# produce all combinations of n choose k in binary

28 views (last 30 days)
NA on 13 Mar 2020
Commented: Stephen23 on 17 Mar 2020
I have
n=5;
k=3;
I want to have combinations of n choose k in binary
result should be (5 choose 3=10) (the order does not matter)
A=[1 0 0 1 1;
1 1 1 0 0;
1 0 1 0 1;
1 0 1 1 0;
0 0 1 1 1;
0 1 1 1 0;
0 1 1 0 1;
1 1 0 0 1;
0 1 1 0 1;
0 1 0 1 1;
]

John D'Errico on 13 Mar 2020
This is easier than you may think, as long as you think outside of the box just a bit.
n = 5;
k = 3;
dec2bin(sum(nchoosek(2.^(0:n-1),k),2)) - '0'
ans =
0 0 1 1 1
0 1 0 1 1
1 0 0 1 1
0 1 1 0 1
1 0 1 0 1
1 1 0 0 1
0 1 1 1 0
1 0 1 1 0
1 1 0 1 0
1 1 1 0 0
The nice thing is, you do not need to generate a long list of all binary numbers, then keep only those that have exactly three bits turned on.
The above scheme should work for up to 52 bit results, since MATLAB can encode integers as large as 2^53-1 in a double. And, of course, if k is at all large, then things will get nasty for n>52. Anyway, I don't think dec2bin will be successful in more than 52 bits though, even if you tried to use uint64.
Stephen23 on 17 Mar 2020
"Is there any way to produce combination one by one?"
Search FEX, e.g.:
Of course you should expect to be waiting a while for any result.

Alex Mcaulley on 13 Mar 2020
One option (Probably a better implementation can be done, but this one should work):
n = 5;
k = 3;
comb = nchoosek(1:n,k);
res = cell2mat(arrayfun(@(x) myfun(x,n,comb),(1:size(comb,1))','uni',0));
function sol = myfun(i,n,comb)
sol = zeros(1,n);
sol(comb(i,:)) = 1;
end
res =
1 1 1 0 0
1 1 0 1 0
1 1 0 0 1
1 0 1 1 0
1 0 1 0 1
1 0 0 1 1
0 1 1 1 0
0 1 1 0 1
0 1 0 1 1
0 0 1 1 1

Akira Agata on 13 Mar 2020
Edited: Akira Agata on 13 Mar 2020
Another solution:
n = 5;
k = 3;
A = unique(perms([zeros(1,n-k) ones(1,k)]),'rows');

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