目录
- Task 1: Frequency Analysis
- Task 2: Encryption using Different Ciphers and Modes
- Task 4: Padding
- Task 5: Error Propagation – Corrupted Cipher Text
- Task 6: Initial Vector (IV) and Common Mistakes
- Task 6.1. IV Experiment
- Task 6.2. Common Mistake: Use the Same IV
- Task 6.3. Common Mistake: Use a Predictable IV
- Task 7: Programming using the Crypto Library
Task 1: Frequency Analysis
大致原理：单表替换密码（将明文中的字母按照对照表映射成密文），会将明文中的统计规律带到密文中，如明文中出现最多的是A，则密文中出现最多的就是A在对照表下的密文。因此我们可以通过频率分析的方法找到对照表，破解密文。
代码思想：读入密文ciphertext.txt（只有小写的字母+空格+回车的密文），然后统计其中各个字母出现的次数并对统计结果进行排序。然后我们还有一个已知的排序结果words_frequency（这个是百度搜到的公开单字母频率），这两个排序结果的对应关系我们就认为是对照表了，接着使用这个对照表翻译密文得到明文out.txt
vim task1.py
sudo python3 task1.py
def get_frequency(path):
res = {}
for i in range(ord('a'),ord('z')+1):
res[chr(i)]=0
with open(path) as f:
for line in f.readlines():
for ch in line.strip():
if ch.isalpha():
res[ch.lower()] +=1
return sorted(res.items(),key=lambda s:s[1],reverse=True)
def main():
ciphertext_frequency= get_frequency('./ciphertext.txt')
words_frequency= "ETAOINRSHDCLMPUFGWYBKJVXQZ";
trans = {}
for i in range(26):
trans[ciphertext_frequency[i][0]] = words_frequency[i].lower()
with open('./out.txt',mode='a') as f1:
with open('./ciphertext.txt') as f2:
for line in f2.readlines():
for ch in line.strip():
if ch.isalpha():
f1.write(trans[ch.lower()])
else:
f1.write(ch)
if __name__ == '__main__':
main()
Task 2: Encryption using Different Ciphers and Modes
用AES加密算法（128位密钥，CBC模式）加密一个文件并进行解密。
openssl enc -aes-128-cbc -e -in ./plaintext.txt -out cipher.bin -K 00112233445566778889aabbccddeeff -iv 0102030405060708
openssl enc -aes-128-cbc -d -in cipher.bin -out out.txt -K 00112233445566778889aabbccddeeff -iv 0102030405060708
用AES加密算法（128位密钥），分别采用ECB和CBC模式加密一个bmp图片，查看加密后的图片。
openssl enc -aes-128-ecb -in ./pic_original.bmp -out ./pic_ecb.bmp -e -K 00112233445566778889aabbccddeeff
openssl enc -aes-128-cbc -in ./pic_original.bmp -out ./pic_cbc.bmp -e -K 00112233445566778889aabbccddeeff -iv 0102030405060708
Task 4: Padding
分别创建大小为5,10,15字节的文件
echo -n "01234" > 5.txt
echo -n "0123456789" > 10.txt
echo -n "012345678999999" > 15.txt
先用AES加密算法（128位密钥，CBC模式）加密这些文件，然后解密这些密文（选择解密时不去除padding的选项），观察解密后各个文件的尺寸大小
注：我们在解密的时候添加了参数-nopad，表示不去除padding
openssl enc -aes-128-cbc -e -in ./5.txt -out ./cbc/5-cbc.txt -K 00112233445566778889aabbccddeeff -iv 0102030405060708
openssl enc -aes-128-cbc -e -in ./10.txt -out ./cbc/10-cbc.txt -K 00112233445566778889aabbccddeeff -iv 0102030405060708
openssl enc -aes-128-cbc -e -in ./15.txt -out ./cbc/15-cbc.txt -K 00112233445566778889aabbccddeeff -iv 0102030405060708
openssl enc -aes-128-cbc -d -in ./cbc/5-cbc.txt -out ./cbc/5-cbc-out.txt -K 00112233445566778889aabbccddeeff -iv 0102030405060708 -nopad
openssl enc -aes-128-cbc -d -in ./cbc/10-cbc.txt -out ./cbc/10-cbc-out.txt -K 00112233445566778889aabbccddeeff -iv 0102030405060708 -nopad
openssl enc -aes-128-cbc -d -in ./cbc/15-cbc.txt -out ./cbc/15-cbc-out.txt -K 00112233445566778889aabbccddeeff -iv 0102030405060708 -nopad
用xxd命令查看各个解密文件中padding的内容
xxd 5-cbc-out.txt
xxd 10-cbc-out.txt
xxd 15-cbc-out.txt
用AES加密算法（128位密钥，ECB模式，CFB模式，OFB模式）进行类似上述（2）和（3）的操作，查看各个解密文件中padding的内容。
注：观察发现，ECB、CBC出现填充，而CFB、OFB没有填充。因为CFB、OFB可以将DES转换为流密码。
ecb
openssl enc -aes-128-ecb -e -in ./5.txt -out ./ecb/5-ecb.txt -K 00112233445566778889aabbccddeeff
openssl enc -aes-128-ecb -e -in ./10.txt -out ./ecb/10-ecb.txt -K 00112233445566778889aabbccddeeff
openssl enc -aes-128-ecb -e -in ./15.txt -out ./ecb/15-ecb.txt -K 00112233445566778889aabbccddeeff
openssl enc -aes-128-ecb -d -in ./ecb/5-ecb.txt -out ./ecb/5-ecb-out.txt -K 00112233445566778889aabbccddeeff -nopad
openssl enc -aes-128-ecb -d -in ./ecb/10-ecb.txt -out ./ecb/10-ecb-out.txt -K 00112233445566778889aabbccddeeff -nopad
openssl enc -aes-128-ecb -d -in ./ecb/15-ecb.txt -out ./ecb/15-ecb-out.txt -K 00112233445566778889aabbccddeeff -nopad
xxd 5-ecb-out.txt
xxd 10-ecb-out.txt
xxd 15-ecb-out.txt
注：xxd的第三列表示的是这一行原本的内容，发现多了一些...
