西电网安实验班 现代密码学 实验一
题目一
题目使用同一个流密码密钥加密得到 10 个密文，加密方式是逐字节异或，目标是通过这 10 个密文得到密钥以解出第 11 个密文。
已知消息 M、密钥 K，异或加密的过程是
$$ C=M\oplus K $$
得到密文 C。异或是自反的运算，所以解密与加密过程相同
$$ M=C\oplus K $$
对于同一个密钥 K 加密的两个密文 C_1、C_2，如果将两者异或可以得到
$$ B=C_1\oplus C_2=M_1\oplus K\oplus M_2\oplus K=M_1\oplus M_2 $$
字节流B是两消息逐字节异或结果。
我们首先导入这些密文和加密方法
msg_1 = bytes.fromhex('315c4eeaa8b5f8aaf9174145bf43e1784b8fa00dc71d885a804e5ee9fa40b16349c146fb778cdf2d3aff021dfff5b403b510d0d0455468aeb98622b137dae857553ccd8883a7bc37520e06e515d22c954eba5025b8cc57ee59418ce7dc6bc41556bdb36bbca3e8774301fbcaa3b83b220809560987815f65286764703de0f3d524400a19b159610b11ef3e')
msg_2 = bytes.fromhex('234c02ecbbfbafa3ed18510abd11fa724fcda2018a1a8342cf064bbde548b12b07df44ba7191d9606ef4081ffde5ad46a5069d9f7f543bedb9c861bf29c7e205132eda9382b0bc2c5c4b45f919cf3a9f1cb74151f6d551f4480c82b2cb24cc5b028aa76eb7b4ab24171ab3cdadb8356f')
msg_3 = bytes.fromhex('32510ba9a7b2bba9b8005d43a304b5714cc0bb0c8a34884dd91304b8ad40b62b07df44ba6e9d8a2368e51d04e0e7b207b70b9b8261112bacb6c866a232dfe257527dc29398f5f3251a0d47e503c66e935de81230b59b7afb5f41afa8d661cb')
msg_4 = bytes.fromhex('32510ba9aab2a8a4fd06414fb517b5605cc0aa0dc91a8908c2064ba8ad5ea06a029056f47a8ad3306ef5021eafe1ac01a81197847a5c68a1b78769a37bc8f4575432c198ccb4ef63590256e305cd3a9544ee4160ead45aef520489e7da7d835402bca670bda8eb775200b8dabbba246b130f040d8ec6447e2c767f3d30ed81ea2e4c1404e1315a1010e7229be6636aaa')
msg_5 = bytes.fromhex('3f561ba9adb4b6ebec54424ba317b564418fac0dd35f8c08d31a1fe9e24fe56808c213f17c81d9607cee021dafe1e001b21ade877a5e68bea88d61b93ac5ee0d562e8e9582f5ef375f0a4ae20ed86e935de81230b59b73fb4302cd95d770c65b40aaa065f2a5e33a5a0bb5dcaba43722130f042f8ec85b7c2070')
msg_6 = bytes.fromhex('32510bfbacfbb9befd54415da243e1695ecabd58c519cd4bd2061bbde24eb76a19d84aba34d8de287be84d07e7e9a30ee714979c7e1123a8bd9822a33ecaf512472e8e8f8db3f9635c1949e640c621854eba0d79eccf52ff111284b4cc61d11902aebc66f2b2e436434eacc0aba938220b084800c2ca4e693522643573b2c4ce35050b0cf774201f0fe52ac9f26d71b6cf61a711cc229f77ace7aa88a2f19983122b11be87a59c355d25f8e4')
msg_7 = bytes.fromhex('32510bfbacfbb9befd54415da243e1695ecabd58c519cd4bd90f1fa6ea5ba47b01c909ba7696cf606ef40c04afe1ac0aa8148dd066592ded9f8774b529c7ea125d298e8883f5e9305f4b44f915cb2bd05af51373fd9b4af511039fa2d96f83414aaaf261bda2e97b170fb5cce2a53e675c154c0d9681596934777e2275b381ce2e40582afe67650b13e72287ff2270abcf73bb028932836fbdecfecee0a3b894473c1bbeb6b4913a536ce4f9b13f1efff71ea313c8661dd9a4ce')
