if i have two vector how to perform single crossover?
7 views (last 30 days)
Show older comments
Firas Al-Kharabsheh
on 9 May 2016
Commented: ALA SHARAF
on 9 May 2019
if i have
p1 = [ 1 0 1 1 1 0 1 0 1 0 1 0 1 0 1 0 0 0 0 0 1 0 1 1]
p2 = [ 0 0 0 0 1 1 1 1 0 0 1 1 0 1 0 1 1 1 0 0 0 1 0 0]
how to perform single crossover
c1 = [ 1 0 1 1 1 0 1 0 1 0 1 0 0 1 0 1 1 1 0 0 0 1 0 0]
c2 = [ 0 0 0 0 1 1 1 1 0 0 1 1 1 0 1 0 0 0 0 0 1 0 1 1]
0 Comments
Accepted Answer
Star Strider
on 9 May 2016
This is how I do it in genetic algorithm programming:
p1 = [ 1 0 1 1 1 0 1 0 1 0 1 0 1 0 1 0 0 0 0 0 1 0 1 1];
p2 = [ 0 0 0 0 1 1 1 1 0 0 1 1 0 1 0 1 1 1 0 0 0 1 0 0];
CrossoverIndex = 12; % This Would Usually Be Random
c1 = [p1(1:CrossoverIndex) p2(CrossoverIndex+1:end)];
c2 = [p2(1:CrossoverIndex) p1(CrossoverIndex+1:end)];
Producing:
c1 = 1 0 1 1 1 0 1 0 1 0 1 0 0 1 0 1 1 1 0 0 0 1 0 0
c2 = 0 0 0 0 1 1 1 1 0 0 1 1 1 0 1 0 0 0 0 0 1 0 1 1
4 Comments
Md. Kawsar Ahmed Asif
on 26 May 2018
Thanks a lot once again Star Strider. Now i've got it... But what about for 2-point crossover like Partially Mapped or Matched Crossover (PMX). Suppose i have two parents as following, (here i'm taking two crossover point at 3rd and 4th position)
P1 = [3 1 2 5 4]
P2 = [2 1 5 4 3] , will produce the 1st Offspring as
C1 = [4 1 2 5 3]
Again,
P2 = [2 1 5 4 3]
P1 = [3 1 2 5 4] , wili produce the remaining 2nd Offspring as
C2 = [3 1 5 4 2]
Now how can i get these Two Offspring proceeding by 'Position Wise Exchanging'? (Where two parents are mapped to each other).
Thank You
ALA SHARAF
on 9 May 2019
Is this two point crossover. Here actually no crossover between parents. only swapping first and last element of a parent and producing a new child.
More Answers (1)
Hassan Sarfraz
on 6 Sep 2018
Thanks... but how to get these parents from these childs again?
0 Comments
See Also
Categories
Find more on Genetic Algorithm in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!