Real coded genetic algorithm in matlab

For create a vector of population with best solution , must randomly select two individuals P1 and P2, it also assumes that P2 is not worse than P1.

How can i do that using this code:

    CostFunction=@(x) sum(x.^2);;     % Cost Function
    nVar=10;             % Number of Decision Variables
    VarSize=[1 nVar];   % Decision Variables Matrix Size
    VarMin=-10;         % Lower Bound of Variables
    VarMax= 10;         % Upper Bound of Variables
    %%Initialization
    empty_individual.Position=[];
    empty_individual.Cost=[];
    pop=repmat(empty_individual,nPop,1);
    for i=1:nPop
    % Initialize Position
    pop(i).Position=unifrnd(VarMin,VarMax,VarSize);

    % Evaluation
    pop(i).Cost=CostFunction(pop(i).Position);
    end

Answer 1

[Walter Roberson]

I would suggest that instead of continually trying to force P2 to not be worse than P1, that you just go ahead and generate and then sort by fitness afterwards.

... Especially if you are trying to generate unique locations. After all, it could be the case that P1 happened to be the single global minimum, in which case the only way generating P2 to be no worse would require P2 to be the same point.

[Sandi J]

What do you mean by 'P2 to be the same point'?

[Walter Roberson]

Suppose

f = @(x) sum(x.^2)

This has a unique global minimum at (0, 0).

Now imagine that you randomly generate (0, 0) as P1.

Now generate P2 such that f(P2) is no worse than f(P1).

"No worse" would require that f(P2) < f(P1) or else that f(P2) == f(P1).

Now, with real valued inputs, can the sum of squares of values ever be less than 0? No. So there is no possible P2 such that f(P2) < f([0,0]) which is 0.

So for "no worse" we must have f(P2) == f(P1)

Now, for f([0,0]) = 0, is there any P2 that is not the same as [0,0] that gives the same result, 0? No, not unless you take into account underflow to 0, such as if the pair of inputs was realmin because the square of realmin is 0 even though realmin is not 0.

Therefore when you require that f(P2) be no worse than f(P1), you could be accidentally requiring that P2 be exactly the same as P1 or at least that it computes the exact same after round-off.

Now suppose that f is defined as being the absolute value of the difference between its input and the sum of the unique factors of the number, including one but excluding the number itself. Also define the result for even values to be 1, and define the result at 1 to be 0. Now let P1 be 1. The value of the function is 0 because you defined it that way. The value at all even numbers is 1, which is a higher value. Now require that the function at P2 be no worse than at P1, and also said that P2 cannot be 1.

What do you have? Well what you now have is that P2 must be an odd "perfect number". A perfect number is a number that is equal to the sum of its unique factors excluding itself. For example 6 = 1+2+3 so 6 is a perfect number.

Now here's the thing: no one knows whether any odd perfect numbers exist. People have put in long effort, but they have not been able to either find an odd perfect number or else prove that none exists. But your algorithm that requires that P2 be no worse than P1 would force the routine to solve that question, to do what hundreds of years of number theorists have not been able to do.

I think you will agree that it is much much easier to just generate some entries independently of each other and then sort the results in descending order. The second sorted result will be no worse than the first sorted result, without any need for any elaborate search for a no worse result that might not exist.

[Sandi J]

Good explanation, it's clear enough, thanks.

Answer 2

[Aaina]

Hello, may I ask why my code has error:

Error using Crossover
Too many input arguments.

Error in nsga2 (line 83)
         [popc(k,1).Position,popc(k,2).Position]=Crossover(p1.Position,p2.Position);

the code for Crossover:

function [y1,y2]=Crossover(x1,x2)

    alpha=rand(size(x1));
    
    y1=alpha.*x1+(1-alpha).*x2;
    y2=alpha.*x2+(1-alpha).*x1;
    
end

[Aaina]

the line of the error:

% Use Tournament Selection method
             i1=TournamentSelection(pop,TournamentSize);
             i2=TournamentSelection(pop,TournamentSize);
               
        % Select Parents
        p1=pop(i1);
        p2=pop(i2);
        
        % Apply Crossover
         [popc(k,1).Position,popc(k,2).Position]=Crossover(p1.Position,p2.Position);

[Aaina]

Code for Tournament Selection:

function i=TournamentSelection(pop,m)
    nPop=numel(pop);
    S=randsample(nPop,m);    
    spop=pop(S);    
    scosts=[spop.Cost];    
    [~, j]=min(scosts);    
    i=S(j);
end

[Walter Roberson]

i1 and i2 are vectors, so pop(i1) and pop(i2) are not scalars. pop is a non-scalar structure array, so p1 = pop(i1) and p2 = pop(i2) are non-scalar structure arrays. When you use p1.Position, you are getting structure expansion, so p1.Position and p2.Position each expand to many arguments.

I suggest

         [p1, p2] = Crossover(vertcat(p1.Position), vertcat(p2.Position));
         p1c = num2cell(p1,2); p2c = num2cell(p2,2);
         [popc(k,1).Position] = p1c{:};
         [popc(k,2).Position] = p2c{:};

You are switching back and forth between fields of separate structure array entries, or treating everything as a group. Fields of separate structure array entries are effectively separate variables as far as storing is concerned in MATLAB, so it is not possible to just output two variables the way you were doing.

[Aaina]

wowwwww.... u solved my problem, thak you soooo much

[Aaina]

Error using vertcat

Dimensions of matrices being concatenated are not consistent.

Error in final_run_withdg (line 70)

[loc1_data, loc2_data] = cross_over_process(vertcat(loc1_data.Position),

vertcat(loc2_data.Position),minval1,maxval1,minval2,maxval2);

what is that mean?

minval1=1;

maxval1=14;

minval2=0.05;

maxval2=0.2;

[Walter Roberson]

You have position data that is not the same number of elements for each entry . It is possible to deal with it but it will require rewriting the crossover function . But before doing that I have to ask how it can be meaningful to do genetic function work when the position vectors are not the same size as each other ? The randomized selection you use makes it almost certain that at some point you will be asking to cross over a p1 with aa corresponding p2 that is aa different length .

[Aaina]

well, im trying to change the selection method(random selection) to tournament method in nsga II coding,

[Walter Roberson]

At some point you will try to tournament two positions of different length at the same time and will try to crossover them . How do you want to define the result?