Условие
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:
- Selection: parent1 wins tournament
tournaments[2k]and parent2 winstournaments[2k + 1]. A tournament[i, j]is won by whichever ofpopulation[i]andpopulation[j]has more 1s; on a tie,population[i]wins. - 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.
Ограничения
2 ≤ N ≤ 200 (N even), 2 ≤ L ≤ 64
0 ≤ number of mutations ≤ 1,000
Примеры
Ввод:
mutations = [
[0, 0]
]
population = ["0000", "1111", "1100", "0011"]
tournaments = [
[0, 1],
[2, 3],
[1, 2],
[0, 3]
]
crossover_points = [2, 1]
Ожидаемый вывод:
expected = ["0100", "1111", "1011", "0111"]Ввод:
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};
Ожидаемый вывод:
std::vector<std::string> expected = {"0100", "1111", "1011", "0111"};Ввод:
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};
Ожидаемый вывод:
string[] expected = new string[] {"0100", "1111", "1011", "0111"};Пояснение
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".
Ввод:
mutations = []
population = ["10", "01"]
tournaments = [
[0, 1],
[1, 0]
]
crossover_points = [1]
Ожидаемый вывод:
expected = ["11", "00"]Ввод:
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};
Ожидаемый вывод:
std::vector<std::string> expected = {"11", "00"};Ввод:
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};
Ожидаемый вывод:
string[] expected = new string[] {"11", "00"};Пояснение
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".
Как подступиться
Сначала попробуйте сами, затем открывайте подсказки по одной.
Подсказка 1
Build children in pairs by first choosing two parents from each tournament, then combining their bit strings at one cut point.
Подсказка 2
Selection depends on fitness only, with ties always favoring the first contestant in the tournament pair.
Подсказка 3
Apply mutations after all children are created, and make sure each flip targets the current child string.
Ожидаемая сложность
O(N * L + M) time, O(N * L) extra space
Ссылки
Это задание адаптировано для CodeUtility. Следуйте условию и ограничениям выше.