# Refuting Random CSPs

Lecturer: Sam Hopkins Scribe: Benjamin Qi

Alert: these notes are a work in progress, and have not been subjected to the usual scrutiny reserved for formal publications!


We’ve already seen how SOS can provide better worst-case guarantees for max-cut. Today we’ll take our first foray into random instances.

## Constraint Satisfication Problems

For this lecture, we will consider the hypercube over $$\{-1,1\}^n$$ rather than $$\{0,1\}^n$$.

Definition (Constraint): We define a constraint to be a predicate of the form $$\phi \colon \{\pm 1\}^k\to \{0,1\}.$$ The constraint is said to be satisfied if it evaluates to $$1$$. Throughout this lecture we assume that $$k$$ is constant, and that the constraint is non-trivial: there exists $$z\in \{\pm 1\}^k$$ such that $$\phi(z)=0$$.

Definition (Constraint Satisfication Problem). An instance $$\vphi\in \text{CSP}_{\phi}^{n,m}$$ on $$n$$ variables $$x_1,\dots,x_n\in \{\pm 1\}$$ and $$m$$ constraints can be written as $$\vphi(x)=\sum_{i=1}^m\phi(x_{S_i}\circ y_i).$$ The constraints are represented by tuples $$S_1,\dots, S_m\in [n]^k$$ and $$y_1,\dots, y_m\in \{\pm 1\}^k$$, which specify whether the variables will be negated in each constraint.

Note on Notation: If $$S=(1,2)$$ and $$y=(+1, -1)$$ then $$x_S\circ y=(x_1,-x_2)$$.

We can express interesting combinatorial objects as instances of CSPs. Some examples: - max-cut: $$\phi(x_1,x_2)=\Brac{x_1\neq x_2}$$. - SAT: $$\phi(x_1,\dots,x_k)=\text{OR}(x_1,\dots,x_k)$$ - NAE-SAT (NAE stands for “not all equal”): $$\phi=1$$ if at least one $$x_i$$ is $$1$$ and not all $$x_i$$ are $$1$$.

## Random CSPs

CSPs also give us an interesting class of optimization problems. Specifically, how can we find a satisfying assignment or an assignment that satisfies as many constraints as possible? Formally, given $$\vphi \sim \text{CSP}_{\phi}^{n,m}$$, what is $$\max_{x\in \{\pm 1\}^n}\vphi(x)$$?

In the worst case, CSPs are hard. By the PCP Theorem, there exists $$\e>0$$ such that distinguishing between satisfiable 3SAT formulas and 3SAT formulas where at most $$1-\e$$ of the clauses can be satisfied (Gap-3SAT$$[1-\e, 1]$$) is NP-Hard. But some SAT-solvers (e.g., Z3) do well in practice. Worst-case analysis isn’t telling us the whole story. Sometimes analyzing behavior on random instances is more useful.

Standard model of random CSPs: An instance of CSP with $$m$$ clauses $$S_{1\dots m}, y_{1\dots m}$$, where each $$S_i$$ and $$y_i$$ is chosen and independently and uniformly at random.

The model we consider today is slightly different than the standard one; it makes the math easier but is essentially the same. Specifically, we include every possible $$S\in [n]^k$$ independently with probability $$m/n^k$$. We refer to this probability as the clause density, and it is typically much less than $$1$$.

There are several regimes depending on how $$m$$ compares to $$n$$. - $$m\gg n\implies$$ no “interesting” $$x$$’s. $$\max_{x}\vphi(x)\approx \E_{\vphi}\max_x\vphi(x)$$. - $$m\ll n \implies$$ many $$x$$’s such that $$\vphi(x)=m$$. Use local search (not particularly interesting). - $$m=\Theta(n)$$: studied heavily in statistical physics. There are interesting algorithmic ideas (belief propagation and survey propagation, for instance).

## Refutations

Definition (Refutation): An algorithm $$\text{ALG}$$ that given $$\vphi\in \text{CSP}_{\phi}^{n,m}$$, outputs $$\text{ALG}(\vphi)\in [0,m]$$ such that:

1. $$\text{ALG}(\vphi)\ge \max_x\vphi(x)$$ for any $$\vphi$$
2. $$\E\Brac{\text{ALG}(\vphi)} \le (1-\delta)m$$ where $$\delta=\Omega(1)$$. This is known as strong refutation.

This is interesting in the regime where $$m\gg n$$.