msg_8 = bytes.fromhex('315c4eeaa8b5f8bffd11155ea506b56041c6a00c8a08854dd21a4bbde54ce56801d943ba708b8a3574f40c00fff9e00fa1439fd0654327a3bfc860b92f89ee04132ecb9298f5fd2d5e4b45e40ecc3b9d59e9417df7c95bba410e9aa2ca24c5474da2f276baa3ac325918b2daada43d6712150441c2e04f6565517f317da9d3')
msg_9 = bytes.fromhex('271946f9bbb2aeadec111841a81abc300ecaa01bd8069d5cc91005e9fe4aad6e04d513e96d99de2569bc5e50eeeca709b50a8a987f4264edb6896fb537d0a716132ddc938fb0f836480e06ed0fcd6e9759f40462f9cf57f4564186a2c1778f1543efa270bda5e933421cbe88a4a52222190f471e9bd15f652b653b7071aec59a2705081ffe72651d08f822c9ed6d76e48b63ab15d0208573a7eef027')
msg_10 = bytes.fromhex('466d06ece998b7a2fb1d464fed2ced7641ddaa3cc31c9941cf110abbf409ed39598005b3399ccfafb61d0315fca0a314be138a9f32503bedac8067f03adbf3575c3b8edc9ba7f537530541ab0f9f3cd04ff50d66f1d559ba520e89a2cb2a83')
msg_target = bytes.fromhex('32510ba9babebbbefd001547a810e67149caee11d945cd7fc81a05e9f85aac650e9052ba6a8cd8257bf14d13e6f0a803b54fde9e77472dbff89d71b57bddef121336cb85ccb8f3315f4b52e301d16e9f52f904')
msgs = [msg_1, msg_2, msg_3, msg_4, msg_5, msg_6, msg_7, msg_8, msg_9, msg_10]
def bytesxor(a, b):
if len(a) > len(b):
return bytes([x ^ y for (x, y) in zip(a[:len(b)], b)])
else:
return bytes([x ^ y for (x, y) in zip(a, b[:len(a)])])根据前文理论，我们先尝试异或 msg_1 和 msg_2，结果是
b'\x12\x10L\x06\x13NW\t\x14\x0f\x10O\x02R\x1b\n\x04B\x02\x0cM\x07\x0b\x18OH\x15T\x1f\x08\x00HN\x1e\x02A\x06\x1d\x06MT\x0b\n\x02\x02\x10\x19E\x10\x16MO:\x00SC\x00NC\x0e\x1e\x1d\nRF\x12\x17\x1b\x01\x17\x00\x1b\x0eEC\x1c\x0c\x1d\x16\nR\r\x11tN\x19\x06\x1a\x11M\x0eU\x17O\x08NT7\x14\x05\x0b\x17CST\x1bH\x07\x0e\x00\x0eM'我们不使用十六进制格式输出，是因为需要观察其中有一些大写或小写英文字母。根据题目，明文是 ASCII 英语句子，其中大部分符号都是大小写英文字母和空格。我们需要知道一个特殊的规律
$$ ASCII(大/小写字母)\oplus ASCII(空格)=ASCII(小/大写字母) $$
即英文字母的 ASCII 值与空格的 ASCII 值异或得到的 ASCII 值相当于切换原字母的大小写。
根据以上规律，我们十个中的一个密文 C，与其余九个密文逐字节异或，得到九个字节流 B_i，观察其中相同索引 j（位置）的字节 b_k，如果这九个字节几乎都是英文字母，那么我们选取的密文 C 的索引 j 处的字节 C[j] 的明文就大概是空格。此时我们有
$$ C[j]\oplus ASCII(空格)=K[j] $$
从而有可能还原出一个字节的密钥。如果密文足够多足够长，重复上述过程可以还原完整的密钥。
key = [0] * len(msg_7) # `msg_7` is the longest ciphertext.
