On learning k-parities with and without noise

TL;DR

This paper advances understanding of learning k-parities by improving time complexity bounds in mistake-bound models and proposing faster algorithms for noisy settings when the noise rate is sufficiently small.

Contribution

It improves the mistake-bound model tradeoff for k-parities and introduces a faster algorithm for learning k-parities under low noise conditions.

Findings

Enhanced the time complexity for online mistake-bound learning of k-parities.

Demonstrated that for small noise rates, the learning problem can be solved faster than the known ${n race k/2}$ barrier.

Provided a polynomial-time algorithm for noisy k-parity learning with improved runtime under specific noise conditions.

Abstract

We first consider the problem of learning $k$-parities in the on-line mistake-bound model: given a hidden vector $x \in \{0,1\}^n$ with $|x|=k$ and a sequence of "questions" $a_1, a_2, ...\in \{0,1\}^n$, where the algorithm must reply to each question with $< a_i, x> \pmod 2$, what is the best tradeoff between the number of mistakes made by the algorithm and its time complexity? We improve the previous best result of Buhrman et al. by an $\exp(k)$ factor in the time complexity. Second, we consider the problem of learning $k$-parities in the presence of classification noise of rate $\eta \in (0,1/2)$. A polynomial time algorithm for this problem (when $\eta > 0$ and $k = \omega(1)$) is a longstanding challenge in learning theory. Grigorescu et al. showed an algorithm running in time ${n \choose k/2}^{1 + 4\eta^2 +o(1)}$. Note that this algorithm inherently requires time ${n \choose k/2}$ even when the noise rate $\eta$ is polynomially small. We observe that for sufficiently small noise rate, it is possible to break the $n \choose k/2$ barrier. In particular, if for some function $f(n) = \omega(1)$ and $\alpha \in [1/2, 1)$, $k = n/f(n)$ and $\eta = o(f(n)^{- \alpha}/\log n)$, then there is an algorithm for the problem with running time $poly(n)\cdot {n \choose k}^{1-\alpha} \cdot e^{-k/4.01}$.