The Projection Games Conjecture and the Hardness of Approximation of super-SAT and related problems

TL;DR

This paper proves the NP-hardness of approximating the Super-SAT problem and related problems within certain factors, assuming the Projection Games Conjecture, thereby establishing strong hardness results for several computational problems.

Contribution

It proves the Super-SAT conjecture assuming the Projection Games Conjecture and reduces SSAT to other problems, establishing their NP-hardness of approximation.

Findings

Proves Super-SAT NP-hardness of approximation assuming PGC.

Reduces SSAT to NCP and LHP, showing their NP-hardness of approximation.

Establishes polynomial factor hardness for SVP and CVP assuming PGC.

Abstract

The Super-SAT or SSAT problem was introduced by Dinur et al.(2002,2003) to prove the NP-hardness of approximation of two popular lattice problems - Shortest Vector Problem(SVP) and Closest Vector Problem(CVP). They conjectured that SSAT is NP-hard to approximate to within a factor of $n^c$ ($c>0$ is constant), where $n$ is the size of the SSAT instance. In this paper we prove this conjecture assuming the Projection Games Conjecture(PGC), given by Moshkovitz (2012). This implies hardness of approximation of SVP and CVP within polynomial factors, assuming PGC. We also reduce SSAT to the Nearest Codeword Problem(NCP) and Learning Halfspace Problem(LHP), as considered by Arora et al.(1997). This proves that both these problems are NP-hard to approximate within a factor of $N^{c'/\log\log n}$($c'>0$ is constant) where $N$ is the size of the instances of the respective problems. Assuming PGC these problems are proved to be NP-hard to approximate within polynomial factors.

Equations

$${\mathcal{R}}_y({\psi }_b)=\{(x_1,\dots ,x_{D_B}):x_j\in {\pi }_e^{-1}(y)\text{ where }e=(a_j,b)\text{ and }{a}_j\text{ is the }{j}^{th}\text{ neighbor of }b\}$$

$$\forall a\in \mathcal{F}:{\pi }_{\mathbf{x}}(\overrightarrow{S(\psi )})[a]=\sum _{r\in {\mathcal{R}}_{\psi }:r{|}_{\mathbf{x}}=a}\overrightarrow{S(\psi )}[r]$$

$${\pi }_{\mathbf{x}}(\overrightarrow{S({\psi }_i)})={\pi }_{\mathbf{x}}(\overrightarrow{S({\psi }_j)})$$

$$\mathcal{I}=⟨\Psi =\{{\psi }_1,\dots ,{\psi }_n\},V=\{{\mathbf{v}}_1,\dots ,{\mathbf{v}}_m\},\{{\mathcal{R}}_{{\psi }_1},\dots ,{\mathcal{R}}_{{\psi }_n}\}⟩$$

$${\mathcal{R}}_y({\psi }_b)=\{(x_1,\dots ,x_{D_B}):x_j\in {\pi }_e^{-1}(y)\text{ where }e=(a_j,b)\text{ and }{a}_j\text{ is the }{j}^{th}\text{ neighbor of }b\}$$

$$\mathbb{E}[L_{\mathbf{a}}]\le N_{\mathbf{a}}+D_A\cdot \frac{g_1}{D_A}\le 2g_1$$

$$P_1(r)\le \sum _{k=0}^{D_B-1}\left(\genfrac{}{}{0ex}{}{D_B-1}{k}\right){\left(1-\frac{g_1}{D_A}\right)}^{D_B-1-k}{\left(\frac{2g_1}{{\ell }^{\prime }}\left(1-\frac{g_1^2}{D_A^2}\right)\right)}^k$$

$$P_1(r)\le {\left(1-\frac{g_1}{D_A}+\frac{2g_1}{{\ell }^{\prime }}-\frac{2g_1^3}{D_A^2{\ell }^{\prime }}\right)}^{D_B-1}$$

$$P_1(r_2|{\psi }_1)\le {\left(1-\frac{g_1}{D_A}+\frac{2g_1}{{\ell }^{\prime }}-\frac{2g_1^3}{D_A^2{\ell }^{\prime }}\right)}^{D_B-1}$$

$$P_{\geq 2}(\Psi^{\prime})=\prod_{\psi_{i}\in\Psi^{\prime}}P_{\geq 2}(\psi_{i}|\psi_{1},\ldots,\psi_{i-1})\geq\Big{(}\frac{c^{\prime\prime}g_{1}}{D_{A}}\Big{)}^{s_{list}|B|}>0.$$

$$\sum _{i=1}^n{a}_i{x}_i-cy+\delta >0,\sum _{i=1}^n{a}_i{x}_i-cy-\delta <0$$

$$x_i-2y<0,x_i+2y>0$$

$$x_i+\delta >0,x_i-\delta <0$$

$${\mathcal{R}}_y({\psi }_b)=\{(x_1,\dots ,x_{D_B}):x_j\in {\pi }_e^{-1}(y)\text{ where }e=(a_j,b)\text{ and }{a}_j\text{ is the }{j}^{th}\text{ neighbor of }b\}$$

$$\forall x\in {\pi }_e^{-1}(y):{\pi }_{{\mathbf{a}}_{\mathbf{i}}}\overrightarrow{(S(\psi ))}[x]=\sum _{x_1,\dots ,x_{i-1},x_{i+1},\dots ,x_{D_B}}{M}_{x_1,\dots ,x_{i-1},x,x_{i+1},\dots ,x_{D_B}}^y$$

$$\Vert \overrightarrow{M^y}\Vert =\sum _{x_1,\dots ,x_{D_B}}|M_{x_1,\dots ,x_{D_B}}|$$

$$\Vert \overrightarrow{S(\psi )}\Vert =\sum _{y\in {\Sigma }_B}\Vert \overrightarrow{M^y}\Vert$$

$${\pi }_{{\mathbf{a}}_1}[x]=\sum _{x_2,\dots ,x_{D_B}}{M}_{x,x_2,\dots ,x_{D_B}}^y=0.$$

$$\sum _{x_1,\dots ,x_{D_B}}{M}_{x_1,\dots ,x_{D_B}}^y=\sum _{x_1}\sum _{x_2,\dots ,x_{D_B}}{M}_{x_1,x_2,\dots ,x_{D_B}}^y=0.$$

$$\sum _{x_2,\dots x_{D_B}}{M}_{i,x_2,\dots ,x_{D_B}}=w_i,i=1,\dots ,x$$

$$\sum _{r\in {\mathcal{R}}_{\psi }}\overrightarrow{S(\psi )}[r]=1$$

$$\sum _{r:r{|}_x=f}\overrightarrow{S({\psi }_i)}[r]+\sum _{r:r{|}_x\ne f}\overrightarrow{S({\psi }_j)}[r]=1$$

$$\sum _{r:r{|}_x=f}\overrightarrow{S({\psi }_i)}[r]=\sum _{r:r{|}_x=f}\overrightarrow{S({\psi }_j)}[r]$$

Full text

††journal: Journal of Computer and System Sciencesmyfootnotemyfootnotefootnotetext: Much of this work was done while the author was in Centre for Quantum Technologies, National University of Singapore.

1 Introduction

1.1 SSAT and lattice problems

The Super-SAT or SSAT problem was introduced by Dinur et al. [1] to prove the hardness of approximation of two popular lattice problems - the Shortest Vector Problem (SVP) and the Closest Vector Problem (CVP). An $n$-dimensional lattice $\mathcal{L}$ is the set of integral linear combinations of $n$ linearly independent vectors in $\operatorname{\mathbb{R}}^{d}$, called the basis of the lattice. The goal of SVP is to find a shortest non-zero lattice vector. Given a target vector $\overrightarrow{t}$, CVP aims at finding a closest lattice vector to it.

Algorithms for these lattice problems are well-studied and have applications in factoring polynomials over rationals [5], integer programming [6, 7], cryptanalysis [8], checking the solvability by radicals [9], solving low-density subset-sum problems [10], cryptography [11, 12, 13, 14, 15, 16].

The SSAT problem is the gap version of $\textsf{SAT}[F]$, which is defined as follows : An instance of $\textsf{SAT}[F]$ consists of a set of constraints or Boolean functions, called tests. The variables in each test take values from a finite set $F$ and each test has a set of satisfying assignments for its variables. The goal is to attach one assignment to each test such that consistency is maintained i.e. each variable gets the same value in all the tests in which it appears. The cardinality of the set of satisfying assignments is bounded by a polynomial in the number of variables.

In SSAT, we attach integer weights to each assignment and call it a super-assignment. It is consistent if for each variable the sum of weights on each value is the same in all the tests with this variable. Values which get non-zero net weight are said to be simultaneously assigned to the variable. If each variable gets at least one value assigned we call it a non-trivial super-assignment. If for at least one test there exists at least one non-zero weighted assignment then we call it not-all-zero super-assignment. This also gives rise to the notion of norm of a super-assignment and accordingly two variants of SSAT has been defined - the one in [1] for $\ell_{1}$ norm and another by Dinur in [2] for $\ell_{\infty}$ norm. An instance is accepted (YES) if each variable gets a single value everywhere giving a consistent super-assignment of norm $1$. The rejection criteria is slightly different in the two variants. Roughly, an instance is rejected (NO) if every consistent super-assignment satisfying some conditions, has norm greater than $g$. If after minimizing the norm of such a consistent super-assignment we get a value between $1$ and $g$, then the instance may be accepted or rejected i.e. any outcome (YES/NO) is fine. A more detailed explanation of these concepts have been given in Section 2.2. The following hardness results have been proved in [1, 2]. Suppose $n$ is the size of the SSAT instance, which is the encoding size of the number of variables, tests and satisfying assignments and this is polynomial in the number of variables (Section 2.2).

SSAT** Theorem [1]** : SSAT is NP-hard for $g=n^{c/\log\log n}$ for some constant $c>0$.

$\textsf{SSAT}_{\infty}$** Theorem [2]** : $\textsf{SSAT}_{\infty}$ is NP-hard for $g=n^{c/\log\log n}$ for some constant $c>0$.

An approximation factor preserving reduction from SSAT to $\textsf{CVP}_{p}$ (where distance is measured in $\ell_{p}$ norm), for $1\leq p<\infty$ was given in [1] and a similar reduction from $\textsf{SSAT}_{\infty}$ to $\textsf{CVP}_{\infty}$ and $\textsf{SVP}_{\infty}$ (where distance or length are measured in $\ell_{\infty}$ norm) was given in [2]. Thus the authors conjectured that the SSAT problems are hard within a polynomial factor, which would imply NP-hardness of the above mentioned lattice problems within polynomial approximation factor.

Conjecture 1.1**.**

SSAT* is NP-hard for $g=n^{c}$ for some constant $c>0$.*

Conjecture 1.2**.**

$\textsf{SSAT}_{\infty}$* is NP-hard for $g=n^{c}$ for some constant $c>0$.*

1.2 Label Cover (LC) and Projection Games Conjecture (PGC)

An instance of a Label Cover (LC) problem (also referred to as Projection Games) consists of (i) a bipartite graph $G=(A,B,E)$; (ii) finite alphabets $\Sigma_{A},\Sigma_{B}$ from which each vertex of $A$ and $B$ (respectively) are assigned a label; (iii) constraints or projections $\pi_{e}:\Sigma_{A}\rightarrow\Sigma_{B}$ for each $e\in E$. Given a labeling or assignment to the vertices, $\varphi_{A}:A\rightarrow\Sigma_{A}$ and $\varphi_{B}:B\rightarrow\Sigma_{B}$, we say an edge $e=(a,b)$ is satisfied if the corresponding projection constraint holds, i.e. $\pi_{e}(\varphi_{A}(a))=\varphi_{B}(b)$. In the optimization version of this problem the task is to find a labeling that maximizes the number of satisfied edges. The decision version of this problem is of interest to us and by Label Cover (LC) we denote this problem of distinguishing between the (YES) case that all edges are satisfied and the (NO) case when at most $s$ (soundness error) fraction of the edges are satisfied. There are other variants of this problem (e.g. [4]) but in this paper we work with this one.