Note that if $$\text{ALG}$$ runs in polynomial time and $$\text{ALG}(\vphi)\le (1-\Omega(1))m$$, then $$\text{ALG}$$ is a “proof system” for $$\vphi$$. Specifically, the computation trace of $$\text{ALG}$$ (i.e., the list of Turing machine moves when executed on $$\vphi$$) provides a proof (a $$\text{poly}(|\vphi|)$$-length string) that $$\text{ALG}(\vphi)\le (1-\Omega(1))m$$, and hence that $$\max_x \phi(x) \leq (1-\Omega(1))m$$. So if a polynomial-time $$\text{ALG}$$ exists, then proofs for most $$\vphi$$ will exist.

Question: How big does $$m=m(n)$$ need to be so that such $$\text{ALG}$$ exists? The intution is that the more clauses there are, the easier it should be to refute the existence of highly satisfying assignments.

Potential $$\text{ALG}$$: For $$d\in \N$$: find the least $$c$$ such that $$\proves_d \phi(x)\le c$$ in $$n^{O(d)}$$ time.

Theorem: $$\forall \phi$$, if $$m\gg n^{k/2}\cdot (\log n)^{O(1)}$$, $$\exists \delta>0$$ such that whp, $$\proves_{O(k)}\vphi(x)\le (1-\delta)m$$.

Proof: We will show that we can reduce general $$\phi$$ to the case of $$k$$-XOR. $$k$$-XOR is defined as

$\phi(z)=\begin{cases} +1 & \prod_{i=1}^kz_i=1 \\ 0 & \prod_{i=1}^kz_i=-1 \end{cases}.$

Note that when $$\phi$$ represents $$k$$-XOR, $$\phi(x_{S_i}\circ y_i)$$ depends only on $$\prod_{j\in S_i} x_j\cdot \prod_{j\le k}y_{ij}$$, so we can replace the list of signs $$y_{i,1\dots k}$$ with a single sign $$a_i\in \{\pm 1\}$$.

Definition: Assume that $$\phi$$ represents $$k$$-XOR (where $$k>0$$). Given $$\vphi\in \text{CSP}_{\phi}^{n,m}$$, define $$\psi(x)\triangleq \sum_{i=1}^ma_i \prod_{j\in S_i}x_j=(\#\text{ sat})-(\#\text{ unsat})$$.

Now let’s return to the case of general $$\phi$$. We can decompose $$\phi$$ into its fourier coefficients: $$\phi(z)=\sum_{T\subseteq [k]}\hat{\phi}_T\cdot z_T$$. Given any instance $$\vphi\sim \text{CSP}_{\phi}^{n,m}$$, we may decompose it into $$2^k$$ $$k'$$-XOR instances with $$0\le k'\le k$$ like so:

$\vphi(x)=\sum \phi(x_{S_i}\circ y_i)=\sum_{T\subseteq [k]}\hat{\phi}_T\sum_{i}\prod_{j\in T}y_{ij}\prod_{j\in T}(x_{S_i})_j=\hat{\phi}_{\emptyset}+\sum_{\emptyset\neq T\subseteq [k]}\hat{\phi}_T\psi_T(x).$

Observe that each $$\psi_T$$ is a random instance of $$k'$$-XOR for $$0< k'\le k$$. If we can show $$\proves_{O(k)}\psi_T(x)\le \e \cdot m$$ and $$\proves_{O(k)}-\psi_T(x)\le \e \cdot m$$ for all $$T$$, then

\begin{align*} \proves_{O(k)}\vphi(x)&\le \hat{\phi}_{\emptyset}+\sum_{T\subseteq [k], T\neq \emptyset}\hat{\phi}_T \psi_T(x)\\ &\le (\hat{\phi}_{\emptyset}+\O(2^k\e))m. \end{align*}

If $$\e \ll 2^{-k}$$, we’ve refuted $$\vphi(x)$$, since $$\hat{\phi_\emptyset} < 1$$ by our assumption that $$\phi(z) = 0$$ for some choice of $$z \in \{ \pm 1\}^k$$.

Note: if the max degree over all nonzero coefficients in the Fourier expansion of $$\phi$$ is $$k'<k$$, then we only need $$m\ge \tilde{\Omega}\paren{n^{k'/2}}$$ rather than $$m\ge \tilde{\Omega}\paren{n^{k/2}}$$. For example, 3-NAE-SAT has $$\hat{\phi}_{\{1,2,3\}}=0$$, so it can be refuted with $$m\le \TO(n^{2/2})=\TO(n)$$. Here, the tilde hides factors of $$\log n$$.

