Rating:
# Pragyan CTF 2019 "Decode This" writeup
This problem is solved by hi120ki.
## check problem
```
Ram has something to show you, which can be of great help. It definitely contains the piece of text "pctf" and whatever follows it is the flag. Can you figure out what it is?
Note: Enclose it in pctf{}
```
There is "pctf" character in secret.txt.
> 「pctf」が secret.txt の中のどこかにあるそうです。
```python
import random
file = open("secret.txt","r")
secret = file.read()
# convert flag to only alphabets
flag = ""
for i in secret:
if i.isalpha():
flag += i
l = len(flag)
# generate random number
key = [[int(random.random()*10000) for e in range(2)] for e in range(2)]
i = 0
ciphertext = ""
while i <= (l-2):
# x and y is 0..25
x = ord(flag[i]) - 97
y = ord(flag[i+1]) - 97
z = (x*key[0][0] + y*key[0][1])%26 + 97
w = (x*key[1][0] + y*key[1][1])%26 + 97
ciphertext = ciphertext + chr(z) + chr(w)
i = i+2
cipherfile = open('ciphertext.txt','w')
cipherfile.write(ciphertext)
```
Flag is converted to only alphabets.
This crypto is called Affin cipher.
<https://en.wikipedia.org/wiki/Affine_cipher>
But this code has some difference.
> flag 文字列はアルファベットのみ取り出されています。
>
> これはアフィン暗号とも呼ばれますが少しカスタマイズされています。
## solve problem
First, key can be thought as key%26.
> まず key は定数であり、mod26 を取り、0 から 25 のどれかの値として考えることができます。
```python
a = key[0][0]%26 # a is 0..25
b = key[0][1]%26 # b is 0..25
c = key[1][0]%26 # c is 0..25
d = key[1][1]%26 # d is 0..25
```
Second, There is "pctf" character in secret.txt.
I try all key, and encrypt "pctf", and make sure ciphertext matches.
If it maches, I decode all ciphertext.
> 次に secret.txt の中に「pctf」が含まれていることから、まず全ての key で pctf を暗号化し、ciphertext と一致するか調べます。
>
> そしてもし一致する部分がある場合、その key を使って ciphertext を復号化します。
```python
cipher_txt = open("ciphertext.txt").read()
cipher_num = []
for i in range(72):
cipher_num.append(ord(cipher_txt[i]) - 97)
p_num = ord("p") - 97
c_num = ord("c") - 97
t_num = ord("t") - 97
f_num = ord("f") - 97
# try all key
for a in range(26):
for b in range(26):
for c in range(26):
for d in range(26):
for i in range(34):
# make sure ciphertext matches
res1 = (a * p_num + b * c_num) % 26
res2 = (c * p_num + d * c_num) % 26
res3 = (a * t_num + b * f_num) % 26
res4 = (c * t_num + d * f_num) % 26
if (
(cipher_num[i * 2] == res1)
and (cipher_num[i * 2 + 1] == res2)
and (cipher_num[i * 2 + 2] == res3)
and (cipher_num[i * 2 + 3] == res4)
):
print("\nkey is", a, b, c, d)
# decode all ciphertext
for item in range(36):
for x in range(26):
for y in range(26):
if (
(cipher_num[item * 2] == (a * x + b * y) % 26)
and (cipher_num[item * 2 + 1] == (c * x + d * y) % 26)
):
print(chr(x + 97), chr(y + 97), end=" ")
# key is 15 10 18 17
# r a m h a s a l i t t l e s e c r e t f o r y o u r i g h t h e r e i t i s p c t f i l i k e c l i m b i n g h i l l s w h a t a b o u t y o u
```
Ram has a little secret for you right here it is pctf i like climbing hills what about you
I get flag : pctf{ilikeclimbinghillswhataboutyou}