A PCP Theorem gives the hardness of LC as follows [[17, 18, 19]]:

Given an input of size $N$ for LC with alphabet size $k$, it is NP-hard to distinguish between the case where all edges can be satisfied and the case where at most $s$ fraction of the edges can be satisfied. ($s$ and $k$ may be functions of $N$.)

Equivalently it can be stated that there is a reduction from (exact) SAT to LC. Raz and Moshkovitz [20] proved the following result.

Theorem 11 in [20] (re-phrased): * There exists $c>0$ such that for every $s\geq 1/N^{c}$, SAT on input of size $n$ can be reduced to LC of size $N$ for $N=n^{1+o(1)}\text{poly}(1/s)$. The LC is over an alphabet of size exponential in $1/s$ and has soundness error $s$. The reduction can be computed in time linear in the size and the alphabet size of the LC instance. The LC is on a bi-regular graph whose degrees are $\text{poly}(1/s)$. *

PCPs which achieve an LC instance of size $N=n^{1+o(1)}\text{poly}(1/s)$ are called almost-linear size PCP because of the exponent of $n$. The soundness error $s$ is at least $1/N$. Assuming $\text{P}\neq\text{NP}$ it can be shown that the alphabet size is at least $1/s$. Certain PCP constructions manage to have an alphabet size of $\text{poly}(1/s)$ at the cost of some other parameters [19]. Thus Moshkovitz [3] conjectured that a similar alphabet size may be achieved in the reduction in [20].

Conjecture 1.3** (Projection Games Conjecture (PGC) [3]).**

There exists $c>0$ such that for every $s\geq 1/N^{c}$, SAT on input of size $n$ can be efficiently reduced to LC of size $N=n^{1+o(1)}\text{poly}(1/s)$ over an alphabet of size $\text{poly}(1/s)$ and has soundness error $s$.

Here again $s$ is at least $1/N$.

1.3 Related work

The Label Cover (LC) problem was introduced by Arora et al. [4] but with a slightly different formulation than what has been stated in this paper. Roughly, in their variant there is a “cost” attached to the labeling of each vertex. The approximation factor is given by the ratio of this cost between the NO and YES case. The authors proved this variant of LC is NP-hard up to an approximation factor of $2^{\log^{0.5-\varepsilon}n}$ where $\varepsilon>0$ is some constant and $n$ is the size of LC instance, under the assumption that $\text{NP}\nsubseteq\text{DTIME}(n^{\text{poly}(\log n)})$ . They gave an approximation factor preserving reduction from LC to a number of other problems like CVP, $\textsf{SVP}_{\infty}$, Nearest Codeword Problem (NCP), Min-Unsatisfy problem and Learning Halfspace Problem (LHP). They also proved that the above problems were NP-hard for constant approximation factors by a reduction from Set Cover.

In 2012, Moshkovitz [3] reduced LC (the variant stated in Section 1.2) to Set Cover and proved that the latter is NP-hard to approximate within $(1-\alpha)\ln n$ ($n$ being the instance size) for arbitrarily small $\alpha>0$. She applied the Projection Games Conjecture to the reduction in [4] and concluded polynomial approximation factors are hard for CVP. Here the conjecture has also been used to study the behavior of CSPs around their approximability threshold.

The first NP hardness result for CVP in all $\ell_{p}$ norms and SVP in the $\ell_{\infty}$ norm was given by van Emde Boas [21]. SVP was proven to be NP-hard to approximate within a constant factor in [22, 23, 24]. Khot [25] and later Haviv and Regev [26] improved the approximation factor to $2^{\log^{1-\varepsilon}n}$ under the assumption that $\text{NP}\nsubseteq\text{RTIME}(n^{\text{poly}(\log n)})$. In [27] it has been shown that for almost all $p\geq 1$, CVP in the $\ell_{p}$ norm cannot be solved in $2^{n(1-\varepsilon)}$ time under the Strong Exponential Time Hypothesis [28]. A similar hardness result has also been obtained for SVP [29].

However, there are barriers for showing stronger inapproximability results. For example, a $\sqrt{n/O(\log n)}$ factor NP-hardness result would imply $\text{coNP}\subseteq\text{AM}$ [30], a factor $\sqrt{n}$ NP-hardness for SVP would imply $\text{NP}=\text{coNP}$ [31], and thus the polynomial hierarchy collapses in all these cases.

1.4 Our results and techniques

Hardness of SSAT and $\textsf{SSAT}_{\infty}$

In this paper we prove Conjecture 1.1 and 1.2, assuming the Projection Games Conjecture (PGC). Specifically we prove the following:

Theorem 1.1**.**

Assuming the PGC, SSAT is NP-hard for $g=n^{c}$ for some constant $c>0$.

Theorem 1.2**.**

Assuming the PGC, $\textsf{SSAT}_{\infty}$ is NP-hard for $g=n^{c}$ for some constant $c>0$.

We give a reduction to both SSAT and $\textsf{SSAT}_{\infty}$ from a variant of Label Cover, introduced by Moshkovitz in [3], let us call it the list-Label Cover, which we explain very briefly here. Here we can assign each $A$-vertex a list of labels. Two vertices $a_{1},a_{2}\in A$ agree on a label $y\in\Sigma_{B}$ for a vertex $b\in B$ if there exists at least one label in their respective lists such that these map to $y$ under the respective edge constraint functions. That is, there exists $x_{1}\in\varphi_{A}(a_{1})$ and $x_{2}\in\varphi_{A}(a_{2})$ such that $\pi_{e_{1}}(x_{1})=\pi_{e_{2}}(x_{2})=y$, where $e_{1}=(a_{1},b),e_{2}=(a_{2},b)\in E$. For this variant of LC a notion of list agreement soundness has been defined which gives the fraction of $B$-vertices on which the $A$-vertices totally disagree (i.e. no two $A$-vertices agree on any one label). A more detailed explanation of these concepts have been given in Section 2.1.

In our construction, each $A$-vertex corresponds to a variable in the SSAT instance and each $B$-vertex is a test. By a result of Moshkovitz [3], we can bound the $B$-degree (degree of each $B$-vertex) by some large enough constant prime power. For simplicity and without much loss of generality, we assume that $\Sigma_{A}$ is in bijective correspondence to some finite field and the variables take values from $\Sigma_{A}$. For each test $\psi_{b}$ (corresponding to some $b\in B$) consider a $y\in\Sigma_{B}$ such that it has at least one pre-image in each of $b$’s neighbor in $A$, consider the following tuples:

[TABLE]

Then the total set of satisfying assignments for $\psi_{b}$ is : $\quad\mathcal{R}(\psi_{b})=\bigcup_{y\in\Sigma_{B}}\mathcal{R}_{y}(\psi_{b})\quad$. Since we assume that the PGC (Conjecture 1.3) is true, so $|\Sigma_{A}|,|\Sigma_{B}|$ and hence $|\mathcal{R}(\psi_{b})|$ is bounded by some polynomial in the number of variables.

Next, we argue that a YES case of LC maps to a YES case of SSAT. If there exists a labeling that satisfies all edges then we can construct an assignment such that each variable is assigned a single value in every test, yielding a consistent super-assignment of norm 1 (Lemma 3.1).

For the soundness proof we need to show that a NO instance of LC maps to a NO case of SSAT. Instead, we give a contrapositive argument. In the $\ell_{1}$ norm (Lemma 3.3) we show that if there exists a consistent non-trivial super-assignment of norm less than $g$ then there exists a labeling of the vertices such that a certain fraction of the $B$-vertices do not have total disagreement. We give a kind of rejection sampling procedure List-Construction, by which every variable (or A-vertex) is assigned a list including all assigned values and some non-assigned values from the list of non-zero satisfying assignments of the tests in which it appears (or B-neighbors). We ensure that the list size of each A-vertex is same. Then we prove that there exists a test or B-vertex $\psi_{b}$ on which at least two of its A-neighbors agree i.e. it has a non-zero weighted satisfying assignment $r_{b}=(x_{1},\ldots,x_{D_{B}})$ such that at least two of the values in this tuple are in the list of their corresponding variables.

We show a similar result in the $\ell_{\infty}$ norm (Lemma 3.5) if there exists a consistent non-trivial (which is obviously not-all-zero) super-assignment. If a consistent super-assignment of norm at most $g$ is not-all-zero but not non-trivial i.e. some tests have all assignments with zero weight, then also such a labeling exists if some conditions hold. Here we must observe that if the norm of a not-all-zero consistent super-assignment is somewhere between $1$ and $g$ then any answer is acceptable. By this contrapositive argument we show that in some cases we get a YES reply from the oracle. So this is sufficient to prove the soundness in the $\ell_{\infty}$ norm.

The SSAT problem played a central role in proving the $n^{c/\log\log n}$ factor NP-hardness of CVP and $\textsf{SVP}_{\infty}$ [2, 1]. But not much work has been done in probing the complexity of this problem. To the best of our knowledge, we first relate this problem to other non-lattice problems. In [2, 1] the authors reduced a PCP instance to an SSAT instance. The variables are embedded into a geometric domain and then this domain is recursively encoded by multiple new domains, adding new variables along the way. The construction relies on strong error correcting properties of low degree fucntions to decode any consistent low norm super-assignment into a satisfying assignment for the original set of tests. We are not going into the detail of this construction. Interested readers are encouraged to refer to the original papers [2, 1]. Our reduction is relatively much simpler because of the PGC. We reduce a variant of the LC problem to SSAT, as described very briefly before (more detail in Section 3). In [3] the author shows how the complexity of this variant of LC relates to the “traditionally defined” LC problem, and hence to SAT. This is sufficient to derive new complexity results for SSAT.

Hardness of lattice problems and some related problems

Dinur et.al. [1, 2] gave approximation factor preserving reduction from SSAT and $\textsf{SSAT}_{\infty}$ to some lattice problems like $\textsf{CVP},\textsf{CVP}_{\infty},\textsf{SVP}_{\infty}$ and a related problem - Short Integer Solution (SIS). As a corollary of Theorem 1.1 and 1.2 we prove these problems are NP-hard to approximate within polynomial factors, provided Projection Games Conjecture holds (Corollary 4.1,4.2,4.3). These problems have been defined explicitly in Section 2.4.

Hardness of Learning Halfspace Problem (LHP)

We briefly describe the Learning Halfspace Problem (LHP) in the presence of malicious errors, as defined in [4]. It has been defined more precisely in Section 2.4. Roughly, the input to this problem is a set of points which are labeled by “+” or “-”, according to the side of a hyperplane they lie in a finite-dimensional space. Since this problem arises in the context of training a perceptron (a learning model) [32], so we say that the input is given to a “learner”, which has to output a hypothesis (i.e. a hyperplane) that correctly classifies as many points as possible. The error of an algorithm is the number of misclassifications by its hypothesis, and the noise of the sample (set of points) is the minimum error achievable by any algorithm. The failure ratio of an algorithm is the ratio of its error to noise.

We take the formulation of LHP given in [4] and give a reduction from SSAT, thus proving that approximating the minimum failure ratio of LHP is NP-hard upto a factor of $n^{c/\log\log n}$ (Theorem 4.2). Assuming PGC this factor can be improved to $n^{c}$ for some constant $c>0$ (Corollary 4.5).

Hardness of Nearest Codeword Problem (NCP)

The input to the Nearest Codeword Problem (NCP) is the generator matrix of an error-correcting code of length $n$ over a $q$-ary alphabet. Given a target vector the goal is to find a codeword that is nearest to the target vector in Hamming distance. More detail description has been given in Section 2.4. Note that this problem is not exactly equivalent to CVP.

