# Can NP-hard statements be proved by PCPs that only involve reading 2 bits?

For non-negative integers q, let PCP(q) denote the set of promise problems
that have polynomial-length probabalistically checkable proofs
over the binary alphabet in which the verifier only reads q bits.
(with no restriction on the amount of randomness that the verifier can use)

Let sPCP(q) denote the analogous class in which the verifier has
(sPCP stands for "strong PCP", a term I made up for this question)

Obviously, $\:$PCP(0) = promiseBPP$\:$.
It is slightly less obvious, but still easy to show, that $\:$PCP(1) = promiseBPP$\:$.
It is known, though highly non-trivial, that $\:$NP $\subseteq$ sPCP(3)$\:$.
Since $\:$2SAT $\in$ P$\:$, $\:$ one has $\:$sPCP(2) $\subseteq$ promise-coRP$\:$ by reasoning similar to the PCP(1) equality.
However, that does yield any comparison between PCP(2) and NP.

Is either $\;\;\;$NP $\subseteq$ PCP(2) $\:$ or $\:$ PCP(2) $\subset$ NP$\;\;\;$ known to hold under plausible assumptions,
such as the promise versions of $\:$ P = BPP $\subset$ NP = MA ≠ coNP $\:$ ?

• doesn't the hardness of approximation of MaxCUT give a 2-bit PCP for NP with constant gap between completeness and soundness? – Sasho Nikolov Jan 10 '14 at 4:19
• I think one would need a way to approximate how large of a cut there $\hspace{1.61 in}$ should be assuming the instance is true. $\:$ – user6973 Jan 10 '14 at 4:36
• Not sure what you mean: the NP-hardness of approximating MaxCut is proved via a gap problem, i.e. it is NP-hard to distinguish between a $c$ fraction of the edges being cut vs. $s$ fraction being cut. $c$ is the completeness, $s$ is the soundness, and bounds on both are given by the reduction. see e.g. theorem 5.14. in Hastad's paper: csce.uark.edu/~dapon/lib/Hastad.pdf – Sasho Nikolov Jan 10 '14 at 5:43
• After reading your first comment, I was looking at that paper and trying to find if there was anything in it which would work. $\:$ How does one go from that theorem to the NP-hardness of such a gap problem? $\:$ (It is not clear to me how one would find the relevant $c$ and $s$ to give a gap problem.) $\;\;\;$ – user6973 Jan 10 '14 at 5:54
• in the reduction, each equation over 3 variables is replaced by 16 equations over 2 variables, so that if the original equation is satisfiable, 12 of the new ones are satisfiable, and if it is not, 10 of them are. since we start with equations given by Hastad's 3-bit PCP where $c = 1-\delta$, $s \leq \frac{1}{2} + \delta$, after the reduction we have $c = \frac{3}{4}-\frac{1}{8}\delta$ and $s \leq \frac{11}{16} + \frac{1}{8}\delta$. – Sasho Nikolov Jan 10 '14 at 13:25

A binary CSP with arity 2 constraints is given by a family $\Pi$ of arity 2 relations on $\{0, 1\}^n$. An instance is given by a set of constraints. The GapCSP$_\Pi$($c$,$s$) problem for the CSP is the promise problem of distinguishing between CSP instances where at least a fraction $c$ of the constraints can be satisfied and instances where at most a fraction $s$ of the constraints can be satisfied. If such a GapCSP$_\Pi$($c$, $s$) problem is NP-hard for constants $c > s$, then we get a 2-bit PCP for NP in the usual way: the proof is a variable assignment, and can be verified by sampling a constraint and checking if it is satisfied.
In Some Optimal Inapproximability Results, Håstad shows that the GapCSP$_\Pi$($\frac{3}{4} - \delta$, $\frac{11}{16} + \delta$) problem is NP-hard for any constant $\delta$ when $\Pi$ is the family of constraints of the type $x_i + x_j = c \pmod 2$. There are two components to the reduction:
• the famous optimal hardness of 3-variable linear equations mod 2: $\text{GapCSP}{_\Pi}$($1 - \delta'$, $\frac{1}{2} + \delta'$) is NP-hard for any constant $\delta'$ and $\Pi$ the family of constraints of the type $x_i + x_j + x_k = c \pmod 2$ (equivalently, this is a result about optimal 3-bit PCP for NP)
You can get a better ratio between $c$ and $s$ if you assume the Unique Games conjecture: Khot, Kindler, O'Donnel, and Mossel showed that for $\Pi$ the max cut constraints $x_i \neq x_j$, solving GapCSP$_\Pi$($\frac{1}{2}(1 - \rho)$, $\frac{1}{\pi}\arccos(\rho) - \delta$) is as hard as Unique Games for any $\rho \in (-1, 1)$ and any constant $\delta > 0$. Optimizing over $\rho$ gives $\frac{s}{c}$ arbitrarily close to the Goemans-Williamson constant.