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)]