The fact that this problem may be related to SSAT was hinted in [33]. However, no proof was given there. We give a reduction from SSAT to NCP which proves that it is NP-hard to approximate the latter within a factor $n^{c/\log\log n}$ (Theorem 4.1), which can be improved to $n^{c}$ ($c>0$ is a constant) assuming the Projection Games Conjecture (Corollary 4.4).

1.5 Future directions

One obvious direction would be to prove the hardness of SSAT and $\textsf{SSAT}_{\infty}$ (Conjecture 1.1 and 1.2) without assuming the Projection Games Conjecture. Alternately, one may prove the PGC (Conjecture 1.3). Either of these will imply improved NP-hard approximation factors for the problems considered in this paper, without any other assumptions.

To the best of our knowledge, apart from this work, the SSAT problem has only been studied to prove the hardness of approximation of lattice problems. Relating this problem to other problems might give interesting hardness results as well as algorithms for them. It might even throw some light on the complexity of SSAT itself.

1.6 Overview of this paper

We give all necessary preliminary definitions and notations in Section 2. The reduction from LC to SSAT appears in Section 3, while the reduction from SSAT to other computational problems are in Section 4.

2 Preliminaries

Notations

We write $\ln$ for natural logarithm and $\log$ for logarithm to the base $2$. $\operatorname{\mathbb{R}},\mathbb{Q},\mathbb{Z}$ denote the set of real numbers, rational numbers and integers respectively. $\mathbb{F}_{p}$ denotes a field of order $p$. We denote variables by bold letters. We denote arrays by letters (lower case letters for $1$-dimensional arrays or vectors) with overhead arrow, e.g. $\overrightarrow{v}^{n}$ and $\overrightarrow{M}^{\ell\times m\times n}$ (or $\overrightarrow{M}_{\ell\times m\times n}$). We may drop the dimension in the superscript (or subscript) whenever it is clear from the context. The $i^{th}$ co-ordinate of $\overrightarrow{v}$ is denoted by $v_{i}$ or $(\overrightarrow{v})_{i}$. The $(i,j,k)^{th}$ entry of $\overrightarrow{M}$ is denoted by $M_{ijk}$ or $\overrightarrow{M}[i,j,k]$. Sometimes we represent a matrix ($2$-dimensional array) as a vector of column (vectors) (e.g. $\overrightarrow{M^{\prime}}^{m\times n}=[\overrightarrow{m^{\prime}}_{1}\overrightarrow{m^{\prime}}_{2}\ldots\overrightarrow{m^{\prime}}_{n}]$ where each $\overrightarrow{m^{\prime}}_{i}$ is an $m-$length vector).

2.1 Label Cover

Definition 2.1** (Label Cover (LC)).**

An instance of LC consists of (i) a bipartite graph $G=(A,B,E)$; (ii) finite alphabets $\Sigma_{A},\Sigma_{B}$, such that each vertex in $A$ and $B$ is assigned a label from $\Sigma_{A}$ and $\Sigma_{B}$ respectively; (iii) a set $\Pi$ of constraints consisting of projections $\pi_{e}:\Sigma_{A}\rightarrow\Sigma_{B},\quad\forall e\in E$. Given a labeling to the vertices $\varphi_{A}:A\rightarrow\Sigma_{A}$ and $\varphi_{B}:B\rightarrow\Sigma_{B}$, an edge $e=(a,b)$ is satisfied if $\pi_{e}(\varphi_{A}(a))=\varphi_{B}(b)$. (With a slight abuse of notation, we sometimes drop the labelings and simply write $\pi_{e}(a)=b$.)

We work with the promise problem where in a YES instance there exists a labeling that satisfies all edges and in a NO instance, for all possible labeling to the vertices at most $s$ fraction of the edges can be satisfied. Such an instance is said to have soundness error $s$. The goal is to distinguish between these two cases. The size of the label cover is $N=|A|+|B|+|E|$ and the size of the alphabet is $\max\{|\Sigma_{A}|,|\Sigma_{B}|\}$.

Feige [34] defined a variant of LC (using the structure obtained from parallel repetition) where the soundness is determined by the fraction of $B$-vertices that have at least two neighbors from $A$ that agree on a label for them. To be more precise, we define the following terms.

Definition 2.2** **(Total disagreement).

Let $(G=(A,B,E),\Sigma_{A},\Sigma_{B},\Pi)$ be an LC instance. $\varphi_{A}:A\rightarrow\Sigma_{A}$ is a labeling to the $A$-vertices. We say that the $A$-vertices totally disagree on a vertex $b\in B$ if there are no two neighbors $a_{1},a_{2}\in A$ of $b$, for which $\quad\pi_{e_{1}}(\varphi_{A}(a_{1}))=\pi_{e_{2}}(\varphi_{A}(a_{2})),\quad$ where $e_{1}=(a_{1},b),\quad e_{2}=(a_{2},b)$.

Definition 2.3** **(Agreement soundness).

Let $\mathcal{G}=(G=(A,B,E),\Sigma_{A},\Sigma_{B},\Pi)$ be an LC instance. We say that $\mathcal{G}$ has agreement soundness error $s_{agr}$, if the following holds: In the YES instance, there exists a labeling $\varphi_{A}:A\rightarrow\Sigma_{A}$, $\varphi_{B}:B\rightarrow\Sigma_{B}$ that satisfies all edges and in the NO instance, for any labeling $\varphi_{A}:A\rightarrow\Sigma_{A}$, the $A$-vertices are in total disagreement on at least $1-s_{agr}$ fraction of the $b\in B$.

Moshkovitz [3] considered a variant of LC where each vertex can be assigned a list of $\ell$ labels and an agreement is interpreted as agreement on one of the labels in the list. We define the following terms related to this variant, which we call the list-Label Cover (list-LC).

Definition 2.4** **(List total disagreement).

Let $\mathcal{G}=(G=(A,B,E),\Sigma_{A},\Sigma_{B},\Pi)$ be an LC instance. Let $\ell\geq 1$ and $\widehat{\varphi_{A}}:A\rightarrow\binom{\Sigma_{A}}{\ell}$ is a labeling that assigns each $A$-vertex $\ell$ alphabet symbols. We say that the $A$-vertices totally disagree on a vertex $b\in B$ if there are no two neighbors $a_{1},a_{2}\in A$ of $b$, for which there exists $\sigma_{1}\in\widehat{\varphi_{A}}(a_{1}),\quad\sigma_{2}\in\widehat{\varphi_{A}}(a_{2})$, such that $\quad\pi_{e_{1}}(\sigma_{1})=\pi_{e_{2}}(\sigma_{2}),\quad$ where $e_{1}=(a_{1},b),\quad e_{2}=(a_{2},b)\in E$.

Definition 2.5** (List agreement soundness).**

Let $\mathcal{G}=(G=(A,B,E),\Sigma_{A},\Sigma_{B},\Pi)$ be an LC instance. We say that $\mathcal{G}$ has list agreement soundness error $(\ell,s_{list})$, if the following holds: In the YES instance, there exists a labeling $\varphi_{A}:A\rightarrow\Sigma_{A}$, $\varphi_{B}:B\rightarrow\Sigma_{B}$ that satisfies all edges and in the NO instance, for any labeling $\widehat{\varphi_{A}}:A\rightarrow\binom{\Sigma_{A}}{\ell}$, the $A$-vertices are in total disagreement on at least $1-s_{list}$ fraction of the $b\in B$.

The following result relates agreement soundness and list agreement soundness.

Lemma 2.1** ([3]).**

Let $\ell\geq 1$ and $0<s_{agr}<1$. An LC with agreement soundness error $s_{agr}$ has list agreement soundness error $(\ell,s_{agr}\ell^{2})$.

2.2 SuperSAT (SSAT)

An SSAT instance has a set $\Psi=\{\psi_{1},\ldots,\psi_{n^{\prime}}\}$ of tests over variables $V=\{\mathbf{v}_{1},\ldots,\mathbf{v}_{m}\}$ which take values from a field $\mathcal{F}$, called range of the variables. An assignment $\mathcal{A}_{\Psi}:V\rightarrow\mathcal{F}$ maps each variable to a value in $\mathcal{F}$. For convenience, we can think an assignment as a tuple of field values (for its variables) and it is satisfying if these values evaluate to some required value for the test. Each test $\psi$ has a list $\mathcal{R}_{\psi}$ of satisfying assignments for its variables and we attach some “weight” to each such assignment. For any assignment $r$ we use $r|_{\mathbf{x}}$ to denote the value of variable $\mathbf{x}$ in this assignment.

Definition 2.6** (Super-assignment to tests).**

A super-assignment is a function $S$ mapping each $\psi\in\Psi$ to a value from $\mathbb{Z}^{\mathcal{R}_{\psi}}$. Thus $\overrightarrow{S(\psi)}$ is a vector of integer coefficients, one for each $r\in\mathcal{R}_{\psi}$.

Denote $\overrightarrow{S(\psi)}[r]$ as the $r^{th}$ co-ordinate of $\overrightarrow{S(\psi)}$. A natural super-assignment assigns each $\psi\in\Psi$ a unit vector $\overrightarrow{e_{r}}\in\mathbb{Z}^{\mathcal{R}_{\psi}}$ with a $1$ in the $r^{th}$ co-ordinate (i.e. $\overrightarrow{S(\psi)}[r]=1$ and $\overrightarrow{S(\psi)}[r^{\prime}]=0$ for all $r^{\prime}\neq r$). A super-assignment is not-all-zero if there is at least one test $\psi\in\Psi$ for which $\overrightarrow{S(\psi)}\neq\overrightarrow{0}$.

Definition 2.7** (Projection).**

Given a super-assignment $S:\Psi\rightarrow\bigcup_{\psi}\mathbb{Z}^{\mathcal{R}_{\psi}}$, the projection of $\overrightarrow{S(\psi)}$ on a variable $\mathbf{x}$ of $\psi,\quad\pi_{\mathbf{x}}(\overrightarrow{S(\psi)})\in\mathbb{Z}^{\mathcal{F}}$, is defined as follows:

[TABLE]

The notion of projection facilitates the definition of consistency between tests.

Definition 2.8** (Consistency).**

A super-assignment $S$ to the tests in $\Psi$ is consistent if the projections of two tests on each mutual variable are equal, i.e. for every pair of tests $\psi_{i}$ and $\psi_{j}$ with a common variable $\mathbf{x}$,

[TABLE]

$S$ is non-trivial if for every variable $\mathbf{x}\in V$ there is at least one test $\psi\in\Psi$ that is not cancelled on $\mathbf{x}$, i.e. $\pi_{\mathbf{x}}(\overrightarrow{S(\psi)})\neq\overrightarrow{0}$. For a variable $\mathbf{x}$ we think of all the values $a\in\mathcal{F}$ receiving non-zero coefficients in $\pi_{\mathbf{x}}(\overrightarrow{S(\psi)})$ (i.e. values for which $\pi_{\mathbf{x}}(\overrightarrow{S(\psi)})[a]\neq 0$) as being “simultaneously” assigned to $\mathbf{x}$ by $\psi$. The non-triviality requirement implies each variable must be assigned at least one value.

To formally define the SSAT problem we have to introduce the notion of norm of a super-assignment. Here we note that in [1] the problem was defined for $\ell_{1}$ norm (though the derived results work for all $\ell_{p}$ norm where $1\leq p<\infty$). Dinur [2] introduced a related problem called the $\textsf{SSAT}_{\infty}$, where some definitions like norm of a super-assignment were modified.

Definition 2.9** **(Norm of a Super-Assignment).

For the problem SSAT the norm of a super-assignment $S$ is the average norm of its individual assignments : $\quad\|S\|=\frac{1}{|\Psi|}\sum_{\psi\in\Psi}\|\overrightarrow{S(\psi)}\|_{1}$.

