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# Kannan's theorems

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Recall that the following existence of hard function by Shannon.

Threre exists a boolean function $f$ such that $f\notin\SIZE[2^n/10n]$.

However, there are two caveats of Shannon’s theorem. First, it does not say which complexity class contains this hard function. Second, the hard function is not constructive. One of the most important jobs for complexity theorists is then to answer these two questions.

Kannan in the early 80s made some pioneering work towards this direction. Concretely, he proved the following theorems.

$\Sigma^{\EXP}_2\not\in\SIZE[2^n/10n]$ and the first function not in $\SIZE[2^n/10n]$ lies in $\Sigma^{\EXP}_3$.

$\NEXP^{\NP}\not\subseteq\Ppoly$.

Note that as $\Ppoly$ is closed under complement, the second theorem actually implies $\NEXP^{\NP}\cap\coNEXP^{\NP}\not\subseteq\Ppoly$.

Both theorems are actually easy to prove as follows. Note that the proof here might be different from the original proof.

### Proof for the first theorem

To show $\Sigma^{\EXP}_2\not\in\SIZE[2^n/10n]$, it suffices to describe a $\Sigma^{\EXP}_2$ algorithm to find a function not in $\SIZE[2^n/10n]$. The algorithm is the following.

1. Nondeterministically guess the truth table of the hard function $f$. This takes $2^n$ nondeterministical time.
2. Co-nondeterministically enumerate all the circuit $C$ of size less than $2^n/10n$. This takes one alternation and $2^n$ (co-)nondeterministical time.
3. Check if there exists $x$ such that $f(x)\neq C(x)$. This takes deterministic $2^n$ time.

To show the first function not in $\SIZE[2^n/10n]$ lies in $\Sigma^{\EXP}_3$, the following algorithm suffices.

1. Nondeterministically guess the truth table of the hard function $f$. This takes $2^n$ nondeterministical time.
2. Co-nondeterministically enumerate all the circuit $C$ of size less than $2^n/10n$ and all the truth table of function $g$ which has smaller lexicogorical order than $f$. This takes one alternation and $2^n$ (co-)nondeterministical time.
3. Nondeterministically find a circuit $C’$ of size less than $2^n/10n$.
4. Check if there exists $x$ such that $f(x)\neq C(x)$ and if for all $x$ g(x)=C(x). This takes deterministic $2^n$ time.

### Proof for the second theorem

We also prove the theorem in an algorithmic way as follows.

1. Nondeterministically guess the truth table of a hard function $f$.
2. Pick the $\NP$ oracle to be the minimum circuit size problem, i.e., $\MCSP$, where one input a truth table of a function $g$ and an integer $s$, output yes if there exists a circuit for $g$ of size at most $s$.
3. Query the $\MCSP$ oracle with the truth table guessed in the first step and integer $2^n/10n$. If the ouput is no, then output $f$.