#!/usr/bin/env python3 # # Polymero # # Imports import os # Local imports with open("flag.txt",'rb') as f: FLAG = f.read() f.close() HDR = r"""| | __________ ______________ _____ | ___ ____/_____ _________ /__ ____/___________ ___________ /_ | __ /_ _ __ `/_ ___/ __/ / __ ___/_ / / /__ __ \ __/ | _ __/ / /_/ /_(__ )/ /_ / /___ _ / _ /_/ /__ /_/ / /_ | /_/ \__,_/ /____/ \__/ \____/ /_/ _\__, / _ .___/\__/ | the Fastest Encryptor in the Digital West /____/ /_/ |""" MSG = r"""| | | This is a free trial version of FastCrypt(tm). | To unlock the full version please provide your license key. | | To show FastCrypt(tm) is safe, we include an encrypted license key under a random key:""" # Class class FastCrypt: # Box created from PI decimals (in byte form), so nothing up our sleeve. PI_BOX = [[36, 63, 106, 136, 133, 163, 8, 211, 19, 25, 138, 46, 3, 112, 115, 68], [164, 9, 56, 34, 41, 159, 49, 208, 250, 152, 236, 78, 108, 137, 69, 40], [33, 230, 119, 190, 84, 102, 207, 52, 233, 12, 192, 172, 183, 201, 124, 80], [221, 132, 213, 181, 71, 23, 146, 22, 217, 121, 251, 27, 209, 11, 166, 223], [47, 253, 114, 219, 26, 184, 225, 175, 237, 38, 126, 150, 186, 144, 241, 44], [127, 153, 161, 179, 145, 247, 1, 242, 226, 142, 252, 99, 105, 32, 216, 113], [87, 88, 254, 244, 147, 61, 13, 149, 116, 143, 182, 139, 205, 130, 21, 74], [238, 123, 29, 194, 90, 89, 156, 48, 57, 42, 96, 197, 176, 35, 240, 202], [65, 24, 239, 220, 58, 14, 158, 30, 62, 215, 193, 189, 75, 39, 120, 218], [85, 92, 37, 243, 170, 171, 148, 72, 98, 232, 20, 64, 16, 180, 204, 17], [206, 134, 111, 188, 43, 169, 93, 246, 155, 135, 214, 51, 122, 50, 83, 129], [59, 107, 185, 196, 191, 97, 128, 117, 177, 2, 235, 101, 15, 109, 131, 66], [4, 200, 31, 94, 198, 104, 154, 103, 81, 160, 210, 167, 110, 228, 118, 67], [140, 125, 165, 195, 224, 86, 54, 6, 55, 10, 18, 234, 73, 7, 212, 222], [227, 76, 151, 79, 178, 82, 203, 168, 95, 0, 248, 60, 173, 5, 187, 255], [91, 45, 77, 174, 199, 100, 28, 229, 162, 245, 157, 53, 231, 249, 70, 141]] PI_BOX = [i for j in PI_BOX for i in j] def __init__(self): # Key for encrypting a valid license key self.secret_key = os.urandom(16) self.license = 'License key: ' + os.urandom(8).hex() def pad(self, msg): if type(msg) == str: msg = msg.encode() return msg + chr(32-len(msg)%32).encode() * (32-len(msg)%32) def unpad(self, msg): numpad = msg[-1] if numpad > 32: return False pdn = msg[-numpad:] msg = msg[:-numpad] if pdn != numpad*chr(numpad).encode(): return False return msg def sub(self, block): return [self.PI_BOX[i] for i in block] def perm(self, block): bitblock = list('{:0256b}'.format(int(bytes(block).hex(),16))) permbits = [bitblock[self.PI_BOX[i]] for i in range(256)] return [int(''.join(permbits[i:i+8]),2) for i in range(0,256,8)] def encrypt_block(self, block, keybits): for keybit in keybits: if keybit == 1: block = self.sub(block) if keybit == 0: block = self.perm(block) return block def encrypt(self, msg, key=None, iv=None): if key is None or key == '': key = self.secret_key assert len(key) == len(self.secret_key) if iv is None: iv = os.urandom(len(self.secret_key)//4) keybits = [int(i) for i in list('{:0{nk}b}'.format(int(key.hex(),16),nk=len(self.secret_key)*8))] ivindex = [int(i,16) for i in list(iv.hex())] keybits += [keybits[ivindex[i]+16*i] for i in range(len(iv)*2)] blocks = self.pad(msg) blocks = [list(blocks[i:i+32]) for i in range(0,len(blocks),32)] ciphertext = [] for b,block in enumerate(blocks): ciphertext += self.encrypt_block(block, keybits) return (iv + bytes(ciphertext)).hex() # Challenge FC = FastCrypt() print(HDR) print(MSG) print("| {}".format(FC.encrypt(FC.license))) while True: try: print("|\n| Menu:") print("| [1] Encrypt (trial)") print("| [2] Enter license key") choice = input("|\n| >> ") if choice == '1': print("|\n| What would you like to encrypt? [enter no iv to use a random iv]:") iv = input("|\n| >> Iv : ") msg = input("| >> Msg: ") try: cip = FC.encrypt(msg.encode(), iv=bytes.fromhex(iv)) except: cip = FC.encrypt(msg.encode()) print("|\n| Encrypted message:\n| {}".format(cip)) if choice == '2': print("|\n| Enter a valid license key:") license = input("|\n| >> ") if license == FC.license[-16:]: print("|\n| Thank you for purchasing FastCrypt(tm), please enjoy this welcome gift!") print('| {}'.format(FLAG)) print('|') exit(0) else: print("|\n| Invalid license key!") except KeyboardInterrupt: print("\n|\n| Thank you for using FastCrypt(tm)!\n|\n") exit(0) except: print("|\n|\n| Something went wrong, please try again.")
keybitsの値によってブロックの暗号化の様子が変わるkeybitsの前半の値はkeyによって決まり、後半の値はivによってきまるやってることはsubとpermで、subが非線形な操作
1ブロックは32バイト
目的は
E(x)からxを求めること。xは known 13 bytes + unknown 16bytes + known 4 bytes からなる
暗号化は何回でもできる
なんかivまわりがバグってて、ivを空にすると単に空のivがつかわれる（
bytes.fromhex("")はb""なので）例外が発生したかどうかを判断できる
