2022 | UACTF | Misc

Lighthouse

#modular equation

题目

Hey, could you help us? It's dark in here.

The flag is of the form UACTF{the minimum amount of turns required per dial for the check to be successful.}

There is a custom character encoding to convert the number of turns for each dial to the character representation.

0 -> A
1 -> B
2 -> C
3 -> D
4 -> E
5 -> F
6 -> G
7 -> H
8 -> I
9 -> J
10 -> K
11 -> L
12 -> M
13 -> N
14 -> O
15 -> P
16 -> Q
17 -> R
18 -> S
19 -> T
20 -> U
21 -> V
22 -> W
23 -> X
24 -> Y
25 -> Z
26 -> _
27 -> !
28 -> *

For example:

0 Turns to dial 1
13 Turns to Dial 2 etc...
UACTF{0 13 10 11 8 3 4 9}
UACTF{ANKLIDEJ}

lighthouse.zip

解题思路

• index.html 有 \(8\) 个转盘，点击任意转盘会改变所有转盘指向的数字，求满足要求每个转盘最少点击的次数
• setup.js 指出每个转盘为通过检查应指向的数字为 8, 11, 22, 4, 14, 26, 3, 21，ki.js 分别对应每个转盘点击后对自己及其它转盘的影响，其中 k2.js 给出了每个转盘初始指向的数字
• 类似 灯，等灯等灯 - Level 0，解一个模 \(29\) 的线性方程组

• 矩阵乘法的结果矩阵第 \(i\) 行第 \(j\) 列元素等于前一个矩阵第 \(i\) 行元素和后一矩阵第 \(j\) 列相应元素乘积的和，因而矩阵乘法的前后顺序很重要

  import numpy as np
  
  N, F = 8, Zmod(29)
  k = [
      [1, 6, 28, 40, 16, 42, 46, 37],
      [46, 1, 14, 25, 44, 17, 27, 27],
      [48, 30, 1, 47, 46, 27, 20, 25],
      [40, 11, 16, 1, 50, 12, 27, 26],
      [48, 49, 26, 16, 1, 6, 16, 2],
      [11, 9, 13, 3, 11, 1, 10, 35],
      [19, 34, 23, 10, 31, 27, 1, 32],
      [12, 10, 36, 6, 19, 24, 8, 1]
  ]
  init = [18, 12, 4, 8, 17, 2, 15, 8]
  target = [8, 11, 22, 4, 14, 26, 3, 21]
  
  L = Matrix(F, N, N)
  for i in range(N):
      for j in range(N):
          L[i, j] = k[i][j]
  d = [chr(ord('A') + i) for i in range(26)]
  d.extend(['_', '!', '*'])
  # use A.solve_left(Y) to solve for X in XA = Y
  for i in L.solve_left(vector(F, np.subtract(target, init))):
      print(d[i], end='')
  

• 也可以使用 Modular Arithmetic Solver - Congruence Calculator - Online 来解 =ω=

Flag

UACTF{Y_Z*MB!E}

参考资料

• Base class for matrices, part 2 — Matrices and Spaces of Matrices

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最后更新: 2022年8月11日 21:28:00

Contributors: YanhuiJessica
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