### Impagliazzo Hard-Core Sets via "Finitary Ergodic-Theory"

In the Impagliazzo hard-core set theorem we are a given a function $g:\{ 0, 1 \}^n \rightarrow \{ 0,1\}$ such that every algorithm in a certain class makes errors at least a $\delta$ fraction of the times when given a random input. We think of $\delta$ as small, and so of $g$ as exhibiting a weak form of average-case complexity. We want to find a large set $H\subseteq \{ 0,1 \}^n$ such that $g$ is average-case hard in a stronger sense when restricted to $H$. This stronger form of average-case complexity will be that no efficient algorithm can make noticeably fewer errors while computing $g$ on $H$ than a trivial algorithm that always outputs the same value regardless of the input. The formal statement of what we are trying to do (see also the discussion in this previous post) is:

Our approach will be to look for a "regular partition" of $\{0,1\}^n$. We shall construct a partition $P= (B_1,\ldots,B_m)$ of $\{0,1\}^n$ such that: (i) given $x$, we can efficiently compute what is the block $B_i$ that $x$ belongs to; (ii) the number $m$ of blocks does not depend on $n$; (iii) $g$ restricted to most blocks $B_i$ behaves like a random function of the same density. (By "density" of a function we mean the fraction of inputs on which the function evaluates to one.)

In particular, we will use the following form of (iii): for almost all the blocks $B_i$, no algorithm has advantage more than $\epsilon$ over a constant predictor in computing $g$ in $B_i$.

Let $M_0$ be the union of all majority-0 blocks (that is, of blocks $B_i$ such that $g$ takes the value 0 on a majority of elements of $B_i$) and let $M_1$ be the union of all majority-1 blocks.

I want to claim that no algorithm can do noticeably better on $M_0$ than the constant algorithm that always outputs 0. Indeed, we know that within (almost) all of the blocks that compose $M_0$ no algorithm can do noticeably better than the always-0 algorithm, so this must be true for a stronger reason for the union. The same is true for $M_1$, with reference to the constant algorithm that always outputs 1. Also, if the partition is efficiently computable, then(in a non-uniform setting) $M_0$ and $M_1$ are efficiently recognizable. It remains to argue that either $M_0$ or $M_1$ is large and not completely unbalanced.

Recalling that we are in a non-uniform setting (where by "algorithms" we mean "circuits") and that the partition is efficiently computable, the following is a well defined efficient algorithm for attempting to compute $g$:

(The majority values of $g$ in the various blocks are just a set of $m$ bits that can be hard-wired into the circuit.)

We assumed that every efficient algorithm must make at least a $\delta$ fraction of errors. The set of $\geq \delta 2^n$ inputs where the Local Majority algorithm makes mistakes is the union, over all blocks $B_i$, of the "minority inputs" of the block $B_i$. (If $b$ is the majority value of $g$ in a block $B$, then the "minority inputs" of $B$ are the set of inputs $x$ such that $g(x) = 1-b$.)

Let $E_0$ be the set of minority inputs (those where our algorithm makes a mistake) in $M_0$ and $E_1$ be the set of minority inputs in $M_1$. Then at least one of $E_0$ and $E_1$ must have size at least $\frac {\delta}{2} 2^n$, because the size of their union is at least $\delta 2^n$. If $E_b$ has size at least $\frac {\delta}{2} 2^n$, then $M_b$ has all the properties of the set $H$ we are looking for.

It remains to construct the partition. We describe an iterative process to construct it. We begin with the trivial partition $P = (B_1)$ where $B_1 = \{ 0,1\}^n$. At a generic step of the construction, we have a partition $P = (B_1,\ldots,B_m)$, and we consider $M_0, M_1,E_0,E_1$ as above. Let $b$ be such that $E_b \geq \frac 12 \delta 2^n$. If there is no algorithm that has noticeable advantage in computing $g$ over $M_b$, we are done. Otherwise, if there is such an algorithm $f$, we refine the partition by splitting each block according to the values that $f$ takes on the elements of the block.

After $k$ steps of this process, the partition has the following form: there are $k$ functions $f_1,\ldots,f_k$ and each of the (at most) $2^k$ blocks of the partition corresponds to a bit string $b_1,\ldots,b_k$ and it contains all inputs $x$ such that $f_1(x)=b_1,\ldots,f_k(x)=b_k$. In particular, the partition is efficiently computable.

We need to argue that this process terminates with $k=poly(1/\epsilon,1/\delta)$. To this end, we define a potential function that measures the "imbalance" of $g$ inside the blocks the partition

$\Psi(B_1,\ldots,B_m) := \sum_{i=1}^m \frac {|B_i|}{2^n} \left( Pr_{x\in B_i} [g(x) = 1] \right)^2 $

and we can show that this potential function increases by at least $poly(\epsilon,\delta)$ at each step of the iteration. Since the potential function can be at most 1, the bound on the number of iterations follows.

A reader familiar with the proof of the Szemeredi Regularity Lemma will recognize the main ideas of iterative partitioning, of using a "counterexample" to the regularity property required of the final partition to do a refinement step, and of using a potential function argument to bound the number of refinement steps.

In which way can we see them as "finitary ergodic theoretic" techniques? As somebody who does not know anything about ergodic theory, I may not be in an ideal position to answer this question. But this kind of difficulty has not stopped me before, so I may attempt to answer this question in a future post.

Impagliazzo Hard-Core Set Theorem, "Constructive Version"

Let $g:\{0,1\}^n \rightarrow \{0,1\}$ be a boolean function, $s$ be a size parameter, $\epsilon,\delta>0$ be given. Then there is a size parameter $s' = poly(1/\epsilon,1/\delta) \cdot s + exp(poly(1/\epsilon,1/\delta))$ such that the following happens.

