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From: Ozge Taskan on 9 Jul 2010 23:25
"Matt Fig" <spamanon(a)yahoo.com> wrote in message <hsrrsd$o20$1(a)fred.mathworks.com>...
> I see. If Jos is correct, then you can also do this with a slight modification of the solution I gave earlier.
>
>
> [m,n] = size(A);
> Rr = npermutek(1:n,m); % The "Row-Rank" matrix.
> G = A((Rr-1)*m+repmat(1:m,n^m,1));
>
> IIRC, COMBN and NPERMUTEK do the same thing using different methods.

Hi,
I would like to ask something related to same thing.
If we have 9 numbers and we would like to assign one of the numbers below how can we look at each possibilities for example
1st number can be 1 or 4
2nd number can be 1 or 4
3rd number can be 4
4th number can be 1 or 2 or 3
5th number can be 1 or 2 or 3
6th number can be 1 or 2 or 3
7th number can be 3
8th number can be 2 or 3
9th number can be 1 or 2 or 3
how can we construct 2*2*1*3*3*3*1*2*3(648) numbers?

thank you very much in advance.
From: Ozge Taskan on 10 Jul 2010 00:51
"Ozge Taskan" <lordgy(a)yahoo.com> wrote in message <i18p6g$kri$1(a)fred.mathworks.com>...
> "Matt Fig" <spamanon(a)yahoo.com> wrote in message <hsrrsd$o20$1(a)fred.mathworks.com>...
> > I see. If Jos is correct, then you can also do this with a slight modification of the solution I gave earlier.
> >
> >
> > [m,n] = size(A);
> > Rr = npermutek(1:n,m); % The "Row-Rank" matrix.
> > G = A((Rr-1)*m+repmat(1:m,n^m,1));
> >
> > IIRC, COMBN and NPERMUTEK do the same thing using different methods.
>
> Hi,
> I would like to ask something related to same thing.
> If we have 9 numbers and we would like to assign one of the numbers below how can we look at each possibilities for example
> 1st number can be 1 or 4
> 2nd number can be 1 or 4
> 3rd number can be 4
> 4th number can be 1 or 2 or 3
> 5th number can be 1 or 2 or 3
> 6th number can be 1 or 2 or 3
> 7th number can be 3
> 8th number can be 2 or 3
> 9th number can be 1 or 2 or 3
> how can we construct 2*2*1*3*3*3*1*2*3(648) combinations?
>
> thank you very much in advance.
From: Roger Stafford on 10 Jul 2010 02:22
"Ozge Taskan" <lordgy(a)yahoo.com> wrote in message <i18p6g$kri$1(a)fred.mathworks.com>...
> Hi,
> I would like to ask something related to same thing.
> If we have 9 numbers and we would like to assign one of the numbers below how can we look at each possibilities for example
> 1st number can be 1 or 4
> 2nd number can be 1 or 4
> 3rd number can be 4
> 4th number can be 1 or 2 or 3
> 5th number can be 1 or 2 or 3
> 6th number can be 1 or 2 or 3
> 7th number can be 3
> 8th number can be 2 or 3
> 9th number can be 1 or 2 or 3
> how can we construct 2*2*1*3*3*3*1*2*3(648) numbers?
>
> thank you very much in advance.
- - - - - - - - - -
You can accomplish that using numbers with mixed bases. That is, start with the number 0 and count by ones up to 648-1. For each one of these numbers, compute the nine digits that would represent that number with the nine different bases 2, 2, 1, 3, 3, 3, 1, 2, 3. If 1 is added to each of these digits, they can be used as indices into a cell array containing all the number sets possible for each digit: the first digit can be 1 or 4, the second one also 1 or 4, the third one only 4, the fourth 1, 2, or 3, etc.

For example, the number N = 556 can be represented in the mixed bases above by: 1 1 0 1 0 2 0 1 1 since

556 = ((((((((1)*2+1)*1+0)*3+1)*3+0)*3+2)*1+0)*2+1)*3+1

Suppose v = [2 2 1 3 3 3 1 2 3] contains the nine bases, that is the nine lengths in the cell array. Suppose N = 556. You can obtain these nine digits with a for-loop:

q = N;
for k = 9:-1:1
r = mod(q,v(k));
q = (q-r)/v(k);
% At this point use r+1 as an index into the k-th cell array
% of possible numbers and place result in a matrix, X(N+1,k)
end

You would have an outer for-loop going through all the possible values of N from 0 to 647. I'll let you work out the rest of the details.

The only difference between the above and converting a number to decimal or binary digits is that the bases are variable in this situation.

Roger Stafford
From: Bruno Luong on 10 Jul 2010 05:03
"Ozge Taskan" <lordgy(a)yahoo.com> wrote in message <i18p6g$kri$1(a)fred.mathworks.com>...
> "Matt Fig" <spamanon(a)yahoo.com> wrote in message <hsrrsd$o20$1(a)fred.mathworks.com>...
> > I see. If Jos is correct, then you can also do this with a slight modification of the solution I gave earlier.
> >
> >
> > [m,n] = size(A);
> > Rr = npermutek(1:n,m); % The "Row-Rank" matrix.
> > G = A((Rr-1)*m+repmat(1:m,n^m,1));
> >
> > IIRC, COMBN and NPERMUTEK do the same thing using different methods.
>
> Hi,
> I would like to ask something related to same thing.
> If we have 9 numbers and we would like to assign one of the numbers below how can we look at each possibilities for example
> 1st number can be 1 or 4
> 2nd number can be 1 or 4
> 3rd number can be 4
> 4th number can be 1 or 2 or 3
> 5th number can be 1 or 2 or 3
> 6th number can be 1 or 2 or 3
> 7th number can be 3
> 8th number can be 2 or 3
> 9th number can be 1 or 2 or 3
> how can we construct 2*2*1*3*3*3*1*2*3(648) numbers?
>

You can use NDGRID as following:

c={[1 4] [1 4] 4 [1 2 3] [1 2 3] [1 2 3] 3 [2 3] [1 2 3]}

n = length(c);
[c{:}]=ndgrid(c{:});
c=cat(n+1,c{:});
c = reshape(c,[],n) % 648 combinations of 9 numbers

% Bruno
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