Generating all combinations containing at least one element of a given set in Matlab

Question

I use combnk to generate a list of combinations. How can I generate a subset of combinations, which always includes particular values. For example, for co

Answer 1

For a brute force solution, you can generate all your combinations with COMBNK then use the functions ANY and ISMEMBER to find only those combinations that contain one or more of a subset of numbers. Here's how you can do it using your above example:

v = 1:10;            %# Set of elements
vSub = [3 5];        %# Required elements (i.e. at least one must appear in the
                     %#   combinations that are generated)
c = combnk(v,2);     %# Find pairwise combinations of the numbers 1 through 10
rowIndex = any(ismember(c,vSub),2);  %# Get row indices where 3 and/or 5 appear
c = c(rowIndex,:);   %# Keep only combinations with 3 and/or 5

---

EDIT:

For a more elegant solution, it looks like Steve and I had a similar idea. However, I've generalized the solution so that it works for both an arbitrary number of required elements and for repeated elements in v. The function SUBCOMBNK will find all the combinations of k values taken from a set v that include at least one of the values in the set vSub:

function c = subcombnk(v,vSub,k)
%#SUBCOMBNK   All combinations of the N elements in V taken K at a time and
%#   with one or more of the elements in VSUB as members.

  %# Error-checking (minimal):

  if ~all(ismember(vSub,v))
    error('The values in vSub must also be in v.');
  end

  %# Initializations:

  index = ismember(v,vSub);  %# Index of elements in v that are in vSub
  vSub = v(index);           %# Get elements in v that are in vSub
  v = v(~index);             %# Get elements in v that are not in vSub
  nSubset = numel(vSub);     %# Number of elements in vSub
  nElements = numel(v);      %# Number of elements in v
  c = [];                    %# Initialize combinations to empty

  %# Find combinations:

  for kSub = max(1,k-nElements):min(k,nSubset)
    M1 = combnk(vSub,kSub);
    if kSub == k
      c = [c; M1];
    else
      M2 = combnk(v,k-kSub);
      c = [c; kron(M1,ones(size(M2,1),1)) repmat(M2,size(M1,1),1)];
    end
  end

end

You can test this function against the brute force solution above to see that it returns the same output:

cSub = subcombnk(v,vSub,2);
setxor(c,sort(cSub,2),'rows')   %# Returns an empty matrix if c and cSub
                                %#   contain exactly the same rows

I further tested this function against the brute force solution using v = 1:15; and vSub = [3 5]; for values of N ranging from 2 to 15. The combinations created were identical, but SUBCOMBNK was significantly faster as shown by the average run times (in msec) displayed below:

N  | brute force | SUBCOMBNK
---+-------------+----------
2  |     1.49    |    0.98
3  |     4.91    |    1.17
4  |    17.67    |    4.67
5  |    22.35    |    8.67
6  |    30.71    |   11.71
7  |    36.80    |   14.46
8  |    35.41    |   16.69
9  |    31.85    |   16.71
10 |    25.03    |   12.56
11 |    19.62    |    9.46
12 |    16.14    |    7.30
13 |    14.32    |    4.32
14 |     0.14    |    0.59*   #This could probably be sped up by checking for
15 |     0.11    |    0.33*   #simplified cases (i.e. all elements in v used)

Answer 2

Well, in your specific example, choosing two integers from the set {1, ..., 10} such that one of the chosen integers is 3 or 5 yields 9+9-1 = 17 known combinations, so you can just enumerate them.

In general, to find all of the n-choose-k combinations from integers {1, ..., n} that contain integer m, that is the same as finding the (n-1)-choose-(k-1) combinations from integers {1, ..., m-1, m+1, ..., n}.

In matlab, that would be

combnk([1:m-1 m+1:n], k-1)

(This code is still valid even if m is 1 or n.)

Answer 3

Just to improve Steve's answer : in your case (you want all combinations with 3 and/or 5) it will be

• all k-1/n-2 combinations with 3 added
• all k-1/n-2 combinations with 5 added
• all k-2/n-2 combinations with 3 and 5 added

Easily generalized for any other case of this type.