We call $\|\overrightarrow{S(\psi)}\|_{1}=\sum_{r\in\mathcal{R}_{\psi}}|\overrightarrow{S(\psi)}[r]|$ as the norm of a test $\psi$.

For the problem $\textsf{SSAT}_{\infty}$ the norm of a super-assignment $S$ is defined as : $\quad\|S\|_{\infty}=\max_{\psi\in\Psi}\|\overrightarrow{S(\psi)}\|_{1}$.

We now formally define both the $g-\textsf{SSAT}$ and $g-\textsf{SSAT}_{\infty}$ problem. The parameter $g$ is an approximation factor for the norm of a super-assignment.

Definition 2.10** ($g-\textsf{SSAT}$ (and $g-\textsf{SSAT}_{\infty}$)).**

The input instance

[TABLE]

consists of (i) a set $\Psi=\{\psi_{1},\ldots,\psi_{n}\}$ of tests over a common set $V=\{\mathbf{v}_{1},\ldots,\mathbf{v}_{m}\}$ of variables that take values in a field $\mathcal{F}$ ; (ii) for each test $\psi\in\Psi$ a list $\mathcal{R}_{\psi}$ of satisfying assignments to its variables, called the range of the test $\psi$. The size of an instance is the encoding size of the number of variables, tests and satisfying assignments. The parameters $m,|\mathcal{F}|$ and $|\mathcal{R}_{\psi}|$ are always bounded by some polynomial in $n$ and hence the size of an instance is also bounded by some polynomial in $n$.

This is a promise problem where in the YES instance there is a consistent natural super-assignment for $\Psi$. In the NO instance of SSAT, for every non-trivial consistent super-assignment $S$ for $\Psi$, $\|S\|\geq g$. While in the NO instance of $\textsf{SSAT}_{\infty}$, for every not-all-zero consistent super-assignment $S$ for $\Psi$, $\|S\|_{\infty}\geq g$. The goal is to distinguish between the YES and NO instance of the respective problems.

2.3 Lattice problems

Definition 2.11**.**

A lattice $\mathcal{L}$ is a discrete additive subgroup of $\operatorname{\mathbb{R}}^{d}$. Each lattice has a basis $\overrightarrow{B}=[\overrightarrow{b}_{1},\overrightarrow{b}_{2},\ldots,\overrightarrow{b}_{n}]$ where $\overrightarrow{b}_{i}\in\operatorname{\mathbb{R}}^{d}$ and \mathcal{L}=\mathcal{L}(\overrightarrow{B})=\Big{\{}\sum_{i=1}^{n}x_{i}\overrightarrow{b}_{i}:x_{i}\in\mathbb{Z}\quad\text{ for }\quad 1\leq i\leq n\Big{\}}.

We call $n$ the rank of $\mathcal{L}$ and $d$ as the dimension.

We consider the following lattice problems. From here on, in all the definitions $c\geq 1$ is some arbitrary approximation factor (usually specified as subscript), which can be a constant or a function of any parameter of the lattice (usually rank). For exact versions of the problems (i.e. $c=1$) we drop the subscript. Typically, we define length in terms of the $\ell_{p}$ norm for some $1\leq p\leq\infty$. Thus $\|\overrightarrow{x}\|_{p}:=(|x_{1}|^{p}+|x_{2}|^{p}+\cdots+|x_{d}|^{p})^{1/p}$ for finite $p$ and $\|\overrightarrow{x}\|_{\infty}:=\max|x_{i}|\;.$

Definition 2.12** (Shortest Vector Problem ($\textsf{SVP}_{c}^{(p)}$)).**

Given a lattice $\mathcal{L}$ with rank $n$ the goal is to find a shortest non-zero vector in the lattice.

In the promise version, usually denoted as $\textsf{GapSVP}_{c}^{(p)}$ the goal is to distinguish between the YES instance when $\exists\overrightarrow{v}\in\mathcal{L}\setminus\{\overrightarrow{0}\}$ such that $\|\overrightarrow{v}\|_{p}\leq r$ (for some positive real $r$ given as input) and the NO instance when all non-zero vectors in the lattice have length greater than $c\cdot r$.

Definition 2.13** (Closest Vector Problem ($\textsf{CVP}_{c}^{(p)}$)).**

Given a lattice $\mathcal{L}$ with rank $n$ and a target vector $\overrightarrow{t}\in\operatorname{\mathbb{R}}^{d}$ the goal is to find a closest lattice vector to $\overrightarrow{t}$.

In the promise version, usually denoted as $\textsf{GapCVP}_{c}^{(p)}$, the goal is to distinguish between the YES instance when $\exists\overrightarrow{v}\in\mathcal{L}$ such that $\|\overrightarrow{v}-\overrightarrow{t}\|_{p}\leq r$ (for some positive real $r$ given as input) and the NO instance when $\forall\overrightarrow{v}\in\mathcal{L},\quad\|\overrightarrow{v}-\overrightarrow{t}\|_{p}>c\cdot r$.

In this paper, with a slight abuse of notation we denote both the optimization and the promise versions of the above problems by the same notation, i.e. SVP and CVP respectively.

2.4 Other computational problems

Shortest Integer Solution

Definition 2.14** (Shortest Integer Solution ($\textsf{SIS}_{c}$) [1]).**

Given (i) a matrix $\overrightarrow{B^{\prime}}\in\mathbb{Z}^{m\times n}$; (ii) target vector $\overrightarrow{t}\in\mathbb{Z}^{m}$ such that $\overrightarrow{t}\in\{\overrightarrow{B^{\prime}x}:\overrightarrow{x}\in\mathbb{Z}^{n}\}$ and (iii) $d\in\mathbb{Z}$, the goal is to distinguish between the YES instance when $\exists\overrightarrow{z}\in\mathbb{Z}^{n}$ such that $\overrightarrow{B^{\prime}}\overrightarrow{z}=\overrightarrow{t}$ and $\|\overrightarrow{z}\|_{p}\leq d$ and the NO instance when $\forall\overrightarrow{z}\in\mathbb{Z}^{n}$ where $\overrightarrow{B^{\prime}}\overrightarrow{z}=\overrightarrow{t}$ we have $\|\overrightarrow{z}\|_{p}>c\cdot d$.

Nearest Codeword Problem

An error-correcting code $\mathcal{A}$ of block length $n$ over a $q$-ary alphabet $\Sigma$ ($=\mathbb{F}_{q}$) is a collection of strings (vectors) from $\Sigma^{n}$, called codewords. A linear code $\mathcal{A}$ is a linear subspace of $\mathbb{F}_{q}^{n}$ over base field $\mathbb{F}_{q}$ and it can be compactly represented by a generator matrix $\overrightarrow{A}\in\mathbb{F}_{q}^{m\times n}$ such that $\mathcal{A}=\{\overrightarrow{Ax}:\overrightarrow{x}\in\mathbb{F}_{q}^{n}\}$.

For any $\overrightarrow{v}\in\Sigma^{m}$, the Hamming weight of $\overrightarrow{v}$ is denoted by $\text{wt}(\overrightarrow{v})=|\{i:v_{i}\neq 0\}|$. The Hamming distance between two vectors $\overrightarrow{u},\overrightarrow{v}\in\Sigma^{m}$ is $\|\overrightarrow{u}-\overrightarrow{v}\|_{H}=\text{wt}(\overrightarrow{u}-\overrightarrow{v})$.

Definition 2.15** (Nearest Codeword Problem($\textsf{NCP}_{c}$) [4]).**

Given (i) a matrix $\overrightarrow{A}$ over $\mathbb{F}_{q}^{m\times n}$; (ii) a target vector $\overrightarrow{t}\in\mathbb{F}_{q}^{m}$ and (iii) an integer $d$, the goal is to distinguish between the YES instance when $\exists\overrightarrow{z}\in\mathbb{F}_{q}^{n}$ such that $\|\overrightarrow{A}\overrightarrow{z}-\overrightarrow{t}\|_{H}\leq d$, and the NO instance when $\forall\overrightarrow{z}\in\mathbb{F}_{q}^{n},\quad\|\overrightarrow{A}\overrightarrow{z}-\overrightarrow{t}\|_{H}>c\cdot d$.

Learning Halfspace Problem

We consider a popular problem in learning theory : learning a halfspace in the presence of malicious errors, as described in [4].

The input to the learner consists of a set of $k$ points in $\operatorname{\mathbb{R}}^{m}$, each labeled with a $+$ (positive examples of a concept) or $-$ (negative examples of a concept). The learner’s output is a hyperplane, $\braket{\overrightarrow{a},\overrightarrow{x}}=b$, where $\overrightarrow{a}\in\operatorname{\mathbb{R}}^{m}$ and $b\in\operatorname{\mathbb{R}}$. The hypothesis correctly classifies a point $\overrightarrow{y}$ marked $+$ (or $-$) if it satisfies $\braket{\overrightarrow{a},\overrightarrow{y}}>b$ (or $\braket{\overrightarrow{a},\overrightarrow{y}}<b$ respectively). Else, it misclassifies the point.

Finding a hypothesis that minimizes the number of misclassifications is the open hemispheres problem, which is NP-hard. The error of the algorithm is the number of misclassifications by its hypothesis, and the noise of the sample is the error of the best possible algorithm. The failure ratio of the algorithm is the ratio of the error to noise.

The LHP can be formulated in the following way.

Definition 2.16** (Learning Halfspace Problem ($\textsf{LHP}_{c}$) [4]).**

Given a set of linear inequalities in $n$ variables over $\operatorname{\mathbb{R}}$ and an integer $d$, distinguish between the YES instance when there exists an assignment to the $n$ variables which does not satisfy at most $d$ inequalities, and the NO instance when every assignment to the $n$ variables does not satisfy at least $c\cdot d$ inequalities.

3 Hardness result for SSAT

We now prove the hardness of SSAT (Theorem 1.1) and $\textsf{SSAT}_{\infty}$ (Theorem 1.2). In Section 3.1 we reduce Label Cover (LC) to SSAT and then appropriately adopt this reduction for $\textsf{SSAT}_{\infty}$ in Section 3.2.

Let $\mathcal{G}^{\prime}=(G^{\prime}=(A^{\prime},B^{\prime},E^{\prime}),\Sigma_{A},\Sigma_{B},\Pi^{\prime})$ be an LC instance obtained after applying the Projection Games Conjecture (Conjecture 1.3). Assume it has size $N^{\prime}$, soundness $s\geq 1/N^{\prime\beta}$ (for some $0<\beta<1$) and alphabet size $\text{poly}(1/s)$. We assume without loss of generality that the LC instance is bi-regular [20, 3]. That is, every $A$ vertex has the same degree $D_{L}$ (which we call the left degree or A-degree) and every $B$ vertex has the same degree $D_{R}$ (which we call the right degree or B-degree).

We use the following result from [3] which relates the soundness error to agreement soundness error.

Proposition 3.1** ([3] (re-phrased)).**

Let $D_{B}\geq 2$ be a prime power, $D_{R}$ be a power of $D_{B}$ and $\varepsilon>0$. From an LC instance with soundness error $\varepsilon^{2}D_{B}^{2}$ and $B$-degree $D_{R}$, we can construct an LC instance with agreement soundness error $2\varepsilon D_{B}^{2}$ and $B$-degree $D_{B}$. The transformation preserves the alphabets. The size is raised to a constant power.

