Collapsing and Separating Completeness Notions Under Average-Case and Worst-Case Hypotheses

Abstract

This paper presents the following results on sets that are complete for NP.

• (i) If there is a problem in NP that requires $2^{n^{\Omega(1)}}$ time at almost all lengths, then every many-one NP-complete set is complete under length-increasing reductions that are computed by polynomial-size circuits.

• (ii) If there is a problem in co-NP that cannot be solved by polynomial-size nondeterministic circuits, then every many-one NP-complete set is complete under length-increasing reductions that are computed by polynomial-size circuits.

• (iii) If there exist a one-way permutation that is secure against subexponential-size circuits and there is a hard tally language in NP∩co-NP, then there is a Turing complete language for NP that is not many-one complete.

Our first two results use worst-case hardness hypotheses whereas earlier work that showed similar results relied on average-case or almost-everywhere hardness assumptions. The use of average-case and worst-case hypotheses in the last result is unique as previous results obtaining the same consequence relied on almost-everywhere hardness results.

Appendix

Here we provide a proof of Lemma 4.5

Definition

Let x,y∈{0,1}^m. $\Delta(x,y) = \frac{1}{m}|\{ i \mid x[i]\neq y[i] \} |$. We say that E:{0,1}ⁿ→{0,1}^m is an error correcting code with distance δ if for every x≠y∈{0,1}ⁿ, Δ(E(x),E(y))≥δ.

Theorem 5.1

(Johnson Bound [27])

If E:{0,1}ⁿ→{0,1}^m is an ECC with distance at least $\frac{1}{2}-\epsilon$, then for every x∈{0,1}^m, and $\eta \geq \sqrt{\epsilon}$, there exist at most l≤1/(2η ²) strings y ₁,…,y _l ∈E({0,1}ⁿ) such that $\Delta(x, y_{i})\leq \frac{1}{2}-\eta$ for every i∈[l].

Consider an ECC $E:\{0,1\}^{2n\log n}\rightarrow\{0,1\}^{n^{4}}$ with distance 1/2−/2n. An explicit example of such ECC is the code obtained by concatenating the Walsh-Hadamard code with the Reed-Solomon code. We will view strings of length 2nlogn and n ⁴ truth tables of languages at certain lengths. For this, both 2nlogn and n ⁴ must be powers of 2. If n is of the form $2^{2^{k-1}}$, then both 2nlogn and n ⁴ are powers of two.

For any language L″, let L′ be fold(L″). It is clear that L″∈NEEE∩coNEEE if and only if L′∈NEEE∩coNEEE. The following fact is easy to verify.

Lemma 5.2

If L″∉EEE/O(logn), then L′∉EEE/O(logn) and for every EEE/O(logn) algorithm $\mathcal{A}$, there are infinitely many lengths n of the form 2^k+1+k such that $\mathcal{A}$ fails to decide L′ correctly at length n.

Lemma 4.5

If there is a language in L″∈NEEE∩coNEEE−EEE/O(logn), then there is a language L∈NEEE∩coNEEE such that for every EEE/O(logn) algorithm $\mathcal{A}$, there exist infinitely many n and $\mathcal{A}$ correctly decides L on at most $\frac{1}{2}+\frac{1}{n}$ fraction of strings from Σⁿ.

Proof

Let L″∈NEEE∩coNEEE−EEE/O(log). Let L′=fold(L″). Let L be the language such that for every k∈ℕ, $L_{2^{k+1}} = E(L'_{2^{k-1}+k})$ where E is the WH-RS code. On other lengths, we set L to be empty. It is clear that L∈NEEE∩coNEEE since we can actually compute the truth table of L′ in NEEE∩coNEEE.

Now, for the sake of contradiction, assume that there is an EEE/O(logn) algorithm $\mathcal{A}$ that correctly decides L on more than $\frac{1}{2}+\frac{1}{n}$ fraction of inputs at every length n. We will first show that this yields a EEE/O(logn) algorithm for L. Since L is empty at lengths that are not of the form 2^k+1, we can decide L easily at those lengths. Thus we will concentrate on lengths of the form 2^k+1.

Consider n of the form 2^k+1. By the definition of L, we know that $L_{n} = E(L'_{2^{k-1}+k})$. Let N=2ⁿ. Recall that the distance of the code E is $\frac{1}{2}-\epsilon$, where $\epsilon= \frac{1}{2\sqrt[4]{N}}$. Let $\eta= \frac{1}{n}=\frac{1}{\log N}> \sqrt{\epsilon}$.

By running $\mathcal{A}$ on all strings of length n compute a candidate sequence, x∈{0,1}^N, for L _n . By our assumption, A is correct on at least $\frac{1}{2}+\frac{1}{n}$ of strings from Σⁿ. Thus $\Delta(x, L_{n}) \leq\frac{1}{2}-\frac{1}{n}$. By the Johnson bound 5.1, there exist at most $l \leq \frac{n^{2}}{2}$ strings y ₁,…,y _l in $E(\Sigma^{2^{k-1}+k})$ such that $\Delta(x, y_{i}) \leq\frac{1}{2}-\frac{1}{n}$. By running E on all strings from $\Sigma^{2^{k-1}+k}$, we can compute these y _i s. Since WH-RS code is computable in polynomial time, this computation can be done in triple exponential time. Since $L_{n} =E(L'_{2^{k-1}+k})$, there exists a j, 1≤j≤l such that L _n =y _j . Thus logl=O(logn) bits of information is enough to identify j for which L _n =y _j . Thus L∈EEE/O(logn).

Recall that $L_{n} = E(L'_{2^{k-1}+k}$) when n=2^k. Since L∈EEE/O(logn), by inverting the WH-RS code E, we can decide L′ correctly on all lengths of the form 2^k+1+k, using an EEE/O(logn) algorithm. This contradicts Lemma 5.2. □