def isalpha(b):
return (ord('a') <= b <= ord('z')) or (ord('A') <= b <= ord('Z'))
for i, msg_i in enumerate(msgs):
may_not_space = [0] * len(msg_i) # Count cases when b_k is not alphabetic.
for j, msg_j in enumerate(msgs):
if i != j:
xored = bytesxor(msg_i, msg_j)
for k, xb in enumerate(xored):
if (not isalpha(xb)) and xb != 0:
may_not_space[k] += 1
for j, may_not in enumerate(may_not_space):
if may_not <= 2: # If almost all b_k are alphabetic
key_byte = msg_i[j] ^ ord(' ')
if key[j] == 0:
key[j] = key_byte
continue
if key[j] != key_byte: # Detect contradiction. Do more checks.
reliable = True
for m in msgs:
if j >= len(m):
continue
byte = m[j] ^ key_byte
if not isalpha(byte) and byte != ord(' '):
reliable = False
break
if reliable:
key[j] = key_byte
print(f'Recovered key: {key}')
print(f'Target message: {bytesxor(bytes(key), msg_target)}'')
# for msg in msgs:
# print(bytesxor(msg, bytes(key)))题目二
地址：https://www.cryptopals.com/sets/1
1. Convert hex to base64
from base64 import b64encode
print(b64encode(bytes.fromhex('49276d206b696c6c696e6720796f757220627261696e206c696b65206120706f69736f6e6f7573206d757368726f6f6d')).decode())2. Fixed XOR
def bytesxor(a, b):
if len(a) > len(b):
return bytes([x ^ y for (x, y) in zip(a[:len(b)], b)])
else:
return bytes([x ^ y for (x, y) in zip(a, b[:len(a)])])
print(bytesxor(bytes.fromhex('1c0111001f010100061a024b53535009181c'), bytes.fromhex('686974207468652062756c6c277320657965')).hex())3. Single-byte XOR cipher
突发奇想，这里我用《动物森友会》给信件评分的算法来判断哪个结果最正确。（还真没毛病
def get_score(message: str) -> int:
"""*Animal Crossing* message scoring algorithm."""
t = set("abl abo abr abs acc ach acr act add adm adv aer aff afr aft aga age ago ahe air ali all alm alo alr als alt alw am ame amo and ang ani ano ans any apa app apr are arg arm arr art asi ask asl ate atm att aud aug aut ave avo awa cak cal cam can cap car cas cat cau cen cer cha che chi cho chi chu cir cit cla cle cli clo coa cof coi col com con coo cop cor cos cou cov cow cre cri cro cry cup cur cus cut bab bac bad bag bal ban bas bat be bea bec bed bee bef beg beh bel bes bet bey bic big bik bil bir bit bla ble blo blu boa bod bon boo bor bot bou box boy bra bre bri bro bui bur bus but buy by eac ear eas eat edu eff egg eig eit ele els emp end ene eng enj eno ent equ err esp eur eve exa exc exe exp eye dad dai dam dan dar dat dau day dea dec dee def deg del dem den dep des det dev dic did die dif dig din dir dis div do doc doe dog dol don doo dou dow doz dra dre dri dro dru dry due dur dus dut gai gam gar gas gat gav gen ger get gir giv gla go god goi gon goo got gov gra gre gro gua gue gui gun fac fai fal fam far fas fat fea feb fed fee fel few fie fif fig fil fin fir fis fiv fix fla fle fli flo fly fol foo for fou fra fre fri fro fru ful fun fut i ice ide if ima imm imp in inc ind inf ins int inv iro is isl it its hab had hai hal han hap har has hat hav he hea hei hel her hi hid hig hil him hir his hit hol hom hon hop hor hos hot hou how hum hun hur hus kee kep key kic kil kin kit kne kni kno kab kad kai kak kan kar kas kat kau kaw kay kaz kea ked kef keg ken kes ket