The running time of this reduction is polynomial in the size of the initial LC instance. Thus after applying this lemma, we can assume we have an LC instance $\mathcal{G}=(G=(A,B,E),\Sigma_{A},\Sigma_{B},\Pi)$ with size $N=N^{\prime\gamma}$ (for some constant $\gamma>1$), right degree $D_{B}$ (constant prime power), left degree $D_{A}$ and agreement soundness error $s_{agr}=\frac{2s}{\varepsilon}=2D_{B}\sqrt{s}$, where $\varepsilon=\frac{\sqrt{s}}{D_{B}}$. Expressing in terms of $N$, we can write $s\geq 1/N^{\beta/\gamma}$. Here we note that $0<\beta<1$ and $\gamma>1$, thus $s\geq 1/N^{c}$ for some $0<c<1$. Let for each edge $e\in E$, $\pi_{e}$ is a $p-to-1$ projection where $p\leq|\Sigma_{A}|$.

Our proof works even without the bi-regularity condition. What is crucial is the fact that degree of each $B$-vertex is bounded by some large enough constant. However, for convenience, we assume we have a bi-regular graph.

3.1 Reduction from LC to SSAT

We reduce the above LC instance $\mathcal{G}$ to a SSAT instance $\mathcal{I}=(V,\Psi,\mathcal{R}_{\Psi})$ as follows. To each $A$-vertex $a$ we associate a variable $\mathbf{a}$ i.e. $|V|=|A|$. To each $B$-vertex $b$ we associate a test $\psi_{b}$ i.e. $|\Psi|=|B|$. The variables in a test $\psi_{b}$ are the neighbors of $b$ in $A$. Thus each test has $D_{B}$ variables and each variable appears in $D_{A}$ tests.

**Values of variables : ** Without much loss of generality we assume that $\Sigma_{A}$ is in bijective correspondence with some field $\mathcal{F}$, which is the range of the variables in $V$. We use the letters $x$ and $y$ (with subscript and superscript as required) for the elements of $\Sigma_{A}$ (or $\mathcal{F}$) and $\Sigma_{B}$ respectively.

**Satisfying assignments for tests : ** Consider a $\psi_{b}\in\Psi$. For each label $y\in\Sigma_{B}$ such that it has at least one pre-image in each of $b$’s neighbors in $A$, consider the following tuples:

[TABLE]

Thus the total set of satisfying assignments for $\psi_{b}$ is : $\quad\mathcal{R}(\psi_{b})=\bigcup_{y\in\Sigma_{B}}\mathcal{R}_{y}(\psi_{b}).\quad$ And cardinality of this set is at most $|\Sigma_{B}|p^{D_{B}}$, which is polynomially bounded by $|V|$, since we assumed the Projection Games Conjecture is true. Each satisfying assignment $r$ is a tuple of size $D_{B}$ consisting of values of some variables, as explained in Section 2.2. So we sometimes use the word “co-ordinate” which implies the corresponding variable. From here on, when we say that a value in $r$ is assigned (or non-assigned) we imply that the value is assigned (or non-assigned) to its respective variable.

Completeness

Lemma 3.1**.**

If there exists a labeling that satisfies all the edges in $G$ then there exists a consistent natural super-assignment satisfying all the tests in $\Psi$.

Proof.

Suppose there is a labeling $\varphi_{A}:A\to\Sigma_{A}$, and $\varphi_{B}:B\to\Sigma_{B}$ that satisfies all edges. Then for any $b\in B$, let $\quad r_{b}:=(\varphi_{A}(a_{1}),\ldots,\varphi_{A}(a_{D_{B}})),\quad$ where $a_{1},\ldots,a_{D_{B}}$ are the neighbors of $b$. Since the labeling ($\varphi_{A}$, $\varphi_{B}$) satisfies all edges, we have that $r_{b}\in\mathcal{R}_{\varphi_{B}(b)}(\psi_{b})$, and hence $r_{b}\in\mathcal{R}(\psi_{b})$.

Consider the super-assignment for the resulting SSAT instance that sets for all $b\in B$, $\overrightarrow{S(\psi_{b})}[r_{b}]=1$ and $\overrightarrow{S(\psi_{b})}[r^{\prime}]=0$ for all $r^{\prime}\neq r_{b}$. This assignment is natural by definition and it is consistent since for each $\mathbf{a}\in V$ and for each neighbor $b$ of $a$, we have that $\pi_{\mathbf{a}}(\overrightarrow{S(\psi_{b})})[\varphi_{A}(a)]=1$ and $\pi_{\mathbf{a}}(\overrightarrow{S(\psi_{b})})[\alpha]=0$ for any $\alpha\neq\varphi_{A}(a)$. Thus we get a consistent natural super-assignment.

∎

Soundness

We can make some observations about the structure of the SSAT instance we constructed. These are not essential for our soundness proof, so we have moved these to A and only state the following result.

Lemma 3.2** (Corollary 1 in A).**

For each test with non-zero norm, in the set of non-zero weighted assignments either there exists at least one assignment such that it has at least two variables with assigned values or all its assignments have exactly one variable with assigned value.

Here we note that the non-triviality condition does not guarantee the existence of at least one variable with assigned value in “each” assignment.

Lemma 3.3**.**

Let $D_{B}$ is a constant prime power such that $N$ is a power of $D_{B}$ and $0<c<1$. Let $s\geq 1/N^{c}$ and $\ell^{\prime}<\frac{1}{\sqrt{2D_{B}}s^{d}}$, where $d<1/4$. Assume $s_{list}=\sqrt{s}D_{B}\ell^{\prime 2}$ 222Here we note that we have an instance of LC with agreement soundness error $s_{agr}=\frac{2s}{\varepsilon}=2D_{B}\sqrt{s}$. By Lemma 2.1, for a list size $\tilde{\ell}$, it has list agreement soundnes $(\tilde{\ell},s_{list})$ where $s_{list}=s_{agr}\tilde{\ell}^{2}$. The bounds in the lemma have been chosen accordingly..

If $\mathcal{G}$ has list agreement soundness error $(\ell^{\prime},s_{list})$ then every non-trivial consistent super-assignment for $\mathcal{I}$ has norm at least $g=N^{c^{\prime}}(1-2s_{list})$, where $c^{\prime}<d$.

Proof.

We prove the contrapositive. Let $\mathcal{I}$ has a non-trivial consistent super-assignment $S$ of norm at most $g$. Then we prove that there exists a labeling such that for at least $s_{list}$ fraction of $B$-vertices, the $A$-vertices do not totally disagree.

If the average norm is at most $g$ then by Markov’s inequality there exists at least $s_{list}$ fraction of tests for which the norm, and hence the number of non-zero weight satisfying assignments, is at most $g_{1}=N^{c^{\prime}}(1-s_{list})$. Let us denote this set of tests by $\Psi^{\prime}(\subseteq\Psi)$. For any variable $\mathbf{a}$ we denote its set of assigned values by $\mathcal{G}_{\mathbf{a}}$ and the non-assigned values by $\overline{\mathcal{G}}_{\mathbf{a}}$. For any assignment $r$ if $r|_{\mathbf{a}}\in\mathcal{G}_{\mathbf{a}}$ then we call the corresponding co-ordinate “good”. Since $S$ is non-trivial each test must have at least one non-zero weight assignment with at least one good co-ordinate.

We now define an assignment $\widehat{\phi_{A}}:A\rightarrow\binom{\Sigma_{A}}{\ell^{\prime}}$ to the $A$-vertices. Let us call this the procedure of List-Construction.

For each variable we include all its assigned values in its list. 2. 2.

For any test $\psi\in\Psi^{\prime}$ if there are no assignment with at least two good co-ordinates, then we consider any one assignment with one good co-ordinate (existence of such an assignment is guaranteed by the non-triviality condition) and include each of the non-assigned values with probability $p=\min\{1,\frac{g_{1}}{D_{A}}\}$ in the lists of the corresponding variables.

(Lemma 3.2 exerts that in such a case each assignment will have exactly one good co-ordinate.)

Expected list size

Consider a variable $\mathbf{a}$ (or corresponding vertex) and let $N_{\mathbf{a}}=|\mathcal{G}_{\mathbf{a}}|(\leq g_{1})$ be the number of values assigned to it. By step (2) of List-Construction, from each of the $D_{A}$ test that $\mathbf{a}$ appears in, one non-assigned value can be included in the list with probability $p$. If $L_{\mathbf{a}}$ is the variable for the size of the list of labels for $a$, then

[TABLE]

This is true for every variable. Thus expected list size is at most $2g_{1}$ and with high probability it remains bounded by $\ell^{\prime}$. By Markov’s inequality, the probability of a variable (or corresponding vertex) having list size more than $\ell^{\prime}$ is at most $\frac{2g_{1}}{\ell^{\prime}}$. This probability is $\text{poly}(\frac{1}{N})$ for small enough $c^{\prime}$. Let $\mathcal{V}$ be the set of vertices with list size bounded by $\ell^{\prime}$ and $\mathcal{V}^{\prime}$ is the set of those vertices with list size more than $\ell^{\prime}$. Each $\mathbf{v}\in\mathcal{V}^{\prime}$ has $N_{\mathbf{v}}$ assigned values and rest non-assigned values. From each of its $D_{A}$ neigbours each such vertex can get at most one non-assigned value (by step 2). So the maximum number of non-assigned values is at most $D_{A}$. We select any $\ell^{\prime}-N_{\mathbf{v}}>g_{1}$ of these values and discard the rest. This ensures that the list size for each vertex remains bounded by $\ell^{\prime}$. Now for each vertex $\mathbf{v}\in\mathcal{V}^{\prime}$, probability ($p_{v}$) of including each non-assigned value is at least $\frac{g_{1}}{D_{A}}\cdot\frac{g_{1}}{D_{A}}>\frac{g_{1}^{2}}{D_{A}^{2}}$. For the lists with size shorter than $\ell^{\prime}$ we add dummy variables. This ensures that every list has size equal to $\ell^{\prime}$.

Now consider a test $\psi_{1}\in\Psi^{\prime}$ with $\mathcal{R}_{\neq 0}(\psi_{1})=\{r_{1},r_{2},\ldots r_{g^{\prime}}\}$ as the set of non-zero weighted assignments, where $g^{\prime}\leq g_{1}$. Let $P_{0}(\psi_{1})$ be the probability that for each $r\in\mathcal{R}_{\neq 0}(\psi_{1})$, not a single value appears in the list of its corresponding variable. By Lemma 3.2, we can say $P_{0}(\psi_{1})=0$ by Step 1 of List-Construction. (We can come to similar conclusion simply by noting the non-triviality condition, without using Lemma 3.2.) Let $P_{\geq 2}(\psi_{1})$ is the probability of the event that for any $r\in\mathcal{R}_{\neq 0}(\psi_{1})$ there exists at least two values that appear in the list of their variables. Here there can be two cases.

In Case (i) there exists at least one $r\in\mathcal{R}_{\neq 0}(\psi_{1})$ such that it has at least two good co-ordinates. Then $P_{\geq 2}(\psi_{1})=1$, by Step 1 of List-Construction.

In Case (ii) for each $r\in\mathcal{R}_{\neq 0}(\psi_{1})$ there is at most one good co-ordinate. Then by Step 2 of List-Construction we know that there is an $r\in\mathcal{R}_{\neq 0}(\psi_{1})$ from which we took some non-assigned values. The probability that only one value is in the list of its corresponding variable is equal to the probability that none of the $D_{B}-1$ non-assigned values have been included and we only took the assigned value. This probability is given as follows.

[TABLE]

In the above equation we take all possible combinations where $k$ ($=0,\ldots,D_{B}-1$) variables are from $\mathcal{V}^{\prime}$ and the rest from $\mathcal{V}$, as described before. Thus

[TABLE]

and $P_{\geq 2}(\psi_{1})\geq 1-P_{1}(r)>\frac{c^{\prime\prime}g_{1}}{D_{A}}$, where $c^{\prime\prime}$ is a positive constant.

