Rating:
# Grille_me - Crypto/net (150)
How well do you know classic ciphers? Some coding may be required...
PEE151R4HEL.LCC8.2TFECE.EOT.901OTHN.ANT23P2RRAG.SNO0.O0TULE.
## Part One
The encoding given above is only the first part of the challenge. After staring at it for a while, I started to notice that the numbers given were the same as the IP address used in previous challenges along with `1204`, the month-day convention for ports previously used. With that it wasn't too hard to see that the characters could just be arranged in a grid and the first message read off the columns:
```
PEE151R4HEL.
LCC8.2TFECE.
EOT.901OTHN.
ANT23P2RRAG.
SNO0.O0TULE.
pleaseconnectto18.205.93.120port1204forthetruechallenge
```
## Part Two
The second part of the challenge comes when you connect to the service above. All you're given is a jumble of characters followed by a prompt. If you enter something wrong, the only response is then 'Invalid password'. After requesting several times, it becomes clear that some of the characters stay constant. A selection of different jumbles is shown below.
```
YPWI2ZORASODSW4E9USR:9N
YPWI2TORASODS6F3DUSR:U6
YPWI1QORASODSGNX3USR:KZ
YPWI46ORASODSL2KAUSR:WH
YPWIDGORASODSDSLKUSR:YK
YPWIX6ORASODS4DPHUSR:0A
YPWIXJORASODSY4Q8USR:BB
YPWIJHORASODSIQDWUSR:1N
YPWIOXORASODSMSZDUSR:C2
YPWIR3ORASODSWHTSUSR:HB
YPWI83ORASODS543RUSR:HF
YPWI70ORASODS8DJ3USR:81
YPWIWIORASODSLNCDUSR:NH
YPWIT1ORASODSOGKVUSR:Z5
YPWIJ7ORASODSLMHYUSR:CG
YPWI9XORASODSGNV3USR:02
YPWIC3ORASODSKXHTUSR:YF
YPWIIUORASODSQVXSUSR:0F
```
From here, I isolated the characters that were constant (`YPWIORASODSUSR:`), and anagramed them to what seems like the logical phrase `YOURPASSWORDIS:`. At this point, I was hopelessly lost on how to un-jumble the remaining 8 characters. Since there were only 8 characters, I chose the brute-force approach and wrote a script to attempt all `8! = 40320` possibilities.
```python
if __name__ == '__main__':
invalid_ids = []
ans = b''
for index in range(40320):
try:
print(index)
positions = [4,5,13,14,15,16,21,22]
s = socket.socket()
s.settimeout(1)
s.connect(('18.205.93.120',1204))
val = s.recv(25)
count = 0
while len(val) != 25 and count < 5:
count += 1
val = val + s.recv(25-len(val))
time.sleep(1)
if count >= 10:
raise Exception('recv error')
if len(val) != 25:
print('error, got:', val)
break
pw = []
x = index
ans_positions = []
while len(positions) > 0:
i = x % len(positions)
x = x // len(positions)
pw.append(val[positions[i]])
ans_positions.append(positions[i])
positions = positions[:i] + positions[i+1:]
s.send(bytes(pw)+b'\n')
res = s.recv(200)
print(val,bytes(pw),res)
if not res.startswith(b'Invalid'):
print(ans_positions)
break
s.close()
except Exception as e:
print(e)
invalid_ids.append(index)
print('invalid_ids =', invalid_ids)
time.sleep(5)
print('invalid ids:', invalid_ids)
```
With this script running, I went to sleep and awoke to find that it hit on the ordering `[13, 4, 14, 21, 15, 5, 16, 22]`.
With the benefit of retrospect, it seems a little more obvious what the actual ordering was. With the numbers arranged in three rows, the order simply reads off as a diagonal pattern as shown below:
```
0, 6 or 10, 17, 7 or 19, 1, 8, 9 or 12 or 18, 9 or 12 or 18, 2, 6 or 10, 7 or 19, 11, 3, 9 or 12 or 18, 20, 13, 4, 14, 21, 15, 5, 16, 22
0 . . . 1 . . . 2 . . . 3 . . . 4 . . . 5 . .
. 6 . 7 . 8 . 9 .10 .11 .12 .13 .14 .15 .16 .
. .17 . . .18 . . .19 . . .20 . . .21 . . .22
```
## Part Three
After entering the correct password to part two, we're greeted with something we expected all along - a [Grille Cipher](https://en.wikipedia.org/wiki/Grille_(cryptography)) (see the challenge name). The response we're given looks like the following, but changes on every new iteration:
```
### #
#### #
## ###
### #
### ##
# ## #
EPEPAL
RUS9E2
OSAOOW
SNBOEE
MROER8
DN:WTE
```
It turns out that this is a similar sentence, but the "password" this time is all contained in the last rotation of the grille. Knowing this, I wrote a quick function to solve the cipher and submitting the answer gives the flag.
```python
def solve_chart(chart):
print(chart.decode('ascii'))
arr = chart.split(b'\n')
pw = []
for i in range(6):
for j in range(6):
print(i,j,chr(arr[j][5-i]),chr(arr[7+i][j]))
if arr[j][5-i] == ord(' '):
pw.append(arr[7+i][j])
print()
return bytes(pw)+b'\n'
```
And my program output:
```
0 0 E
0 1 # P
0 2 # E
0 3 # P
0 4 # A
0 5 # L
1 0 # R
1 1 U
1 2 # S
1 3 9
1 4 # E
1 5 2
2 0 O
2 1 # S
2 2 # A
2 3 # O
2 4 O
2 5 # W
3 0 # S
3 1 # N
3 2 B
3 3 # O
3 4 # E
3 5 # E
4 0 # M
4 1 # R
4 2 # O
4 3 # E
4 4 # R
4 5 8
5 0 # D
5 1 # N
5 2 # :
5 3 W
5 4 # T
5 5 # E
b'EU92OOB8W\n'
b'Have a flag: AOTW{m33t_m3_at_grillby_s}\n'
```
