收敛到相同种群的遗传算法的交叉

Question

I am doing a project for the OneMax algorithm and am having a problem with the crossover.

Throughout its iterations, I take the top 'division' of people and loop through them.

For even numbers i assign 5 odd number people who they will crossover with (ie 0,4,8,12 & 16 all match with 1,2,3,5,7 & 9 and then 2,6,10,14,18 all match with 11, 13, 15, 17, 19). This is used in order to ensure there are no duplicates.

I then choose a random cross point and split the lists, then the lists are split and returned as 2 split lists for parent A & B, which can then be used to make child A and B.

My problem is the fact that after a few iterations, the population converge to be the exact same bitstring.

Any help would be greatly appreciated!

Code:

Crossover

def crossover(top20):
    end = len(top20) - 1

    newPop = []

    # Person A crosses over with a list of 5 people

    for i in range(0, end, 2):
        crossPoint = (random.randint(1, len(range(stringLen - 2))))
        crossPoint2 = stringLen - crossPoint

        if i % 4 == 0:
            peopleB = [1, 3, 5, 7, 9]
        else:
            peopleB = [11, 13, 15, 17, 19]

        personA = pop[top20[i]]

        for j in range(len(peopleB)):
            personB = pop[top20[peopleB[j]]]

            sizes = [crossPoint, crossPoint2]
            parA1, parA2 = splitList(sizes, personA)
            parB1, parB2 = splitList(sizes, personB)

            childA = list(chain(parA1, parB2))
            childB = list(chain(parB1, parA2))
            newPop.append(childA)
            newPop.append(childB)

return newPop

Split List

def splitList(sizes, ls):
    par1 = []
    par2 = []

    for s in range(stringLen):
        if s < sizes[0]:
            par1.append(ls[s])
        else:
            par2.append(ls[s])

    return par1, par2

Heres a link to the gist of the file

Edit: Another thing is the fact that it always stops at a number in exactly the hundrets for example. Like it would stop at an average of exactly 16.00 and the sum of the top chosen would be exactly 8000

Answer 1

Here is my final product for the OneMax problem with 3 methods, One being the regular one max problem (ie try to get to all 1s), 2 being Evolving to a target string, and number 3 being a Deceptive Landscape. It can found here in this gist