Now consider test $\psi_{2}\in\Psi^{\prime}$ with $\mathcal{R}_{\neq 0}(\psi_{2})$ being the set of non-zero weighted assignments. We calculate a bound on $P_{\geq 2}(\psi_{2}|\psi_{1})$, which is the probability that $\exists r_{2}\in\mathcal{R}_{\neq 0}(\psi_{2})$ such that at least two of its values appear in the list of their corresponding variables, conditioned on the fact that such an event happens for $\psi_{1}$. Here we observe that there is a correlation between the tests in terms of the variables shared. But to calculate the required probability what is important is the fact that how the values have been selected in the list of the variables. This is determined by the List-Construction procedure. Due to consistency the assigned values of the shared variables are same and we include all of them in the list. Thus if $\exists r_{2}\in\mathcal{R}_{\neq 0}(\psi_{2})$ such that it has at least two good co-ordinates then $P_{\geq 2}(\psi_{2}|\psi_{1})=1$.

Now let us consider the case when for each $r_{2}\in\mathcal{R}_{\neq 0}(\psi_{2})$ there is at most one good co-ordinate. By Step 2 of List-Construction we know $\exists r_{2}\in\mathcal{R}_{\neq 0}(\psi_{2})$ from which we took one assigned value and few non-assigned values. Probability of including these non-assigned values in this step is not affected by other tests. We sort of “look into” each test separately and do the sampling independently. Even during discarding values from longer (than $\ell^{\prime}$) lists we do not consider the tests from which these values have been taken. So the probability, $P_{1}(r_{2}|\psi_{1})$, that only one value is in the list of its corresponding variable can be written as follows.

[TABLE]

Thus $P_{\geq 2}(\psi_{2}|\psi_{1})\geq 1-P_{1}(r_{2}|\psi_{1})>\frac{c^{\prime\prime}g_{1}}{D_{A}}$, where $c^{\prime\prime}$ is a positive constant. Similar arguments and bound hold for $P_{\geq 2}(\psi_{i}|\psi_{1},\ldots,\psi_{i-1})$.

Let $P_{\geq 2}(\Psi^{\prime})$ be the probability that for each test $\psi\in\Psi^{\prime}$ there exists at least one assignment such that at least two of its values appear in the list of their corresponding variable. Then

[TABLE]

Thus there exists a labeling of the $A$-vertices (variables) such that for every $b\in\Psi^{\prime}$ (tests), we have at least two vertices $a_{1},a_{2}$ in $A$ such that $\sigma_{1}\in\widehat{\phi_{A}}(a_{1}),\quad\sigma_{2}\in\widehat{\phi_{A}}(a_{2})$ and $\pi_{e_{1}}(\sigma_{1})=\pi_{e_{2}}(\sigma_{2})$, where $e_{1}=(a_{1},b),e_{2}=(a_{2},b)\in E$, i.e. both the vertices agree (values appear in the same assignment). Hence we do not have an agreement soundness of $(\ell,s_{list})$.

∎

Thus we prove Theorem 1.1:     Assuming the Projection Games Conjecture SSAT is NP-hard for $g=n^{c}$ for some constant $c>0$.

Remark 3.1*.*

By Lemma 3.3 the constant $c$ in the above theorem is less than $1/2$. This is because the given LC instance has $s_{list}\propto s_{agr}\ell^{2}$ and $s_{agr}\propto\sqrt{s}$. Since $s_{list}<1$, so $g<\sqrt{N}$, where $N$ is the size of the LC instance.

3.2 Reduction from LC to $\textsf{SSAT}_{\infty}$

Given an LC instance $\mathcal{G}$ we construct an $\textsf{SSAT}_{\infty}$ instance $\mathcal{I}$, as done in Section 3.1. The completeness proof is also similar to Lemma 3.3, so we do not repeat it again here.

Soundness

We need to show that a “NO” instance of LC, i.e. when it has a list-agreement soundness of $(\ell^{\prime},s_{list})$, maps to a “NO” instance of $\textsf{SSAT}_{\infty}$, i.e. when every consistent not-all-zero super-assignment has norm at least $g$. When the norm is between $1$ and $g$ then any outcome (YES/NO) is fine.

If we assume that the constructed $\textsf{SSAT}_{\infty}$ instance has a consistent non-trivial solution of norm at most $g$, then similar to Lemma 3.3 we can prove that there exists a labeling such that the given LC instance does not have a list agreement soundness error $(\ell^{\prime},s_{list})$. The only difference is that since here $\ell_{\infty}$ norm is at most $g$, so norm of every test remains bounded by $g$. Note that a non-trivial solution is also not-all-zero.

Lemma 3.4**.**

Let $D_{B}$ is a constant prime power such that $N$ is a power of $D_{B}$ and $0<c<1$. Let $s\geq 1/N^{c}$ and $\ell^{\prime}<\frac{1}{\sqrt{2D_{B}}s^{d}}$, where $d<1/4$. Assume $s_{list}=\sqrt{s}D_{B}\ell^{\prime 2}$.

If $\mathcal{I}$ has a consistent non-trivial super-assignment of $\ell_{\infty}$ norm at most $g=N^{c^{\prime}}$ then $\mathcal{G}$ has list agreement soundness error $(\ell^{\prime},s_{list})$, where $c^{\prime}<d$.

Now suppose the constructed $\textsf{SSAT}_{\infty}$ instance of norm at most $g$ does not have a non-trivial solution, i.e. every variable does not have at least one value assigned, but the solution is consistent and not-all-zero. Let $\Psi^{\prime}\subseteq\Psi$ is the set of tests with non-zero norm and $V^{\prime}\subset V$ is the set of variables with no assigned values, i.e. for every $\mathbf{a}\in V^{\prime}$ we have $\mathcal{G}_{\mathbf{a}}=\emptyset$. When a value has projection weight [math], it can be due to the weights of the super-assignments (and how they cancelled) or because the value never appeared in any super-assignment (e.g. if the projection functions are partial functions). Thus in the $\ell_{\infty}$ case even if consistency is maintained, for variables with no assigned values, we cannot say that one particular value appears in all its tests. Then we can change the List-Construction procedure of Lemma 3.3 as follows. Lemma 3.2 remains true even in the $\ell_{\infty}$ norm.

For each variable we include all its assigned values in its list. 2. 2.

For any test in $\psi\in\Psi^{\prime}$ if each of its non-zero assignment has exactly one good co-ordinate, then we consider any one such assignment and include each of the non-assigned values with probability $p=\min\{1,\frac{g}{D_{A}}\}$ in the list of the corresponding variables. Once values have been included we consider the tests as “marked”. 3. 3.

If $\Psi\setminus\Psi^{\prime}\neq\emptyset$ then we do the following sequentially. Consider any variable in $V^{\prime}$ (say $\mathbf{a_{1}}$) and let $\Psi^{\prime}_{\mathbf{a_{1}}}\subseteq\Psi\setminus\Psi^{\prime}$ be the tests in which it appear. Start with any test in $\Psi^{\prime}_{\mathbf{a_{1}}}$ and take any one of its assignment. We include the value of $\mathbf{a_{1}}$ in its list and the values of the other variables (in the lists of the respective variables) with probability $p$. We consider this test “marked” and go to the next one in $\Psi^{\prime}_{\mathbf{a_{1}}}$. If there are any assignment such that the value of $\mathbf{a_{1}}$ has already been included in the list, then we consider it. Else we take any assignment and include the value of $\mathbf{a_{1}}$ in its list. The values of the other variables are included in their lists with probability $p$. This continues till we have taken at most $g$ values of $\mathbf{a_{1}}$ (we call the values obtained from this step as “marked values”). Let $\Psi^{\prime\prime}\subset\Psi\setminus\Psi^{\prime}$ be the set of tests marked in this way. We repeat this procedure again with another variable in $V^{\prime}$ and the remaining “unmarked” tests. This continues till we exhaust all variables in $V^{\prime}$.

The expected list size for variables with assigned values remain at most $2g$, similar to Lemma 3.3. For each variable $\mathbf{a_{1}}\in V^{\prime}$ there can be at most $g$ “marked” values. In each of the $D_{A}$ tests that $\mathbf{a_{1}}$ appears one “unmarked” value (apart from the marked one) can be in the list with probability $p$. So expected list size is at most $g+D_{A}\cdot\frac{g}{D_{A}}=2g$. Similar to Lemma 3.3 here also we separate the variables into two sets $\mathcal{V}$, where the list size is bounded by $\ell^{\prime}$ and $\mathcal{V}^{\prime}$ where the list size is more than $\ell^{\prime}$. We can discard some non-assigned or unmarked values and ensure that the list size remains equal to $\ell^{\prime}$ for each variable.

In Step (3) of List-Construction we see that each variable $\mathbf{a_{1}}\in V^{\prime}$ can mark at least $g$ tests in $\Psi^{\prime}$. Each variable with assigned values can mark $D_{A}$ tests each. By the condition of not-all-zero there is at least one variable with assigned values. Thus at least $g|A|$ tests can be marked. So $s_{list}=\frac{g|A|}{|B|}=\frac{gD_{B}}{D_{A}}$. Next, we can argue in the same way as in Lemma 3.3 that for at least $s_{list}$ fraction of the $B$-vertices there will not be total disagreement.

Now, one crucial thing here is, for the parameters to make sense, $gD_{A}\approx\frac{1}{\sqrt{s}}$. Or, the projection functions are such that more tests can be marked in Step (3). For example if they are total functions. Or they are partial functions but there exists a set of variables $V^{\prime\prime}\subseteq V$ such that we can find a set of values $\mathcal{V}_{\mathbf{a}}$ ($\mathbf{a}\in V^{\prime\prime}$) of cardinality at most $g$ such that for at least $s_{list}$ fraction of tests there exists at least one pre-image in that set i.e. for each such test there exists some assignment $r$ such that $r|_{\mathbf{a}}\in\mathcal{V}_{\mathbf{a}}$.

Thus we can state the following lemma.

Lemma 3.5**.**

Let $D_{B}$ is a constant prime power such that $N$ is a power of $D_{B}$ and $0<c<1$. Let $s\geq 1/N^{c}$ and $\ell^{\prime}<\frac{1}{\sqrt{2D_{B}}s^{d}}$, where $d<1/4$. Assume $s_{list}=\sqrt{s}D_{B}\ell^{\prime 2}$.

*If $\mathcal{I}$ has a consistent not-all-zero (but not non-trivial) super-assignment of $\ell_{\infty}$ norm at most $g=N^{c^{\prime}}$ ($c^{\prime}<d$), then $\mathcal{G}$ has list agreement soundness error $(\ell^{\prime},s_{list})$ if either

(i) $gD_{A}\approx\frac{1}{\sqrt{s}}$; OR     (ii) For most variables we can find a set of values of cardinality at most $g$ such that for at least $s_{list}$ fraction of tests (with that variable) there exists at least one assignment with values in that set.*

This proves Theorem 1.2:     Assuming the Projection Games Conjecture $\textsf{SSAT}_{\infty}$ is NP-hard for $g=n^{c}$ for some constant $c>0$. Again from Lemma 3.5 and the preceding discussion we get $c<1/2$.

4 Applications : Reduction from SSAT to other problems

4.1 Complexity of lattice problems

Dinur et.al. [1] reduced $g-\textsf{SSAT}$ to $g-\textsf{SIS}$ and $g-\textsf{CVP}$. In a separate paper Dinur [2] reduced $g-\textsf{SSAT}_{\infty}$ to $g-\textsf{SVP}_{\infty}$. Thus using Theorem 1.1 and 1.2 we have the following corollaries.

Corollary 4.1**.**

Assuming the Projection Games Conjecture SIS is NP-hard to approximate within a factor $g=N^{c}$, for some constant $c>0$.

Corollary 4.2**.**

Assuming the Projection Games Conjecture CVP on $N$-dimensional lattice is NP-hard to approximate within a factor $g=N^{c}$, for some constant $c>0$.

Corollary 4.3**.**

Assuming the Projection Games Conjecture $\textsf{SVP}_{\infty}$ on $N$-dimensional lattice is NP-hard to approximate within a factor $g=N^{c}$, for some constant $c>0$.

