One Generation of a Genetic Algorithm प्रोग्रामिंग चैलेंज

समस्या

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 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.
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.

सीमाएँ

2 ≤ N ≤ 200 (N even), 2 ≤ L ≤ 64

0 ≤ number of mutations ≤ 1,000

उदाहरण

उदाहरण 1

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

अपेक्षित आउटपुट:
expected = ["0100", "1111", "1011", "0111"]

इनपुट:
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};

अपेक्षित आउटपुट:
std::vector<std::string> expected = {"0100", "1111", "1011", "0111"};

इनपुट:
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};

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

व्याख्या

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".

उदाहरण 2

इनपुट:
mutations = []
population = ["10", "01"]
tournaments = [
    [0, 1],
    [1, 0]
]
crossover_points = [1]

अपेक्षित आउटपुट:
expected = ["11", "00"]

इनपुट:
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};

अपेक्षित आउटपुट:
std::vector<std::string> expected = {"11", "00"};

इनपुट:
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};

अपेक्षित आउटपुट:
string[] expected = new string[] {"11", "00"};

व्याख्या

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".

संदर्भ

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One Generation of a Genetic Algorithm

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समस्या

Input

Output

सीमाएँ

उदाहरण

व्याख्या

व्याख्या

संदर्भ

समुदाय समाधान