我正在尝试在python中创建一个数独检查器：
ill_formed = [[5,3,4,6,7,8,9,1,2],
[6,7,2,1,9,5,3,4,8],
[1,9,8,3,4,2,5,6,7],
[8,5,9,7,6,1,4,2,3],
[4,2,6,8,5,3,7,9], # <---
[7,1,3,9,2,4,8,5,6],
[9,6,1,5,3,7,2,8,4],
[2,8,7,4,1,9,6,3,5],
[3,4,5,2,8,6,1,7,9]]
easy = [[2,9,0,0,0,0,0,7,0],
[3,0,6,0,0,8,4,0,0],
[8,0,0,0,4,0,0,0,2],
[0,2,0,0,3,1,0,0,7],
[0,0,0,0,8,0,0,0,0],
[1,0,0,9,5,0,0,6,0],
[7,0,0,0,9,0,0,0,1],
[0,0,1,2,0,0,3,0,6],
[0,3,0,0,0,0,0,5,9]]
我期待这样的输入 - 一个包含9个列表的列表。零表示用户尚未填写的数字。它们可以连续，列或3x3多次出现。
def check_sudoku(grid):
if len(grid) == 9:
numsinrow = 0
for i in range(9):
if len(grid[i]) == 9:
numsinrow += 1
if numsinrow == 9:
for i in range(9):
rowoccurence = [0,0,0,0,0,0,0,0,0,0]
for j in range(9):
rowoccurence[grid[i][j]] += 1
temprow = rowoccurence[1:10]
if temprow == [1,1,1,1,1,1,1,1,1]:
return True
else:
return False
else:
return False
else:
return False
我显然需要检查是否存在9x9列表（网格），并且每行，列和3x3小方块都没有重复项。在代码中，我首先检查是否有适当的行数（应该有9个）。然后我检查每行中有9个元素（使用ill_formed示例，您会发现情况并非如此）。然后我尝试检查每一行中的重复项，但我遇到了一些麻烦。我以为我可以遍历每一行并循环遍历该行中的每个元素，并将1添加到一个int列表中（rowoccurence）。例如，如果第一个数字是2，那么rowoccurence [2]应该等于1.零处于rowoccurence [0]并且不被检查（我有一个临时列表，它应该采取除第一个元素之外的所有内容 - 零 - 因为连续可能有多于零，并且网格仍然是合法的。我尝试检查临时列表（基本上是rowoccurence）与正确值的参考列表，但它似乎没有工作。你能帮我检查一下这个数独检查器中的重复行吗？非常感谢你！
答案 0 :(得分：12)
请记住，您不是在搜索重复项 - 只是非零重复项。总结一组适用于此。您还可以同时检查行/列的合法性：
def sudoku_ok(line):
return (len(line) == 9 and sum(line) == sum(set(line)))
def check_sudoku(grid):
bad_rows = [row for row in grid if not sudoku_ok(row)]
grid = list(zip(*grid))
bad_cols = [col for col in grid if not sudoku_ok(col)]
squares = []
for i in range(9, step=3):
for j in range(9, step=3):
square = list(itertools.chain(row[j:j+3] for row in grid[i:i+3]))
squares.append(square)
bad_squares = [square for square in squares if not sudoku_ok(square)]
return not (bad_rows or bad_cols or bad_squares)
答案 1 :(得分：4)
你return True太早了，所以你永远不会参加你希望看到失败的考试：
if temprow == [1,1,1,1,1,1,1,1,1]:
return True # <-- this is the culprit
else:
return False
其他注释：一种简单的方法可以确保某些向量的所有元素都等于某个常量：
all(i == const for i in vector)
另一个，更简单：如果（不好主意，请参阅下面的评论。）vec[1:10]全部为1，那么sum(vec[1:10])必须为9。
答案 2 :(得分：2)
参考@llb 的答案，它没有检查缺失值或零，这是我的解决方案，适用于负值、零和缺失值
def line_ok(e):
if len(set(e)) != 9: return False
for i in range(len(e)):
if e[i] not in range(1,10): return False
return True
def checker(grid):
bad_rows = [False for row in grid if not line_ok(row)]
grid = list(zip(*grid))
bad_cols = [False for col in grid if not line_ok(col)]
squares = []
for i in range(0,9,3):
for j in range(0,9,3):
square = list(itertools.chain.from_iterable(row[j:j+3] for row in grid[i:i+3]))
squares.append(square)
bad_squares = [False for sq in squares if not line_ok(sq)]
