Exact Cover Problem via Dancing Links

sage.combinat.dlx.AllExactCovers(M)

Use A. Ajanki’s DLXMatrix class to solve the exact cover problem on the matrix M (treated as a dense binary matrix).

EXAMPLES:

sage: M = Matrix([[1,1,0],[1,0,1],[0,1,1]])  #no exact covers
sage: for cover in AllExactCovers(M):
....:     print(cover)
sage: M = Matrix([[1,1,0],[1,0,1],[0,0,1],[0,1,0]]) #two exact covers
sage: for cover in AllExactCovers(M):
....:     print(cover)
[(1, 1, 0), (0, 0, 1)]
[(1, 0, 1), (0, 1, 0)]

class sage.combinat.dlx.DLXMatrix(ones, initialsolution=None)

Bases: object

Solve the Exact Cover problem by using the Dancing Links algorithm described by Knuth.

Consider a matrix M with entries of 0 and 1, and compute a subset of the rows of this matrix which sum to the vector of all 1’s.

The dancing links algorithm works particularly well for sparse matrices, so the input is a list of lists of the form: (note the 1-index!):

[
 [1, [i_11,i_12,...,i_1r]]
 ...
 [m,[i_m1,i_m2,...,i_ms]]
]

where M[j][i_jk] = 1.

The first example below corresponds to the matrix:

1110
1010
0100
0001

which is exactly covered by:

1110
0001

and

1010
0100
0001

EXAMPLES:

sage: from sage.combinat.dlx import *
sage: ones = [[1,[1,2,3]]]
sage: ones+= [[2,[1,3]]]
sage: ones+= [[3,[2]]]
sage: ones+= [[4,[4]]]
sage: DLXM = DLXMatrix(ones,[4])
sage: for C in DLXM:
....:      print(C)
[4, 1]
[4, 2, 3]

Note

The 0 entry is reserved internally for headers in the sparse representation, so rows and columns begin their indexing with 1. Apologies for any heartache this causes. Blame the original author, or fix it yourself.

next()

Search for the first solution we can find, and return it.

Knuth describes the Dancing Links algorithm recursively, though actually implementing it as a recursive algorithm is permissible only for highly restricted problems. (for example, the original author implemented this for Sudoku, and it works beautifully there)

What follows is an iterative description of DLX:

stack <- [(NULL)]
level <- 0
while level >= 0:
    cur <- stack[level]
    if cur = NULL:
        if R[h] = h:
            level <- level - 1
            yield solution
        else:
            cover(best_column)
            stack[level] = best_column
    else if D[cur] != C[cur]:
        if cur != C[cur]:
            delete solution[level]
            for j in L[cur], L[L[cur]], ... , while j != cur:
                uncover(C[j])
        cur <- D[cur]
        solution[level] <- cur
        stack[level] <- cur
        for j in R[cur], R[R[cur]], ... , while j != cur:
            cover(C[j])
        level <- level + 1
        stack[level] <- (NULL)
    else:
        if C[cur] != cur:
            delete solution[level]
            for j in L[cur], L[L[cur]], ... , while j != cur:
                uncover(C[j])
        uncover(cur)
        level <- level - 1

sage.combinat.dlx.OneExactCover(M)

Use A. Ajanki’s DLXMatrix class to solve the exact cover problem on the matrix M (treated as a dense binary matrix).

EXAMPLES:

sage: M = Matrix([[1,1,0],[1,0,1],[0,1,1]])  # no exact covers
sage: OneExactCover(M)

sage: M = Matrix([[1,1,0],[1,0,1],[0,0,1],[0,1,0]]) # two exact covers
sage: OneExactCover(M)
[(1, 1, 0), (0, 0, 1)]