Gaussian elimation - Linear equation matrices, algorithm

Question

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)

My question is: With these values as input (and the lenght of matrix 3x7),

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 3x7 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!

Answer 1

For an array of 10x15 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 4x4 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 

Answer 2

You can use singular value decomposition to compute a non zero least squares solution to a system of linear homogeneous (and non homogeneous) equations in matrix form.

For a quick overview see:

http://campar.in.tum.de/twiki/pub/Chair/TeachingWs05ComputerVision/3DCV_svd_000.pdf

You should first write out your systems as a matrix equation in the form Ax = b, where x is the 21 unknowns as a column vector, and A is the 28 x 21 matrix that forms the linear system when multiplied out. You essentially need to a compute the matrix A of linear equations, compute the singular value decomposition of A and plug the results into the equation as shown in equation 9.17

There are plenty of libraries that will compute the SVD for you in C, so you only need to formulate the matrix and perform the computations in 9.17. The most difficult part is probably understanding how it all works, with a library SVD function there is relatively little code needed.

To get you started on how to form the equation of linear systems, consider a simple 3 x 3 case.

Suppose that our system is a matrix of the form

1 0 1
0 1 0
1 0 1

We would have the following inputs to the linear system:

2 1 2             (sum of rows                 - row)
2 1 2             (sum of colums               - col)
1 0 3 0 1         (sum of first diagonal sets  - t2b)
1 0 3 0 1         (sum of second diagonal sets - b2t)

so now we create a matrix for the linear system

 A            a1 a2 a3 b1 b2 b3 c1 c2 c3     unknowns (x)    =   result (b)
sum of row 1 [ 1  1  1  0  0  0  0  0  0 ]      [a1]                [2]       
sum of row 2 [ 0  0  0  1  1  1  0  0  0 ]      [a2]                [1]
sum of row 3 [ 0  0  0  0  0  0  1  1  1 ]      [a3]                [2]
sum of col 1 [ 1  0  0  1  0  0  1  0  0 ]      [b1]                [2]
sum of col 2 [ 0  1  0  0  1  0  0  1  0 ]      [b2]                [1]
sum of col 3 [ 0  0  1  0  0  1  0  0  1 ]      [b3]                [2]
sum of t2b 1 [ 1  0  0  0  0  0  0  0  0 ]      [c1]                [1]
sum of t2b 2 [ 0  1  0  1  0  0  0  0  0 ]      [c2]                [0]
sum or t2b 3 [ 0  0  1  0  1  0  1  0  0 ]      [c3]                [3]
sum of t2b 4 [ 0  0  0  0  0  1  0  1  0 ]                          [0]
sum of t2b 5 [ 0  0  0  0  0  0  0  0  1 ]                          [1]
sum of b2t 1 [ 0  0  0  0  0  0  1  0  0 ]                          [1]
sum of b2t 2 [ 0  0  0  1  0  0  0  1  0 ]                          [0]
sum of b2t 3 [ 1  0  0  0  1  0  0  0  1 ]                          [3]
sum of b2t 4 [ 0  1  0  0  0  1  0  0  0 ]                          [0]
sum of b2t 5 [ 0  0  1  0  0  0  0  0  0 ]                          [1]

When you multiply out Ax, you see that you get the linear system of equations. For example if you multiply out the first row by the unkown column, you get

a1 + a2 + a3 = 2

All you have to do is put a 1 in any of the colums that appear in the equation and 0 elsewhere.

Now all you have to do is compute the SVD of A and plug the result into equation 9.17 to compute the unknowns.

I recommend SVD because it can be computed efficiently. If you would prefer, you can augment the matrix A with the result vector b (A|b) and put A in reduced row echelon form to obtain the result.

Answer 3

How about this as another variation

Count the amount of unknown squares each sum passes through   
While there are unsolved cells
   Solve all the cells which are passed through by a sum with only one unknown square
       Cells are solved by simply subtracting off all the known cells from the sum
   Update the amount of unknown squares each sum passes through

No boundary cases but very similar to the previous answer. This would first solve all the corners, then those adjacent to the corners, then those one step more interior from that, and so on...

Edit: Also zero out any paths that have a sum of zero, that should solve any that are solvable (I think)

Answer 4

Start with the first column. You know the top and bottom values (from the first values of the red & yellow lists). Subtract the sum of these two from the first in the green list, and now you have the middle value as well.

Now just work to the right.

Subtract the first column's middle value from the next value in the red list, and you have the second column's top value. Subtract that same middle value from the next value in the yellow list, and you have the second column's bottom value. Subtract the sum of these two from the next value in the green list, and now you have the middle value for the second column.

et cetera

If you're going to code this up, you can see that the first two columns are a special case, and that'll make the code ugly. I'd suggest using two "ghost" columns of all zeros to the left so that you can use a single method for determining the top, bottom, and middle values for each column.

This is also easily generalizable. You'll just have to use (#rows)-1 ghost columns.

Enjoy.