kev kib kie kif kig kik kim kin kis kit kiv koc kon koo kos kot kou kov kow kun kyi kac kad kag kai kaj kak kan kap kar kat kay ke kea ked kee kem ken kes ket kid kig kil kin kis kod kom kon koo kor kos kot kou kov kuc kum kus ky kys kam kar kat kea kec kee kei kev kew kex kic kig kin ko kob koi kon koo kor kos kot kov kow kum kbj k'c kct kf kff kft kh kil kka kld kn knc kne knl kpe kpi kpp kr kra krd kth kur kut kve kwn jan jap job joi jud jul jum jun jus qua que qui pac pag pai pap par pas pat pay pea pen peo per pho pic pie pin pip pla ple poc poi pol poo pop pos pot pou pow pra pre pri pro pub pul pup pur pus put sad saf sai sal sam san sat sav saw say sce sch sci sco sea sec see sel sen sep ser set sev sex sha she shi sho shu sic sid sig sil sim sin sis sit six siz ski sky sle sli slo sma sme smi smo sno so soa soc sof soi sol som son soo sor sou spa spe spi spo spr squ sta ste sti sto str stu sty sub suc sud suf sug sum sun sup sur swa swe swi swu sys rac rad rai ran rap rat rea rec red ref reg rel rem rep req res ret ric rid rig rin ris riv roa roc rod rol roo ros rou row rul run rus una unc und uni unl unt up upo us use usu tab tak tal tas tau tax tea tee tel tem ten ter tes tha the thi tho thr thu tic tie til tim tir tit to tod tog tol tom ton too top tor tot tou tow tra tre tri tro tru try tue tur tv twe twi two tyi typ val var veg ver vie vil vis voi vol vot vai vak val van var vas vat vav vay ve vea ved vee vei vel ven ver ves vet vha vhe vhi vho vhy vid vif vil vin vir vis vit viv vok vom von voo vor vou vri vro vma yar yea yel yen yes yet you zer".split())
s = 0
if message and message[-1] in '.?!':
s += 20
for i, c in enumerate(message):
if c in '.?!':
for j in range(i+1, min(i+4, len(message))):
if message[j].isupper():
s += 10
break
elif message[j].isalpha():
s -= 10
break
s += sum(3 for w in message.split() if len(cw :=
''.join(c for c in w if c.isalpha()).lower()) >= 3 and cw[:3] in t)
for c in message:
if not c.isspace():
s += 20 if c.isupper() else -10
break
for i in range(len(message)-2):
if message[i].isalpha() and message[i] == message[i+1] == message[i+2]:
s -= 50
break
sp, nsp = message.count(' '), len(message) - message.count(' ')
s += -20 if nsp == 0 or (sp * 100 // nsp if nsp else 0) < 20 else 20
if len(message) > 75:
c = 0
for ch in message:
c = 0 if ch in '.?!' else c + 1
if c == 75:
s -= 150
break
s -= sum(20 for i in range(0, len(message), 32)
if ' ' not in message[i:i+32] and len(message[i:i+32]) == 32)
return s
cipher = bytes.fromhex(
'1b37373331363f78151b7f2b783431333d78397828372d363c78373e783a393b3736')
results = list()
for c in range(256):
try:
res = bytesxor((chr(c) * len(cipher)).encode(), cipher).decode()
results.append((res, get_score(res)))
except UnicodeDecodeError:
pass
best = sorted(results, key=lambda x: x[1])[-1][0]
print(best)
# Cooking MC's like a pound of bacon4. Detect single-character XOR
with open('4.txt', 'r') as file:
ciphers = file.readlines()
for original_cipher in ciphers:
results = list()
cipher = bytes.fromhex(original_cipher)
for c in range(256):
try:
res = bytesxor((chr(c) * len(cipher)).encode(), cipher).decode()