## Refutation for XOR

It remains to tackle the case of $$k$$-XOR. Weakly refuting $$k$$-XOR is actually easy – treating a $$k$$-XOR instance as a set of linear equations over $$\mathbb{F}_2$$, you can use Gaussian elimination to check if there is an assignment $$x \in \{\pm 1\}^n$$ which satisfies all the equations; if there isn’t, you know that $$\psi(x) \leq m - 1$$. But we need a much stronger refutation: $$\psi(x) \leq \e m$$ for some tiny $$\e$$. That is, we want a proof that no assignment satisfies more than a $$(1/2 + \e)$$-fraction of the clauses – note that this is the best we can hope for, since a random assignment satisfies $$1/2$$ of the clauses.

The main tools we use are as follows:

1. Spectral SoS certificates.

Claim: Let $$d$$ be even and let $$f \, : \, \{-1,1\}^n \rightarrow \R$$. Suppose $$f(x)$$ has $$f(x) = (x^{\otimes d/2})^T M x^{\otimes d/2}$$ over $$x\in \{\pm 1\}^n$$, where $$M$$ is a symmetric matrix. Then $$\proves_d f(x)\le n^{d/2}\vmag{M}$$.

Proof:

\begin{align*} \vmag{M}I-M\succeq 0 \implies 0&\preceq (x^{\otimes d/2})^T(\vmag{M}I-M)x^{\otimes d/2} \\ &=\vmag{M}\cdot \vmag{x}_2^d-f(x)\\ &=\vmag{M}n^{d/2}-f(x). \end{align*} Here we have used the notation that $$g(x) \succeq 0$$ for a polynomial $$g$$ if it is a sum of squares.

2. Matrix Bernstein Inequality.

Aside: The Ordinary Bernstein Inequality is roughly as follows. Given independent random variables $$a_1,\dots,a_n$$ such that $$\E a_i=0$$ and $$|a_i|\le R$$, $$\sum a_i$$ is approximately Gaussian with variance $$\E[\sum a_i^2]$$ close to the origin.

The matrix version is as follows. Given independent random symmetric matrices $$A_1,\dots, A_n\in \R^{d\times d}$$ with $$\E A_i=0$$ and $$\vmag{A_i}\le R$$ with probability 1,

$\E\vmag{\sum A_i}\le \O\paren{\vmag{\E \sum A_i^2}^{1/2}\cdot \sqrt{\log d}+R\log d}.$

### Part 1: $$k=2$$

Let $$a_{ij}$$ equal $$\pm 1$$ (whichever sign is appropriate) if $$\phi(\pm x_i,\pm x_j)$$ appears as a clause in $$\vphi$$, or $$0$$ otherwise. Then

$\psi(x)=\sum_{ij}a_{ij}x_ix_j=x^T\paren{\sum a_{ij} E_{ij}}x.$

Here $$E_{ij}$$ is the $$n\times n$$ matrix $$M$$ with $$M_{ij}=M_{ji}=\frac{1}{2}$$ if $$i\neq j$$ or $$M_{ii}=1$$ if $$i=j$$, and zeros everywhere else.

Random matrix rule of thumb: Let $$M$$ be a $$D\times D$$ symmetric matrix. It is well-known that $\vmag{M}_F=(\ell_2\text{ norm of }M)=\sqrt{\sum \lambda_i^2}.$

If $$M$$ is “unstructured” (not a well defined notion), then we can hope that $$M$$’s largest squared eigenvalue behaves like the average squared eigenvalue, in which case we would have that $$\max \lambda_i\approx \frac{\vmag{M}_F}{\sqrt D}$$.

If this occurs for the matrix $$\sum a_{ij} E_{ij}$$, we can see that the spectral norm would be around $$\|\sum_{aij} E_{ij}\| \approx \sqrt{m/n}$$. The calculation below shows that this is correct, at least when $$m \gg n$$.

