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From: Vinh Le on 20 Mar 2010 19:02
I have a binary matrix A (3x7):
A =

0 0 0 0 0 0 0
0 0 0 1 1 1 1
0 0 1 0 0 1 1

I would like to write a code to generate a matrix with all possible alternative of n element(s) in each row of A.

For example,
============================================
Row 1 of A is
0 0 0 0 0 0 0
*if n = 1, then I will have matrixRow1
matrixRow1 =

0 0 0 0 0 0 0
0 0 0 0 0 0 1
0 0 0 0 0 1 0
........
1 0 0 0 0 0 0
The result matrix S should have all possible alternative elements of all rows (24 by 8 matrix).
----------------------------------------------------------------------------------------------------
*if n = 2, the I will have matrixRow1 with BOTH 1 OR 2 elements changed
matrixRow1 =

0 0 0 0 0 0 0
0 0 0 0 0 0 1
0 0 0 0 0 1 0
........
1 0 0 0 0 0 0
0 0 0 0 1 1 0
0 0 0 0 1 0 1
0 0 0 0 1 1 0
........
1 1 0 0 0 0 0
==============================================
From: Vinh Le on 20 Mar 2010 19:10
sorry same typing errors
the first result matrix should be 24x7 in dimension
the first row has value 0 0 0 0 1 1 0 should be 0 0 0 0 0 1 1
From: Vinh Le on 20 Mar 2010 19:16
I tried the very small one - but it does not work
A = [ 0 0 0; 0 1 1; 1 1 1];
[r, c] = size(A);
for i=1:r
row = A(i,:)
for j=1:r
addedRow = row;
saveRow = row;
row(1,j) = '1';
row = saveRow
resultMatrix = [row,addedRow];
end
end
==============
really need assistance, please help
From: Vinh Le on 20 Mar 2010 19:32
I fixed my little code but still does not work well
================
A = [ 0 0 0; 0 1 1; 1 1 1]
[r, c] = size(A)
resultMatrix = []
for i=1:r
row = A(i,:)
for j=0:r-1
saveRow = row
if row(1,c-j) == 1
row(1,c-j) = 0
else
row(1,c-j) = 1
end
addedRow = row
row = saveRow
resultMatrix = [resultMatrix;addedRow]
end
end
=================
resultMatrix =

0 0 1
0 1 0
1 0 0
0 1 0
0 0 1
1 1 1
1 1 0
1 0 1
0 1 1

should be
0 0 1
0 1 0
1 0 0
From: Vinh Le on 20 Mar 2010 19:36
Sorry, the result matrix should be
resultMatrix =

0 0 0
0 0 1
0 1 0
1 0 0

0 1 1
0 1 0
1 1 0
1 0 1

1 1 1
1 1 0
1 0 1
0 1 1
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