Crypto Challenge Set 1解题报告

1、Convert hex to base64

题意：给出一个hex编码过的字符串，将它进行base64加密

解题关键：直接利用base64库函数实现

1 import base64
2 str1="49276d206b696c6c696e6720796f757220627261696e206c696b65206120706f69736f6e6f7573206d757368726f6f6d".decode("hex")
3 me = base64.b64encode(str1)
4 print me

2、Fixed XOR

题意：将两个16进制字符串进行异或

解题关键：将16进制字符串解码，对每个字符分别进行异或，最后编码成16进制即可。

 1 #coding=utf-8
 2 import base64
 3 import re
 4 #异或操作无论对什么数都是以二进制的形式实现，所以无所谓进制
 5 #str1=long("1c0111001f010100061a024b53535009181c",16)
 6 #str2=long("686974207468652062756c6c277320657965",16)
 7 str1="1c0111001f010100061a024b53535009181c".decode('hex')
 8 str2="686974207468652062756c6c277320657965".decode('hex')
 9 str3=[]
10 for i in range(0,len(str1)):
11     str3+=[chr(ord(str1[i])^ord(str2[i]))]
12 str3="".join(str3)
13 print str3.encode('hex')
14 
15 #str=str1^str2
16 #print hex(str)
17 #print str

3、Single-byte XOR cipher

题意：字符串被某个单字符加密过，找出这个字符

解题关键：1、字符所处的ASCII码范围为0-255，暴力搜索，得出字符串字母数量最多的即为解。

                  2、判断的方式还可以按照英文中各字母出现的频率计算。

法一：

 1 #coding=utf-8
 2 import re
 3 str="1b37373331363f78151b7f2b783431333d78397828372d363c78373e783a393b3736"
 4 score=0
 5 for i in range(0,129):
 6     tmp=[]
 7     for j in re.findall(".{2}",str):#任意两个字符的字符串
 8         tmp += chr(i^int(j,16))
 9     tmpstr = "".join(tmp)
10     num=0
11     for j in range(0,len(tmpstr)):
12         if tmpstr[j]>='a'and tmpstr[j]<='z':#or tmpstr[j]>='A'and tmpstr[j]<='Z':
13             num+=1
14     if num>score:
15         #print tmpstr
16         score=num#用于更新用
17         ansstr=tmpstr
18         key=chr(i)
19 print key
20 print ansstr

法二：

 1 #coding=utf-8
 2 import re
 3 def english_test(sentence):
 4     score = 0
 5     freqs = {
 6         'a': 0.0651738, 'b': 0.0124248, 'c': 0.0217339,
 7         'd': 0.0349835, 'e': 0.1041442, 'f': 0.0197881,
 8         'g': 0.0158610, 'h': 0.0492888, 'i': 0.0558094,
 9         'j': 0.0009033, 'k': 0.0050529, 'l': 0.0331490,
10         'm': 0.0202124, 'n': 0.0564513, 'o': 0.0596302,
11         'p': 0.0137645, 'q': 0.0008606, 'r': 0.0497563,
12         's': 0.0515760, 't': 0.0729357, 'u': 0.0225134,
13         'v': 0.0082903, 'w': 0.0171272, 'x': 0.0013692,
14         'y': 0.0145984, 'z': 0.0007836, ' ': 0.1918182}
15     for x in sentence.lower():
16         if x in freqs:
17             score += freqs[x]
18     return score
19 
20 str="1b37373331363f78151b7f2b783431333d78397828372d363c78373e783a393b3736"
21 score=0
22 for i in range(0,129):
23     tmp=[]
24     for j in re.findall(".{2}",str):#任意两个字符的字符串
25         tmp += chr(i^int(j,16))
26     tmpstr = "".join(tmp)
27     num=english_test(tmpstr)
28     if num>score:
29         #print tmpstr
30         score=num#用于更新用
31         ansstr=tmpstr
32         key=chr(i)
33 print key
34 print ansstr

 