Now let’s try to bound $$\vmag{\sum a_{ij}E_{ij}}$$. Observe that $$\E \Brac{\sum_{ij} a_{ij}^2 E_{ij}^2}\approx \sum_{ij} \frac{m}{n^2}E_{ij}^2$$ and

$E_{ij}^2=\begin{pmatrix} 0 & 1/2 \\ 1/2 & 0 \end{pmatrix}^2=\begin{pmatrix} 1/4 & 0\\ 0 & 1/4 \end{pmatrix}=\frac{1}{4}\paren{E_{ii}+E_{jj}}.$

So

\begin{align*} \sum_{ij} \frac{m}{n^2}E_{ij}^2 &\preceq \O\paren{\frac{m}{n^2}\sum_{i,j}\paren{E_{ii}+E_{jj} }}\\ &= \O\paren{m/n^2\cdot n\cdot I_n}\\ &= \O(m/n\cdot I_n). \end{align*}

Now by Matrix Bernstein, $\E \vmag{\sum a_{ij}E_{ij}}\le \O\paren{\sqrt{m/n}\cdot \sqrt{\log n}+\log n}=\O\paren{\sqrt{m/n}\cdot \sqrt{\log n}}.$

Furthermore, by Markov’s inequality, we can adjust the constant hidden by the $$\O$$ such that $$\vmag{\sum a_{ij}E_{ij}}\le \O\paren{\sqrt{m/n}\cdot \sqrt{\log n}}$$ holds with probability at least $$1-\e$$ for any constant $$\e>0$$ of our choice.

So we’ve shown that whp,

\begin{align*} \proves_{2}\psi(x)&=x^T\paren{\sum a_{ij} E_{ij}}x\\ &\le (\sqrt n)^2 \vmag{\paren{\sum a_{ij} E_{ij}}} \\ &\le \O\paren{n\cdot \sqrt{m/n}\cdot \sqrt{\log n}}. \end{align*}

For this to be $$\ll \e m$$, we need:

$\sqrt n \sqrt{\log n}\ll \e \sqrt m \implies m\gg \frac{n\log n}{\e^2}.$

As long as the inequality on the right is true, then whp degree-2 SOS refutes 2-XOR.

### $$k=4$$

$\psi(x)=\sum a_{ijkl}\cdot x_ix_jx_kx_l=\paren{x^{\otimes 2}}^T\overbrace{\paren{\sum a_{ijkl}E_{ij,kl}}}^{n^2\times n^2}x^{\otimes 2}$

Using a similar heuristic argument as $$k=2$$:

\begin{align*} \vmag{\sum a_{ijkl}E_{ij,kl}}_F^2 \le \O(m) &\implies \vmag{\sum a_{ijkl}E_{ij,kl}}\lesssim \sqrt{\frac{m}{n^2}}=\frac{\sqrt m}{n} \\ &\implies \proves_{4}\psi(x)\le \TO\paren{n\sqrt m}. \end{align*}

We conclude that $$\psi(x)$$ is much less than $$m$$ if $$m\gg n^2$$. The reasoning for larger even $$k$$ is similar.

### $$k=3$$

This case is trickier than the previous two. Firstly, we can write

\begin{align*} \psi(x)&=\sum_{ijk} a_{ijk}x_ix_jx_k\\ &=x^T \cdot \paren{\overbrace{\sum_{ijk} a_{ijk}E_{i,jk}}^{A}} \cdot x^{\otimes 2}, \end{align*}

where $$A$$ is an $$n\times n^2$$ matrix. Naively, we could rewrite this expression as:

$=(x,x^{\otimes 2})^T\cdot \overbrace{\begin{bmatrix}0_{n\times n} & A/2 \\ A^T/2 & 0_{n^2\times n^2} \end{bmatrix}}^{M} \cdot (x,x^{\otimes 2})$

where $$M$$ is an $$(n+n^2)\times (n+n^2)$$ symmetric matrix, and try to use the same approach as in the even-$$k$$ case, using the maximum eigenvalue of $$M$$. But the maximum eigenvalue of $$M$$ is much larger than we would like, because even though $$M$$ is $$(n+n^2) \times (n+n^2)$$, it has rank at most $$2n$$.

Another approach: using Cauchy-Schwarz,

\begin{align*} \psi(x)^2&=\paren{\sum a_{ijk}x_ix_jx_k}^2\\ &=\paren{\sum_i x_i\cdot \sum_{j,k}a_{ijk}x_jx_k}^2\\ &\le \paren{\sum_i x_i^2}\cdot \paren{\sum_i \paren{\sum_{j,k}a_{ijk}x_jx_k}^2} \\ &\triangleq n\cdot \paren{\sum_i (x^TA_ix)^2}, \end{align*}

where $$A_i$$ is the symmetric $$n\times n$$ matrix such that $$A_i(j,k)=\frac{a_{ijk}+a_{ikj}}{2}$$. The next step is to define a matrix $$A'$$ such that

