I have a matrix A, with the coefficients of a binary equation system.

In order to solve it, I need to know if all of those equations are linearly independent. However, rank(A) doesn't work here, because MATLAB doesn't know they're binary equations.

Here's an example:

>> A
A =

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

>> rank(A)
ans =  3

It's important to mention that it's a $Ax = 0 \pmod{2}$ type system.

As you can see, if you add rows 2 and 3, the result is the first one. This means the three of them aren't linearly independent, while rank(A) is saying they are.

How can I solve this?

  • $\begingroup$ The result of adding row 2 and row 3 is (1,2,1,1), which is not (1,0,1,1). Computing the determinant of the $3\times 3$ leading principal submatrix it is clear that this submatrix has full rank and hence the rank of $A$ is $3$. $\endgroup$
    – K. Miller
    Dec 10 '16 at 1:31
  • $\begingroup$ @K.Miller Forgot to mention that the answer is $0 \pmod{2}$ (I've just updated the question). Wouldn't that require to make a bitwise XOR instead of a normal addition? $\endgroup$
    – Hewbot
    Dec 10 '16 at 11:02

rank(A) of Matlab works under the assuption that the matrix has elements in the field of complex numbers, so this won't work for you. But, the rank detection of a binary system is really easy (only algebra no numerical noise considerations). Just apply Gaussian algorithm with rank detection (i.e., in general with row and column permutations).

In your example the first non-trivial elimination step is adding the first equation to the last. In this way you get: \begin{align*} A_2=\pmatrix{1&0&1&1\\0&1&1&1\\0&1&1&1} \end{align*} The next elimination step consists of adding the second row to the just created modified third row. In this way you get just a zero row: \begin{align*} A_2=\pmatrix{1&0&1&1\\0&1&1&1\\0&0&0&0} \end{align*} You 've got two non-zero rows of the staircase system. Therefore, $\operatorname{rank}(A)=2$.

By applying the transformation steps also to the right-hand side you can use the algorithm to solve your system if it is solvable.

There is code for Gauss' algorithm at Mathworks. For completing the answer there follows a code for Octave. Maybe it runs in Matlab too:

## Usage: [x,r,LU,pr,pc,res] = gfGauss(A,y,p,pr,pc)
## Solve system Ax=y in the Galois Field GF(p).
## @param[in] A system matrix
## @param[in] y right-hand side
## @param[in] p size of field (must be prim)
## @param[in] pr optional row permutation; when specified A is already decomposed
## @param[in] pc optional column permutation
## @param[in] r optional rank; only required when A is already decomposed
## @return x solution
## @return r rank of the system
## @return LU LU-factorization of A
## @return pr row permutation vector
## @return pc column permutation vector
## @return res residuum The system cannot be solved if this is non-zero.
function [x,r,LU,pr,pc,res] = gfGauss(A,y,p,pr,pc,r)
  if ~exist("y","var")
    y = [];
  if ~exist("p","var")
    p = 2;
  n = min(size(A));
  nc = size(A,2);
  ny = size(y,2);
  if exist("pr","var") ## System already decomposed
    ## Solve L*z = y (z stored in y)
    LU = A;
    if ~exist("r","var")
      r = n;
    x = zeros(nc,ny);
    y = y(pr,:);
    for iDiag=1:r
      y(iDiag,:) = y(iDiag,:) - LU(iDiag,1:(iDiag-1))*y(1:(iDiag-1),:);
    x = [y(1:r,:);zeros(nc-r,ny)];
  else ## decomposition required
    ## Decompose L*U = A with permuation
    ## and simultaneously solve L*z = y (z stored in y)
    LU = mod([A,y],p);
    r = n;
    pr = 1:size(A,1);
    pc = 1:size(A,2);
    for iDiag=1:n
      [ir,ic] = find(LU(iDiag:end,iDiag:nc)); # Efficiency: search could stop at first non-zero element
      if length(ir)==0
    r = iDiag-1;
    ir = ir(1)+iDiag-1;
    ic = ic(1)+iDiag-1;
      ## row/column permutation
      pr([ir,iDiag]) = pr([iDiag,ir]);
      LU([ir,iDiag],:) = LU([iDiag,ir],:);
      pc([ic,iDiag]) = pc([iDiag,ic]);
      LU(:,[ic,iDiag]) = LU(:,[iDiag,ic]);
      ## elimination step
      pivInv = mod(LU(iDiag,iDiag)^(p-2),p); ## multiplicative inverse of the pivot in GF(p); only works for pime p
      LU((iDiag+1):end,iDiag) = mod(pivInv.*LU((iDiag+1):end,iDiag),p);
      LU((iDiag+1):end,(iDiag+1):end) = mod(LU((iDiag+1):end,(iDiag+1):end)-LU((iDiag+1):end,iDiag)*LU(iDiag,(iDiag+1):end),p);
    x = [LU(1:r,(nc+1):end);zeros(nc-r,ny)];
  if ny > 0
    ## back-substitution
    for iDiag=r:-1:1
      pivInv = mod(LU(iDiag,iDiag)^(p-2),p);
      x(iDiag) = mod(pivInv.*(x(iDiag)-LU(iDiag,(iDiag+1):r)*x((iDiag+1):r)),p);
    x(pc,:) = x;
    x = [];
  LU = LU(:,1:nc);
  if isargout(6)
    res = mod(A*x-y,p);

The following code segment shows the application of the algorithm to the problem in the question for deducing the rank of $A$:

A = [1   0   1   1
     0   1   1   1
     1   1   0   0];

If the size p of the Galois field is not included in the argument list it defaults to GF(2). That means the computation is based on dual numbers. If the right-hand side is omitted an empty vector x=[] is returned. Nevertheless, the LU-decomposition and the rank-detection are carried out.

The output of the program is:

>>     [x,r,LU,pr,pc]=gfGauss(A)
x = [](0x0)
r =  2
LU =

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

pr =

   1   2   3

pc =

   1   2   3   4

The rank information r=2 that is included in the output is most interesting in this example with respect to the question.

  • $\begingroup$ Thank you for your answer! How can I do this programatically? Because if I have to try and add all the possible combinations of rows in the matrix, it would take a lot of time. $\endgroup$
    – Hewbot
    Dec 10 '16 at 11:28
  • $\begingroup$ @Hewbot Am I right that you do not know the Gaussian algorithm with row/column-permutation for systems with real matrices? (Just to know where one can build on.) $\endgroup$
    – Tobias
    Dec 10 '16 at 11:32
  • $\begingroup$ Right you are, I'm afraid... I've used MATLAB's rref(), but I guess it has the same problem as rank(). $\endgroup$
    – Hewbot
    Dec 10 '16 at 11:34
  • $\begingroup$ @Hewbot Note, there is a modrank function that does exactly what you want (not tested by me). The base is also Gaussian algorithm with row and column pivoting. $\endgroup$
    – Tobias
    Dec 10 '16 at 11:42
  • $\begingroup$ I had just found it as you mentioned the Gaussian algorithm. Just tested it and works like a charm, thank you very much! $\endgroup$
    – Hewbot
    Dec 10 '16 at 11:58

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