How can I generate all permutations of a matrix, in which value "1" cannot be repeated in any column or row? Rest of the elements are identical, and a number between 0 and 1.

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Bidsitlov
Bidsitlov 2016-5-21
评论： Roger Stafford 2016-5-21
Each row must have at least an element with value 1.
Say we have 3 by 4 matrix. Then an acceptable arrangement would be:
x x 1 x
x x x 1
x 1 x x
and there will be:
4! / (4-3)! = 24
different allowed permutations of a 3 by 4 matrix.
Say for a 4 by 5 matrix, violations would be
1 x x x 1
x x x x x
x x x 1 x
x x x 1 x
where first row has more than one values of 1, and column 4 has the same violation. Lastly, there is no 1 value in row 2.
Similarly number of allowed permutations would be:
5! / (5-4)! = 120
that is, the number of different matrices.
My attempt is through different for loops, which is cumbersome for larger elements. Any smarter way to make this happen?
Thank you.

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采纳的回答

Roger Stafford
Roger Stafford 2016-5-21
编辑：Roger Stafford 2016-5-21
I don’t know how you want to arrange all the matrices that are to be generated so I leave that aspect to you. Suppose you wish to generate all m-by-n matrices using the value x where m<=n. First create the matrix A:
A = toeplitz([1,repmat(x,1,m-1)],[1,repmat(x,1,n-1)]);
Next do this:
P = perms(1:m);
T = nchoosek(1:n,m);
C = zeros(size(T,1),n);
for k = 1:size(C,1)
C(k,[T(k,:),setdiff(1:n,T(k,:))]) = 1:n; % <-- Corrected
end
Now for every combination of ix in 1:size(P,1) and jx in 1:size(C,1) create
A(P(ix,:),C(jx,:))
There will be n!/(n-m)! of these altogether. If m>n, do this the other way around.
  2 个评论
Bidsitlov
Bidsitlov 2016-5-21
I tweaked your answer a bit so that it completely generates a list:
function [matrixlist] = allpermsALT(dim1,dim2,J)
%m=dim1 n=dim2
howmuch = factorial(dim2) / factorial(dim2 - dim1);
matrixlist = cell(howmuch,1);
offdiag = exp(-J);
A = toeplitz([1,repmat(offdiag,1,dim1-1)],[1,repmat(offdiag,1,dim2-1)])
P = perms(1:dim1);
T = nchoosek(1:dim2,dim1);
C = zeros(size(T,1),dim2);
for k = 1:size(C,1)
C(k,:) = [T(k,:),setdiff(1:dim2,T(k,:))];
end
counter = 1;
for i = 1:size(P,1)
for j = 1:size(C,1)
matrixlist{counter,1} = A(P(i,:),C(j,:));
counter = counter + 1;
end
end
end
It was instrutive to see an alternative way. Thank you for your answer!
Roger Stafford
Roger Stafford 2016-5-21
I’m afraid I made an error on that code for C. I have corrected it in my answer. As it stood it would duplicate certain matrices and omit others.

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更多回答（1 个）

Bidsitlov
Bidsitlov 2016-5-21
Alright, so personal solution is below:
function [matrixlist] = allperms(dim1,dim2,J)
howmuch = factorial(dim2) / factorial(dim2 - dim1);
matrixlist = cell(howmuch,1);
trfm = trfmgod(dim1,dim2,J);
blank = linspace(1,dim2,dim2);
diff = dim1 + 1;
for i=diff:dim2
if blank(i) >= diff
blank(i) = diff;
end
end
perm_list = unique(perms([blank]),'rows');
listlength = length(perm_list);
for j = 1:listlength
matrixlist{j,1} = trfm(:,perm_list(j,:));
end
end
Function 'trfmgod' is a function, which initially yields, say for 3 by 5 matrix:
1 x x x x
x 1 x x x
x x 1 x x
x is a number generated by a J value, and, value 'howmuch' is calculated mathematically, giving number of allowed permutations.
Since in initial matrix, columns after the 'square portion' keeps repeating itself, parameter 'blank' should be [1 2 3 4 4] in our case.
Finally permutation list is generated by the 'blank' parameter, and a for loop generates all possible permutations.
If you have a better answer that would be informative.
Thank you.

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