$\sum_i (x^TA_ix)^2=(x^{\otimes 2})^TA'x^{\otimes 2}.$

We need to choose $$A'$$ such that $$\psi(x)^2=n\paren{x^{\otimes 2}}^TA'x^{\otimes 2} \ll m^2$$ when $$m\gg n^{1.5}$$. There are several possible ways to define $$A'$$ given $$A_i$$:

1. $$A'\triangleq \sum_iA_i^f(A_i^f)^T$$
• Here, $$A_i^f\in \R^{n^2}$$ is a flattened version of $$A_i$$ where $$A_i^f(jk)=A_i(j,k)$$. Then each $$A_i^f(A_i^f)^T$$ will have rank $$1$$, so $$A'$$ will have rank at most $$n$$. As in the naive approach, due to $$A'$$ having low rank, it can have relatively large eigenvalues, which is bad.
1. $$A'\triangleq \sum_i A_i\otimes A_i$$
• Let’s proceed with this way.

Define the $$n^2\times n^2$$ matrix $$B_i$$ to be equal to $$A_i\otimes A_i$$, except with $$B_i(jj',kk')=0$$ whenever $$\{j,k\}=\{j',k'\}$$. Informally, $$B_i$$ ignores all terms in the summation $$\sum_{i}(x^TA_ix)^2$$ that are definitely non-negative (that is, of the form $$A_i(j,k)^2x_j^2x_k^2$$). Then

$A'=\sum_i B_i+\sum_i (A_i\otimes A_i-B_i).$

The second summation is easy to deal with; for all $$x \in \{\pm 1\}^{n}$$,

$\proves_4 \paren{x^{\otimes 2}}^T\sum_i (A_i\otimes A_i-B_i)\paren{x^{\otimes 2}}\le \O\paren{\#\text{ nonzero }a_{ijk}} \le \O\paren {m}$

whp, where we used $$\proves_4 x_i^2 x_j^2 \leq 1$$.

It remains to deal with the first summation. We can bound $$\E\vmag{\sum_{i\le n}B_i}$$ using the Matrix Bernstein Inequality. Firstly, note that $$\E[B_i]=0$$ for all $$i$$, because all entries of $$B_i$$ that weren’t zeroed out are equal to the product of two independently distributed entries of $$A_i$$, each with mean zero. Secondly, we bound the operator norm of $$B' \triangleq \E\sum_{i\le n}B_i^2$$. By definition of $$B'$$, \begin{align*} B'(jj',kk')&=\sum_{i,l,l'\le n}\E \Brac{B_i(jj',ll')B_i(ll',kk')}\\ &=\E\sum_{i,l,l'\le n, \{j,l\}\neq \{j',l'\}, \{l,k\}\neq \{l',k'\}}\Brac{A_i({j,l})A_i(j',l')A_i(l,k)A_i(l',k')}\\ &=n\cdot \begin{cases} \O\paren{(m/n^3)^2\cdot n^2} & \{j,j'\}=\{k,k'\} \\ 0 & \text{otherwise } \\ \end{cases} \\ \end{align*} so $$\vmag{B'}\le \O\paren{n\cdot (m/n^3)^2\cdot n^2}=\O\paren{m^2/n^3}$$. Next, by Matrix Bernstein, $\vmag{\sum_{i\le n} B_i}\le \O\paren{\sqrt{m^2/n^3}\sqrt{\log n}+R\log n}.$ From our work for $$k=2$$, $$\vmag{A_i}\le \TO\paren{\sqrt{m/n^2}+1}$$ whp, which in turn implies $$\vmag{A_i\otimes A_i}\le \TO\paren{m/n^2+1}$$ whp. Therefore we can set $$R=\TO\paren{m/n^2+1}$$. It remains to verify that $\psi(x)^2=\TO\paren{n\paren{n^2\vmag{\sum_{i\le n} B_i}+m+n^2}} \ll m^2$ when $$m\ge \tilde\Omega\paren{n^{1.5}}$$, which is true.

So we’ve shown that $$\proves_6 \psi(x)^2\ll m^2$$. By the following lemma we may conclude that $$\proves_6 \psi(x)\ll m$$.

Fact: Let $$B$$ be a positive constant. $$\proves_d f^2\le B\implies \proves_d f\le \sqrt B$$.

Proof: Clearly $\proves_d (f/B^{1/4}-B^{1/4})^2\ge 0,$

as the LHS is a square of degree at most $$d$$. Expanding the LHS, we get $\proves_d f^2/\sqrt B-2f+\sqrt B\ge 0$

$\implies \proves_d f\le \frac{f^2/\sqrt B+\sqrt B}{2}.$ Since $$\proves_d f^2\le B$$ by assumption, the inequality above simplifies to $$\proves_d f\le \sqrt B$$, as desired.

