c ++遗传算法变异错误 [英] c++ Genetic Algorithm Mutation error

问题描述

我的遗传算法中的变异函数有问题.我也不太清楚我在做什么错.我看过这段代码已经有一段时间了，我认为逻辑是正确的，只是没有产生我想要的结果.

I Have a problem with the mutation function within my genetic Algorithm. I can't quite see what I am doing wrong either. I've looked at this code for a while and I think the logic is correct, it's just not producing the results i want.

问题 当我输出位于子结构中的Binary数组时，如果在任何位上发生了变异，则将更改一个随机数，而不是应该更改的数.

The problem When i output the Binary array located in the Child Struct, If mutation has occured on any of the bits, then a random number will be changed, and not the one that should be.

例如

• 0000000是二进制字符串
• 第二次发生了变异 位
• 0001000将是结果

• 0000000 is the binary string
• mutation has occured on the second bit
• 0001000 would be the result

此部分位于主体内.

for (int Child = 0; Child < ParentNumberInit; Child++)
{
    cout << endl;
    mutation(child[Child],Child);
}

这是变异函数

void mutation(struct Parent Child1,int childnumber)
{
    int mutation; // will be the random number generated

    cout << endl << "Child " << (childnumber+1) << endl;

    //loop through every bit in the binary string
    for (int z = 0; z < Binscale; z++)
    {
        mutation = 0;   // set mutation at 0 at the start of every loop
        mutation = rand()%100;      //create a random number

        cout << "Generated number = " << mutation << endl;

        //if variable mutation is smaller, mutation occurs
        if (mutation < MutationRate)
        {
            if(Child1.binary_code[z] == '0')
                Child1.binary_code[z] = '1';
            else if(Child1.binary_code[z] == '1')
                Child1.binary_code[z] = '0';
        }
    }
}

像这样在主体中输出

    for (int childnumber = 0; childnumber < ParentNumberInit; childnumber++)
    {
        cout<<"Child "<<(childnumber+1)<<" Binary code = ";
        for (int z = 0; z < Binscale; z ++)
        {
        cout<<child[childnumber].binary_code[z];
        }
        cout<<endl;
     }

推荐答案

您无法通过这种方式来限制转换率.您需要将发生突变的位与发生突变的可能性区分开来.

You can't throttle the multation rate this way. You need to separate the mutated bit from the probability of the mutation occuring.

for (int z = 0; z < Binscale; z++)     
{         
    if (rand() % 100 < MutationRate)        
    {
        // flip bit             
        Child1.binary_code[z] += 1; 
        Child1.binary_code[z] %= 2;
    }
} 

更简单的翻转位方式:

Child1.binary_code[z] ^= 1;

这篇关于c ++遗传算法变异错误的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！