cfb
openssl enc -aes-128-cfb -e -in ./5.txt -out ./cfb/5-cfb.txt -K 00112233445566778889aabbccddeeff -iv 0102030405060708
openssl enc -aes-128-cfb -e -in ./10.txt -out ./cfb/10-cfb.txt -K 00112233445566778889aabbccddeeff -iv 0102030405060708
openssl enc -aes-128-cfb -e -in ./15.txt -out ./cfb/15-cfb.txt -K 00112233445566778889aabbccddeeff -iv 0102030405060708
openssl enc -aes-128-cfb -d -in ./cfb/5-cfb.txt -out ./cfb/5-cfb-out.txt -K 00112233445566778889aabbccddeeff -iv 0102030405060708 -nopad
openssl enc -aes-128-cfb -d -in ./cfb/10-cfb.txt -out ./cfb/10-cfb-out.txt -K 00112233445566778889aabbccddeeff -iv 0102030405060708 -nopad
openssl enc -aes-128-cfb -d -in ./cfb/15-cfb.txt -out ./cfb/15-cfb-out.txt -K 00112233445566778889aabbccddeeff -iv 0102030405060708 -nopad
xxd 5-cfb-out.txt
xxd 10-cfb-out.txt
xxd 15-cfb-out.txt
ofb
openssl enc -aes-128-ofb -e -in ./5.txt -out ./ofb/5-ofb.txt -K 00112233445566778889aabbccddeeff -iv 0102030405060708
openssl enc -aes-128-ofb -e -in ./10.txt -out ./ofb/10-ofb.txt -K 00112233445566778889aabbccddeeff -iv 0102030405060708
openssl enc -aes-128-ofb -e -in ./15.txt -out ./ofb/15-ofb.txt -K 00112233445566778889aabbccddeeff -iv 0102030405060708
openssl enc -aes-128-ofb -d -in ./ofb/5-ofb.txt -out ./ofb/5-ofb-out.txt -K 00112233445566778889aabbccddeeff -iv 0102030405060708 -nopad
openssl enc -aes-128-ofb -d -in ./ofb/10-ofb.txt -out ./ofb/10-ofb-out.txt -K 00112233445566778889aabbccddeeff -iv 0102030405060708 -nopad
openssl enc -aes-128-ofb -d -in ./ofb/15-ofb.txt -out ./ofb/15-ofb-out.txt -K 00112233445566778889aabbccddeeff -iv 0102030405060708 -nopad
xxd 5-ofb-out.txt
xxd 10-ofb-out.txt
xxd 15-ofb-out.txt
Task 5: Error Propagation – Corrupted Cipher Text
创建一个至少1000字节大的文本文件
#!/bin/bash
f=""
for (( i = 1; i < 1024; ++i)); do f=$f"6"; done
echo $f >> f.txt
用AES加密算法（128位密钥，ECB模式）加密这个文件
openssl enc -aes-128-ecb -e -in ./f.txt -out ./ecb/f-ecb.txt -K 00112233445566778889aabbccddeeff
用bless工具修改密文中第55个字节中的1个bit
bless ./ecb/f-ecb.txt
注：bless工具貌似是不支持直接修改的，只能删除、复制和黏贴。比如我想将DF修改成EF，做法就是选中EF（是鼠标拖动，让ED块变红），然后右键复制，选中DF右键删除，然后再右键黏贴。
解密被上述修改后的密文，查看解密后的明文有什么变化
openssl enc -aes-128-ecb -d -in ./ecb、f-ecb.txt -out ./ecb/f-ecb-out.txt -K 00112233445566778889aabbccddeeff
vim ./ecb/f-ecb-out.txt
用AES加密算法（128位密钥，CBC模式，OFB模式）进行类似上述（2）-（4）的操作，查看对应解密后的明文有什么变化
注：不同的模式下，密文改变带来的影响不同，具体怎么个不同法和采用的模式有关。
cbc模式
openssl enc -aes-128-cbc -e -in ./f.txt -out ./cbc/f-cbc.txt -K 00112233445566778889aabbccddeeff -iv 0102030405060708
bless ./cbc/f-cbc.txt
openssl enc -aes-128-cbc -d -in ./cbc/f-cbc.txt -out ./cbc/f-cbc-out.txt -K 00112233445566778889aabbccddeeff -iv 0102030405060708
vim ./cbc/f-cbc-out.txt
ofb模式