return not any([bad_rows, bad_cols, bad_squares])
print(checker(sudoku_correct))
PS：由于代表较少，无法发表评论。希望谁需要它，找到它:)
答案 3 :(得分：1)
定义一个函数来验证没有重复项，然后您可以使用它来检查行，列和3x3网格。如果不满足某些条件，则可以通过提前返回来减少嵌套块，例如，行数大于9.如果没有任何检查失败，则只在函数的最末端返回true。
from collections import Counter
def check_dups(l):
counts = Counter()
for cell in l:
if cell != 0: counts[cell] += 1
if cell > 9 or counts[cell] > 1: return False
return True
def check_sudoku(grid):
if len(grid) != 9: return False
if sum(len(row) == 9 for row in grid) != 9: return False
for row in grid:
if not check_dups(row): return False
return True
答案 4 :(得分：1)
我只发布这个，因为大多数其他解决方案虽然可能非常有效，但几乎无法读取。对于那些刚刚尝试学习的人，我相信下面的代码是有用且可读的。希望这有助于任何想要学习如何创建数独检查器的人。
def check_sudoku(grid):
for row in range(len(grid)):
for col in range(len(grid)):
# check value is an int
if grid[row][col] < 1 or type(grid[row][col]) is not type(1):
return False
# check value is within 1 through n.
# for example a 2x2 grid should not have the value 8 in it
elif grid[row][col] > len(grid):
return False
# check the rows
for row in grid:
if sorted(list(set(row))) != sorted(row):
return False
# check the cols
cols = []
for col in range(len(grid)):
for row in grid:
cols += [row[col]]
# set will get unique values, its converted to list so you can compare
# it's sorted so the comparison is done correctly.
if sorted(list(set(cols))) != sorted(cols):
return False
cols = []
# if you get past all the false checks return True
return True
答案 5 :(得分：0)
如果要检查重复行，而不是
rowoccurence = [0,0,0,0,0,0,0,0,0,0]
for j in range(9):
rowoccurence[grid[i][j]] += 1
temprow = rowoccurence[1:10]
if temprow == [1,1,1,1,1,1,1,1,1]:
return True
else:
return False
数：
b = True
for i in range(9):
grid[i].count(grid[i][j]) > 1:
b = False
return b
您的方法不仅仅是重复检查，还需要注意只有一位数字存在，否则会引发超出范围的异常
答案 6 :(得分：0)
如何用以下方法检查每一行/列：
sorted(row) == range(1,10)
或用于python 3
sorted(row) == list(range(1,10))
我认为运行时间主要是创建一个新列表（无论是直方图方法还是排序方法）所以log（n）的额外因子不应该是显而易见的
为了检查每一行，每列和子方，我建议使用提取器方法从矩阵中获取第n行，列和子方（即不要尝试将它们全部放在一个方法中）。
例如：
getSubSquare(m, i):
subrow = (i // 3) * 3
subcol = (i % 3) * 3
v = [0] * 9
for j in range(9):
subrj = j // 3
subcj = j % 3
v[j] = m[subrow + subrj][subcol + subcj]
return v
getRow(m,i):
return m[i]
getCol(m,i):
return [m[j][i] for j in range(9)]
答案 7 :(得分：0)
我认为您的代码崩溃的原因是因为您的缩进。你应该这样做：
for j in range(9):
rowoccurence[grid[i][j]] += 1
temprow = rowoccurence[1:10]
if temprow == [1,1,1,1,1,1,1,1,1]:
return True
else:
return False
而不是：
for j in range(9):
rowoccurence[grid[i][j]] += 1
temprow = rowoccurence[1:10]
if temprow == [1,1,1,1,1,1,1,1,1]:
return True
else:
return False
或使用Counter：
from collections import Counter
...
