## Onlude### Challenge> Encryption and Decryption could be really easy, while we expected the decryption to be harder!
```python#!/usr/bin/env sage
from sage.all import *from flag import flag
global p, alphabetp = 71alphabet = '=0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ$!?_{}<>'
flag = flag.lstrip('CCTF{').rstrip('}')assert len(flag) == 24
def cross(m):    return alphabet.index(m)
def prepare(msg):    A = zero_matrix(GF(p), 11, 11)    for k in range(len(msg)):        i, j = 5*k // 11, 5*k % 11        A[i, j] = cross(msg[k])    return A
def keygen():    R = random_matrix(GF(p), 11, 11)    while True:        S = random_matrix(GF(p), 11, 11)        if S.rank() == 11:            _, L, U = S.LU()            return R, L, U
def encrypt(A, key):    R, L, U = key    S = L * U    X = A + R    Y = S * X    E = L.inverse() * Y    return E
A = prepare(flag)key = keygen()R, L, U = keyS = L * UE = encrypt(A, key)print(f'E = \n{E}')print(f'L * U * L = \n{L * U * L}')print(f'L^(-1) * S^2 * L = \n{L.inverse() * S**2 * L}')print(f'R^(-1) * S^8 = \n{R.inverse() * S**8}')```
```pythonE = [25 55 61 28 11 46 19 50 37  5 21][20 57 39  9 25 37 63 31 70 15 47][56 31  1  1 50 67 38 14 42 46 14][42 54 38 22 19 55  7 18 45 53 39][55 26 42 15 48  6 24  4 17 60 64][ 1 38 50 10 19 57 26 48  6  4 14][13  4 38 54 23 34 54 42 15 56 29][26 66  8 48  6 70 44  8 67 68 65][56 67 49 61 18 34 53 21  7 48 32][15 70 10 34  1 57 70 27 12 33 46][25 29 20 21 30 55 63 49 11 36  7]L * U * L = [50  8 21 16 13 33  2 12 35 20 14][36 55 36 34 27 28 23 21 62 17  8][56 26 49 39 43 30 35 46  0 58 43][11 25 25 35 29  0 22 38 53 51 58][34 14 69 68  5 32 27  4 27 62 15][46 49 36 42 26 12 28 60 54 66 23][69 55 30 65 56 13 14 36 26 46 48][25 48 16 20 34 57 64 62 61 25 62][68 39 11 40 25 11  7 40 24 43 65][54 20 40 59 52 60 37 14 32 44  4][45 20  7 26 45 45 50 17 41 59 50]L^(-1) * S^2 * L = [34 12 70 21 36  2  2 43  7 14  2][ 1 54 59 12 64 35  9  7 49 11 49][69 14 10 19 16 27 11  9 26 10 45][70 17 41 13 35 58 19 29 70  5 30][68 69 67 37 63 69 15 64 66 28 26][18 29 64 38 63 67 15 27 64  6 26][ 0 12 40 41 48 30 46 52 39 48 58][22  3 28 35 55 30 15 17 22 49 55][50 55 55 61 45 23 24 32 10 59 69][27 21 68 56 67 49 64 53 42 46 14][42 66 16 29 42 42 23 49 43  3 23]R^(-1) * S^8 = [51  9 22 61 63 14  2  4 18 18 23][33 53 31 31 62 21 66  7 66 68  7][59 19 32 21 13 34 16 43 49 25  7][44 37  4 29 70 50 46 39 55  4 65][29 63 29 43 47 28 40 33  0 62  8][45 62 36 68 10 66 26 48 10  6 61][43 30 25 18 23 38 61  0 52 46 35][ 3 40  6 45 20 55 35 67 25 14 63][15 30 61 66 25 33 14 20 60 50 50][29 15 53 22 55 64 69 56 44 40  8][28 40 69 60 28 41  9 14 29  4 29]```
### Solution
In this challenge, we are given the source code of an encryption scheme that uses matrix operations for encryption and decryption, the corresponding encrypted flag and some hints.
