Complete set of combinations combining 3 set

Question

I need to generate the complete set of combinations obtained combining three different subset:

• Set 1: choosing any 4 numbers from a vector of 13 elements.
• Set 2: choosing any 2 numbers from a vector of 3 elements.
• Set 3: choosing any 2 numbers from a vector of 9 elements.

Example: sample 3 from vector of 4 (H=3 and L=4) for the Set A, H=2 L=3 for the Set B and H=2 L=4 for the Set B:

           4    4   4
           3    4   4
           3    3   4
           3    3   3
           2    4   4
           2    3   4
           2    3   3
           2    2   4
Set A =    2    2   3
           2    2   2
           1    4   4
           1    3   4
           1    3   3
           1    2   4
           1    2   3
           1    2   2
           1    1   4
           1    1   3
           1    1   2
           1    1   1

           3    3
           2    3
Set B =    2    2
           1    3
           1    2
           1    1

           4    4
           3    4
           3    3
           2    4
Set C =    2    3
           2    2
           1    4
           1    3
           1    2
           1    1

[Set A] = [20 x 3], [Set B] = [6 x 2], [Set C] = [10 x 2]. Then I need to obtain all possible combinations from these three sets: AllComb = [Set A] x [Set B] x [Set C] = [1200 x 8]. The AllComb matrix will be like this:

4   4   4   |   3   3   |   4   4
4   4   4   |   3   3   |   3   4
4   4   4   |   3   3   |   3   3
4   4   4   |   3   3   |   2   4
4   4   4   |   3   3   |   2   3
4   4   4   |   3   3   |   2   2
4   4   4   |   3   3   |   1   4
4   4   4   |   3   3   |   1   3
4   4   4   |   3   3   |   1   2
4   4   4   |   3   3   |   1   1
4   4   4   |   2   3   |   4   4
4   4   4   |   2   3   |   3   4
4   4   4   |   2   3   |   3   3
4   4   4   |   2   3   |   2   4
4   4   4   |   2   3   |   2   3
                 .
                 .
                 .
1   1   1   |   1   1   |   1   1

Unfortunately I can not use the same number for the sets since I need to substitute the numbers like that:

• For Set A: 1 = 10, 2 = 25, 3 = 30 and 4 = 45
• For Set B: 1 = 5, 2 = 20 and 3 = 35
• For Set C: 1 = 10, 2 = 20, 3 = 30 and 4 = 50

Any ideas? Real case sets will often lead to an AllComb matrix ~[491 400 x 8] so vectorized solutions will be gladly accepted.

Note: Each set is obtained with the following code:

a = combnk(1:H+L-1, H);
b = cumsum([a(:,1)  diff(a,[],2) - 1],2);

What is H and L?

H are the hoppers of a MultiheadWeigher (MHW) machines. I have a MHW with H=8 and I need to deliver in each of these hoppers some materials. If I need to deliver just one type of material all possibile combinations are (L+H-1)!/(H!(L-1)!) and i compute them with the code write above (a and b). Now, suppose to have 3 different product then we have 4 hoppers for product A, 2 for B and 2 for C. Product A in the first 4 hoppers can assume values 10:10:130, Product B 10:10:30 and c 10:10:90. Then the number of steps are for A L=13, B L=3 and C L=9

Answer 1

I guess this could be further optimized but this generates you AllComb:

H=3;
L=4;
a = combnk(1:H+L-1, H);
b = cumsum([a(:,1)  diff(a,[],2) - 1],2);
H=2;
L=3;
c = combnk(1:H+L-1, H);
d = cumsum([c(:,1)  diff(c,[],2) - 1],2);
H=2;
L=4;
e = combnk(1:H+L-1, H);
f = cumsum([e(:,1)  diff(e,[],2) - 1],2);
u=[];
for k=1:10
u=vertcat(u,d);
end
u=sortrows(u,[1 2]);
v=[];
for k=1:6
v= vertcat(v,f);
end
w= [u,v];
v=[];
for k=1:20
 v= vertcat(v,w);
end
u=[];
for k=1:60
 u = vertcat(u,b);
end
u=sortrows(u,[1 2 3]);
AllComb= [u,v];

Here b,d and f are your 3 sets. Then i loop over the numbers of permutation in d and f and replicate them so that all possibilities are constructed. One of them is sorted and then i write them in a new matrix w. THis process is repeated with Set A (b) and this new constructed matrix. Resulting in the end in AllComb.

Answer 2

You basically need to find

1. Combinations with repetition for each set;
2. "Multi-variations" (I don't know the correct name for this) of the results of stage 1.

Both stages can be solved with more or less the same logic, taken from here.

%// Stage 1, set A
LA = 4;
HA = 3;
SetA = cell(1,HA);
[SetA{:}] = ndgrid(1:LA);
SetA = cat(HA+1, SetA{:});
SetA = reshape(SetA,[],HA);
SetA = unique(sort(SetA(:,1:HA),2),'rows');

%// Stage 1, set B
LB = 3;
HB = 2;
SetB = cell(1,HB);
[SetB{:}] = ndgrid(1:LB);
SetB = cat(HB+1, SetB{:});
SetB = reshape(SetB,[],HB);
SetB = unique(sort(SetB(:,1:HB),2),'rows');

%// Stage 1, set C
LC = 4;
HC = 2;
SetC = cell(1,HC);
[SetC{:}] = ndgrid(1:LC);
SetC = cat(HC+1, SetC{:});
SetC = reshape(SetC,[],HC);
SetC = unique(sort(SetC(:,1:HC),2),'rows');

%// Stage 2
L = 3; %// number of sets
result = cell(1,L);
[result{:}] = ndgrid(1:size(SetA,1),1:size(SetB,1),1:size(SetC,1));
result = cat(L+1, result{:});
result = reshape(result,[],L);
result = [ SetA(result(:,1),:) SetB(result(:,2),:) SetC(result(:,3),:) ];
result = flipud(sortrows(result)); %// put into desired order

This gives

result =
     4     4     4     3     3     4     4
     4     4     4     3     3     3     4
     4     4     4     3     3     3     3
     4     4     4     3     3     2     4
     4     4     4     3     3     2     3
     4     4     4     3     3     2     2
     4     4     4     3     3     1     4
     4     4     4     3     3     1     3
     4     4     4     3     3     1     2
     4     4     4     3     3     1     1
     4     4     4     2     3     4     4
     4     4     4     2     3     3     4
     4     4     4     2     3     3     3
     4     4     4     2     3     2     4
     4     4     4     2     3     2     3
     4     4     4     2     3     2     2
     4     4     4     2     3     1     4
     4     4     4     2     3     1     3
     4     4     4     2     3     1     2
     ...