4、 Detect single-character XOR

题意： 文本中有一个字符串被单字符加密过，找出这个字符串

解题关键：对file中所有的字符串进行暴力匹配字符，最终得到小写字母数量最多的字符串和秘钥字符即为解（以下进行异或匹配的时候，判断方式均可运用频率函数）

 1 #coding=utf-8
 2 import re
 3 #with open("ex4.txt") as fp:
 4 #wenben=[i for i in open("ex4.txt").readlines()]
 5 wenben=[]
 6 for i in open("ex4.txt","r").readlines():
 7     wenben+=[i.replace("\n","")] #序列相加
 8 score=0
 9 for k in wenben:
10     for i in range(0,129):
11         tmp=[]
12         for j in re.findall(".{2}",k):#任意两个字符的字符串
13             tmp += chr(i^int(j,16))
14         tmpstr = "".join(tmp)
15         num=0
16         num=len(re.findall(r'[a-zA-Z ]',tmpstr))#一定注意不要落下空格
17         if num>score:
18             score=num#用于更新用
19             ansstr=tmpstr
20             c=k
21             key=chr(i)
22 print c
23 print key
24 print ansstr

5、Implement repeating-key XOR

题意：将某个字符串使用秘钥进行重复异或

解题关键： 将秘钥扩展，单字节异或即可，最后编码成16进制

1 import re
2 str1=re.findall('.{2}',"Burning 'em, if you ain't quick and nimble I go crazy when I hear a cymbal".encode('hex'))
3 str2=re.findall('.{2}',("ICE"*200).encode('hex'))
4 str3=[]
5 for i in range(0,len(str1)):
6     str3 +=[(chr(int(str1[i],16)^int(str2[i],16)))]
7 print "".join(str3).encode('hex')
 1 import re
 2 str1="Burning 'em, if you ain't quick and nimble I go crazy when I hear a cymbal"
 3 str2="ICE"*200
 4 str3=[]
 5 for i in range(0,len(str1)):
 6     str3 +=[(chr(ord(str1[i])^ord(str2[i])))]
 7 print "".join(str3).encode('hex')
 8 #for i in range(0,len(str1)):
 9 #    str3 +=[(hex(ord(str1[i])^ord(str2[i])))]
10 #print "".join(str3)

 

6、Break repeating-key XOR

题意：base64编码后的字符串使用某串key加密过，解出明文

解题关键：首先按照可能的keysize对密文进行分块，取前4个进行两两求汉明距离，若分块的长度等于keysize，则应具有最小的汉明距离，（字母之间的汉明距离小），得到keysize之后，对分块的第一位、第二位分别进行匹配，最终得到解。

