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Asked: 2026-05-28T18:57:14+00:00

Let’s assume that we have a simple matrix 3rows x 7cols. The matrix includes

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Let’s assume that we have a simple matrix 3rows x 7cols.
The matrix includes only zeros (0) and (1) like:

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

Senario:
If we know the sum of non-zeros in each row,

(in first row is 4, in second row is 2, in third row is 3.) (blue line)

additional, if we know the sum of each col (1 , 0, 3, 2, 2, 1, 0) (green line)

also if we know the sum of each diagonal from the top-left to bottom-right (1,0,1,2,3,0,1,1,0)(red lines) anti-clockwise

and finally we know the sum of each diagonal from the bottom-left to top-right (0,0,2,1,3,2,1,0,0) (yellow lines)

enter image description here

My question is:
With these values as input (and the lenght of matrix 3×7),

4, 2, 3
1, 0, 3, 2, 2, 1, 0
1, 0, 1, 2, 3, 0, 1, 1, 0
0, 0, 2, 1, 3, 2, 1, 0, 0

How we can draw the first matrix?
After a lot of thoughts I came to the conclusion that this is a linear equation system with 3×7 unknown values and some equations.
Right?

How can I make an algorithm in C, or whatever, to solve these equations?
Should I use a method like gausian equation?

Any help would be greatly appreciated!

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1 Answer

  1. Editorial Team

    For an array of 10×15 ones and zeros, you would be trying to find 150 unknowns and have 10+15+2*(10+15-1) = 73 equations if you ignore that the values are limited to being either one or zero. Obviously you can’t create a linear system on that basis which has a unique solution.

    So is that constraint enough to give a unique solution?

    For a 4×4 matrix with the following sums there are two solutions:

    - 1 1 1 1
| 1 1 1 1
\ 0 1 1 0 1 1 0
/ 0 1 1 0 1 1 0

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

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

    So I wouldn’t expect there to be a unique solution for larger matrices – the same symmetry would exist in many places:

    - 1 1 0 0 1 1
| 1 1 0 0 1 1
\ 0 1 0 0 1 0 1 0 0 1 0
/ 0 1 0 0 1 0 1 0 0 1 0

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

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