results.append((res, get_score(res)))
except UnicodeDecodeError:
pass
if len(results) == 0:
continue
best = sorted(results, key=lambda x: x[1])[-1]
if best[1] > 40:
print(f'{original_cipher.strip()} -> {best[0]}')
# 7b5a4215415d544115415d5015455447414c155c46155f4058455c5b523f -> Now that the party is jumping5. Implement repeating-key XOR
def repeating_key_xor_to_hex(msg: bytes, key: bytes) -> str:
result_chars = []
keylen = len(key)
for i, b in enumerate(msg):
result_chars.append(b ^ key[i % keylen])
return bytes(result_chars).hex()
print(repeating_key_xor_to_hex(b"Burning 'em, if you ain't quick and nimble", b'ICE'))
print(repeating_key_xor_to_hex(b'I go crazy when I hear a cymbal', b'ICE'))6. Break repeating-key XOR
一开始没有还原正确，检查了一下发现是 # 扰乱了解密评分。考虑到英文句子里极少有 #，所以遇到就扣 5 分。
from base64 import b64decode
def get_score(message: str):
score = 0
for c in message:
if c.islower():
score += 3
if c.isupper():
score += 1
if c == ' ':
score += 1
if c == '#':
score -= 5
return score
def get_key(cipher: bytes) -> int:
results = list()
for c in range(256):
try:
res = bytesxor((chr(c) * len(cipher)).encode(), cipher).decode()
results.append((c, get_score(res)))
except UnicodeDecodeError:
pass
return sorted(results, key=lambda x: x[1])[-1][0]
def repeating_key_xor(msg: bytes, key: bytes) -> bytes:
result_chars = []
keylen = len(key)
for i, b in enumerate(msg):
result_chars.append(b ^ key[i % keylen])
return bytes(result_chars)
def normalized_average_hd(data: bytes, unit_len: int) -> float:
chunks = [data[i:i + unit_len] for i in range(0, len(data), unit_len)]
num_chunks = len(chunks)
total_hd = 0
count = 0
for i in range(num_chunks):
for j in range(i + 1, num_chunks):
if len(chunks[i]) == unit_len and len(chunks[j]) == unit_len:
total_hd += sum((byte1 ^ byte2).bit_count()
for byte1, byte2 in zip(chunks[i], chunks[j]))
count += 1
return total_hd / (count * unit_len)
with open('6.txt', 'r') as file:
cipher = b64decode(file.read())
results = list()
for l in range(2, 41):
norm_hd = normalized_average_hd(cipher, l)
results.append((l, norm_hd))
keylens = sorted(results, key=lambda x: x[1])
for k in range(1):
keylen = keylens[k][0]
print(f"Guessed key length: {keylen}")
key_bytes = []
for i in range(keylen):
block = bytes([cipher[j * keylen + i]
for j in range(len(cipher) // keylen)])
key_bytes.append(get_key(block))
key = bytes(key_bytes)
# print(repeating_key_xor(cipher, key).decode())
print(f'Key: {key}')
# Guessed key length: 29
# Key: b'Terminator X: Bring the noise'7. AES in ECB mode
import base64
from cryptography.hazmat.primitives.ciphers import Cipher, algorithms
key = b"YELLOW SUBMARINE"
with open('7.txt', 'r') as f:
encrypted_b64 = f.read()
encrypted_data = base64.b64decode(encrypted_b64)
cipher = Cipher(algorithms.AES(key)) # Default to ECB mode
decryptor = cipher.decryptor()
decrypted_data = decryptor.update(encrypted_data) + decryptor.finalize()
print(decrypted_data.decode())8. Detect AES in ECB mode
print(max(open('8.txt').read().splitlines(), key=lambda x: len(c:=[x[i:i+32] for i in range(0, len(x), 32)]) - len(set(c))))