문제
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.
제약 조건
2 ≤ N ≤ 200 (N even), 2 ≤ L ≤ 64
0 ≤ number of mutations ≤ 1,000
예시
입력:
mutations = [
[0, 0]
]
population = ["0000", "1111", "1100", "0011"]
tournaments = [
[0, 1],
[2, 3],
[1, 2],
[0, 3]
]
crossover_points = [2, 1]
예상 출력:
expected = ["0100", "1111", "1011", "0111"]입력:
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};
예상 출력:
std::vector<std::string> expected = {"0100", "1111", "1011", "0111"};입력:
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};
예상 출력:
string[] expected = new string[] {"0100", "1111", "1011", "0111"};설명
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".
입력:
mutations = []
population = ["10", "01"]
tournaments = [
[0, 1],
[1, 0]
]
crossover_points = [1]
예상 출력:
expected = ["11", "00"]입력:
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};
예상 출력:
std::vector<std::string> expected = {"11", "00"};입력:
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};
예상 출력:
string[] expected = new string[] {"11", "00"};설명
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".
접근 방법
먼저 스스로 풀어 본 뒤 힌트를 하나씩 열어 보세요.
힌트 1
Build children in pairs by first choosing two parents from each tournament, then combining their bit strings at one cut point.
힌트 2
Selection depends on fitness only, with ties always favoring the first contestant in the tournament pair.
힌트 3
Apply mutations after all children are created, and make sure each flip targets the current child string.
예상 복잡도
O(N * L + M) time, O(N * L) extra space
참고 자료
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