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WRITEUP.md

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Challenge name: One True Problem

Description:

Two of my friends were arguing about which CTF category is the best, but they encrypted it because they didn't want anyone to see. Lucky for us, they reused the same key; can you recover it?

Here are the ciphertexts:

213c234c2322282057730b32492e720b35732b2124553d354c22352224237f1826283d7b0651

3b3b463829225b3632630b542623767f39674431343b353435412223243b7f162028397a103e

Solution:

There is an algorithm named One-Time-Pad. You generate a true random key with the size as long as plaintext size. Then XOR them, bit by bit.

In this problem, I try something else. We know the flag start with utflag{. So, if we XOR first seven bytes with it, we can find first seven bytes of key:

def nonrepeatxor(text, key):
    sk = len(key)
    for i in range(0, 2 * sk, 2):
            x = int(text[i:i + 2], 16)
            y = ord(key[i // 2])
            if (x ^ y > 31):
                    print(chr(x ^ y), end='')
            else:
                    print('\\x%x' % (x ^ y), end='')

c1 = '213c234c2322282057730b32492e720b35732b2124553d354c22352224237f1826283d7b0651'
c2 = '3b3b463829225b3632630b542623767f39674431343b353435412223243b7f162028397a103e'

nonrepeatxor(c1, 'utflag{)

Output

THE BES

It seems, the key starts with "THE BEST". With a look at challenge description, next word could be "CTF" or "Category".

Now we can XOR cipher with key and get the plaintext. Let's open commandline after running script and use this function and updating key for catching plaintext. First I tried "Category" for next and failed! Then tried "CTF" and it was okey.

def run():
    st = len(c1)
    sk = len(key)

    nonrepeatxor(c1, key)
    print ((st - sk) * '?')
    nonrepeatxor(c2, key)
    print ((st - sk) * '?')

Output

The key is: THE BES
Commandline:
>>> run()
utflag{?????????????????????????????????????????????????????????????????????
os\x3\x18kg\x8?????????????????????????????????????????????????????????????????????
>>> key += "T CTF CATEGORY"
>>> run()
utflag{tw0_tim3_p4ds}???????????????????????????????????????????????????????
os\x3\x18kg\x8b\x12 _\x12\x6`7+| \xbcm???????????????????????????????????????????????????????
>>>

final script.py

The Flag

utflag{tw0_tim3_p4ds}

P.S: 2 day after CTF, i wonder why other characters wasn't printable, so I thinked a little more and realized, the flag, wasn't plaintext (should realize sooner! as the challenge description said!), but it was the key! With this function. We can use this function and find plaintext:

def repeatxor(text, key):
	st = len(text)
	sk = len(key)
	empty=""
	for i in range(0, st, 2):
		x = int(text[i:i + 2], 16)
		y = ord(key[(i // 2) % sk])
		print(chr(x ^ y), end='')

Output

THE BEST CTF CATEGORY IS CRYPTOGRAPHY!
NO THE BEST ONE IS BINARY EXPLOITATION