Conclusion: $$\proves_6 \psi_T(x)\le \e m$$ whp if $$m\ge n^{1.5}\text{poly}(\log n,1/\e)$$.

## Application: Tensor Completion

Related to learning theory.

Matrix Completion Problem (AKA Netflix Problem). Consider an $$n\times m$$ matrix $$M$$ with rows indexed by $$n$$ users and $$m$$ columns indexed by movies. You know only a few entries of the matrix, each corresponding to a user rating a movie. For example, $$+1$$ could represent a user strongly liking a movie, and $$-1$$ could represent a user strongly disliking a movie. The goal is to recover the rest of $$M$$.

Question: Assume that $$M$$ has low rank $$r$$; that is, $$M=\sum_{i=1}^r u_iv_i^T$$. How many random entries of $$M$$ need to be revealed in order to recover the rest of $$M$$?

It turns out that we can do so with $$\O(n+m)$$ entries, given $$r\le \O(1)$$ and some other assumptions (including bounds on the magnitudes of $$u_i$$ and $$v_i$$).

Tensor Completion Problem (AKA Yelp Problem). Consider an $$n\times n\times n$$ tensor $$T$$ indexed by users, restaurants, and time of day. Again, assume $$T$$ has a low-rank decomposition $$\sum_{i=1}^ru_i\otimes v_i\otimes w_i$$. The task is the same: recover $$T$$ from as few entries as possible.

Naive solution: Flatten the tensor into an $$n\times n^2$$ matrix: $$T=\sum u_i (v_i\otimes w_i)^T$$. Then run matrix completion. This requires $$\Omega(n^2)$$ entries to be revealed.

Theorem (Boaz Barak, Ankur Moitra): Consider an $$n\times n\times n$$ tensor $$T$$ that can be written in the form:

$T=\sum_{i=1}^r u_i\otimes u_i\otimes u_i, \vmag{u_i}_{\infty}\le 1.$

Then there exists a polynomial time algorithm to complete $$T$$ using $$\TO(n^{1.5})$$ observed entries. The set of observed entries is given by $$\Omega\subseteq [n]^3$$, where $$\E |\Omega|=m$$ and $$\Omega$$ was chosen by sampling each entry of $$T$$ independently with probability $$\frac{m}{n^3}$$. The algorithm outputs $$X\in \R^{n\times n\times n}$$ such that $$\E\vmag{X-T}_2^2\le \text{poly}(r)\cdot n^3\cdot \paren{\frac{n^{1.5}\log n}{m}}^{\Omega(1)} + r^{O(1)} n^2$$.

Remark: Interpreting the guarantees here, notice that the entries of $$T$$ are of order $$r$$, so when $$m \gg n^{1.5} \log n$$ we are finding $$X$$ such that for typical $$i,j,k$$, $$(X_{ijk} - T_{ijk})^2 \ll T_{ijk}^2$$.

The $$\infty$$-norm assumption, or something like it, is necessary in this context, where it is often referred to as incoherence. The $$r^{O(1)} n^2$$ term can be removed with a bit of additional work – it shows up here because we won’t worry about correctly completing the entries $$T_{ijk}$$ where $$i,j,k$$ are not all distinct.

Approach: Let us associate to $$T$$ the following probability distribution $$\mu$$ on $$\{ \pm 1\}^n$$. First, sample $$i \sim [r]$$ uniformly at random. Then, for each $$j \leq n$$, independently sample $$x_j$$ such that $$\E x_j = u_i(j)$$.

For each tuple of indices $$(i,j,k)$$ with $$i,j,k$$ all distinct, $$\frac 1 r T_{ijk} = \E_{x \sim \mu} x_i x_j x_k$$. A natural approach to completing $$T$$ would be to find the degree $$3$$ moments of a probability distribution that agrees with the revealed entries of $$T$$. We don’t know how to search for such moments in polynomial time, but we can search for pseudo-moments instead.

Algorithm: Find a degree 12 $$\pE$$ such that for all $$(i,j,k) \in \Omega$$ with $$i,j,k$$ distinct, $$\pE x_i x_j x_k = \frac 1 r T_{ijk}$$. Output $$\pE x^{\otimes 3}$$.

It turns out that the success of this algorithm depends on SOS refuting random 3-XOR.