if numsinrow == 9:
for i in range(9):
count = Counter(grid[i])
return False if max(count.values()) > 1 else True
答案 8 :(得分：0)
def check_sudoku(grid):
if len(grid) == 9:
numsinrow = 0
for i in range(9):
if len(grid[i]) == 9:
numsinrow += 1
if numsinrow == 9:
if checkrow(grid):
if checkcol(grid):
return True
else:
return False
else:
return False
else:
return False
else:
return False
def checkrow(grid):
for i in range(9):
rowoccurence = [0,0,0,0,0,0,0,0,0,0]
for j in range(9):
rowoccurence[grid[i][j]] += 1
temprow = rowoccurence[1:10]
for q in range(9):
if temprow[q] == 1 or temprow[q] == 0:
continue
else:
return False
return True
def checkcol(grid):
for num in range(9):
coloccurence = [0,0,0,0,0,0,0,0,0,0]
for i in range(9):
coloccurence[grid[i][num]] += 1
tempcol = coloccurence[1:10]
for q in range(9):
if tempcol[q] == 1 or tempcol[q] == 0:
continue
else:
return False
return True
好的家伙我回来的功能是检查哪些行有效。非常感谢你们所有的广泛帮助。我有点像这样的菜鸟，所以我不明白一些答案，但我意识到我过早地回归真实。我也意识到如果连续存在多个零，那么rowoccurence / temp列表中就会出现一些数字。这就是我必须在rowoccurence / temp列表中检查1和0的原因。我还写了一个类似的功能来检查列。再次感谢！
答案 9 :(得分：0)
写了一个简单的类来模拟（完成的）数独板。没有什么棘手的，但对于9x9电路板来说是一个简单的解决方案。
class SudokuBoard(object):
DIGITS = set(range(1, 10))
def __init__(self, values):
self.values = values
def row(self, n):
return self.values[n]
def rows(self):
return self.values
def cols(self):
return [self.col(i) for i in xrange(9)]
def col(self, n):
return [self.values[i][n] for i in xrange(len(self.values))]
def groups(self):
return [self.group(i) for i in xrange(9)]
def group(self, n):
start_r = (n / 3) * 3
start_c = n * 3 % 9
values = []
for row in xrange(start_r, start_r + 3):
for col in xrange(start_c, start_c + 3):
values.append(self.values[row][col])
return values
def is_correct(self):
for row in self.rows():
if self.DIGITS - set(row):
return False
for col in self.cols():
if self.DIGITS - set(col):
return False
for group in self.groups():
if self.DIGITS - set(group):
return False
return True
答案 10 :(得分：0)
我在https://github.com/loghmanb/daily-coding-problem
上编写了代码from collections import defaultdict def solution_valid(board) #check row and column is distict no or not?! rows = defaultdict(int) columns = defaultdict(int) squares = defaultdict(int) for i in range(9): rows.clear() columns.clear() squares.clear() for j in range(9): if board[i][j] is not None: columns[board[i][j]] += 1 if columns[board[i][j]]>1: return False if board[j][i] is not None: rows[board[j][i]] += 1 if rows[board[j][i]]>1: return False new_j = (i*3 + j%3)%9 new_i = (i//3)*3 + j//3 if squares[board[new_i][new_j]] is not None: squares[board[new_i][new_j]] += 1 if squares[board[new_i][new_j]]>1: return False return True
答案 11 :(得分：0)
valid_solution= lambda board: not any([sorted(row)!=list(range(1,10)) for row in board]) and not any([sorted(list(col))!=list(range(1,10)) for col in zip(*board)]) and not any([sorted(board[i][j:j+3]+board[i+1][j:j+3]+board[i+2][j:j+3]) !=list(range(1,10)) for i in range(0,9,3) for j in range(0,9,3)])
答案 12 :(得分：0)
import numpy as np
def is_valid(row):
# checks whether a given set of values forms a valid row in sudoku
return len(list(filter(lambda val: type(val) == int and 0 < val < 10, set(row))) == 9
def check_sudoku(grid):
""" Check a sudoku board is correctly completed or not. """
# checks whether the grid has 9 rows
if len(grid) != 9:
return False
# checks whether the grid has 9 columns
for i in range(9):
if len(grid[i]) != 9:
return False
# turns grid from list to numpy array
grid = np.array(grid)
# checks whether the grid is filled with integers
if grid.dtype != np.int:
return False
for i in range(9):
# checks for repetition in rows
if not is_valid(grid[i, :]):
return False
# checks for repetition in columns
if not is_valid(grid[:, i]):
return False
# checks for repetition in squares
if not is_valid(grid[i//3*3:i//3*3+3, j%3*3:j%3*3+3]):
return False
# returns true if none of the conditions reached
return True