python数独伪代码回溯法_数独#回溯算法#CTF

1. intro&＃xff1a;巅峰极客的一道逆向

刷巅峰极客2020里的rev题fu!kpy&＃xff0c;复杂得不行但是看到if d[1][0] !&＃61; &＃39;8&＃39; or d[1][7] !&＃61; &＃39;2&＃39;和if check(h1) !&＃61; 45 or check(h2) !&＃61; 45 or check(h9) !&＃61; 45这种内容&＃xff0c;结合flag相关的部分可以确认是flag{9*9}的数独。

# uncompyle6 version 2.11.5

# Python bytecode 2.7 (62211)

# Decompiled from: Python 2.7.16 (default, Apr 6 2019, 01:42:57)

# [GCC 8.3.0]

# Embedded file name: test233_ol.py

# Compiled at: 2020-03-20 13:22:50

(lambda __g, __print: [ [ (lambda __after: if __name__ &＃61;&＃61; &＃39;__main__&＃39;:

[ (lambda __after: if len(input) !&＃61; 87:

(__print(&＃39;Error len!&＃39;), (exit(), __after())[1])[1]__after())(lambda : [ [ [ [ (lambda __after: if fmt1 !&＃61; &＃39;flag{&＃39; or fmt2 !&＃61; &＃39;}&＃39;:

(__print(&＃39;Error fmt!&＃39;), (exit(0), __after())[1])[1]__after())(lambda : (

d.append(context[0:9]), (d.append(context[9:18]), (d.append(context[18:27]), (d.append(context[27:36]), (d.append(context[36:45]), (d.append(context[45:54]), (d.append(context[54:63]), (d.append(context[63:72]), (d.append(context[72:81]), [ [ [ [ [ [ [ [ [ [ [ [ [ [ [ [ [ [ [ [ [ [ [ [ [ [ [ (lambda __after: if d[0][2] !&＃61; &＃39;5&＃39; or d[0][3] !&＃61; &＃39;3&＃39;:

(__print(&＃39;Error!&＃39;), (exit(), __after())[1])[1]__after())(lambda : (lambda __after: if d[1][0] !&＃61; &＃39;8&＃39; or d[1][7] !&＃61; &＃39;2&＃39;:

(__print(&＃39;Error!&＃39;), (exit(), __after())[1])[1]__after())(lambda : (lambda __after: if d[2][1] !&＃61; &＃39;7&＃39; or d[2][4] !&＃61; &＃39;1&＃39; or d[2][6] !&＃61; &＃39;5&＃39;:

(__print(&＃39;Error!&＃39;), (exit(), __after())[1])[1]__after())(lambda : (lambda __after: if d[3][0] !&＃61; &＃39;4&＃39; or d[3][5] !&＃61; &＃39;5&＃39; or d[3][6] !&＃61; &＃39;3&＃39;:

(__print(&＃39;Error!&＃39;), (exit(), __after())[1])[1]__after())(lambda : (lambda __after: if d[4][1] !&＃61; &＃39;1&＃39; or d[4][4] !&＃61; &＃39;7&＃39; or d[4][8] !&＃61; &＃39;6&＃39;:

(__print(&＃39;Error!&＃39;), (exit(), __after())[1])[1]__after())(lambda : (lambda __after: if d[5][2] !&＃61; &＃39;3&＃39; or d[5][3] !&＃61; &＃39;2&＃39; or d[5][7] !&＃61; &＃39;8&＃39;:

(__print(&＃39;Error!&＃39;), (exit(), __after())[1])[1]__after())(lambda : (lambda __after: if d[6][1] !&＃61; &＃39;6&＃39; or d[6][3] !&＃61; &＃39;5&＃39; or d[6][8] !&＃61; &＃39;9&＃39;:

(__print(&＃39;Error!&＃39;), (exit(), __after())[1])[1]__after())(lambda : (lambda __after: if d[7][2] !&＃61; &＃39;4&＃39; or d[7][7] !&＃61; &＃39;3&＃39;:

(__print(&＃39;Error!&＃39;), (exit(), __after())[1])[1]__after())(lambda : (lambda __after: if d[8][5] !&＃61; &＃39;9&＃39; or d[8][6] !&＃61; &＃39;7&＃39;:

(__print(&＃39;Error!&＃39;), (exit(), __after())[1])[1]__after())(lambda : (lambda __after: if check(h1) !&＃61; 45 or check(h2) !&＃61; 45 or check(h3) !&＃61; 45 or check(h4) !&＃61; 45 or check(h5) !&＃61; 45 or check(h6) !&＃61; 45 or check(h7) !&＃61; 45 or check(h8) !&＃61; 45 or check(h9) !&＃61; 45:

(__print(&＃39;Error!&＃39;), (exit(), __after())[1])[1]__after())(lambda : (lambda __after: if check(l1) !&＃61; 45 or check(l2) !&＃61; 45 or check(l3) !&＃61; 45 or check(l4) !&＃61; 45 or check(l5) !&＃61; 45 or check(l6) !&＃61; 45 or check(l7) !&＃61; 45 or check(l8) !&＃61; 45 or check(l9) !&＃61; 45:

(__print(&＃39;Error!&＃39;), (exit(), __after())[1])[1]__after())(lambda : (lambda __after: if check(k1) !&＃61; 45 or check(k2) !&＃61; 45 or check(k3) !&＃61; 45 or check(k4) !&＃61; 45 or check(k5) !&＃61; 45 or check(k6) !&＃61; 45 or check(k7) !&＃61; 45 or check(k8) !&＃61; 45 or check(k9) !&＃61; 45:

(__print(&＃39;Error!&＃39;), (exit(), __after())[1])[1]__after())(lambda : (lambda __after: if check1(h1) !&＃61; 1 or check1(h2) !&＃61; 1 or check1(h3) !&＃61; 1 or check1(h4) !&＃61; 1 or check1(h5) !&＃61; 1 or check1(h6) !&＃61; 1 or check1(h7) !&＃61; 1 or check1(h8) !&＃61; 1 or check1(h9) !&＃61; 1:

(__print(&＃39;Error!&＃39;), (exit(), __after())[1])[1]__after())(lambda : (lambda __after: if check1(l1) !&＃61; 1 or check1(l2) !&＃61; 1 or check1(l3) !&＃61; 1 or check1(l4) !&＃61; 1 or check1(l5) !&＃61; 1 or check1(l6) !&＃61; 1 or check1(l7) !&＃61; 1 or check1(l8) !&＃61; 1 or check1(l9) !&＃61; 1:

(__print(&＃39;Error!&＃39;), (exit(), __after())[1])[1]__after())(lambda : (lambda __after: if check1(k1) !&＃61; 1 or check1(k2) !&＃61; 1 or check1(k3) !&＃61; 1 or check1(k4) !&＃61; 1 or check1(k5) !&＃61; 1 or check1(k6) !&＃61; 1 or check1(k7) !&＃61; 1 or check1(k8) !&＃61; 1 or check1(k9) !&＃61; 1:

(__print(&＃39;Error!&＃39;), (exit(), __after())[1])[1]__after())(lambda : (

__print(&＃39;Yes! You got it!&＃39;), __after())[1]))))))))))))))) for __g[&＃39;k9&＃39;] in [context[60] &＃43; context[61] &＃43; context[62] &＃43; context[69] &＃43; context[70] &＃43; context[71] &＃43; context[78] &＃43; context[79] &＃43; context[80]] ][0] for __g[&＃39;k8&＃39;] in [context[57] &＃43; context[58] &＃43; context[59] &＃43; context[66] &＃43; context[67] &＃43; context[68] &＃43; context[75] &＃43; context[76] &＃43; context[77]] ][0] for __g[&＃39;k7&＃39;] in [context[54] &＃43; context[55] &＃43; context[56] &＃43; context[63] &＃43; context[64] &＃43; context[65] &＃43; context[72] &＃43; context[73] &＃43; context[74]] ][0] for __g[&＃39;k6&＃39;] in [context[33] &＃43; context[34] &＃43; context[35] &＃43; context[42] &＃43; context[43] &＃43; context[44] &＃43; context[51] &＃43; context[52] &＃43; context[53]] ][0] for __g[&＃39;k5&＃39;] in [context[30] &＃43; context[31] &＃43; context[32] &＃43; context[39] &＃43; context[40] &＃43; context[41] &＃43; context[48] &＃43; context[49] &＃43; context[50]] ][0] for __g[&＃39;k4&＃39;] in [context[27] &＃43; context[28] &＃43; context[29] &＃43; context[36] &＃43; context[37] &＃43; context[38] &＃43; context[45] &＃43; context[46] &＃43; context[47]] ][0] for __g[&＃39;k3&＃39;] in [context[6] &＃43; context[7] &＃43; context[8] &＃43; context[15] &＃43; context[16] &＃43; context[17] &＃43; context[24] &＃43; context[25] &＃43; context[26]] ][0] for __g[&＃39;k2&＃39;] in [context[3] &＃43; context[4] &＃43; context[5] &＃43; context[12] &＃43; context[13] &＃43; context[14] &＃43; context[21] &＃43; context[22] &＃43; context[23]] ][0] for __g[&＃39;k1&＃39;] in [context[0] &＃43; context[1] &＃43; context[2] &＃43; context[9] &＃43; context[10] &＃43; context[11] &＃43; context[18] &＃43; context[19] &＃43; context[20]] ][0] for __g[&＃39;l9&＃39;] in [context[8] &＃43; context[17] &＃43; context[26] &＃43; context[35] &＃43; context[44] &＃43; context[53] &＃43; context[62] &＃43; context[71] &＃43; context[80]] ][0] for __g[&＃39;l8&＃39;] in [context[7] &＃43; context[16] &＃43; context[25] &＃43; context[34] &＃43; context[43] &＃43; context[52] &＃43; context[61] &＃43; context[70] &＃43; context[79]] ][0] for __g[&＃39;l7&＃39;] in [context[6] &＃43; context[15] &＃43; context[24] &＃43; context[33] &＃43; context[42] &＃43; context[51] &＃43; context[60] &＃43; context[69] &＃43; context[78]] ][0] for __g[&＃39;l6&＃39;] in [context[5] &＃43; context[14] &＃43; context[23] &＃43; context[32] &＃43; context[41] &＃43; context[50] &＃43; context[59] &＃43; context[68] &＃43; context[77]] ][0] for __g[&＃39;l5&＃39;] in [context[4] &＃43; context[13] &＃43; context[22] &＃43; context[31] &＃43; context[40] &＃43; context[49] &＃43; context[58] &＃43; context[67] &＃43; context[76]] ][0] for __g[&＃39;l4&＃39;] in [context[3] &＃43; context[12] &＃43; context[21] &＃43; context[30] &＃43; context[39] &＃43; context[48] &＃43; context[57] &＃43; context[66] &＃43; context[75]] ][0] for __g[&＃39;l3&＃39;] in [context[2] &＃43; context[11] &＃43; context[20] &＃43; context[29] &＃43; context[38] &＃43; context[47] &＃43; context[56] &＃43; context[65] &＃43; context[74]] ][0] for __g[&＃39;l2&＃39;] in [context[1] &＃43; context[10] &＃43; context[19] &＃43; context[28] &＃43; context[37] &＃43; context[46] &＃43; context[55] &＃43; context[64] &＃43; context[73]] ][0] for __g[&＃39;l1&＃39;] in [context[0] &＃43; context[9] &＃43; context[18] &＃43; context[27] &＃43; context[36] &＃43; context[45] &＃43; context[54] &＃43; context[63] &＃43; context[72]] ][0] for __g[&＃39;h9&＃39;] in [context[72:81]] ][0] for __g[&＃39;h8&＃39;] in [context[63:72]] ][0] for __g[&＃39;h7&＃39;] in [context[54:63]] ][0] for __g[&＃39;h6&＃39;] in [context[45:54]] ][0] for __g[&＃39;h5&＃39;] in [context[36:45]] ][0] for __g[&＃39;h4&＃39;] in [context[27:36]] ][0] for __g[&＃39;h3&＃39;] in [context[18:27]] ][0] for __g[&＃39;h2&＃39;] in [context[9:18]] ][0] for __g[&＃39;h1&＃39;] in [context[0:9]] ][0])[1])[1])[1])[1])[1])[1])[1])[1])[1])

for __g[&＃39;d&＃39;] in [[]] ][0]

for __g[&＃39;context&＃39;] in [input[5:-1]] ][0]

for __g[&＃39;fmt2&＃39;] in [input[-1]] ][0]

for __g[&＃39;fmt1&＃39;] in [input[0:5]] ][0])

for __g[&＃39;input&＃39;] in [raw_input(&＃39;Input your flag:&＃39;)] ][0]__after())(lambda : None) for __g[&＃39;check1&＃39;], check1.__name__ in [(lambda arg: (lambda __l: [ (lambda __after: if len(list(set(__l[&＃39;arg&＃39;]))) !&＃61; 9:

01)(lambda : None) for __l[&＃39;arg&＃39;] in [arg] ][0])({}), &＃39;check1&＃39;)] ][0] for __g[&＃39;check&＃39;], check.__name__ in [(lambda arg: (lambda __l: [ sum(map(int, __l[&＃39;arg&＃39;])) for __l[&＃39;arg&＃39;] in [arg] ][0])({}), &＃39;check&＃39;)] ][0])(globals(), __import__(&＃39;__builtin__&＃39;, level&＃61;0).__dict__[&＃39;print&＃39;])

熟练的用numpy画出格子&＃xff0c;然后发现这道题太难了我不会做。这个时候需要搜一下……

[[0. 0. 5. 3. 0. 0. 0. 0. 0.]