We haven’t discussed how to implement this kind of algorithm in polynomial time; for today we will take it on faith that finding pseudoexpectations which satisfy some linear equations can be done in polynomial time (up to small numerical error) via semidefinite programming.

Analysis: We will combine some standard tricks from statistical learning theory with our theorem on SOS refutation of $$k$$-XOR to analyze this algorithm.

We want to bound

$(*) := \sup_{\pE}\Brac{ \E_{ijk}\paren{\tfrac 1 rT_{ijk}-\pE x_i x_j x_k }^2-\frac{1}{\E |\Omega|}\sum_{ijk\sim \Omega}(\tfrac 1 r T_{ijk}-\pE x_ix_jx_k)^2}.$

Suppose we can show that $$(*)$$ is at most $$\delta$$. The $$\pE$$ found by our algorithm will have the property that $$\pE x_i x_j x_k - \frac 1 r T_{ijk} = 0$$ for $$ijk \in \Omega$$ all distinct. So for this $$\pE$$,

\begin{align*} \frac{1}{\E |\Omega|}\E_\Omega \sum_{ijk \sim \Omega} (\tfrac 1 r T_{ijk} - \pE x_i x_j x_k)^2 &\leq \O(1) \cdot \frac{1}{\E |\Omega|} \cdot \E_\Omega \sum_{ijk \sim \Omega} 1[i,j,k \text{ not all distinct}] \\ &\leq \O(1/n), \end{align*}

and hence, again for this $$\pE$$,

$\E_{\Omega} \, \E_{ijk} \paren{\tfrac 1 r T_{ijk} - \pE x_i x_j x_j}^2 \leq \O(1/n) + \delta,$

which after rescaling by $$r^2 n^3$$, becomes $$\|X - T\|_2^2 \leq \O(r^2 n^2) + \delta r^2 n^3.$$ Hence, our goal is to obtain $$\delta \leq \paren{\tfrac{n^{1.5} \log n}{m}}^{\Omega(1)}$$.

Next, let’s use “ghost samples;” this is a standard trick in learning theory. Specifically, consider generating a set of samples $$\Omega'$$ in the same way as $$\Omega$$. Then we can rewrite the expected supremum above as

\begin{align*} (*)&= \frac{1}{\E |\Omega|} \E_{\Omega}\sup_{\pE}\Brac{\E_{\Omega'}\sum_{ijk\sim \Omega'}\paren{\tfrac 1 rT_{ijk}-\pE x_i x_j x_k }^2-\sum_{ijk\sim \Omega}\paren{\tfrac 1 rT_{ijk}-\pE x_i x_j x_k }^2}\\ &\le \frac{1}{\E |\Omega|} \E_{\Omega,\Omega'}\sup_{\pE}\Brac{\sum_{ijk\sim \Omega'}\paren{\tfrac 1 rT_{ijk}-\pE x_i x_j x_k }^2-\sum_{ijk\sim \Omega}\paren{\tfrac 1 rT_{ijk}-\pE x_i x_j x_k }^2}\\ &= \frac{1}{\E |\Omega|}\E_{\Omega,\Omega',\sigma}\sup_{\pE}\Brac{\sum_{ijk\sim \Omega'}\sigma_{ijk}\paren{\tfrac 1 rT_{ijk}-\pE x_i x_j x_k }^2-\sum_{ijk\sim \Omega}\sigma_{ijk}\paren{\tfrac 1 rT_{ijk}-\pE x_i x_j x_k }^2}\\ &\le \frac{2}{\E |\Omega|}\E_{\Omega,\sigma}\sup_{\pE}\Brac{\sum_{ijk\sim \Omega}\paren{\tfrac 1 rT_{ijk}-\pE x_i x_j x_k }^2} \\ &\le \O(1)\cdot \E_{\Omega,\sigma}\sup_{\pE}\E_{ijk\sim \Omega}\sigma_{ijk}\paren{\frac{1}{r}T_{ijk}-\pE x_ix_jx_k}^2 \end{align*}

Here, each $$\sigma_{ijk}$$ equals $$\pm 1$$ independently with equal probability; see the note at the end of this document for details. The first inequality follows from Jensen’s inequality, the second follows from the triangle inequality, and the last holds because $$|\Omega|$$ is within a constant factor of $$\E |\Omega|$$ whp.

\begin{align*} \E_{\Omega,\sigma}&\sup_{\pE}\E_{ijk\sim \Omega} \sigma_{ijk}\paren{\frac{1}{r}T_{ijk}-\pE x_ix_jx_k}^2\\ &\le \E_{\Omega,\sigma}\Brac{\E_{ijk\sim \Omega}\sigma_{ijk}\paren{\frac{1}{r}T_{ijk}}^2+\sup_{\pE}\E_{ijk\sim \Omega}\sigma_{ijk}\paren{\pE x_ix_jx_k}^2+2\sup_{\pE}\E_{ijk\sim \Omega}\paren{\sigma_{ijk}\cdot \frac{T_{ijk}}{r}\pE x_ix_jx_k}} \end{align*}

We can deal with the summands independently.

