how to add some value to random matrix with specific condition?

if i have A represent the number of group ones in matrix in the row

A = [ 1 2 0
      2 1 1 
      3 1 1 ]

i want to generate matrix like this depend on A which between the group of ones at least one zero

F = [1 0 1 1 0 0 0 
     1 1 0 1 0 1 0
     1 1 1 0 1 0 1 ]

[Stephen23]

Question is continued here:

http://www.mathworks.com/matlabcentral/answers/281669-i-want-a-function-to-compute-some-value-from-matrix-and-then-set-this-value-on-a-new-matrix

Answer 1

[Stephen23]

A = [ 1 2 0
      2 1 1 
      3 1 1 ]
%
N = max(sum(1+A,2));
C = arrayfun(@(n){[ones(1,n),0]},A);
C = cellfun(@(c){[c{:}]},num2cell(C,2));
C = cellfun(@(v){[v,zeros(1,N-numel(v))]},C);
F = vertcat(C{:});
F(:,end) = []

prints the output in the command window:

F =
 1   0   1   1   0   0   0
 1   1   0   1   0   1   0
 1   1   1   0   1   0   1

[Firas Al-Kharabsheh]

when i apply this on

   A = [   1     0     0
           3     0     0
           5     0     0
           7     0     0
           9     0     0
          11     0     0
          13     0     0
          13     0     0
          15     0     0
          15     0     0
          15     0     0
          6     1     6
          4     1     4
          3     0     0
          5     0     0  ]

 the solution will be 

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

 and this is wrong where the solution will be 

   [ 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0
      0 0 0 0 0 0 1 1 1 0 0 0 0 0 0
      0 0 0 0 0 1 1 1 1 1 0 0 0 0 0
      0 0 0 0 1 1 1 1 1 1 1 0 0 0 0
      0 0 0 1 1 1 1 1 1 1 1 1 0 0 0
      0 0 1 1 1 1 1 1 1 1 1 1 1 0 0
      0 1 1 1 1 1 1 1 1 1 1 1 1 1 0
      0 1 1 1 1 1 1 1 1 1 1 1 1 1 0
      1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
      1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
      1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
      1 1 1 1 1 1 0 1 0 1 1 1 1 1 1
      0 1 1 1 1 0 0 1 0 0 1 1 1 1 0
      0 0 0 0 0 0 1 1 1 0 0 0 0 0 0
      0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 ]

[Stephen23]

My answer is not wrong, because you actually changed the requirements afterwards.

Please explain your new requirements clearly:

1. exactly how many leading zeros there need to be, or
2. where the ones should be.

and then explain how this new requirement matches your original example:

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

Why are these ones not centered ?

[Firas Al-Kharabsheh]

i want to write another question and i will explain what i need clearly

[Guillaume]

If you start another question, then you still should accept Stephen's answer considering he did answer the current question correctly.