[8. 0. 0. 0. 0. 0. 0. 2. 0.]

[0. 7. 0. 0. 1. 0. 5. 0. 0.]

[4. 0. 0. 0. 0. 5. 3. 0. 0.]

[0. 1. 0. 0. 7. 0. 0. 0. 6.]

[0. 0. 3. 2. 0. 0. 0. 8. 0.]

[0. 6. 0. 5. 0. 0. 0. 0. 9.]

[0. 0. 4. 0. 0. 0. 0. 3. 0.]

[0. 0. 0. 0. 0. 9. 7. 0. 0.]]

2. 知识点引入&＃xff1a;backtracking

回溯算法(backtracking)是暴力搜索算法的一种。正是由于回溯算法具有“强大的”暴力搜索的能力&＃xff0c;它被应用于一些游戏问题&＃xff0c;例如&＃xff1a;N 皇后、解数独、祖玛游戏、24 点游戏、走迷宫、生成迷宫。许多复杂的、规模较大的问题都可以使用回溯搜索算法得到所有可行解&＃xff0c;进而得到最优解&＃xff0c;因此回溯算法有“通用解题方法”的美称&＃xff0c;回溯算法也是经典的人工智能的基础算法。

由于回溯算法的时间复杂度很高&＃xff0c;因此&＃xff0c;在遍历的时候&＃xff0c;如果能够提前知道这一条分支不能搜索到满意的结果&＃xff0c;这一分支就可以跳过&＃xff0c;这一步操作就是在一棵树上剪去一个枝叶&＃xff0c;被人们很形象地称之为剪枝。

结论&＃xff0c;backtracking可以较好地解数独题。(喂)

2.1. 算法入门-全排列

「力扣」上第 46 号问题&＃xff1a;“全排列”。

题目描述&＃xff1a;给定一个没有重复数字的序列&＃xff0c;返回其所有可能的全排列。

在一个树形问题上做一次深度优先遍历。为了理解回溯的过程画个流程图如下。

graph TB;

0--1:depth&＃43;&＃61;1,state&＃61;0b001-->A(1)

A--2:depth&＃43;&＃61;1,state&＃61;0b011-->B("[1,2]")

B--3:depth&＃43;&＃61;1,state&＃61;0b111-->C("[1,2,3]")

C--4:depth-&＃61;1,state&＃61;0b011-->B

B--5:-depth-&＃61;1,state&＃61;0b001-->A

A--6:depth&＃43;&＃61;1,state&＃61;0b101-->D("[1,3]")

D--7:depth&＃43;&＃61;1,state&＃61;0b111-->E("[1,3,2]")

E--8:depth-&＃61;1,state&＃61;0b101-->D

D--9:depth-&＃61;1,state&＃61;0b001-->A

0(start)

image.png

抄的py3代码如下()。

def permute(nums):

def dfs(nums, size, depth, path, state, res):

# depth满足时递归终止

if depth &＃61;&＃61; size:

res.append(path)

return

for i in range(size):

#print(i,bin(state),path,res)

#state判断第i位的元素是否已经使用过&＃xff0c;位运算优化

if ((state >> i) & 1) &＃61;&＃61; 0:

dfs(nums, size, depth &＃43; 1, path &＃43; [nums[i]], state ^ (1 <

size &＃61; len(nums)

if size &＃61;&＃61; 0:

return []

res &＃61; []

dfs(nums, size, 0, [], 0, res)

return res

permute([1,2,3])

2.2. 解数独

乐扣37题&＃xff0c;解数独。空白格用 &＃39;.&＃39; 表示。

数独的规则&＃xff1a;

- 数字 1-9 在每一行只能出现一次。

- 数字 1-9 在每一列只能出现一次。

- 数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。

和原py题一样按照「行优先」的顺序依次枚举每一个空白格中填的数字&＃xff0c;通过递归 &＃43; 回溯的方法枚举所有可能的填法。

初始化state[row,line,palace]&＃xff0c;表明哪些数字已经被使用过了。

尝试去填充数组。如果行、列、 或 3*3 的方格内出现已经被使用过的数字则不填充&＃xff0c;否则尝试填充。

填充失败则回溯。

深度满足则递归结束。

为了理解流程&＃xff0c;先来个3*3的数独。没有小框框

[[1. 2. 0.]

