[英]How can I improve performance of Sudoku solver?
我无法提高以下 Sudoku Solver 代码的性能。 我知道这里有 3 个循环,它们可能会导致性能下降,但我找不到更好/更有效的方法。 “board”随着递归的每次迭代而发生变化——如果没有零剩下,我只需要退出递归。
我试图将“板”与突变隔离开来,但它并没有改变性能。 我还尝试对前 2 个“for”循环使用列表理解(即仅循环使用零的行和列),尝试查找所有零的坐标,然后通过它们使用单个循环到 go - 没有帮助.
我认为我在这里用递归做一些根本错误的事情 - 关于如何使解决方案更快的任何建议或建议?
def box(board,row,column):
start_col = column - (column % 3)
finish_col = start_col + 3
start_row = row - (row % 3)
finish_row = start_row + 3
return [y for x in board[start_row:finish_row] for y in x[start_col:finish_col]]
def possible_values(board,row,column):
values = {1,2,3,4,5,6,7,8,9}
col_values = [v[column] for v in board]
row_values = board[row]
box_values = box(board, row, column)
return (values - set(row_values + col_values + box_values))
def solve(board, i_row = 0, i_col = 0):
for rn in range(i_row,len(board)):
if rn != i_row: i_col = 0
for cn in range(i_col,len(board)):
if board[rn][cn] == 0:
options = possible_values(board, rn, cn)
for board[rn][cn] in options:
if solve(board, rn, cn):
return board
board[rn][cn] = 0
#if no options left for the cell, go to previous cell and try next option
return False
#if no zeros left on the board, problem is solved
return True
problem = [
[9, 0, 0, 0, 8, 0, 0, 0, 1],
[0, 0, 0, 4, 0, 6, 0, 0, 0],
[0, 0, 5, 0, 7, 0, 3, 0, 0],
[0, 6, 0, 0, 0, 0, 0, 4, 0],
[4, 0, 1, 0, 6, 0, 5, 0, 8],
[0, 9, 0, 0, 0, 0, 0, 2, 0],
[0, 0, 7, 0, 3, 0, 2, 0, 0],
[0, 0, 0, 7, 0, 5, 0, 0, 0],
[1, 0, 0, 0, 4, 0, 0, 0, 7]
]
solve(problem)
您可以做三件事来加快速度:
possible_values = row_candidates[row] & col_candidates[col] & box_candidates[box]
。 这是一个不变的因素,在您的方法中变化很小。由于 53x15 和 kosciej16,结果最终奏效。 不理想或最佳但通过了所需的性能测试:
def solve(board, i_row = 0, i_col = 0):
cells_to_solve = [((rn, cn), possible_values(board,rn,cn)) for rn in range(len(board)) for cn in range(len(board)) if board[rn][cn] == 0]
if not cells_to_solve: return True
min_n_of_values = min([len(x[1]) for x in cells_to_solve])
if min_n_of_values == 0: return False
best_cells_to_try = [((rn,cn),options) for ((rn,cn),options) in cells_to_solve if len(options) == min_n_of_values]
for ((rn,cn),options) in best_cells_to_try:
for board[rn][cn] in options:
if solve(board, rn, cn):
return board
board[rn][cn] = 0
return False
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