問題
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.
制約
2 ≤ N ≤ 200 (N even), 2 ≤ L ≤ 64
0 ≤ number of mutations ≤ 1,000
例
入力：
mutations = [
[0, 0]
]
population = ["0000", "1111", "1100", "0011"]
tournaments = [
[0, 1],
[2, 3],
[1, 2],
[0, 3]
]
crossover_points = [2, 1]
期待される出力：
expected = ["0100", "1111", "1011", "0111"]入力：
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};
期待される出力：
std::vector<std::string> expected = {"0100", "1111", "1011", "0111"};入力：
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};
期待される出力：
string[] expected = new string[] {"0100", "1111", "1011", "0111"};解説
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".
入力：
mutations = []
population = ["10", "01"]
tournaments = [
[0, 1],
[1, 0]
]
crossover_points = [1]
期待される出力：
expected = ["11", "00"]入力：
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};
期待される出力：
std::vector<std::string> expected = {"11", "00"};入力：
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};
期待される出力：
string[] expected = new string[] {"11", "00"};解説
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".
解き方のヒント
まずは自分で考えてから、ヒントを 1 つずつ開いてみましょう。
ヒント 1
Build children in pairs by first choosing two parents from each tournament, then combining their bit strings at one cut point.
ヒント 2
Selection depends on fitness only, with ties always favoring the first contestant in the tournament pair.
ヒント 3
Apply mutations after all children are created, and make sure each flip targets the current child string.
目安の計算量
O(N * L + M) time, O(N * L) extra space
参考資料
この課題はCodeUtility用に調整されています。上記の説明と制約に従ってください。