[2. 0. 1.]

[0. 0. 0.]]

流程图如下。

graph TB;

0--"1:depth&＃43;&＃61;1,state&＃61;[111,011,000,011,010,101]"-->A("[1,2,3],[2,0,1],[0,0,0]")

A--"2:depth&＃43;&＃61;1,state&＃61;[111,111,000,011,110,101]"-->B("[1,2,3],[2,3,1],[0,0,0]")

B--"3:depth&＃43;&＃61;1,state&＃61;[111,111,100,111,110,101]"-->C("[1,2,3],[2,3,1],[3,0,0]")

C--"4:depth&＃43;&＃61;1,state&＃61;[111,111,101,111,111,101]"-->D("[1,2,3],[2,3,1],[3,1,0]")

D--"5:depth&＃43;&＃61;1,state&＃61;[111,111,111,111,111,111]"-->e("[1,2,3],[2,3,1],[3,1,2]")

0("start,depth&＃61;4,state&＃61;[011,011,000,011,010,001]")

image.png

没时间了抄了份别人的哈希表解法&＃xff0c;好清晰啊。以后有空再优化成位运算0.0

def solveSudoku(board):

nums &＃61; {"1", "2", "3", "4", "5", "6", "7", "8", "9"}

row &＃61; [set() for _ in range(9)]

col &＃61; [set() for _ in range(9)]

palace &＃61; [[set() for _ in range(3)] for _ in range(3)] # 3*3

blank &＃61; [] # 存储待填格的位置

# 初始化&＃xff0c;按照行、列、宫 分别存入哈希表

for i in range(9):

for j in range(9):

ch &＃61; board[i][j]

if ch &＃61;&＃61; &＃39;0&＃39;:

#if ch &＃61;&＃61; ".":

blank.append((i, j))

else:

row[i].add(ch)

col[j].add(ch)

palace[i // 3][j // 3].add(ch)

def dfs(n):

if n &＃61;&＃61; len(blank):

return True

i, j &＃61; blank[n]

rst &＃61; nums - row[i] - col[j] - palace[i // 3][j // 3] # 剩余的数字

if not rst:

return False

for num in rst:

board[i][j] &＃61; num

row[i].add(num)

col[j].add(num)

palace[i // 3][j // 3].add(num)

if dfs(n &＃43; 1):

return True

row[i].remove(num)

col[j].remove(num)

palace[i // 3][j // 3].remove(num)

dfs(0)

board&＃61;[[&＃39;0&＃39;,&＃39;0&＃39;,&＃39;5&＃39;,&＃39;3&＃39;,&＃39;0&＃39;,&＃39;0&＃39;,&＃39;0&＃39;,&＃39;0&＃39;,&＃39;0&＃39;]

,[&＃39;8&＃39;,&＃39;0&＃39;,&＃39;0&＃39;,&＃39;0&＃39;,&＃39;0&＃39;,&＃39;0&＃39;,&＃39;0&＃39;,&＃39;2&＃39;,&＃39;0&＃39;]

,[&＃39;0&＃39;,&＃39;7&＃39;,&＃39;0&＃39;,&＃39;0&＃39;,&＃39;1&＃39;,&＃39;0&＃39;,&＃39;5&＃39;,&＃39;0&＃39;,&＃39;0&＃39;]

,[&＃39;4&＃39;,&＃39;0&＃39;,&＃39;0&＃39;,&＃39;0&＃39;,&＃39;0&＃39;,&＃39;5&＃39;,&＃39;3&＃39;,&＃39;0&＃39;,&＃39;0&＃39;]

,[&＃39;0&＃39;,&＃39;1&＃39;,&＃39;0&＃39;,&＃39;0&＃39;,&＃39;7&＃39;,&＃39;0&＃39;,&＃39;0&＃39;,&＃39;0&＃39;,&＃39;6&＃39;]

,[&＃39;0&＃39;,&＃39;0&＃39;,&＃39;3&＃39;,&＃39;2&＃39;,&＃39;0&＃39;,&＃39;0&＃39;,&＃39;0&＃39;,&＃39;8&＃39;,&＃39;0&＃39;]

