A General Dichotomy of Evolutionary Algorithms on Monotone Functions

TL;DR

This paper establishes a broad dichotomy in the efficiency of various mutation-based evolutionary algorithms on monotone functions, showing that their performance sharply depends on specific parameters, with crossover enabling efficiency across all mutation strengths.

Contribution

It generalizes the known dichotomy for the (1+1)-EA to a wide class of algorithms, identifying key parameters that determine their efficiency on monotone functions.

Findings

All considered mutation-based algorithms exhibit a similar parameter-dependent dichotomy.

For the (1+(mbda),mbda)-GA, the dichotomy depends on the product cmma.

Crossover allows genetic algorithms to be efficient regardless of mutation strength when population size is large.

Abstract

It is known that the evolutionary algorithm $(1+1)$-EA with mutation rate $c/n$ optimises every monotone function efficiently if $c<1$, and needs exponential time on some monotone functions (HotTopic functions) if $c\geq 2.2$. We study the same question for a large variety of algorithms, particularly for $(1+\lambda)$-EA, $(\mu+1)$-EA, $(\mu+1)$-GA, their fast counterparts like fast $(1+1)$-EA, and for $(1+(\lambda,\lambda))$-GA. We find that all considered mutation-based algorithms show a similar dichotomy for HotTopic functions, or even for all monotone functions. For the $(1+(\lambda,\lambda))$-GA, this dichotomy is in the parameter $c\gamma$, which is the expected number of bit flips in an individual after mutation and crossover, neglecting selection. For the fast algorithms, the dichotomy is in $m_2/m_1$, where $m_1$ and $m_2$ are the first and second falling moment of the number of bit flips. Surprisingly, the range of efficient parameters is not affected by either population size $\mu$ nor by the offspring population size $\lambda$. The picture changes completely if crossover is allowed. The genetic algorithms $(\mu+1)$-GA and fast $(\mu+1)$-GA are efficient for arbitrary mutations strengths if $\mu$ is large enough.