法一：转化为hex求解

  1 #coding:utf-8
  2 import re
  3 import base64
  4 with open("ex6.txt","r") as fp:
  5     wenben=[base64.b64decode(i) for i in fp.readlines()]
  6 wenben="".join(wenben)
  7 
  8 def english_test(sentence):
  9     score = 0
 10     freqs = {
 11         'a': 0.0651738, 'b': 0.0124248, 'c': 0.0217339,
 12         'd': 0.0349835, 'e': 0.1041442, 'f': 0.0197881,
 13         'g': 0.0158610, 'h': 0.0492888, 'i': 0.0558094,
 14         'j': 0.0009033, 'k': 0.0050529, 'l': 0.0331490,
 15         'm': 0.0202124, 'n': 0.0564513, 'o': 0.0596302,
 16         'p': 0.0137645, 'q': 0.0008606, 'r': 0.0497563,
 17         's': 0.0515760, 't': 0.0729357, 'u': 0.0225134,
 18         'v': 0.0082903, 'w': 0.0171272, 'x': 0.0013692,
 19         'y': 0.0145984, 'z': 0.0007836, ' ': 0.1918182}
 20     for x in sentence.lower():
 21         if x in freqs:
 22             score += freqs[x]
 23     return score
 24 
 25 def hanming(x,y):
 26     num=0
 27     for i in range(0,len(x)):
 28         t=ord(x[i])^ord(y[i])
 29         while t:
 30            if t&1 : num+=1
 31            t>>=1
 32     return num
 33 
 34 def thechar(st1):
 35     score = 0
 36     for i in range(0, 255):
 37         tmp = []
 38         for j in range(0,len(st1)):  # 任意两个字符的字符串
 39             tmp += chr(i ^ int(st1[j],16))
 40         tmpstr = "".join(tmp)
 41 
 42         #num=len(re.findall(r'[a-zA-Z ,\.;?!:]',tmpstr))  #'[a-zA-Z ,\.?!:;]'
 43         num=english_test(tmpstr)
 44         if num > score:
 45             score = num  # 用于更新用
 46             key = chr(i)
 47             #print key,score
 48     return key
 49 
 50 
 51 ans = []
 52 for i in range(1,41):
 53     str1=[]
 54     str2=[]
 55     str3=[]
 56     str4=[]
 57     for j in range(0,i): str1+=[wenben[j]]
 58     for j in range(i,2*i): str2+=[wenben[j]]
 59     for j in range(2*i,3*i): str3+=[wenben[j]]
 60     for j in range(3*i,4*i): str4+=[wenben[j]]
 61     str1="".join(str1)
 62     str2="".join(str2)
 63     str3="".join(str3)
 64     str4="".join(str4)
 65     x1=float(hanming(str1,str2))/i
 66     x2=float(hanming(str2,str3))/i
 67     x3=float(hanming(str3,str4))/i
 68     x4=float(hanming(str1,str4))/i
 69     x5=float(hanming(str1,str3))/i
 70     x6=float(hanming(str2,str4))/i
 71     aa=(x1+x2+x3+x4+x5+x6)/6
 72     ans+=[(i,aa)]
 73 ans.sort(lambda x,y:cmp(x[1],y[1]))
 74 for i in range(len(ans)):
 75     print ans[i][0],ans[i][1]
 76 #print len(wenben)
 77 #print len(wenben)%29
 78 wenben=wenben.encode('hex')
 79 
 80 block=[re.findall(r'(.{2})',z)  for z in re.findall(r'(.{58})',wenben)]
 81 
 82 
 83 
 84 keyy = []
 85 for i in range(0,29):
 86     tmp=[]
 87     for j in range(0,len(block)):
 88         tmp+=[block[j][i]]
 89     keyy+=[thechar(tmp)]
 90 keyy="".join(keyy)
 91 
 92 print keyy
 93 keyy=keyy*10000
 94 
 95 wenben=wenben.decode('hex')
 96 an=[]
 97 for i in range(0,len(wenben)):
 98     an+=[chr(ord(wenben[i])^ord(keyy[i]))]
 99 an="".join(an)
100 print an