From the discussion in Section 3 and keeping in mind the blow-up in size we get from the reduction of SSAT to SIS (B) we can conclude that the constant $c$ in the exponent in each of the above corollaries is less than $1/2$. Thus these results do not violate the polynomial hierarchy [30].

4.2 Complexity of Nearest Codeword Problem (NCP)

Here we give an approximation factor preserving reduction from SSAT to NCP and prove the following using the SSAT Theorem [1].

Theorem 4.1**.**

Approximating NCP within factor $n^{c/\log\log n}$ is NP-hard.

Proof.

Let $\mathcal{I}=\langle\Psi=\{\psi_{1},\ldots,\psi_{n}\},V=\{\mathbf{v}_{1},\ldots,\mathbf{v}_{m}\},\{\mathcal{R}_{\psi_{1}},\ldots,\mathcal{R}_{\psi_{n}}\}\rangle$ be an SSAT instance. From this we efficiently construct a matrix $\overrightarrow{B^{\prime}}$ (SIS matrix) and a target vector $\overrightarrow{t^{\prime}}$, as given in [1]. Here we give a brief description of this matrix and the relevant facts that we require. For completeness we give the reduction from $g-\textsf{SSAT}$ to $g-\textsf{SIS}$ and prove these facts in B.

$\overrightarrow{B^{\prime}}$ is a $0-1$ matrix and its dimension is $n^{\prime}\times m^{\prime}$ where $n^{\prime},m^{\prime}\in\text{poly}(n)$. $\overrightarrow{t}^{\prime}$ is an all-$1$ column vector of length $n^{\prime}$. If the SSAT instance has a natural consistent super-assignment (YES) then there exists a coefficient vector $\overrightarrow{z}$ such that $\overrightarrow{B^{\prime}z}=\overrightarrow{t^{\prime}}$ and $\|\overrightarrow{z}\|_{1}=n$. If every non-trivial consistent super-assignment of the SSAT instance has norm greater than $g$ (NO) then for all coefficient vectors $\overrightarrow{z}$ with $\overrightarrow{B^{\prime}z}=\overrightarrow{t^{\prime}}$ we have $\|\overrightarrow{z}\|_{1}>gn$. Let $D>gn$ be some integer. We construct the matrix $\overrightarrow{A}\in\{0,1\}^{n^{\prime}D+m^{\prime}\times m^{\prime}}$ for NCP as follows.

We can think of $\overrightarrow{A}$ as consisting of two parts - the first $n^{\prime}D$ rows in the upper part ($\overrightarrow{A^{\prime}}$, say) and the rest $m^{\prime}$ rows in the lower part. In the upper part $\overrightarrow{A}[ik,j]=\overrightarrow{B}^{\prime}[i,j]$ for $1\leq i\leq n^{\prime},1\leq k\leq D,1\leq j\leq m^{\prime}$. The lower part is the identity matrix $\overrightarrow{\mathbb{I}}^{m^{\prime}\times m^{\prime}}$.

The target vector $\overrightarrow{t}$ is as follows : $t_{ik}=1$ for $1\leq i\leq n^{\prime},1\leq k\leq D$. Let us call these rows as $\overrightarrow{t}^{\prime\prime}$. The rest of $\overrightarrow{t}$ is [math].

Let $q$ be a prime number greater than $g$ times $\max\{n^{\prime},m^{\prime}\}$. The NCP instance obtained by reduction is $(\overrightarrow{A},\overrightarrow{t},n)$ over $\mathbb{F}_{q}$. For any coefficient vector $\overrightarrow{z}$ we have $\|\overrightarrow{Az}-\overrightarrow{t}\|=\|\overrightarrow{A^{\prime}z}-\overrightarrow{t}^{\prime\prime}\|+\|\overrightarrow{z}\|$.

Completeness : There exists $\overrightarrow{z}$ such that $\overrightarrow{B^{\prime}z}=\overrightarrow{t}^{\prime}$ and $\|\overrightarrow{z}\|_{1}=n$. This implies $\|\overrightarrow{Az}-\overrightarrow{t}\|=\|\overrightarrow{z}\|_{1}=n$ and NCP oracle outputs YES.

Soundness : For all coefficient vectors $\overrightarrow{z}$ where $\overrightarrow{B^{\prime}z}=\overrightarrow{t}$ we have $\|\overrightarrow{z}\|_{1}>gn$. In the NCP instance if $\overrightarrow{A^{\prime}z}-\overrightarrow{t}^{\prime\prime}\neq\overrightarrow{0}$ then it implies that at least $D$ co-ordinates are non-zero, else it implies $\overrightarrow{B^{\prime}z}=\overrightarrow{t}^{\prime}$ In both the cases $\|\overrightarrow{Az}-\overrightarrow{t}\|>gn$ and the NCP oracle outputs NO. ∎

We can get an improved hardness of approximation factor using Theorem 1.1.

Corollary 4.4**.**

Assuming the Projection Games Conjecture, approximation of NCP up to a factor of $n^{c}$ is NP-hard, for some constant $c>0$.

4.3 Complexity of Learning Halfspaces Problem (LHP)

Theorem 4.2**.**

Approximating the minimum failure ratio of Learning Halfspace Problem (LHP) within a factor of $n^{c/\log\log n}$ is NP-hard.

Proof.

From an SSAT instance $\mathcal{I}=\langle\Psi=\{\psi_{1},\ldots,\psi_{n}\},V=\{\mathbf{v}_{1},\ldots,\mathbf{v}_{m}\},\{\mathcal{R}_{\psi_{1}},\ldots,\mathcal{R}_{\psi_{n}}\}\rangle$, we construct an SIS instance $(\overrightarrow{B^{\prime}},\overrightarrow{t^{\prime}},d)$ as in [1] and derive inequalities, somewhat simiar to [4]. The theorem then follows from the SSAT Theorem [1].

A brief description of the SIS-matrix $\overrightarrow{B^{\prime}}$ and the relevant facts for this proof have been given in Theorem 4.1. More details can be found in B. Observe that an SIS instance $(\overrightarrow{B}^{\prime},\overrightarrow{t^{\prime}},d)$ can be viewed as a set of linear equations $\overrightarrow{B^{\prime}z}-\overrightarrow{t^{\prime}}=\overrightarrow{0}$ for some variable vector $\overrightarrow{z}$. Note that while learning a halfplane $\braket{\overrightarrow{a},\overrightarrow{x}}=b$, the unknown variables are $\overrightarrow{a}$ and $b$. So it is a homogeneous system and the coefficient of one of the variables (here $b$) is always $\pm 1$. We use a standard technique for homogenization of the SIS equations- use a new variable $y$ and replace every constant $c$ by $cy$.

Let $U=n^{c/\log\log n}|\Psi|$ and $d=|\Psi|$. Following are the linear inequalities for LHP instance.

We make $U$ copies of the linear inequality : $\quad\dfrac{-y}{U}<\delta<\dfrac{y}{U}$. 2. 2.

For each equation from SIS of the form $\sum_{i=1}^{n}a_{i}x_{i}=c$, we make $U$ copies of each of the following two inequalities:

[TABLE] 3. 3.

For every variable make $U$ copies of each of the following :

[TABLE] 4. 4.

For every variable make a copy of the following :

[TABLE] 5. 5.

Make $U$ copies of the inequality : $\quad y>0$

**Completeness : ** There exists a $\{0,1\}$ coefficient vector (variables $\overrightarrow{x}$) with norm $d$ that satisfies the linear equations. In the LHP instance assign the same values to the $x_{i}$ variables and put $y=1$ and $\delta$ a very small number tending to [math]. This will satisfy all the inequalities of type $1$, $2$ and $3$. Among inequalities of type $4$ only when $x_{i}=1$ the second inequality will not be satisfied. Thus the number of unsatisfied linear inequalities are $d$.

**Soundness : ** We give a contrapositive argument. Suppose there exists an assignment such that less than $U$ inequalities are unsatisfied. So the assignment must satisfy all inequalities of type $1,2$ and $3$. Now in the SIS instance simply assign the variables the corresponding values divided by value of $y$. This implies the variables in SIS get values from $\{-1,0,1\}$ and satisfies all the SIS equations. Hence we have a solution with norm less than $U$ for SIS. This would imply the existence of a non-trivial consistent super-assignment of norm at most $n^{c/\log\log n}$ for the SSAT instance.

∎

Using Theorem 1.1 we get the following corollary.

Corollary 4.5**.**

Assuming the Projection Games Conjecture, approximating the minimum failure ratio of LHP up to a factor of $n^{c}$ is NP-hard, for some constant $c>0$.

From the discussion in Section 3 the constant $c$ in the exponent in Corollary 4.4 and 4.5 is less than $1/2$.

Acknowledgement

The author would like to thank Divesh Aggarwal and Rajendra Kumar for helpful discussions. The author would like to thank Dana Moshkovitz for clarifying some concepts about Projection Games Conjecture, via personal correspondence with Divesh Aggarwal. The author would like to thank anonymous reviewers for helpful comments. The author was partly funded by National University of Singapore during this research. Research at IQC was supported in part by the Government of Canada through Innovation, Science and Economic Development Canada, Public Works and Government Services Canada and Canada First Research Excellence Fund.

Appendix A Some results about the SSAT instance in Section 3.1

Here we prove some results about the SSAT instance obtained in Section 3.1 by a reduction from an LC instance. We give a quick recap of some relevant facts.

We are given an LC instance $\mathcal{G}=(G=(A,B,E),\Sigma_{A},\Sigma_{B},\Pi)$ with size $N$, right degree $D_{B}$ (constant prime power), left degree $D_{A}$. For each edge $e\in E$, $\pi_{e}$ is a $p-to-1$ projection where $p\leq|\Sigma_{A}|$.

We reduce $\mathcal{G}$ to a SSAT instance $\mathcal{I}=(V,\Psi,\mathcal{R}_{\Psi})$ as follows. To each $A$-vertex $a$, we associate a variable $\mathbf{a}$, i.e. $|V|=|A|$. To each $B$-vertex $b$ we associate a test $\psi_{b}$, i.e. $|\Psi|=|B|$.The variables in a test $\psi_{b}$ are the neighbors of $b$ in $A$.

**Values of variables : ** Without much loss of generality we assume that the variables take values from a field $\mathcal{F}$, which is in bijective correspondence to $\Sigma_{A}$. We use the letters $x$ and $y$ (with subscript and superscript as required) for the elements of $\Sigma_{A}$ (or $\mathcal{F}$) and $\Sigma_{B}$ respectively.

**Satisfying assignments for tests : ** Consider a $\psi_{b}\in\Psi$. For each label $y\in\Sigma_{B}$ such that it has at least one pre-image in each of $b$’s neighbors in $A$, consider the following tuples:

[TABLE]

Thus the total set of satisfying assignments for $\psi_{b}$ is : $\quad\mathcal{R}(\psi_{b})=\bigcup_{y\in\Sigma_{B}}\mathcal{R}_{y}(\psi_{b})$. And cardinality of this set is at most $|\Sigma_{B}|p^{D_{B}}$, which is polynomially bounded by $|V|$, assuming the Projection Games Conjecture is true.

