CryptoHack

Note : A JOURNEY TO GAIN KNOWLEDGE

General

#XOR Properties

import binascii
def byte_xor(ba1, ba2):
    return bytes([_a ^ _b for _a, _b in zip(ba1, ba2)])
s1 = "a6c8b6733c9b22de7bc0253266a3867df55acde8635e19c73313"
s2 = "37dcb292030faa90d07eec17e3b1c6d8daf94c35d4c9191a5e1e"
s3 = "c1545756687e7573db23aa1c3452a098b71a7fbf0fddddde5fc1"
s4 = "04ee9855208a2cd59091d04767ae47963170d1660df7f56f5faf"
s2 = bytes.fromhex(s2)
s3 = bytes.fromhex(s3)
s4 = bytes.fromhex(s4)
key1 = bytes.fromhex(s1)
key2 = b'\x91\x14\x04\xe1?\x94\x88N\xab\xbe\xc9%\x85\x12@\xa5/\xa3\x81\xdd\xb7\x97\x00\xddm\r'
key13 = b'\xf6\x88\xe5\xc4kq\xdf\xe3\x0b]F\x0b\xd7\xe3f@m\xe33\x8a\xdb\x14\xc4\xc4\x01\xdf'
k = b'g\x9c\xe1%T\xe5W\xad\xa0\xe3\x8f.R\xf1&\xe5B@\xb2Wl\x83\xc4\x19l\xd2'
a_list = byte_xor(k,s4)

print(a_list)

The special thing of XOR operator is when you xor with same key you can get the original message .

a ^ b = c => a = b ^ c

(encrypt = message ^ key => message = encrypt ^ key)

crypto{x0r_i5_ass0c1at1v3}

#Favourite byte

key word "a single byte" . Xor each character with integer in range [0,256).

crypto{0x10_15_my_f4v0ur173_by7e}

You either know, XOR you don't

Idea: Find key by xor encrypt message and key_form then xor key with encrypt.

crypto{1f_y0u_Kn0w_En0uGH_y0u_Kn0w_1t_4ll}

#Lemur XOR

What do you think about Adobe?

Adobe Photoshop why not ? :>

crypto{X0Rly_n0t ? }

#Break RSA

Tonelli–Shanks algorithm

Adrien's Signs

Thanks for reading. Have a good day <3

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