问题
Given an m*n binary matrix A, m*p binary matrix B, where n > m what is an efficient algorithm to compute X such that AX=B?
For example:
A =
1 1 0 0 1 1 0 1 0 0
1 1 0 0 1 0 1 0 0 1
0 1 1 0 1 0 1 0 1 0
1 1 1 1 1 0 0 1 1 0
0 1 1 0 1 0 1 1 1 0
B =
0 1 0 1 1 0 1 1 0 1 0 0 1 0
0 0 1 0 1 1 0 0 0 1 0 1 0 0
0 1 1 0 0 0 1 1 0 0 1 1 0 0
0 0 1 1 1 1 0 0 0 1 1 0 0 0
1 0 0 1 0 0 1 0 1 0 0 1 1 0
Note, when I say binary matrix I mean matrix defined over the field Z_2, that is, where all arithmetic is mod 2.
If it is of any interest, this is a problem I am facing in generating suitable matrices for a random error correction code.
回答1:
You can do it with row reduction: Place B to the right of A, and then swap rows (in the whole thing) to get a 1 in row 0, col 0; then xor that row to any other row that has a '1' in column 0, so you have only a single 1 in column 0. Then move to the next column; if [1,1] is zero then swap row 1 with a later row that has a 1 there, then xor rows to make it the only 1 in the column. Assuming 'A' is a square matrix and a solution exists, then you eventually have converted A to unity, and B is replaced with the solution to Ax=B. If n > m, you have a system with more unknowns than equations, so you can solve for some of the unknowns, and set the others to zero. During the row reduction, if there are no values in a column which have a '1' to use (below the rows already reduced) you can skip that column and make the corresponding unknown zero (you can do this at most n-m times).
来源：https://stackoverflow.com/questions/14637478/how-can-i-find-a-solution-of-binary-matrix-equation-ax-b