前言

利用Python來解數(shù)獨(dú)~~~
起因大概是：
自己解數(shù)獨(dú)實(shí)在是太費(fèi)勁了！！！

代碼效果展示

所需工具

python版本： 3.5.4

主要思路

思路很簡單：

將每個(gè)空格可能填入的數(shù)先列舉出來，然后就是深搜來解數(shù)獨(dú)。

數(shù)獨(dú)的規(guī)則為：

每個(gè)謎題都由一個(gè)在不同位置給與提示數(shù)字的9x9網(wǎng)格組成。游戲的目的是將空方格填上數(shù)字，使得每一行，每一列以及每一個(gè)3x3宮都沒有重復(fù)的數(shù)字出現(xiàn)。

代碼實(shí)現(xiàn)

# 點(diǎn)類class point():	def __init__(self, x, y):		self.x = x		self.y = y		self.available = []		self.value = 0# 該空格所在行有哪些數(shù)def rowNum(p ,sudoku):	# set用于去重，因?yàn)?不止一個(gè)！	row = set(sudoku[p.y*9: (p.y+1)*9])	row.remove(0)	return row# 該空格所在列有哪些數(shù)def colNum(p, sudoku):	col = []	length = len(sudoku)	for j in range(p.x, length, 9):		col.append(sudoku[j])	col = set(col)	col.remove(0)	return col# 該空格所在小九宮有哪些數(shù)def blockNum(p, sudoku):	block_x = p.x//3	block_y = p.y//3	block = []	start_point = block_y*3*9 + block_x*3	for j in range(start_point, start_point+3):		block.append(sudoku[j])	for j in range(start_point+9, start_point+9+3):		block.append(sudoku[j])	for j in range(start_point+9+9, start_point+9+9+3):		block.append(sudoku[j])	block = set(block)	block.remove(0)	return block# 初始化，作用為：# 把每個(gè)空格可能的點(diǎn)先列舉出來# 比如空格所在的行和列還有小九宮內(nèi)有數(shù)字1、2、3# 那么空格只能填入4、5、6、7、8、9中的某個(gè)數(shù)def initialize(sudoku):	sudokuList = []	length = len(sudoku)	for index in range(length):		# 找到需要填入的單元，即空格		if sudoku[index] == 0:			p = point(index%9, index//9)			for i in range(1, 10):				# 如果行、列、小九宮中均沒有i這個(gè)數(shù)				if (i not in rowNum(p, sudoku)) and (i not in colNum(p, sudoku)) and (i not in blockNum(p, sudoku)):					p.available.append(i)			sudokuList.append(p)	return sudokuList# 檢驗(yàn)該數(shù)填入空格后是否滿足數(shù)獨(dú)規(guī)則def check(p, sudoku):	if p.value == 0:		return False	if (p.value not in rowNum(p, sudoku)) and (p.value not in colNum(p, sudoku)) and (p.value not in blockNum(p, sudoku)):		return True	else:		return False# 展示數(shù)獨(dú)結(jié)果def showResult(sudoku):	for r in range(9):		for c in range(9):			print("%d " % (sudoku[r*9+c]), end="")		print("")# 深搜來解數(shù)獨(dú)def solve(p, sudoku):	available_Num = p.available	for ava in available_Num:		p.value = ava		if check(p, sudoku):			sudoku[p.y*9+p.x] = p.value			if len(sudokuList) < 1:				showResult(sudoku)				exit()			p_next = sudokuList.pop()			solve(p_next, sudoku)			sudoku[p_next.y*9+p_next.x] = 0			sudoku[p.y*9+p.x] = 0			p_next.value = 0			sudokuList.append(p_next)		else:			passif __name__ == "__main__":	# 0代表需要填入的單元，即空格	sudoku = [				0, 0, 0, 0, 0, 0, 0, 0, 0,				0, 9, 3, 6, 2, 8, 1, 4, 0,				0, 6, 0, 0, 0, 0, 0, 5, 0,  				0, 3, 0, 0, 1, 0, 0, 9, 0,  				0, 5, 0, 8, 0, 2, 0, 7, 0,  				0, 4, 0, 0, 7, 0, 0, 6, 0,  				0, 8, 0, 0, 0, 0, 0, 3, 0,  				0, 1, 7, 5, 9, 3, 4, 2, 0,  				0, 0, 0, 0, 0, 0, 0, 0, 0, 			]	sudokuList = initialize(sudoku)	print("數(shù)獨(dú)題目為：/n")	showResult(sudoku)	print("/n數(shù)獨(dú)的解為：/n")	p_first = sudokuList.pop()	solve(p_first, sudoku)