  法二：直接对字符串进行求解：debug了两天，原来是正则表达式的.无法匹配换行符的原因，换了个写法就过了

  1 #coding:utf-8
  2 import re
  3 import base64
  4 with open("ex6.txt","r") as fp:
  5     wenben=[base64.b64decode(i) for i in fp.readlines()]
  6 print len(wenben[2])
  7 wenben="".join(wenben)
  8 
  9 def english_test(sentence):
 10     score = 0
 11     freqs = {
 12         'a': 0.0651738, 'b': 0.0124248, 'c': 0.0217339,
 13         'd': 0.0349835, 'e': 0.1041442, 'f': 0.0197881,
 14         'g': 0.0158610, 'h': 0.0492888, 'i': 0.0558094,
 15         'j': 0.0009033, 'k': 0.0050529, 'l': 0.0331490,
 16         'm': 0.0202124, 'n': 0.0564513, 'o': 0.0596302,
 17         'p': 0.0137645, 'q': 0.0008606, 'r': 0.0497563,
 18         's': 0.0515760, 't': 0.0729357, 'u': 0.0225134,
 19         'v': 0.0082903, 'w': 0.0171272, 'x': 0.0013692,
 20         'y': 0.0145984, 'z': 0.0007836, ' ': 0.1918182}
 21     for x in sentence.lower():
 22         if x in freqs:
 23             score += freqs[x]
 24     return score
 25 
 26 def hanming(x,y):
 27     num=0
 28     for i in range(0,len(x)):
 29         t=ord(x[i])^ord(y[i])
 30         while t:
 31            if t&1 : num+=1
 32            t>>=1
 33     return num
 34 
 35 def thechar(st1):
 36     score = 0
 37     for i in range(0, 255):
 38         tmp = []
 39         for j in range(0,len(st1)):  # 任意两个字符的字符串
 40             #print str1
 41             tmp += chr(i ^ ord(st1[j]))
 42         tmpstr = "".join(tmp)
 43         #num=len(re.findall(r'[a-zA-Z ,\.;?!:]',tmpstr))  #'[a-zA-Z ,\.?!:;]'
 44         num=english_test(tmpstr)
 45         if num > score:
 46             score = num  # 用于更新用
 47             key = chr(i)
 48             #print key,score
 49     return key
 50 
 51 ans = []
 52 for i in range(1,41):
 53     str1=[]
 54     str2=[]
 55     str3=[]
 56     str4=[]
 57     for j in range(0,i): str1+=[wenben[j]]
 58     for j in range(i,2*i): str2+=[wenben[j]]
 59     for j in range(2*i,3*i): str3+=[wenben[j]]
 60     for j in range(3*i,4*i): str4+=[wenben[j]]
 61     str1="".join(str1)
 62     str2="".join(str2)
 63     str3="".join(str3)
 64     str4="".join(str4)
 65     x1=float(hanming(str1,str2))/i
 66     x2=float(hanming(str2,str3))/i
 67     x3=float(hanming(str3,str4))/i
 68     x4=float(hanming(str1,str4))/i
 69     x5=float(hanming(str1,str3))/i
 70     x6=float(hanming(str2,str4))/i
 71     aa=(x1+x2+x3+x4+x5+x6)/6
 72     ans+=[(i,aa)]
 73 ans.sort(lambda x,y:cmp(x[1],y[1]))
 74 for i in range(len(ans)):
 75     print ans[i][0],ans[i][1]
 76 #print len(wenben)
 77 #print len(wenben)%29
 78 #wenben=wenben.encode('hex')
 79 #print wenben
 80 #wenben=wenben.decode('utf-8')
 81 block=re.findall(r'[\s\S]{29}',wenben)
 82 #block=[wenben[i:i+29] for i in xrange(0,len(wenben),29)]
 83 #for i in range(0,len(block)):
 84 #    block[i]=block[i]
 85 b1=[]
 86 print block
 87 #print block[0][4].encode('hex')
 88 print block[0]
 89 print len(block[0])
 90 nn=0
 91 for i in range(0,len(block)):
 92     for j in range(0,len(block[i])):
 93         nn+=1
 94 print nn
 95 
 96 print [z.encode('hex')  for z in block[3]]
 97 print block[3][28].encode('hex')
 98 print len(block[3])
 99 print block[3][0].encode('hex')
100 print block[3]
101 #['37', '16', '06', '0c', '1a', '17', '41', '1d', '01', '52', '54', '30', '5f', '00', '20', '13', '0a', '05', '47', '4f', '12', '48', '08', '45', '4e', '65', '3e', '16', '09']
102 
103 keyy = []
104 for i in range(0,29):
105     tmp=[]
106     for j in range(0,len(block)):
107         tmp+=block[j][i]
108     tmp="".join(tmp)
109     keyy+=thechar(tmp)
110 keyy="".join(keyy)
111 
112 print keyy
113 keyy=keyy*10000
114 
115 #wenben=wenben.decode('hex')
116 an=[]
117 for i in range(0,len(wenben)):
118     an+=[chr(ord(wenben[i])^ord(keyy[i]))]
119 an="".join(an)
120 print an

 

 

7、AES in ECB mode

题意：解密进行aes-128加密过的字符串

解题关键：利用pycrypto库，首先将秘钥编码，然后进行AES编码即可。

 1 from Crypto.Cipher import AES
 2 import base64,re
 3 with open("ex7.txt","r") as fp:
 4     C=[base64.b64decode(i.replace("\n","")) for i in fp.readlines()]
 5 C="".join(C)
 6 
 7 key = "YELLOW SUBMARINE"
 8 cipher=AES.new(key,AES.MODE_ECB)
 9 m= cipher.decrypt(C)
10 print m

8、Detect AES in ECB mode

题意：找出被ECB加密过的字符串

解题关键：将文本连接起来，以16字节分块，根据ECB的性质，相同的 16 字节明文经过加密后总会产生相同的 16 字节密文，若存在相同块，即可ECS编码

1 import re
2 with open("ex8.txt","r") as fp:
3     wenben=[i.replace("\n","")  for i in fp.readlines()]
4 for ecb in wenben:
5     block =re.findall(".{16}",ecb)
6     if len(block)-len(set(block)):
7         print ecb

 

posted @ 2017-09-30 18:22  Elpsywk  阅读(1046)  评论(0)  编辑  收藏  举报