Complete set of combinations combining 3 set - arrays

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

You basically need to find
Combinations with repetition for each set;
"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
...

Answer 2

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.