Based on this result, a dichotomy for complex-valued Holant problems with symmetric signatures (the function value depends only on the Hamming weight of the input) was established [14]. For asymmetric signatures, the first result is a dichotomy for a restricted class called $\operatorname{Holant}^{\ast}$ problems where all unary signatures are assumed to be available [15]. Later, it was generalized to (first real-valued [17] and then complex-valued [1, 2]) $\operatorname{Holant}^{c}$ problems where two pinning unary signatures $(1,0)$ and $(0,1)$ are available. In addition, based on the dichotomy for $\operatorname{Holant}^{\ast}$ problems, a dichotomy for non-negative $\operatorname{Holant}$ problems was proved [27] without assuming any auxiliary signatures. Simultaneously, progress has been made for Holant problems parameterized by complex-valued signatures of even arities. The base case is a single 4-ary signature. (The case that all signatures are binary is known to be tractable.) A dichotomy is proved for complex-valued six-vertex models [13] and later it was generalized to complex-valued eight-vertex models [11]. The framework of #EO problems generalizes six-vertex models to higher arities. In this paper, we prove a dichotomy for complex-valued #EO problems with ars, which can be transformed to a class of real-valued Holant problems. Combining with other results, we hope this result will lead to a classification for all real-valued Holant problems. The ultimate goal is definitely a full classification for all complex-valued Holant problems. In order to achieve this, we think a classification for complex-valued #EO problems without assuming ars may serve as a building block.

Recent Development

Most recently, after this paper was accepted, the vision that is enunciated in this paper has been realized. A full complexity dichotomy for all real-valued Holant problems on the Boolean domain has been achieved [32], and the results in the present paper indeed played a crucial role in its proof.

Acknowledgement

We thank the anonymous referees for their very helpful comments. We also thank the editor Prof. Lane Hemaspaandra for his very detailed editorial comments. Their many suggestions have helped us greatly to improve the presentation of the material.

Appendix A An $8$-ary Signature not Satisfying the $\Delta$-Property

Some lemmas in Section 4 (e.g., Lemmas 4.3, 4.4, and 4.5) are subtle both in their precise statements and proofs. In particular, they require high arities. Here, we give a signature of arity $8$ that cannot be handled by using “commutativity of merging operations”, which is the underlying “principle” that is essential for our induction proof (Lemma 4.6, which requires arity at least 10). This illustrates some of the intricacies and why we have to deal with signatures of lower arities separately (Lemma 4.7). For convenience, we use $(ij)$ to denote the binary disequality signature on variables $x_{i}$ and $x_{j}$, and in the table (A.1) below we omit the tensor product symbol $\otimes$ between the binary disequality signatures. Let $f$ be an EO signature satisfying ars on eight variables $x_{1},x_{2},\ldots,x_{8}$. Suppose $f$ has the following structural decompositions in $\mathcal{B}$ under various merging operations. (The precise forms in the following 28 tensor product factorizations in $\mathcal{B}$ are meticulously chosen; embedded in it is a nontrivial symmetry subgroup of $S_{8}$. It is not clear at this point that there are any function that can satisfy all these requirements.)

[TABLE]

There are a total of ${8\choose 2}=28$ binary disequalities $(ij)$, each appears exactly three times in (A.1) (each disequality $(ij)$ appears in a single row in table (A.1)). One can verify that this $f$ does not satisfy the $\Delta$-property. We carefully design these particular factorizations so that for any binary disequality signature $(ij)$, it divides exactly three totally disjoint pairs $\partial_{(s_{1}t_{1})}f$, $\partial_{(s_{2}t_{2})}f$ and $\partial_{(s_{3}t_{3})}f$, which do not form a “triangle”. However (it is tedious but one can verify that) by taking any two disjoint pairs $\{i,j\}$ and $\{u,v\}$, the above tensor factorizations always satisfy $\partial_{(ij)(uv)}f=\partial_{(uv)(ij)}f$. (Of course any actual signature $f$ must satisfy this.) It follows that this case cannot be handled by our induction proof.

The above structural decompositions (A.1) under various merging operations may look like artificially designed and one may wonder whether there exists such a signature that satisfies them all. While not obvious, the answer is yes. We give a signature that does satisfy (A.1).

Consider the following signature $f_{8}$ which has its support set defined by

[TABLE]

The function is defined to be $f_{8}(\alpha)=1$ when $\alpha\in\mathscr{S}(f)$ and $f_{8}(\alpha)=0$ elsewhere. It can be verified that $f_{8}$ is an EO signature with ars. Its support has the following structure: (1) it is on half-weight; (2) the sum of the first four variables is even; (3) the last four variables are either identical to the first four variables, or are the complement of the first four variables.

One can verify that the signature $f_{8}$ satisfies forms (A.1). Thus, there does exist a case that cannot be handled by our induction proof alone. This partially explains some of the intricacies of our proof.