How to represent set of sets in MATLAB?

I want to create a series of sets C that

... ...

So i use the code

N = 10;
C{1} = [];  % C_0 because MATLAB index starts from 1
C{2} = 1:N;

C{3} = zeros(1,length(C{2})*N);
count = 0;
for r = 1:length(C{2})
    ri = C{2}(r);
    for s = ri+1:N
        count = count + 1;
        ri2 = [ri,s]
        C{3}(count) = ri2; % Error: Unable to perform assignment because the left and right sides have a different number of elements.
    end
end
C{3} = C{3}(C{3}~=0);

How can i fix this code and improve efficiency?

Thanks

Answer 1

[Walter Roberson]

Use a single cell array, with one entry for each of the number of levels you want to carry this out to. Each entry will be a numeric array with the same number of columns as (cell index minus 1)

To get level M from level (M-1), take the array from level M-1 and find out how many rows it currently has, R. Now repmat() the content of that previous level vertically N times, and append a new column which is the elements 1:N repeated vertically R times each.

The only explicit loop needed is according to the number of levels you are using. (repmat or repelem might use implicit loops.)

[2020noOlympics]

Thank you for reply so soon!

Do you mean this:

rowNum = size(C{j},1);
a = repmat(C{j},N,1);
b = repelem((1:N)',rowNum,1);
C{j+1} = [a,b];

But in this way, it will create a lot of unwanted rows like: duplicate rows, like (1,2) and (2,1) are the same; and rows with same elements (1,1), (2,2) ...

[Walter Roberson]

Good points about the unwanted entries.

For any given existing row, you can setdiff(1:N, row) to get the list of values that can be appended this time. A faster way to do that might be

N1 = 1:N;

v = true(1,N);
v(row) = false;
avail = N1(v);

[2020noOlympics]

Hi Walter, thank you for detailed replying. However, when I got to three-number sets, how can I avoid duplicate rows like (8,18,2) and (18,2,8) and so on?

Furthermore, with more elements in sets (my target is 10-element set), duplicate rows are way too many to accelerate the algorithm.

Thank you very much for your time.

[Walter Roberson]

N = 10;

C = cell(N,1);
C{1} = [];
C{2} = (1:N).';

for K = 3 : N
    old = C{K-1};
    nold = size(old,1);
    nnew = N-K+2;
    new = cell(nold,1);
    for r = 1 : nold
        avail = old(r,end)+1:N;
        new{r} = [repmat(old(r,:),length(avail),1), avail(:)];
    end
    C{K} = vertcat(new{:});
end

C{N+1} = 1:N;