Suppose that for every function $f:\{0,1\}^n \rightarrow \{0,1\}$ computable by a circuit of size $s'$ we have

$Pr_{x \in \{0,1\}^n} [ f(x) = g(x) ] \leq 1-\delta$

Then there is a set $H$ such that: (i) $H$ is recognizable by circuits of size $\leq s'$; (ii) $|H| \geq \delta 2^n$, and in fact the number of $x$ in $H$ such that $g(x)=0$ is at least $\frac 12 \delta 2^n$, and so is the number of $x$ in $H$ such that $g(x)=1$; and (iii) for every $f$ computable by a circuit of size $\leq s$,

$Pr_{x\in H} [ g(x) = f(x) ] \leq max \{ Pr_{x\in H}[ g(x) = 0] , Pr_{x\in H} [g(x)=1] \} + \epsilon$

Our approach will be to look for a "regular partition" of $\{0,1\}^n$. We shall construct a partition $P= (B_1,\ldots,B_m)$ of $\{0,1\}^n$ such that: (i) given $x$, we can efficiently compute what is the block $B_i$ that $x$ belongs to; (ii) the number $m$ of blocks does not depend on $n$; (iii) $g$ restricted to most blocks $B_i$ behaves like a random function of the same density. (By "density" of a function we mean the fraction of inputs on which the function evaluates to one.)

In particular, we will use the following form of (iii): for almost all the blocks $B_i$, no algorithm has advantage more than $\epsilon$ over a constant predictor in computing $g$ in $B_i$.

Let $M_0$ be the union of all majority-0 blocks (that is, of blocks $B_i$ such that $g$ takes the value 0 on a majority of elements of $B_i$) and let $M_1$ be the union of all majority-1 blocks.

I want to claim that no algorithm can do noticeably better on $M_0$ than the constant algorithm that always outputs 0. Indeed, we know that within (almost) all of the blocks that compose $M_0$ no algorithm can do noticeably better than the always-0 algorithm, so this must be true for a stronger reason for the union. The same is true for $M_1$, with reference to the constant algorithm that always outputs 1. Also, if the partition is efficiently computable, then(in a non-uniform setting) $M_0$ and $M_1$ are efficiently recognizable. It remains to argue that either $M_0$ or $M_1$ is large and not completely unbalanced.

Recalling that we are in a non-uniform setting (where by "algorithms" we mean "circuits") and that the partition is efficiently computable, the following is a well defined efficient algorithm for attempting to compute $g$:

Algorithm. Local Majority

On input $x$:

determine the block $B_i$ that $x$ belongs to;

output $1$ if $Pr_{z\in B_i} [g(z)=1] \geq \frac 12$;

otherwise output $0$

(The majority values of $g$ in the various blocks are just a set of $m$ bits that can be hard-wired into the circuit.)

We assumed that every efficient algorithm must make at least a $\delta$ fraction of errors. The set of $\geq \delta 2^n$ inputs where the Local Majority algorithm makes mistakes is the union, over all blocks $B_i$, of the "minority inputs" of the block $B_i$. (If $b$ is the majority value of $g$ in a block $B$, then the "minority inputs" of $B$ are the set of inputs $x$ such that $g(x) = 1-b$.)

Let $E_0$ be the set of minority inputs (those where our algorithm makes a mistake) in $M_0$ and $E_1$ be the set of minority inputs in $M_1$. Then at least one of $E_0$ and $E_1$ must have size at least $\frac {\delta}{2} 2^n$, because the size of their union is at least $\delta 2^n$. If $E_b$ has size at least $\frac {\delta}{2} 2^n$, then $M_b$ has all the properties of the set $H$ we are looking for.

It remains to construct the partition. We describe an iterative process to construct it. We begin with the trivial partition $P = (B_1)$ where $B_1 = \{ 0,1\}^n$. At a generic step of the construction, we have a partition $P = (B_1,\ldots,B_m)$, and we consider $M_0, M_1,E_0,E_1$ as above. Let $b$ be such that $E_b \geq \frac 12 \delta 2^n$. If there is no algorithm that has noticeable advantage in computing $g$ over $M_b$, we are done. Otherwise, if there is such an algorithm $f$, we refine the partition by splitting each block according to the values that $f$ takes on the elements of the block.

After $k$ steps of this process, the partition has the following form: there are $k$ functions $f_1,\ldots,f_k$ and each of the (at most) $2^k$ blocks of the partition corresponds to a bit string $b_1,\ldots,b_k$ and it contains all inputs $x$ such that $f_1(x)=b_1,\ldots,f_k(x)=b_k$. In particular, the partition is efficiently computable.

We need to argue that this process terminates with $k=poly(1/\epsilon,1/\delta)$. To this end, we define a potential function that measures the "imbalance" of $g$ inside the blocks the partition

$\Psi(B_1,\ldots,B_m) := \sum_{i=1}^m \frac {|B_i|}{2^n} \left( Pr_{x\in B_i} [g(x) = 1] \right)^2 $

and we can show that this potential function increases by at least $poly(\epsilon,\delta)$ at each step of the iteration. Since the potential function can be at most 1, the bound on the number of iterations follows.

A reader familiar with the proof of the Szemeredi Regularity Lemma will recognize the main ideas of iterative partitioning, of using a "counterexample" to the regularity property required of the final partition to do a refinement step, and of using a potential function argument to bound the number of refinement steps.

In which way can we see them as "finitary ergodic theoretic" techniques? As somebody who does not know anything about ergodic theory, I may not be in an ideal position to answer this question. But this kind of difficulty has not stopped me before, so I may attempt to answer this question in a future post.