1. The first summand does not depend on $$\pE$$, so by a Chernoff bound it can be shown to be $$\le \O\paren{m^{-1/2}}$$.

2. The second summand can be written as $\E_{\Omega,\sigma}\sup_{\pE}\E_{ijk\sim \Omega}\sigma_{ijk}\pE \underbrace{x_ix_i'}_{y_i} \underbrace{x_jx_j'}_{y_j}\underbrace{x_kx_{k}'}_{y_k}\le \E_{\Omega,\sigma}\sup_{\pE}\pE \frac{1}{|\Omega|}\sum_{ijk}\sigma_{ijk}y_iy_jy_k.$

Here, on the LHS $$\pE$$ is the degree-$$12$$ pseudoexpectation which results from defining a “product pseudoexpectation” on variables $$x_1,\ldots,x_n,x_1',\ldots,x_n'$$; then we define polynomials $$y_i = x_i x_i'$$ to obtain a degree-$$6$$ pseudoexpectation in variables $$y_1,\ldots,y_n$$. So by our work for refuting 3-XOR it is bounded above by $$\O\paren{\frac{n^{1.5}\log n}{m}}^{\Omega(1)}$$ whp.

3. Similarly as the second summand, the third summand can be written as

$\E_{\Omega,\sigma}\sup_{\pE}\E_{ijk\in \Omega}\sigma_{ijk}\E_{x'\sim \mu}\pE \underbrace{x_ix'_i}_{z_i}\underbrace{x_jx'_j}_{z_j}\underbrace{x_kx'_k}_{z_k}\le \E_{\Omega,\sigma}\sup_{\pE}\pE \frac{1}{|\Omega|}\sum_{ijk}\sigma_{ijk}y_iy_jy_k,$

where $$\mu$$ is the distribution on $$\{ \pm 1\}^n$$ we associated to $$T$$, and $$z_i$$ is a product pseudoexpectation defined as $$z_i=x_ix_i'$$. So it is also bounded above by $$\O\paren{\frac{n^{1.5}\log n}{m}}^{\Omega(1)}$$ whp.

Thus, $$\delta \le (*)\le \O\paren{\frac{n^{1.5}\log n}{m}}^{\Omega(1)}$$, as desired.

Note: The signs $$\sigma_{ijk}$$ can be introduced via an exchangability argument which goes roughly like this.

Suppose we have a family of functions $$f\colon [n]^3 \rightarrow \R$$. (In this case, the functions are given by pseudoexpectations, and the function $$f_{\pE}$$ associated to $$\pE$$ is $$f_{\pE}(i,j,k) = \paren{\tfrac 1r T_{ijk} - \pE x_i x_j x_k}^2$$.)

Look at $$\sup_f \Brac{\sum_{ijk \in \Omega} f(ijk) - \sum_{ijk \in \Omega'} f(ijk)}$$. We can rewrite as

$\sup_f \Brac{\sum_{ijk} f(ijk) 1(ijk \in \Omega) - \sum_{ijk} f(ijk) 1(ijk \in \Omega')} = \sup_f \Brac{\sum_{ijk} f(ijk) (1(ijk \in \Omega) - 1(ijk \in \Omega'))}.$

Now, the random variables $$1(ijk \in \Omega)- 1(ijk \in \Omega')$$ are independent for distinct $$ijk, i'j'k'$$, and identically distributed to $$\sigma_{ijk} (1(ijk \in \Omega) - 1(ijk \in \Omega'))$$, where $$\sigma_{ijk}$$ are independent random signs. So the whole $$\sup$$ is identically distributed to $$\sup_f \sum_{ijk \in \Omega}\Brac{ \sigma_{ijk} f_{ijk} - \sum_{ijk \in \Omega'} \sigma_{ijk} f_{ijk}}$$. Then we can use the triangle inequality to break this into $$2$$ identical terms.