Java 遗传算法竞赛选择

我正在写一个遗传算法，我计划从轮盘赌到锦标赛，但我怀疑我的理解可能有缺陷

如果我只是选择总体中的n/2最佳解决方案，那么我的总体肯定会很快用完吗

我对算法的理解是：

for(Member m in currentPopulation){
    Member randomMember1 = random member of currentPopulation which is then removed from currentPopulation
    Member randomMember2 = as above;
    //Mutate and crossover

    if(randomMember1.getScore() > randomMember2.getScore()){
        nextGeneration.add(randomMember1);
    } else {
        nextGeneration.add(randomMember2);
    }
}

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我理解正确吗？

在锦标赛选择中，所选个体不会从群体中移除。您可以选择同一个人参加多个锦标赛

仔细看了一下你的代码，我发现你还有另一个误解。您通常不会对锦标赛的所有成员进行变异/交叉。相反，您执行一个锦标赛，锦标赛的获胜者被选为个体进行变异/交叉。这意味着，对于变异，您的比赛规模必须至少为2，对于交叉，规模必须至少为3，最好的2人获胜（或者您可以执行2个单独的比赛，以选择每个父母进行交叉）

一些伪代码可能会有所帮助：

while (nextPopulation too small) {
    Members tournament = randomly choose x members from currentPopulation

    if(crossover){
        Member parents = select best two members from tournament
        Member children = crossover(parents)
        nextPopulation.add(children);
    } else {
        Member parent = select best one member from tournament
        Member child = mutate(parent)
        nextPopulation.add(child);
    }
}

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如果你在每一代从你的种群中选择n/2个个体，你最终会达到一个种群为1的点。除了选择之外，你想做的是使用变异或交叉为你的下一代创建新成员，通常是在锦标赛的胜利者身上

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因此，对于每一代，你都有一个n大小的群体——你通过选择将其减少到n/2，然后这些n/2的成员繁殖和/或变异，为你的下一代多产生大约n/2的成员（平均而言，比上一代没有进步的成员“更适合”）.

锦标赛选择：

choose k (the tournament size) individuals from the population at random
choose the best individual from pool/tournament with probability p
choose the second best individual with probability p*(1-p)
choose the third best individual with probability p*((1-p)^2)
and so on...

Matepool=randi(PopLength,PopLength,2);%%select two individuals randomly for tournament and chooose the one with best fitness value
%% number of tournament is equal to the number of population size
for i=1:PopLength
    if Fitness(Matepool(i,1))>= Fitness(Matepool(i,2))
        SelectedPop(i,1:IndLength)=CurrentPop(Matepool(i,1),1:IndLength);
    else
        SelectedPop(i,1:IndLength)=CurrentPop(Matepool(i,2),1:IndLength);
    end
end

• 锦标赛选择是一种从个体群体中选择个体的方法
• 锦标赛选择包括在从人群中随机挑选的几个人中进行几场“锦标赛”
• 选择每场比赛的获胜者（身体状况最佳者）进行交叉
• 当竞赛规模较小时，竞赛选择也给所有个体一个被选择的机会，因此它保持了多样性，尽管保持多样性可能会降低收敛速度
• 但如果比赛规模较大，弱者被选中的机会较小，从而导致多样性丧失

伪代码：

choose k (the tournament size) individuals from the population at random
choose the best individual from pool/tournament with probability p
choose the second best individual with probability p*(1-p)
choose the third best individual with probability p*((1-p)^2)
and so on...

Matepool=randi(PopLength,PopLength,2);%%select two individuals randomly for tournament and chooose the one with best fitness value
%% number of tournament is equal to the number of population size
for i=1:PopLength
    if Fitness(Matepool(i,1))>= Fitness(Matepool(i,2))
        SelectedPop(i,1:IndLength)=CurrentPop(Matepool(i,1),1:IndLength);
    else
        SelectedPop(i,1:IndLength)=CurrentPop(Matepool(i,2),1:IndLength);
    end
end

确定性锦标赛选择在任何锦标赛中选择最佳个人（当p=1时）。单向锦标赛（k=1）选择等同于随机选择。如果需要，可将所选个体从进行选择的群体中移除，否则可为下一代多次选择个体。与（随机）适应度比例选择方法相比，由于缺乏随机噪声，比赛选择在实践中经常被实现

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MatLab中的锦标赛选择：

choose k (the tournament size) individuals from the population at random
choose the best individual from pool/tournament with probability p
choose the second best individual with probability p*(1-p)
choose the third best individual with probability p*((1-p)^2)
and so on...

Matepool=randi(PopLength,PopLength,2);%%select two individuals randomly for tournament and chooose the one with best fitness value
%% number of tournament is equal to the number of population size
for i=1:PopLength
    if Fitness(Matepool(i,1))>= Fitness(Matepool(i,2))
        SelectedPop(i,1:IndLength)=CurrentPop(Matepool(i,1),1:IndLength);
    else
        SelectedPop(i,1:IndLength)=CurrentPop(Matepool(i,2),1:IndLength);
    end
end

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请正确设置代码的格式。哦，对不起！看起来其他人已经有了，下次我会记住这一点。它是如何得到比轮盘赌等选择方法更好的解决方案的？请参阅我的编辑。只有锦标赛的获胜者会经历变异/交叉，并进入下一个群体。也就是说，平均而言（如果我没有弄错的话），如果你做3个比较，66%的人群会经历变异/交叉。非常感谢，这正是我想要的。