For the encryption part:
$$ E = L^{-1}Y = L^{-1}SX = L^{-1}LU(A+R) = U(A+R) $$
The three hints we have is $LUL$, $L^{-1}S^2L$ and $R^{-1}S^8$
Since $S=LU$, we can rewrite the second hint as $ULUL$, then we can get $U$ by dividing hint1.For the third hint $R^{-1}S^8$, we can rewrite it as $R^{-1}(LULU)^4=R^{-1}(h_1U)^4$, then we can get $R$.
With $U$ and $R$ known, we can then recover the flag $A$ from $E$.
```pythonfrom sage.all import *
E = Matrix(GF(71),[[25,55,61,28,11,46,19,50,37,5,21],[20,57,39,9,25,37,63,31,70,15,47],[56,31,1,1,50,67,38,14,42,46,14],[42,54,38,22,19,55,7,18,45,53,39],[55,26,42,15,48,6,24,4,17,60,64],[1,38,50,10,19,57,26,48,6,4,14],[13,4,38,54,23,34,54,42,15,56,29],[26,66,8,48,6,70,44,8,67,68,65],[56,67,49,61,18,34,53,21,7,48,32],[15,70,10,34,1,57,70,27,12,33,46],[25,29,20,21,30,55,63,49,11,36,7]])
hint1 = Matrix(GF(71),[[50,8,21,16,13,33,2,12,35,20,14],[36,55,36,34,27,28,23,21,62,17,8],[56,26,49,39,43,30,35,46,0,58,43],[11,25,25,35,29,0,22,38,53,51,58],[34,14,69,68,5,32,27,4,27,62,15],[46,49,36,42,26,12,28,60,54,66,23],[69,55,30,65,56,13,14,36,26,46,48],[25,48,16,20,34,57,64,62,61,25,62],[68,39,11,40,25,11,7,40,24,43,65],[54,20,40,59,52,60,37,14,32,44,4],[45,20,7,26,45,45,50,17,41,59,50]])
hint2=Matrix(GF(71),[[34,12,70,21,36,2,2,43,7,14,2],[1,54,59,12,64,35,9,7,49,11,49],[69,14,10,19,16,27,11,9,26,10,45],[70,17,41,13,35,58,19,29,70,5,30],[68,69,67,37,63,69,15,64,66,28,26],[18,29,64,38,63,67,15,27,64,6,26],[0,12,40,41,48,30,46,52,39,48,58],[22,3,28,35,55,30,15,17,22,49,55],[50,55,55,61,45,23,24,32,10,59,69],[27,21,68,56,67,49,64,53,42,46,14],[42,66,16,29,42,42,23,49,43,3,23]])
hint3=Matrix(GF(71),[[51,9,22,61,63,14,2,4,18,18,23],[33,53,31,31,62,21,66,7,66,68,7],[59,19,32,21,13,34,16,43,49,25,7],[44,37,4,29,70,50,46,39,55,4,65],[29,63,29,43,47,28,40,33,0,62,8],[45,62,36,68,10,66,26,48,10,6,61],[43,30,25,18,23,38,61,0,52,46,35],[3,40,6,45,20,55,35,67,25,14,63],[15,30,61,66,25,33,14,20,60,50,50],[29,15,53,22,55,64,69,56,44,40,8],[28,40,69,60,28,41,9,14,29,4,29]])
U = hint2/hint1R = (hint3/U/hint1/U/hint1/U/hint1/U/hint1).inverse()A = U.inverse()*E-Ralphabet = '=0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ$!?_{}<>'flag = ''for k in range(24):    i, j = 5*k // 11, 5*k % 11    flag+=alphabet[A[i, j]]print('CCTF{'+flag+'}')```
##### Flag
`CCTF{LU__D3c0mpO517Ion__4L90?}`