All possible combinations of 2 vectors.

Hi everyone.

I have one vector and one number. For example [1 3 5] and 0.

How do I generate all possible combinations? Like this:

 0 3 5
 1 0 5
 1 3 0
 0 0 5
 0 3 0
 1 0 0
 0 0 0

[Matt Fig]

Why is the last row all zeros? It looks like the rule is: take at least one element from each vector, with repetition allowed only for the shorter vector. But then the last row breaks this. So what is the rule?

[Artyom]

The rule is:

1) we have an n - dimensional vector.

2) replace one number with zero and find all combinations

3) replace two numbers with zero and find all combinations

4) ...

5) replace n-1 number with zero and find all combinations

6) replace n number with zero

Answer 1

[Matt Fig]

Here is a solution:

function H = mycomb(V)
% Help
L = length(V);
H = cell(1,L);

for ii = 1:L-1
    C = nchoosek(1:L,L-ii);
    R = cumsum(ones(size(C)));
    M = max(R(:,1));
    H{ii} = zeros(M,L);
    H{ii}(R+(C-1)*M) = V(C);
end

H{L} = zeros(1,L);
H = vertcat(H{:});

Now try it out from the command line:

>> mycomb([4 5 6])

ans =

       4     5     0
       4     0     6
       0     5     6
       4     0     0
       0     5     0
       0     0     6
       0     0     0
  >> mycomb([4 5 6 7])

ans =

       4     5     6     0
       4     5     0     7
       4     0     6     7
       0     5     6     7
       4     5     0     0
       4     0     6     0
       4     0     0     7
       0     5     6     0
       0     5     0     7
       0     0     6     7
       4     0     0     0
       0     5     0     0
       0     0     6     0
       0     0     0     7
       0     0     0     0

Answer 2

[Andrei Bobrov]

variant

t = [1 3 5];
ii = perms([t, zeros(size(t))]);
out = unique(sort(t(:,1:numel(t)),2),'rows');

or

t = [1 3 5];
out = [];
n = numel(t);
for jj = 1:n
    k = nchoosek(t,n - jj);
    out = [out;[zeros(size(k,1),jj),k]];
end

or

k = ones(1,numel(t)) * 2.^(numel(t)-1:-1:0)';
out = bsxfun(@times,t,dec2bin(0:k - 1,numel(t))-'0');

[Matt Fig]

Very nice! (The last one)

Answer 3

[Azzi Abdelmalek]

save this function

function y=arrangement(v,n)
 m=length(v);
 y=zeros(m^n,n);
 for k = 1:n
    y(:,k) = repmat(reshape(repmat(v,m^(n-k),1),m*m^(n-k),1),m^(k-1),1);
 end

then type

x=arrangement([1 3 5 0],3)
out=x(~all(x,2),:)

If you don't need repetition add

 s=arrayfun(@(t) sort(out(t,:)),(1:size(out,1))','un',0)
 out1=unique(cell2mat(s),'rows')

Answer 4

[Matt J]

    t=[1 3 5];
    n=length(t);

    result =  bsxfun(@times, [1,3,5],  dec2bin(2^n-1:-1:0)-'0')