,[&＃39;0&＃39;,&＃39;6&＃39;,&＃39;0&＃39;,&＃39;5&＃39;,&＃39;0&＃39;,&＃39;0&＃39;,&＃39;0&＃39;,&＃39;0&＃39;,&＃39;9&＃39;]

,[&＃39;0&＃39;,&＃39;0&＃39;,&＃39;4&＃39;,&＃39;0&＃39;,&＃39;0&＃39;,&＃39;0&＃39;,&＃39;0&＃39;,&＃39;3&＃39;,&＃39;0&＃39;]

,[&＃39;0&＃39;,&＃39;0&＃39;,&＃39;0&＃39;,&＃39;0&＃39;,&＃39;0&＃39;,&＃39;9&＃39;,&＃39;7&＃39;,&＃39;0&＃39;,&＃39;0&＃39;]]

solveSudoku(board)

from pprint import pprint

pprint(board)

2.3. 来到ctf数独题

其实已经出来了。。只要补一下生成

def solveSudoku(board):

nums &＃61; {"1", "2", "3", "4", "5", "6", "7", "8", "9"} #用set保证唯一

row &＃61; [set() for _ in range(9)]

col &＃61; [set() for _ in range(9)]

palace &＃61; [[set() for _ in range(3)] for _ in range(3)] # 3*3

blank &＃61; [] # 存储待填格的位置

# 初始化&＃xff0c;按照行、列、宫 分别存入哈希表

for i in range(9):

for j in range(9):

ch &＃61; board[i][j]

if ch &＃61;&＃61; &＃39;0&＃39;:

blank.append((i, j))

else:

row[i].add(ch)

col[j].add(ch)

palace[i // 3][j // 3].add(ch)

def dfs(n):

if n &＃61;&＃61; len(blank): # 填满

return True

i, j &＃61; blank[n]

rst &＃61; nums - row[i] - col[j] - palace[i // 3][j // 3] # 剩余的数字

if not rst: # 无解的情况

return False

for num in rst:

board[i][j] &＃61; num

row[i].add(num)

col[j].add(num)

palace[i // 3][j // 3].add(num)

if dfs(n &＃43; 1):

return True

row[i].remove(num)

col[j].remove(num)

palace[i // 3][j // 3].remove(num)

dfs(0)

import numpy as np

d &＃61; np.zeros(81).reshape(9, 9)

d[0][2] &＃61; &＃39;5&＃39;

d[0][3] &＃61; &＃39;3&＃39;

d[1][0] &＃61; &＃39;8&＃39;

d[1][7] &＃61; &＃39;2&＃39;

d[2][1] &＃61; &＃39;7&＃39;

d[2][4] &＃61; &＃39;1&＃39;

d[2][6] &＃61; &＃39;5&＃39;

d[3][0] &＃61; &＃39;4&＃39;

d[3][5] &＃61; &＃39;5&＃39;

d[3][6] &＃61; &＃39;3&＃39;

d[4][1] &＃61; &＃39;1&＃39;

d[4][4] &＃61; &＃39;7&＃39;

d[4][8] &＃61; &＃39;6&＃39;

d[5][2] &＃61; &＃39;3&＃39;

d[5][3] &＃61; &＃39;2&＃39;

d[5][7] &＃61; &＃39;8&＃39;

d[6][1] &＃61; &＃39;6&＃39;

d[6][3] &＃61; &＃39;5&＃39;

d[6][8] &＃61; &＃39;9&＃39;

d[7][2] &＃61; &＃39;4&＃39;

d[7][7] &＃61; &＃39;3&＃39;

d[8][5] &＃61; &＃39;9&＃39;

d[8][6] &＃61; &＃39;7&＃39;

board &＃61; d

board&＃61; np.asarray(board,dtype&＃61;np.int16)

board &＃61; np.asarray(board,dtype&＃61;np.unicode_)

solveSudoku(board)

for i in board:

print(&＃39;&＃39;.join(i),end&＃61;&＃39;&＃39;)

flag{145327698839654127672918543496185372218473956753296481367542819984761235521839764}

3. 总结

机器学习赶紧接着学啊&＃xff01;

继续学A* 算法解迷宫题啊&＃xff01;手动解迷宫你也好意思&＃xff1f;()

现在安全竞赛是不是越来越往算法大赛转了。。。落后专业人才转型迫在眉睫。