How to construct a combinatorics array from two data sets with each combination having a specified number of elements?

Hi I have two data sets. The first one consists of N1(let's say, 10) elements and the second one has N2(i.e., 8) elements. All the elements of the smaller set has its counterpart in the larger set. I need to construct a combinatorics array that will consist of all the possible combinations of the two data sets with each combinations having N1 elements. However, an element from a data set cannot co-exist with its counter-part from the other data-set.

For example, Set1 = [F1 F2 F3] and set2 = [S2 S3] ; (here, N1 = 3, N2 =2) Now the desired set will be: M = [F1 F2 F3; F1 S2 F3; F1 F2 S3; F1 S2 S3]

So, the total number of possible combinations is : 2^N2. I want Matlab to construct the 'M' set. Any suggestions/insight to this problem will be highly appreciated. Thanks

Answer 1

[James Tursa]

E.g., but of course only makes sense for small sizes of your variables

Set1 = [1,2,3];
Set2 = [4 5];
columns = [2 3]; % the columns in Set1 corresponding to Set2
N = 2^numel(Set2); % number of combinations
b2 = logical(dec2bin(0:(N-1))-'0'); % logical locations of the replacements in Set2
b1 = false(N,numel(Set1)); % logical locations of the replacements in Set1
b1(:,columns) = b2; % fill appropriate columns of b1 with b2
result = repmat(Set1,N,1); % result is Set1 by default
Set2rep = repmat(Set2,N,1); % get the replacement values
result(b1) = Set2rep(b2); % replace the appropriate values

The result:

>> result
result =
     1     2     3
     1     2     5
     1     4     3
     1     4     5

[MRR]

Hi James, Thanks a lot. It really helped!

Thanks again.

Answer 2

[Bruno Luong]

Sad, people keep asking unreasonable thing. The number of combinations is

2^80,

Each of those is 100 numbers so the size in byte is

>> 2^80*100*8

ans =

     9.6714e+26 bytes

>> (2^80*100*8)/(1024^4)

ans =

     8.7961e+14 Tera bytes

You can take all the computers on earth, it'ss still impossible to store such quantity, let alone compute it.

[MRR]

Thanks for your kind reply. While the 80 and 100 are just numbers as an example, N1 and N2 are clearly noted in the question that can take any value. Further, in the illustration, N1 and N2 are only 3 and 2 respectively. While the number is arbitrary (be it 80 or 8 or 80000!), the algorithm is not, right? Thanks for your effort though in answering the question! However, the numbers in the questions are modified so that no one get distracted by something that is necessarily not part of the problem.

[John D'Errico]

The point is however, that what you really want to solve is a problem of that size. For 2 or 3 levels, you would never have bothered to ask the question in the first place.

So IF you cannot solve the problem you think that you want to solve, that means you need to focus on how to make the problem manageable, thus using a completely different fundamental algorithm.