Create random matrix (MATLAB)

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high speed
high speed on 15 Jan 2022
Commented: high speed on 20 Jan 2022
Dear,
I have these initial parameters :
numRows = 216;
numCols = 432;
A = zeros(numRows,numCols);
numOnesPerCol = randi(([2,3]),[1,numCols]);
numOnesPerRow = randi(([5,6]),[numRows,1]);
and I want to create a binary matrix with dimensions (numRows*numCols) that has numOnesPerCol and numOnesPerRow.
How can I do that please
  4 Comments
high speed
high speed on 15 Jan 2022
@Tina I will give you a small example to understand the idea
numRows = 5;
numCols = 10;
A = zeros(numRows,numCols);
numOnesPerCol = randi(([2,3]),[1,numCols]); % Number of ones in each column
numOnesPerRow = randi(([5,6]),[numRows,1]); % Number of ones in each Row
I obtain for example this binary matrix
which contain 5 rows and 10 columns, in each column I have 2 or 3 ones. And in each row I have 5 or 6 ones.
high speed
high speed on 15 Jan 2022
@Torsten I will give you a small example to understand the idea
numRows = 5;
numCols = 10;
A = zeros(numRows,numCols);
numOnesPerCol = randi(([2,3]),[1,numCols]); % Number of ones in each column
numOnesPerRow = randi(([5,6]),[numRows,1]); % Number of ones in each Row
I obtain for example this binary matrix
which contain 5 rows and 10 columns, in each column I have 2 or 3 ones. And in each row I have 5 or 6 ones.

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Accepted Answer

Torsten
Torsten on 15 Jan 2022
Edited: Torsten on 20 Jan 2022
Use intlinprog for the program
min: sum_{i=1}^{numRows} (e1_i+ + e1_i-) + sum_{j=1}^{numCols} (e2_j+ + e2_j-)
under the constraints
e1_j+ - e1_j- - sum_{i=1}^{numRows} A(i,j) = -numOnesPerCol(j) (j=1,...,numCols)
e2_i+ - e2_i- - sum_{j=1}^{numCols} A(i,j) = -numOnesPerRow(i) (i=1,...,numRows)
e1_j+, e1_j- >= 0 (j=1,...,numCols)
e2_i+, e2_i- >= 0 (i=1,...,numRows)
A(i,j) in {0,1} for all i,j
  6 Comments
Torsten
Torsten on 20 Jan 2022
Edited: Torsten on 20 Jan 2022
Test whether this code works for your purpose.
If fopt as output from intlinprog is equal to zero, a matrix with the desired properties exists and is written to console.
If you want to run the program for larger values of numRows and numCols, you should construct Aeq as a sparse matrix.
numRows = 5;
numCols = 10;
%numOnesPerCol = randi(([2,3]),[1,numCols]); % Number of ones in each column
%numOnesPerRow = randi(([5,6]),[numRows,1]); % Number of ones in each Row
numOnesPerCol = [3 2 3 3 2 3 2 3 3 2];
numOnesPerRow = [5;6;5;5;5];
%Usually, nothing needs to be changed after this line
f = [ones(2*numCols,1);ones(2*numRows,1);zeros(numRows*numCols,1)];
intcon = 2*numCols+2*numRows+1:2*numRows+2*numCols+numRows*numCols;
Aeq11 = [eye(numCols,numCols),-eye(numCols,numCols),zeros(numCols,numRows),zeros(numCols,numRows)];
Aeq12 = -repmat(eye(numCols,numCols),1,numRows);
Aeq21 = [zeros(numRows,numCols),zeros(numRows,numCols),eye(numRows,numRows),-eye(numRows,numRows)];
Aeq22 = [];
for i=1:numRows
Aeq22 = [Aeq22;zeros(1,(i-1)*numCols),ones(1,numCols),zeros(1,(numRows-i)*numCols)];
end
Aeq22 = -Aeq22;
Aeq = [Aeq11,Aeq12;Aeq21,Aeq22];
beq = [-numOnesPerCol.';-numOnesPerRow];
Aineq = [];
bineq = [];
lb = [zeros(2*numCols,1);zeros(2*numRows,1);zeros(numCols*numRows,1)];
ub = [Inf(2*numCols,1);Inf(2*numRows,1);ones(numCols*numRows,1)];
[xopt,fopt,status,output] = intlinprog(f,intcon,Aineq,bineq,Aeq,beq,lb,ub);
disp(xopt)
disp(fopt)
disp(status)
disp(output)
if abs(fopt) < eps
for i = 1:numRows
for j = 1:numCols
A(i,j) = xopt(2*numRows+2*numCols+(i-1)*numCols+j);
end
end
disp(A)
end
high speed
high speed on 20 Jan 2022
@Torsten It works with your code. Thank you so much
I really appreciate your help

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More Answers (1)

Image Analyst
Image Analyst on 15 Jan 2022
@high speed I think what you need to do is to first construct a Latin Rectangle:
Then mask it with two numbers like
output = latinRectangle == 1 | latinRectangle == 2;
Sorry I don't have Latin Rectangle code but there is Latin Square, and maybe Latin Rectangle, code in the File Exchange.
  1 Comment
high speed
high speed on 15 Jan 2022
@Image Analyst Thank you for your response, but I think that I don't need to work with Latin Rectangle.
Because the idea here is to obtain binary matrix that contains a variety of ones between 2 and 3 in each column and a variety of ones between 5 and 6 in each row as in this example

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