Consider a test $\psi\in\Psi$ consisting of the variables $\mathbf{a_{1}},\mathbf{a_{2}},\ldots,\mathbf{a_{D_{B}}}$. We can partition its set of satisfying assignments $\mathcal{R}(\psi)$ as follows:

For each $y\in\Sigma_{B}$, $\mathcal{R}_{y}(\psi)$ can be viewed as a $D_{B}$-dimensional array $\overrightarrow{M^{y}}$ where the $i^{th}$ dimension or co-ordinate corresponds to variable $\mathbf{a_{i}}$. The number of values of $\mathbf{a_{i}}$ in its dimension is the number of pre-images of $y$. Thus there can be at most $p$ values in each dimension. With a slight abuse of notation, we denote array elements by $M^{y}_{x_{1},\ldots,x_{D_{B}}}$, where $x_{j}\in\pi_{e}^{-1}(y)$, $e=(a_{j},b)$ and $a_{j}$ is the $j^{th}$ neighbor of $b$. This is the weight of the corresponding assignment $(x_{1},\ldots,x_{D_{B}})$. Thus the projection of a super-assignment $\overrightarrow{S(\psi)}$ on a variable $\mathbf{a_{i}}$ for any pre-image of $y$ is :

[TABLE]

where the summation is over all the pre-images of $y$ for each variable $\mathbf{a_{1}},\ldots,\mathbf{a_{i-1}},\mathbf{a_{i+1}},\ldots\mathbf{a_{D_{B}}}$. Note, since each edge constraint are functions, so the values for each variable that appear in one array cannot come in the other. In this case the arrays can be considered as “disjoint” or “non-interfering”. While taking projection on a variable for any value, it is sufficient to consider only the array in which this value appears, and not the other arrays.

We can build these arrays in polynomial time and for convenience we can think of giving these arrays as input to the SSAT oracle, so that it can fill the arrays with weights according to some well-defined criteria. A co-ordinate of an array is good (with respect to it) if at least one of the values is assigned to its corresponding variable, else it is bad.

We define the norm of each array $\overrightarrow{M^{y}}$ as

[TABLE]

The norm of a super-assignment $\overrightarrow{S(\psi)}$ is:

[TABLE]

(If the reduction is to $\textsf{SSAT}_{\infty}$ then $\|\overrightarrow{S(\psi)}\|_{\infty}=\max_{y\in\Sigma_{B}}\|\overrightarrow{M^{y}}\|$. )

Claim 1**.**

With a consistent super-assignment if a test has an array with non-zero norm but all bad co-ordinates, then we can always find a consistent super-assignment with a lesser norm.

Proof.

This follows from the “disjoint”-ness of the arrays, as has been explained above. Since the weights cannot cancel from entries in other arrays, we can put all-zero weight in this array (with all bad co-ordinates). We get a super-assignment of smaller norm, but consistency is maintained. ∎

Claim 2**.**

In any array if any variable has all non-assigned values, i.e. any co-ordinate is bad, then the total sum of all the entries is zero.

Proof.

Let in array $\overrightarrow{M^{y}}$ the first variable $\mathbf{a_{1}}$ has all non-assigned values. Thus for each value $x$ of $\mathbf{a_{1}}$

[TABLE]

The sum of all array entries is

[TABLE]

∎

Claim 3**.**

In an array with not-all bad co-ordinates, there can be either at least two good co-ordinates or one good co-ordinate with at least two values assigned to the corresponding variable.

If there is only one good co-ordinate then a consistent super-assignment can be made with a constant (one should suffice) number of non-assigned values in the other co-ordinates.

Proof.

If possible let there be only one good co-ordinate corresponding to variable $\mathbf{a_{1}}$ (say) and it has only one assigned value, say $x$. Thus $\sum_{x_{2},\ldots,x_{D_{B}}}M_{x,x_{2},\ldots,x_{D_{B}}}=w\neq 0$ and $\sum_{x_{2},\ldots,x_{D_{B}}}M_{x^{\prime},x_{2},\ldots,x_{D_{B}}}=0$, for all $x^{\prime}\neq x$ (which appear in this matrix).

Hence $\sum_{x_{1},\ldots,x_{D_{B}}}M_{x_{1},\ldots,x_{D_{B}}}=w\neq 0$, which cannot be true by Claim 2.

For the second part of the claim, without loss of generality let $\mathbf{a_{1}}$ be the co-ordinate with assigned values $1,2,\ldots,x$ and corresponding projection weights $w_{1},w_{2},\ldots w_{x}$. The following conditions must hold:

[TABLE]

By Claim 2 $\sum_{i}w_{i}=0$.

Thus we can have a super-assignment as follows: Fix $\mathbf{a_{2}},\ldots,\mathbf{a_{D_{B}}}$ to some value $x_{2},\ldots,x_{D_{B}}$ respectively. Assign $M_{ix_{2},\ldots,x_{D_{B}}}=w_{i}$, for all $i=1,\ldots,x$. It is easy to see that consistency is maintained and this gives the minimum norm. ∎

Claim 4**.**

If an array has more than one good co-ordinates then either (i) there exists at least one assignment with at least two assigned values or (ii) each assignment has one assigned value and there are only a constant number of non-assigned values for each variable.

Proof.

Easy to see from Claim 3. ∎

As a corollary of the above claims we can conclude that

Corollary 1**.**

For each test with non-zero norm, in the set of non-zero weighted assignments either there exists at least one assignment such that it has at least two variables with assigned values or all its assignments have exactly one variable with assigned value.

Appendix B $g-\textsf{SSAT}$ to $g-\textsf{SIS}$ reduction

In this section we give a brief outline of the approximation factor preserving reduction from SSAT to SIS given by Dinur et al. [1]. Given a $g-\textsf{SSAT}$ instance $\mathcal{I}=\langle\Psi=\{\psi_{1},\ldots,\psi_{n}\},V=\{\mathbf{v}_{1},\ldots,\mathbf{v}_{m}\},\{\mathcal{R}_{\psi_{1}},\ldots,\mathcal{R}_{\psi_{n}}\}\rangle$ we construct a $g-\textsf{SIS}$ instance $\mathcal{S}=(\overrightarrow{B},\overrightarrow{t},d)$ as follows.

The target vector $\overrightarrow{t}$ is an all-$1$ vector and $d=|\Psi|=n$.

The SIS-matrix $\overrightarrow{B}$ has a column for every pair $(\psi,r)$ where $\psi\in\Psi$ is a test and $r\in\mathcal{R}_{\psi}$ is a satisfying assignment for it. Thus there are $\sum_{i=1}^{n}|\mathcal{R}_{\psi_{i}}|$ columns. It can be divided into two parts : the upper part consists of consistency rows to take care of consistency and the lower part consists of non-triviality rows to take care of non-triviality.

**Non-triviality rows : ** There is a row for each test. In the row for $\psi$ all the columns associated with $\psi$ have $1$, and all the other columns have [math]. Thus there are $|\Psi|$ non-triviality rows.

**Consistency rows : ** There are $|\mathcal{F}|$ rows for each pair of tests $\psi_{i}$ and $\psi_{j}$ and common variable $x$ shared by them. If $a_{ij}$ is the number of variables shared by $\psi_{i}$ and $\psi_{j}$ then the number of consistency rows is $\sum_{i,j}a_{ij}\cdot|\mathcal{F}|$. These rows contain a consistency-ensuring gadget and only the columns for $\psi_{i}$ and $\psi_{j}$ will have non-zero values in these rows.

The consistency-ensuring gadget for pair of tests $\psi_{i}$ and $\psi_{j}$ with common variable $x$ ensures that the super-assignments to these tests are consistent on $x$. It consists of a pair of matrices $\overrightarrow{G_{1}}^{|\mathcal{F}|\times|\mathcal{R}_{\psi_{i}}|}$ and $\overrightarrow{G_{2}}^{|\mathcal{F}|\times|\mathcal{R}_{\psi_{j}}|}$. The $|\mathcal{F}|$ rows of each matrix correspond to the possible assignments for the variable $x$. The $r^{th}$ column in $G_{1}$ is the characteristic function of $r|_{x}$, i.e. it has $1$ on the value of $x$ in $r$, and [math] everywhere else. For $G_{2}$ the $r^{\prime th}$ column is a “sort of” negation of the characteristic function of $r^{\prime}|_{x}$, i.e. it has [math] on the value of $x$ in $r^{\prime}$ and $1$ everywhere else.

Correctness

We show that the YES instance of $g-\textsf{SSAT}$ maps to the YES instance of $g-\textsf{SIS}$.

Lemma 1**.**

If there is a consistent natural super-assignment to the $g-\textsf{SSAT}$ instance $\mathcal{I}$ then there exists a solution of $\ell_{1}$ norm $|\Psi|$ to the $g-\textsf{SIS}$ instance $\mathcal{S}$.

Proof.

Let $S$ be a consistent natural super-assignment. We will construct a solution $\overrightarrow{z}$ to the $g-\textsf{SIS}$ as follows: Note each $\overrightarrow{S(\psi_{i})}$ is a $|\mathcal{R}_{\psi_{i}}|$-length vector. $\overrightarrow{z}$ is a $\sum_{i=1}^{n}|\mathcal{R}_{\psi_{i}}|$-long vector consisting of the concatenation of the vectors $\overrightarrow{S(\psi_{1})},\overrightarrow{S(\psi_{2})},\ldots,\overrightarrow{S(\psi_{n})}$.

Since $S$ is natural, it assigns a $+1$ to exactly one assignment of every test. Thus the target vector is reached in the non-triviality rows.

To show that the target vector is reached in the consistency rows, consider the $|\mathcal{F}|$ rows belonging to a pair of tests $\psi_{i}$ and $\psi_{j}$ with common variable $x$. Let $\overrightarrow{S(\psi_{i})}[r_{1}]$ and $\overrightarrow{S(\psi_{j})}[r_{2}]$ be the single $1$’s in $\overrightarrow{S(\psi_{i})}$ and $\overrightarrow{S(\psi_{j})}$ respectively. Since $S$ is consistent so $r_{1}|_{x}=r_{2}|_{x}$. By the construction of the gadget matrices in $\overrightarrow{B}$ we see that the sum of these two columns gives an all-$1$ vector. So the target vector is reached in the consistency rows as well.

$\|\overrightarrow{z}\|_{1}=\sum_{i=1}^{n}\|\overrightarrow{S(\psi_{i})}\|_{1}=|\Psi|$, since $\|S\|=1$.

∎

Soundness

We need to show that a NO instance of $g-\textsf{SSAT}$ maps to a NO instance of $g-\textsf{SIS}$. Instead we give a contrapositive argument and prove the following.

Lemma 2**.**

If there exists a solution $\overrightarrow{z}$ of the $g-\textsf{SIS}$ instance $\mathcal{S}$ such that $\|\overrightarrow{z}\|_{1}\leq g|\Psi|$, then there exists a non-trivial consistent super-assignment $S$ of norm at most $g$ for the $g-\textsf{SSAT}$ instance $\mathcal{I}$.

Proof.

Given $\overrightarrow{z}$ we construct a super-assignment $S$ as follows : $\overrightarrow{z}$ is of length $\sum_{i=1}^{n}|\mathcal{R}_{\psi_{i}}|$. We break $\overrightarrow{z}$ into $|\Psi|$ pieces of length $|\mathcal{R}_{\psi_{1}}|,\ldots,|\mathcal{R}_{\psi_{n}}|$, one for each test $\psi\in\Psi$. We obtain a super-assignment of norm $\|S\|=\frac{1}{|\Psi|}\|\overrightarrow{z}\|_{1}\leq g$.

Since for each $\psi\in\Psi$, the target vector is reached in the $\psi^{th}$ row of the non-triviality rows, so

[TABLE]

and $S$ is non-trivial.

Let $\psi_{i},\psi_{j}\in\Psi$ with common variable $x$. Consider the $|\mathcal{F}|$ rows that correspond to $\psi_{i},\psi_{j},x$. In each of these rows the sum of the vectors is $1$, i.e. for any $f\in\mathcal{F}$,

[TABLE]

Subtracting Eq.(1) for $\psi_{j}$ from Eq.(2) gives,

[TABLE]

which implies $\pi_{x}(\overrightarrow{S(\psi_{i})})=\pi_{x}(\overrightarrow{S(\psi_{j})})$.

Thus we have a consistent non-trivial super-assignment of norm at most $g$.

∎