As reported here, here and here, Mark Braverman has just announced a proof of a 1990 conjecture by Linial and Nisan.

Mark proves that if {C} is an AC0 boolean circuit (with NOT gates and with AND gates and OR gates of unbounded fan-in) of depth {d} and size {S}, and if {X} is any {k}-wise independent distribution with {k = (\log S)^{O(d^2)}}, then

\displaystyle  {\mathbb E} C(U_n) \approx {\mathbb E} C(X)

that is, {X} “fools” the circuit {C} into thinking that {X} is the uniform distribution {U_n} over {\{0,1\}^n}. Plausibly, this might be true even for {k = O((\log S)^{d-1})}.

Nothing was known for depth 3 or more, and the depth-2 case was settled only recently by Bazzi, with a proof that, as you may remember, has been significantly simplified by Razborov about six months ago.

Mark’s proof relies on approximating {C} via low-degree polynomials. The point is that if {p} is an {n}-variate (real valued) polynomial of degree {\leq k}, and {X} is a {k}-wise independent distribution ranging over {\{0,1\}^n}, then

\displaystyle  {\mathbb E} p(X) = {\mathbb E} p(U_n)

Now if we could show that {p} approximates {C} both under {X} and under {U_n}, in the sense that {{\mathbb E} p(X) \approx {\mathbb E} C(X)}, and also {{\mathbb E} p(U_n) \approx {\mathbb E} C(U_n)}, then we would be done.

The Razborov-Smolenski lower bound technique gives a probabilistic construction of a polynomial {p} such that for every input {x} one has a high probability that {p(x) = C(x)}. In particular, one get one polynomial {p} such that both

\displaystyle   {\mathbb P}_{x\sim X} [p(x) = C(x) ] = 1-o(1) \ \ \ \ \ (1)


\displaystyle   {\mathbb P}_{x\sim U_n} [p(x) = C(x) ] = 1-o(1) \ \ \ \ \ (2)

Unfortunately this is not sufficient, because the polynomial {p} might be very large at a few points, and so even if {p} agrees with {C} with high probability there is no guarantee that the average of {p} is close to the average of {C}.

Using a result of Linial, Mansour and Nisan (developed in the context of learning theory), one can construct a different kind of low-degree approximating polynomial {p}, which is such that

\displaystyle   {\mathbb E}_{x\sim U_n} | C(x) - p(x)|^2 = o(1) \ \ \ \ \ (3)

The Linial-Mansour-Nisan approximation, however, says nothing about the relation between {p} and {C} under the distribution {X}.

Using ideas of Bazzi’s, however, if we had a single polynomial {p} such that properties (1), (2) and (3) are satisfied simultaneously, then we could construct another low-degree polynomial {p'} such that {{\mathbb E} p'(X) \approx {\mathbb E} C(X)}, and also {{\mathbb E} p'(U_n) \approx {\mathbb E} C(U_n)}, giving us that {C} is fooled by {X}.

As far as I understand, Mark constructs a polynomial satisfying properties (1), (2) and (3) by starting from the Razborov-Smolenski polynomial {p}, and then observing that the indicator function {E} of the points on which {p \neq C} is itself a boolean function admitting a Linial-Mansour-Nisan approximation {e}. Defining {p' := p\cdot (1-e)}, we have that {p'} has all the required properties, because multiplying by {1-e} “zeroes out” the points on which {p} is excessively large.

I have been interested in this problem for some time because of a connection with the complexity of 3SAT on random instances.

As I have discussed in the past, if we take a random instance of 3SAT constructed by randomly picking, say, 10n clauses over n variables, then there is an extremely high probability that the formula is unsatisfiable. Certifying that such formulas are unsatisfiable, however, appears to be hard.

Here what we want is a certification algorithm {A} that, on input a boolean formula {\phi}, outputs either unsatisfiable or don’t know; if the algorithm outputs unsatisfiable then the formula is definitely guaranteed to be unsatisfiable; and the algorithm outputs unsatisfiable on at least, say, a inverse polynomial constant fraction of inputs. (Thanks for Albert Atserias for pointing out that the problem is trivial if one requires only inverse polynomial refutation probability.)

The existence of such an algorithm for random 3SAT with {n} variables and {10n} clauses has been ruled out in a variety of bounded models of computation. (And of course the lower bounds extend to higher number of clauses than {10n}.) For example, we know that, except with very small probability, no sub-exponential size “tree-like resolution” proof of unsatisfiability can exist for formulas from this distribution, and we know that when a back-tracking based algorithm finishes without having found a satisfying assignment, its computation defines a tree-like resolution proof. Hence no back-tracking based algorithm can run in sub-exponential time. We also know lower bounds for algorithms based on convex relaxations of Max 3SAT, including algorithms based on Lasserre semidefinite programming relaxations, which are about the most powerful convex programming relaxations that we know how to construct.

Notably, I don’t know of a lower bound showing that no such refutation algorithm can be designed in AC0, even though AC0 is a class for which we are usually able to prove lower bounds. (As far as I know, the question remains open.)

All the above lower bounds, by the way, apply also to random instances of the 3XOR problem with, say, n variables and 10n equations. (3XOR is like 3SAT but every “clause” is a linear equation mod 2 involving three variables.) 3XOR is of course solvable in polynomial time via Gaussian elimination, but in many simplified models of computation it seems to capture the hardness of 3SAT.

So what about refuting random 3XOR in AC0? Consider the following two distributions of instances:

  • Distribution {A} contains random instances with {n} variables, {10n} clauses, and uniformly chosen right-hand-side

  • Distribution {B} contains instances with a random left-hand side involving {n} variables and {10n} clauses, and a right-hand side chosen to be consistent with a random assignment.

Now the left-hand side has the same distribution in {A} and {B}, and for, most left-hand-sides, the distribution of the right-hand side in {B} in {\Omega(n)}-wise independent, while the right-hand side in {A} is always uniform.

From Mark’s result it follows that {A} and {B} are indistinguishable by AC0 circuits, and so no AC0 circuit can refute random instances from distribution {B}. (Otherwise it would incorrectly output unsatisfiable with positive probability given samples from {B}, while all samples from {B} are satisfiable.)

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