Description: Genetic algorithms (GAs) nd approximate solutions to di cult problems by employing the principles of Darwinian evolution. The algorithm maintains a population of candidate solutions that are iteratively improved through a number of generations using reproduction, mutation, and survival of the ttest. Reproduction takes the form of crossover. In crossover, two individuals (parents) are recombined to create some number of new individuals (children). Each child contains some of the genome of each parent. For this assignment, you will implement several crossover functions.
The genomes for this assignment are binary strings with xed length 64. (In general, genomes can be of any size and contain bits, integers, real numbers, etc.) That is, each individual has a list of 64 bits that encodes its solution to the problem.
Details: Details of the crossover operations appear below. Each will be implemented as a Python function that returns two children. You should also write a main program that calls the crossover functions to test them.
1. Uniform crossover: Much like the problem from Quiz 3, uniform crossover ex-changes bits. This function will create a copy of each parent. child1 is a copy of parent1 and child2 is a copy of parent2. Each bit in child1 is replaced with the corresponding bit from parent2 with probability indpb. child2 is similarly altered by parent1.
parameters:
parent1: list of bits
parent2: list of bits
indpb: probability in [0.0, 1.0], default value = 0.2
returns:
child1: list of bits, de ned above
child2: list of bits, de ned above
1
2. One-point crossover: In one-point crossover, a split point is selected in the genome. Assume that position k is chosen (at random) as the split point. Then child1 would inherit genes 0..k from parent1 and genes k+1..n-1 from parent2. child2 is created from the symmetric case.
parameters:
parent1: list of bits
parent2: list of bits
returns:
child1: list of bits, de ned above
child2: list of bits, de ned above
3. Two-point crossover: In two-point crossover, two split points are selected in the genome. Assume that positions j and k are chosen (at random) as the split points. Then child1 would inherit genes 0..k and j+1..n-1 from parent1 and genes k+1..j from parent2. In other words, the middle section of child1 comes from parent2. child2 is formed by the symmetric case.
parameters:
parent1: list of bits
parent2: list of bits
returns:
child1: list of bits, de ned above
child2: list of bits, de ned above
2
