如何避免遗传算法中的无效搜索空间？
我正在为一个学校项目开发一个 GA，我注意到在评估我的适应度函数时，个体相当于它的逆函数。
例如，集合 (1, 1, -1, 1) 相当于 (-1, -1, 1, -1)。为了缩小搜索空间并更有效地找到解决方案，如何避免在搜索空间的后半部分中进行交叉搜索？
I am developing a GA for a school project and I've noticed that upon evaluating my functions for fitness, an individual is equivalent to its inverse.
For example, the set (1, 1, -1, 1) is equivalent to (-1, -1, 1, -1). To shrink my search space and reach a solution more efficiently, how can I avoid my crossovers from searching in this second half of the search space?
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将空间减半。要求第一个元素为非负数，那么如果有 (x_1, .. , x_n) 及其逆元素 (-x_1, .. , -x_n) ，则搜索空间中只有一个元素。 （如果 x_1 = 0 它们是相同的）
顺便说一句，你要解决的问题是什么？
Cut the space in half. Require the first element to be non-negative, then if you have (x_1, .. , x_n) and its inverse (-x_1, .. , -x_n) only one will be in the search space. (if x_1 = 0 they are the same)
BTW, what is the problem you are solving?
尝试重写您的健身函数，以将权重适当地分配给个人 - 不确定您要解决的问题或如何重新分配权重。请记住，最适合的个体应该在池中，并且更有可能被选择进行交叉 - 您不希望池中充满了与良好解决方案具有相同适应度的假人，这违背了遗传算法的目的
Try to re-write your fitness function to assign weights appropriately to individuals - not sure on the problem you're trying to solve or how weight can be re-assigned. Remember, fittest individuals should be in the pool and more likely to be selected for crossover - you don't want a pool full of dummys that have the same fitness as good solutions that defeats the purpose of the genetic algorithm
它的计算成本很高，但类似的东西应该可以完成这项工作。忽略第一代，对于每一代> = 2：
如果通过交叉/变异产生的子代是逆的，则忽略它并生成另一个子代，直到它是“好”子代。
Its computationally expensive, but something like this should do the job. Ignoring the 1st generation, for every generation >=2:
If the children produced via crossover/mutation is an inverse, ignore it and generate another child till it is a "good" child.