Who do I create a sequence of matrices?

2012 11 17
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Hello everybody!
I need to create a sequence of matrices of the following form. For example: From the input:
1 2 0
x = 0 0 0
0 0 0
I need to get:
1 0 0 1 1 0 1 2 0 0 1 0 0 2 0
y = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
That is, I need to get a "combinatoric" sequence of matrices from the one input. The size of the matrix can vary. Is there any suitable function for this? Thank you.

11 个评论
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Azzi Abdelmalek
Azzi Abdelmalek 2012-11-17
What is the relation y=f(x)
Walter Roberson
Walter Roberson 2012-11-17
编辑：Walter Roberson 2012-11-17
Why are multiple 1 allowed? And not multiple 2 ? Why is the first matrix different from the fourth and yet none of the matrices show a 2 in the first column ?
SomeUser
SomeUser 2012-11-17
Because the element y(2,2) can only be 0,1,2. Generaly, y(i,j) can lie between 0 and x(i,j). Only integers allowed.
Azzi Abdelmalek
Azzi Abdelmalek 2012-11-17
编辑：Azzi Abdelmalek 2012-11-17
it's still not clear for me. How x and y are related?
SomeUser
SomeUser 2012-11-17
x if a matrix. y is a 3D array, where y(i,j,:) are matrices like in my first post. The size(y,3) is a number of all such matrices.
the cyclist
the cyclist 2012-11-17
What people are saying here is that we do not understand the rule for generating y from x. Please give more than a one-sentence explanation. Is your example y ALL the values that you would expect from that x? Please trust us when we say it is simply not at all clear what the output should be for general x.
Matt Fig
Matt Fig 2012-11-17
编辑：Matt Fig 2012-11-17
The rule seems to be:
For each non-zero value in x, generate a set of matrices where that value is held fixed while all other values vary from 0 to the value. So you can see there is a 1 and a 2. So hold the 1 fixed and generate matrices by letting the element in the position of the 2 vary from 0 to 2. Then hold the 2 fixed and generate matrices by letting the element in the 1 position vary from 0 to 1.
Or something like that....
So there is a typo in y??
SomeUser
SomeUser 2012-11-17
Yeah, something like that, but i'll better show it. For 2x2 matrix the result should be:
y = zeros(size(x));
for i = 0:x(1,1)
for j = 0:x(1,2)
for k = 0:x(2,1)
for s = 0:x(2,2)
matrix = [[i;k],[j;s]];
y = cat(3, y, matrix);
end
end
end
end
But I need to do it without "fors" cause this should work for matrix of any dimension.
Matt Fig
Matt Fig 2012-11-17
SomeUser, that code produces an error with the x you show....
x = [1 2 0;0 0 0;0 0 0];
Error using cat
CAT arguments dimensions are not consistent.
Error in for_loop3 (line 13)
y = cat(3, y, matrix);
SomeUser
SomeUser 2012-11-17
编辑：SomeUser 2012-11-17
Yeah, because you specified a 3x3 matrix, this code works for 2x2 matrices only. I noted that.
And I want to extend it for, say, my expample from 1 post, which you just tried to follow.
To reproduse my example for x = [1 2 0;0 0 0;0 0 0]; one should add an additional 5 "for" loops to this code. And that is a problem.
Try to restart my code with x = [1 2; 1 0], for example.
Matt Fig
Matt Fig 2012-11-17
Ah, o.k. Thanks.

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Matt Fig
Matt Fig 2012-11-17
编辑：Matt Fig 2012-11-18

0 个投票

This will do the job. Note that all of the matrices are there, but they are in a different order. You can figure out how to re-order them if you work on it. You need this file to make the code work as I have written it: NPERMUTEK.
Note that like most combinatorial problems, things can quickly get out of control with either many elements or a small number of large elements of x.
UPDATE: Simplified Code greatly
function H = combinatoricmat(x)
% help goes here...
L = [size(x) prod(x(x>0)+1)]; % Final array size.
H = npermutek(0:max(x(:)),numel(x));
H = reshape(H(all(bsxfun(@le,H,x(:).'),2),:).',L);
Now test it once it is saved on the MATLAB path:
>> x = [1 2 0;0 0 0;0 0 0];
>> combinatoricmat(x)
ans(:,:,1) =
0 0 0
0 0 0
0 0 0
ans(:,:,2) =
0 1 0
0 0 0
0 0 0
ans(:,:,3) =
0 2 0
0 0 0
0 0 0
ans(:,:,4) =
1 0 0
0 0 0
0 0 0
ans(:,:,5) =
1 1 0
0 0 0
0 0 0
ans(:,:,6) =
1 2 0
0 0 0
0 0 0

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