%% Homework Solution:  Change BinStrings.m to Solve Knapsack.
%
% The Knapsack problem uses the following constants (in order)
KnapNames={'Bug','Stove','Water','Clothes','Food','Aid','Light','Book','Rain','Bag','Tent','Filter'};

%%  Setup the GA parameters

ff='testfunction2'; % objective function file name

maxit=250;        % max number of iterations (for stopping)
maxcost=9999999; %Maximum allowable cost (for stopping)
popsize=20; % set population size (it is constant)
mutrate=0.2; % set mutation rate (a small probability)
lenx=12; % Length of the chromosome (20 here)

%% Create the initial population
pop=round(rand(popsize,lenx)); % random population of 1s 
                               % and 0s (using default, 25x12)
popKept=0.5;  %Fraction of the population to keep
keep=floor(popKept*popsize);  %How many individuals are kept
M=ceil((popsize-keep)/2); % number of matings; 2 mates create 2 offspring
                                    
% Initialize cost and other items to set up the main loop
cost=feval(ff,pop); % calculates population cost using ff

[cost,ind]=sort(cost,'descend'); % max element in first entry
pop=pop(ind,:); % sorts population with max cost first

probs=cost/sum(cost);  %Simple normalization for probabilities. 

% We'll be tracking the following quantities for our plot:
maxc(1)=max(cost); % minc contains min of population
meanc(1)=mean(cost); % meanc contains mean of population


%% MAIN LOOP

iga=0;  % Initalize the variable used in the loop below.
        %  This will keep track of how many iterations we've
        %  used and will get us out of the loop at the max
        
while iga<maxit
    
    iga=iga+1; % increments generation counter
    
% Choose mates

    ma=RandChooseN(probs,M); % mate #1
    pa=RandChooseN(probs,M); % mate #2    
% ma and pa contain the *indices* of the chromosomes that will mate
    
% Select crossover values for each set of parents    
    xp=randi([1,lenx],1,M);  % M integers from 1 to lenx  
                             % In this code, crossover is different
                             %  for each set of parents.
                              
    Temp=pop;  %Just temporary storage as we perform crossover
    Replace=zeros(popsize-keep,lenx);
%Crossover: One offspring will be stored in the odd indices, 
%           the other in the even indices.
    for k=1:M
      Replace(2*k-1,:)=[Temp(ma(k),1:xp(k)) Temp(pa(k),xp(k)+1:end)]; 
      Replace(2*k,:)=[Temp(pa(k),1:xp(k)) Temp(ma(k),xp(k)+1:end)]; 
    end
    pop=[Temp(1:keep,:); Replace];    
% Mutate the population
    nmut=ceil((popsize-1)*lenx*mutrate); % total number of mutations
    mrow=ceil(rand(1,nmut)*(popsize-1))+1; % row to mutate
    mcol=ceil(rand(1,nmut)*(lenx-1))+1; % column to mutate
    for ii=1:nmut
        pop(mrow(ii),mcol(ii))=abs(pop(mrow(ii),mcol(ii))-1); % toggles bits
    end % ii

%% The population is re-evaluated for cost
    cost=feval(ff,pop); % calculates population cost using ff
    [cost,ind]=sort(cost,'descend'); % max element in first entry
    pop=pop(ind,:); % sorts population with max cost first
    
    probs=cost/sum(cost);  %Re-set probabilities
    
% We keep track of some values for graphical output:    
    maxc(iga+1)=max(cost); 
    meanc(iga+1)=mean(cost); 

%% Stopping criteria.  The double bar: || is an "or"
    if iga>maxit || cost(1)>maxcost
        break
    end

end %iga


%% Displays the output
day=clock;
disp(datestr(datenum(day(1),day(2),day(3),day(4),day(5),day(6)),0))
%disp(['optimized function is ' ff])
format short g
disp(['popsize = ' num2str(popsize) ' mutrate = ' num2str(mutrate)]);
disp(['#generations=' num2str(iga) ' best cost=' num2str(cost(1))]);
fprintf('best solution\n%s\n',mat2str(int8(pop(1,:))));
figure(1)
iters=0:length(maxc)-1;
plot(1:(iga+1),maxc,1:(iga+1),meanc);
xlabel('generation');ylabel('cost');
legend('Max cost', 'Mean Cost')
for j=1:12
if pop(1,j)==1
fprintf('Include %s\n',KnapNames{j});
end
end