Create random matrix (MATLAB)

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

[Tina]

Hi there ! Can you be a liitle clear?Do you want all the rows of first col have the numonesperrow data or do you want the first col of all the row have numonepercoldata ??

You cant merge your parameter in any way but you can try

numRows = 216;
numCols = 432;
A = zeros(numRows,numCols);
numOnesPerCol = randi(([2,3]),[215,numCols]);
numOnesPerRow = randi(([5,6]),[1,numCols]);
a=[numOnesPerRow ;numOnesPerCol ]

a = 216×432

5 6 6 5 5 6 6 5 5 6 5 5 5 5 6 5 6 6 5 5 5 6 6 5 6 6 5 6 6 6 3 2 3 2 2 2 2 3 2 3 3 3 2 3 3 2 3 3 2 2 2 3 3 2 2 2 3 2 2 3 3 3 2 2 2 2 2 2 2 2 2 2 2 2 3 2 2 3 3 3 2 2 2 3 2 2 2 3 2 3 2 3 2 2 3 3 3 2 2 2 2 3 2 3 3 2 2 3 2 3 3 2 2 3 2 3 3 2 3 2 3 3 3 3 3 2 2 3 2 2 2 2 3 3 2 3 3 3 3 3 2 3 2 2 2 3 3 3 3 2 2 3 3 2 3 3 2 2 2 2 3 2 2 3 2 2 2 2 2 3 2 2 3 3 2 3 2 2 2 3 2 2 2 3 2 3 3 2 3 3 2 2 2 2 2 2 3 3 2 2 2 2 2 2 2 3 3 2 3 3 3 3 3 2 3 2 2 3 3 3 2 3 2 3 3 3 3 2 3 2 3 3 3 2 3 3 2 2 3 3 3 2 3 3 2 2 3 2 3 2 3 2 2 2 2 3 2 3 2 3 3 3 3 2 3 3 3 2 2 3 3 2 3 2 3 2 3 2 3 2 2 3 3 2 3 2 3 2 3 2 2 2 2 3 2 2 2 2 2 3

or vice versa

[Torsten]

numOnesPerCol is a row vector (1x432) with only 2's and 3's.

numOnesPerRow is a row vector (1x216) with only 5's and 6's.

So what do you mean by

"I want to create a matrix with dimensions (numRows*numCols) that has numOnesPerCol and numOnesPerRow"

[high speed]

@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]

@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.

Answer 1

[Torsten]

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

[high speed]

@Torsten Sorry I didn't understand

[Torsten]

Use the optimizer "intlinprog" and solve the system I defined above for the matrix elements A(i,j).

[Image Analyst]

A solution may not exist. For example if you wanted, in a 5x10 matrix to have 9 or more in each row but only 1 or 2 in each column. There is no solution for that. Solutions exist only for certain combinations.

[Torsten]

Yes, the solver will show up that the problem is not feasible (hopefully).

[Torsten]

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]

@Torsten It works with your code. Thank you so much

I really appreciate your help

Answer 2

[Image Analyst]

@high speed I think what you need to do is to first construct a Latin Rectangle:

https://en.wikipedia.org/wiki/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.

[high speed]

@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