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One Generation of a Genetic Algorithm

Breed the next generation of bit strings with tournament selection, single-point crossover and mutation.

O que você vai aprender

  • Select parents with tournaments
  • Combine solutions with crossover
  • Add variation with mutation

Problema

Ver tradução

A genetic algorithm improves a population of candidate solutions the way evolution does: select fit parents, combine them, and add random mutations. Here the candidates are bit strings and fitness is the number of 1s (the OneMax problem).

To make it testable, every random choice is given. With N strings (N is even), build N children in pairs. For pair k = 0, 1, …, N/2 - 1:

  1. Selection: parent1 wins tournament tournaments[2k] and parent2 wins tournaments[2k + 1]. A tournament [i, j] is won by whichever of population[i] and population[j] has more 1s; on a tie, population[i] wins.
  2. Crossover at c = crossover_points[k]: child 2k = parent1[0:c] + parent2[c:] and child 2k + 1 = parent2[0:c] + parent1[c:].

Then apply each mutation [child, position] in order by flipping that bit of that child.

Implement ga_generation(population, tournaments, crossover_points, mutations).

Input

population: N bit strings of equal length L.

tournaments: N pairs of population indices.

crossover_points: N/2 cut points, each between 1 and L - 1.

mutations: [child, position] pairs (may be empty).

Output

Return the N children in order.

Restrições

2 ≤ N ≤ 200 (N even), 2 ≤ L ≤ 64

0 ≤ number of mutations ≤ 1,000

Exemplos

Exemplo 1
Entrada:
mutations = [
    [0, 0]
]
population = ["0000", "1111", "1100", "0011"]
tournaments = [
    [0, 1],
    [2, 3],
    [1, 2],
    [0, 3]
]
crossover_points = [2, 1]

Saída esperada:
expected = ["0100", "1111", "1011", "0111"]
Entrada:
std::vector<std::vector<long long>> mutations = {
    {0, 0}
};
std::vector<std::string> population = {"0000", "1111", "1100", "0011"};
std::vector<std::vector<long long>> tournaments = {
    {0, 1},
    {2, 3},
    {1, 2},
    {0, 3}
};
std::vector<long long> crossover_points = {2, 1};

Saída esperada:
std::vector<std::string> expected = {"0100", "1111", "1011", "0111"};
Entrada:
long[][] mutations = new long[][] {
    new long[] {0, 0}
};
string[] population = new string[] {"0000", "1111", "1100", "0011"};
long[][] tournaments = new long[][] {
    new long[] {0, 1},
    new long[] {2, 3},
    new long[] {1, 2},
    new long[] {0, 3}
};
long[] crossover_points = new long[] {2, 1};

Saída esperada:
string[] expected = new string[] {"0100", "1111", "1011", "0111"};

Explicação

Pair 0: "1111" beats "0000", and "1100" ties "0011" so the first one wins. Cutting after 2 bits gives "1100" and "1111". Pair 1: parents "1111" and "0011", cut after 1 bit: "1011" and "0111". Finally child 0 has bit 0 flipped: "0100".

Exemplo 2
Entrada:
mutations = []
population = ["10", "01"]
tournaments = [
    [0, 1],
    [1, 0]
]
crossover_points = [1]

Saída esperada:
expected = ["11", "00"]
Entrada:
std::vector<std::vector<long long>> mutations = {};
std::vector<std::string> population = {"10", "01"};
std::vector<std::vector<long long>> tournaments = {
    {0, 1},
    {1, 0}
};
std::vector<long long> crossover_points = {1};

Saída esperada:
std::vector<std::string> expected = {"11", "00"};
Entrada:
long[][] mutations = new long[][] {};
string[] population = new string[] {"10", "01"};
long[][] tournaments = new long[][] {
    new long[] {0, 1},
    new long[] {1, 0}
};
long[] crossover_points = new long[] {1};

Saída esperada:
string[] expected = new string[] {"11", "00"};

Explicação

Both strings have one 1, so each tournament is won by its first index: parent1 = "10" and parent2 = "01". Cutting after 1 bit gives "11" and "00".

Referências

Este desafio foi adaptado para o CodeUtility. Siga o enunciado e as restrições acima.

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