changing position of numbers in a vector

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Brwa
Brwa 2013-5-20
回答： Mian Jehanzaib 2016-12-22
Im new in matlab, i hope someone can help me with my problem. I need to make a code to solve this Cobinatoric example. I have a vector with 6 numbers where 2 of them are 1 and the rest are 0 as you can see below
A=[1 1 0 0 0 0]
I want to make a code which can help me change the position of all (1)s and put them in all the possible positions as you can see bellow without changing it all of them by myself, because my code is not only 5 numbers, i have a huge number.
A=[0 1 1 0 0 0]
A=[0 0 1 1 0 0]
A=[0 0 0 1 1 0]
A=[0 0 0 0 1 1]
A=[1 0 1 0 0 0]
A=[1 0 0 1 0 0]
A=[1 0 0 0 1 0]
A=[1 0 0 0 0 1]
A=[0 1 0 1 0 0]
A=[0 1 0 0 1 0]
A=[0 1 0 0 0 1]
A=[0 0 1 0 1 0]
A=[0 0 1 0 0 1]
A=[0 0 0 1 0 1]

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回答（4 个）

Andrei Bobrov
Andrei Bobrov 2013-5-20
编辑：Andrei Bobrov 2013-5-20
out = unique(perms([1 1 0 0 0 0]),'rows');
or [EDIT]
A = [1 1 1 1 0 0 0 0 0 0];
n = numel(A);
b = nnz(A);
M = 0:ones(1,n)*pow2(n-1:-1:0)';
z = rem(floor(M(:)*pow2(1-n:0)),2);
out = z(sum(z,2) == b,:);
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Brwa
Brwa 2013-6-5
Dear Andrei Bobrov
is it possible to use that code for big numbers such as n => 30 and b => 5 ?
when i tried n = 25 and b =5 i got error said
??? Error using ==> mtimes Out of memory. Type HELP MEMORY for your options.
Error in ==> Position at 5
z = rem(floor(M(:)*pow2(1-n:0)),2);
and when i tried n = 30 i got this error
??? Error using ==> colon Maximum variable size allowed by the program is exceeded.
Error in ==> Position at 4
M = 0:ones(1,n)*pow2(n-1:-1:0)';
I wish you have a solution for this problem
Thanks for your help.
Triveni
Triveni 2015-10-31
@Andrei Bobrov
If A= [-45 -45 -45 -45 -45 -45 0 0 0 0 0 0 0 0 45 45 45 45 45 45]
then how can be write??

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David Sanchez
David Sanchez 2013-5-20
Andrei's answer maybe the best choice, but in case you want to see what's behind:
A=zeros(1,6);
for k=1:size(A,2)
A=zeros(1,6);
A(k) = 1;
for n = k+1:size(A,2)
if n>k+1
A(n-1) = 0;
end
A(n) = 1
end
end
Roger Stafford
Roger Stafford 2013-5-20
This problem is naturally made for Matlab's 'nchoosek' function. This method requires no rejection afterward and will therefore be easiest on your memory. Let A be a row vector of ones and zeros.
n = size(A,2);
k = sum(A==1);
C = nchoosek(1:n,k);
m = size(C,1); % m will equal n!/k!/(n-k)!
B = zeros(m,n);
B(repmat((1-m:0)',1,k)+m*C) = 1; % Place ones according to indices in C
The desired position combinations will appear in the rows of B, with k ones in each row.
Note: Beware of "huge" numbers! If you have, say, 15 ones and 15 zeros in A, the number of possible arrangements of them is an enormous 155,117,520.
Mian Jehanzaib
Mian Jehanzaib 2016-12-22
How about creating one possible combination row at a time in a for loop and not creating a huge matrix at once?
I don't require to store the all possibilities in a matrix rather I need to generate them one at a time. Could you suggest any solution please

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