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arraysmatlabconstraintscombinations
NLed
NLed

Reputation: 1865

Combination of possible values in an array with multiple constraints in Matlab?

I have an array B of :

B=[1 2 3; 10 20 30 ; 100 200 300 ; 1000 2000 3000]

such that 

B =

           1           2           3
          10          20          30
         100         200         300
        1000        2000        3000

I am using the following code to find the possible combination between these variables that are below a certain value (a constraint) -- 2000 in this case :

A=[123; 323; 12 ; 421]
SA=sum(A)
V=cell(length(B),1);
n=1;
for k = 1:length(B)
    for idx = nchoosek(1:length(B), k)'
        B_subset = B(idx);
        if (SA + sum(B_subset) <= 2000)
            V(n)={B_subset(:)}; %store them in a cell
            n=n+1;
        end
    end
end

However I failed to combine them the way I want below.

Objective :

Find the possible combinations from B to be added with SA so that their sum is less than 2000 ?

Constraint 1 :

For example, this is NOT acceptable : [1 2 20] [2 20 30]
This is the correct one : [1 20 100] [3 200 3000]

Constraint 2 : - The answers should be stored in the cell V in only one column (as initialised in the code above).

The cell should have an output similar to the one I'm currently having :

V = 

    [       100]
    [       300]
    [       200]
    [2x1 double]
    [2x1 double]
    [2x1 double]
    [3x1 double]

Upvotes: 1

Views: 1331

Answers (2)

Dedek Mraz
Dedek Mraz

Reputation: 814

Changed your code just a bit and added a test below - if V is unchanged -> there were no combinations -> display that. Edited to allow saving it into cell array V and simultaneously constructing a string for printing.

This code considers every combination of three elements from B, where each element is from a different column.

V=cell(length(B),1);
A=[123; 323; 12 ; 421];
SA=sum(A);
S = 'possible combinations :';
n = 1
for ii=1:4
    for jj=1:4
        if jj == ii
            continue
        end
        for kk=1:4
            if or(kk == jj,kk == ii)
                continue
            end
            B_subset = [B(ii,1), B(jj,2), B(kk,3)];
            if (SA + sum(B_subset) <= 2000)
                S = [S, ' ', mat2str(B_subset)];
                V{n} = B_subset;
                n += 1;
            end
        end
    end
end

if V == 'possible combinations :'
    disp('No possible combinations found')
else
    disp(S)
end

EDIT: adding answer to the new part of the question. In the inner-most loop the different combinations of included rows are calculated.

V = {}
for ii=1:3
    for jj=1:3
        for kk=1:3
            for ll = 1:3
                rows = [ii, jj, kk, ll]
                if isequal(rows, unique(rows))
                    % loop for turning off individual elements
                    result = [B(1,ii), B(2,jj), B(3,kk), B(4,ll)]
                    for mm = 0:1:15
                        % make a binary string - will loop through all combinations of zeros and ones
                        str1 = dec2bin(mm,4)
                        mask = zeros(1,4)
                        for nn = 1:4 % make a numeric vector
                            mask(nn) = str2num(str1(nn))
                        end
                        masked_result = mask.*result
                        V = [V {masked_result}]
                    end
                end
            end
        end
    end
end

Upvotes: 1

Eitan T
Eitan T

Reputation: 32930

Here, this fix should do the trick:

SA = sum(A);
V = cell(numel(B), 1);                  % // Changed 'length' to 'numel'
n = 1;
for k = 1:size(B, 1)                    % // Changed 'length' to 'size'
    for idx = nchoosek(1:numel(B), k)'  %'// Changed 'length' to 'numel'

        %// Ignore the combination if there are two elements from the same row
        rows = mod(idx, size(B, 1));
        if ~isequal(rows, unique(rows))
            continue
        end

        B_subset = B(idx);
        if (SA + sum(B_subset) <= 2000)
            V(n) = {B_subset(:)};
            n = n + 1;
        end
    end
end

Perhaps it's not the most efficient solution there is, but it's short and it works.

Upvotes: 1

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