openssl enc -aes-128-ofb -e -in ./f.txt -out ./ofb/f-ofb.txt -K 00112233445566778889aabbccddeeff -iv 0102030405060708
bless ./ofb/f-ofb.txt
openssl enc -aes-128-ofb -d -in ./ofb/f-ofb.txt -out ./ofb/f-ofb-out.txt -K 00112233445566778889aabbccddeeff -iv 0102030405060708
vim ./ofb/f-ofb-out.txt
Task 6: Initial Vector (IV) and Common Mistakes
Task 6.1. IV Experiment
用AES加密算法（128位密钥，CBC模式），采用相同的密钥和不同的初始化向量加密一个相同的文件，观察加密结果是否相同
echo -n "01234" > f.txt
openssl enc -aes-128-cbc -e -in ./f.txt -out ./f-cbc-iv1.txt -K 123456 -iv 111111
openssl enc -aes-128-cbc -e -in ./f.txt -out ./f-cbc-iv2.txt -K 123456 -iv 222222
cat f-cbc-iv1.txt
cat f-cbc-iv2.txt
用AES加密算法（128位密钥，CBC模式），采用相同的密钥和相同的初始化向量加密一个相同的文件，观察加密结果是否相同
openssl enc -aes-128-cbc -e -in ./f.txt -out ./f-cbc-iv1-same.txt -K 123456 -iv 111111
cat f-cbc-iv1.txt
cat f-cbc-iv1-same.txt
思考为什么IV必须唯一
从实验结果可以发现，在key相同的情况，相同的IV会导致加密结果相同，所以为了信息安全IV必须唯一。
Task 6.2. Common Mistake: Use the Same IV
已知明文攻击：假设采用OFB模式进行加密，加密时所采用的IV是相同的，请破译P2的内容。
Task 6.3. Common Mistake: Use a Predictable IV
鸽
Task 7: Programming using the Crypto Library
大致思想：这个任务说的是，已知key是由words.txt文件中的某一行填充而成，给你明文、密文、iv以及words.txt，能不能找到key
代码思路：核心思想就是把words.txt中的每一个单词当成key，结合已知的iv构造aes，然后使用这个aes去解码密文，如果得到的明文和已知明文相同，那就找到了key。
碰到的问题：python中的Crypto包，把iv定死了成16bite，而任务给的是32bite的iv，没有办法，就自己重新生成了密文，希望老师不介意。（并且自己来做的话，填充函数这边就可以自己发挥了）
import base64
from Crypto.Cipher import AES
class EncryptDate:
def __init__(self, key, iv):
self.length = 16
self.key = self.pad(key).encode("utf-8")
self.iv = iv.encode("utf-8")
self.aes = AES.new(self.key, AES.MODE_CBC, self.iv)
self.unpad = lambda s: s[0:-s[-1]]
def pad(self, text): # 填充函数
count = len(text.encode('utf-8'))
add = self.length - (count % self.length)
entext = text + (chr(add) * add)
return entext
def encrypt(self, encrData): # 加密函数
a = self.pad(encrData)
res = self.aes.encrypt(a.encode("utf-8"))
msg = str(base64.b64encode(res), encoding="utf8")
return msg
def decrypt(self, decrData): # 解密函数
res = base64.decodebytes(decrData.encode("utf-8"))
msg_text = self.aes.decrypt(res)
decrypt_text = self.unpad(msg_text).decode('utf8')
return decrypt_text
def main():
# 使用Syracuse作为key，9999999999999999作为iv对明文进行加密
aes_key = "Syracuse"
aes_iv = "9999999999999999"
text_data = 'This is a top secret.'
encrypt_data = EncryptDate(aes_key, aes_iv).encrypt(text_data)
# 在words.txt中找到key
# 寻找的方式就是使用该单词作为key，配合已知的iv进行解密，若得到的明文和已知的明文相同，则找到了key
with open("words.txt") as f:
for line in f.readlines():
word = line.strip()
try:
decrypt_data = EncryptDate(word, aes_iv).decrypt(encrypt_data)
except Exception as e:
continue
if(decrypt_data == text_data):
print("原来key是："+word)
break;
if __name__ == '__main__':
main()
