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使用python怎么实现一个数独游戏-创新互联
这期内容当中小编将会给大家带来有关使用python怎么实现一个数独游戏，文章内容丰富且以专业的角度为大家分析和叙述，阅读完这篇文章希望大家可以有所收获。
Python主要用来做什么
Python主要应用于：1、Web开发；2、数据科学研究；3、网络爬虫；4、嵌入式应用开发；5、游戏开发；6、桌面应用开发。
import random
import itertools
from copy import deepcopy
def make_board(m = 3):
numbers = list(range(1, m**2 + 1))
board = None
while board is None:
board = attempt_board(m, numbers)
return board
def attempt_board(m, numbers):
n = m**2
board = [[None for _ in range(n)] for _ in range(n)]
for i, j in itertools.product(range(n), repeat = 2):
i0, j0 = i - i % m, j - j % m
random.shuffle(numbers)
for x in numbers:
if(x not in board[i]) and all(row[j] != x for row in board) and all(x not in row[j0:j0+m] for row in board[i0:i]):
board[i][j] = x
break
else:
return None
return board
def print_board(board, m = 3):
numbers = list(range(1, m**2 + 1))
omit = 5
challange = deepcopy(board)
for i, j in itertools.product(range(omit), range(m ** 2)):
x = random.choice(numbers) - 1
challange[x][j] = None
spacer = "++---+---+---++---+---+---++---+---+---++"
print (spacer.replace('-', '='))
for i, line in enumerate(challange):
print("|| {0} | {1} | {2} || {3} | {4} | {5} || {6} | {7} | {8} ||".format(*(cell or ' ' for cell in line)))
if(i + 1) % 3 == 0:
print(spacer.replace('-', '='))
else:
print(spacer)
return challange
def print_answer(board):
spacer = "++---+---+---++---+---+---++---+---+---++"
print(spacer.replace('-','='))
for i, line in enumerate(board):
print("|| {0} | {1} | {2} || {3} | {4} | {5} || {6} | {7} | {8} ||".format(*(cell or ' ' for cell in line)))
if(i + 1) % 3 == 0:
print(spacer.replace('-','='))
else:
print(spacer)
def is_full(challange, m = 3):
for i, j in itertools.product(range(m**2), repeat = 2):
if challange[i][j] is None:
return False
return True
def cal_candidate(challange, x, y, m = 3):
candidate = range(1, m ** 2 + 1)
for i in range(m ** 2):
if challange[x][i] in candidate:
candidate.remove(challange[x][i])
if challange[i][y] in candidate:
candidate.remove(challange[i][y])
for i, j in itertools.product(range(m), repeat = 2):
x0, y0 = x - x % m, y - y % m
if challange[x0 + i][y0 + j] in candidate:
candidate.remove(challange[x0 + i][y0 + j])
return candidate
def least_candidate(challange, m = 3):
least, x, y = m ** 2, -1, -1
for i, j in itertools.product(range(m ** 2), repeat = 2):
if not challange[i][j]:
num = len(cal_candidate(challange, i, j))
if num < least:
least = num
x, y = i, j
return x, y
def solving_soduku(challange, m = 3):
if is_full(challange):
return challange
x, y = least_candidate(challange)
id = x * (m ** 2) + y
result = try_candidate(challange, id)
return result
def try_candidate(challange, id, m = 3):
if is_full(challange):
return challange
x = id / (m ** 2)
y = id % (m ** 2)
while challange[x][y]:
id = (id + 1) % m ** 4
x = id / (m ** 2)
y = id % (m ** 2)
candidate = cal_candidate(challange, x, y)
if len(candidate) == 0:
return False
for i in range(len(candidate)):
challange[x][y] = candidate[i]
result_r = try_candidate(challange, (id + 1) % m ** 4)
if not result_r:
pass
else:
return challange
challange[x][y] = None
return False
#Board = make_board()
#print Board
#challange = print_board(Board)
#print_answer(Board)
#result = solving_soduku(challange)
#print_answer(result)
testing = [[8, None, None, None, None, None, None, None, None],
[None, None, 3, 6, None, None, None, None, None],
[None, 7, None, None, 9, None, 2, None, None],
[None,5 , None, None, None, 7, None, None, None ],
[None, None, None, None, 4, 6, 7, None, None],
[None, None, None, 1, None, None, None, 3, None],
[None, None, 1, None, None, None, None, 6, 8],
[None, None, 8, 5, None, None, None, 1, None],
[None, 9, None, None, None, None, 4, None, None]]
result = solving_soduku(testing)
print_answer(result)上述就是小编为大家分享的使用python怎么实现一个数独游戏了，如果刚好有类似的疑惑，不妨参照上述分析进行理解。如果想知道更多相关知识，欢迎关注创新互联成都网站设计公司行业资讯频道。
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标题名称：使用python怎么实现一个数独游戏-创新互联
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