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.

3 vues (au cours des 30 derniers jours)
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.

Réponse acceptée

Roger Stafford
Roger Stafford le 21 Mai 2016
Modifié(e) : Roger Stafford le 21 Mai 2016
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 commentaires
Bidsitlov
Bidsitlov le 21 Mai 2016
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 le 21 Mai 2016
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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Bidsitlov
Bidsitlov le 21 Mai 2016
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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