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

import base64
str1="49276d206b696c6c696e6720796f757220627261696e206c696b65206120706f69736f6e6f7573206d757368726f6f6d".decode("hex")
me = base64.b64encode(str1)
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。

#coding=utf-8
import base64
import re
#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
#str1=long("1c0111001f010100061a024b53535009181c",16)
#str2=long("686974207468652062756c6c277320657965",16)
str1="1c0111001f010100061a024b53535009181c".decode('hex')
str2="686974207468652062756c6c277320657965".decode('hex')
str3=[]
for i in range(0,len(str1)):
    str3+=[chr(ord(str1[i])^ord(str2[i]))]
str3="".join(str3)
print str3.encode('hex')

#str=str1^str2
#print hex(str)
#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：

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

法二：

#coding=utf-8
import re
def english_test(sentence):
    score = 0
    freqs = {
        'a': 0.0651738, 'b': 0.0124248, 'c': 0.0217339,
        'd': 0.0349835, 'e': 0.1041442, 'f': 0.0197881,
        'g': 0.0158610, 'h': 0.0492888, 'i': 0.0558094,
        'j': 0.0009033, 'k': 0.0050529, 'l': 0.0331490,
        'm': 0.0202124, 'n': 0.0564513, 'o': 0.0596302,
        'p': 0.0137645, 'q': 0.0008606, 'r': 0.0497563,
        's': 0.0515760, 't': 0.0729357, 'u': 0.0225134,
        'v': 0.0082903, 'w': 0.0171272, 'x': 0.0013692,
        'y': 0.0145984, 'z': 0.0007836, ' ': 0.1918182}
    for x in sentence.lower():
        if x in freqs:
            score += freqs[x]
    return score

str="1b37373331363f78151b7f2b783431333d78397828372d363c78373e783a393b3736"
score=0
for i in range(0,129):
    tmp=[]
    for j in re.findall(".{2}",str):#任意两个字符的字符串
        tmp += chr(i^int(j,16))
    tmpstr = "".join(tmp)
    num=english_test(tmpstr)
    if num>score:
        #print tmpstr
        score=num#用于更新用
        ansstr=tmpstr
        key=chr(i)
print key
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）

#coding=utf-8
import re
#with open("ex4.txt") as fp:
#wenben=[i for i in open("ex4.txt").readlines()]
wenben=[]
for i in open("ex4.txt","r").readlines():
    wenben+=[i.replace("
","")] #序列相加
score=0
for k in wenben:
    for i in range(0,129):
        tmp=[]
        for j in re.findall(".{2}",k):#任意两个字符的字符串
            tmp += chr(i^int(j,16))
        tmpstr = "".join(tmp)
        num=0
        num=len(re.findall(r'[a-zA-Z ]',tmpstr))#一定注意不要落下空格
        if num>score:
            score=num#用于更新用
            ansstr=tmpstr
            c=k
            key=chr(i)
print c
print key
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

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

import re
str1="Burning 'em, if you ain't quick and nimble I go crazy when I hear a cymbal"
str2="ICE"*200
str3=[]
for i in range(0,len(str1)):
    str3 +=[(chr(ord(str1[i])^ord(str2[i])))]
print "".join(str3).encode('hex')
#for i in range(0,len(str1)):
#    str3 +=[(hex(ord(str1[i])^ord(str2[i])))]
#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

#coding:utf-8
import re
import base64
with open("ex6.txt","r") as fp:
    wenben=[base64.b64decode(i) for i in fp.readlines()]
wenben="".join(wenben)

def english_test(sentence):
    score = 0
    freqs = {
        'a': 0.0651738, 'b': 0.0124248, 'c': 0.0217339,
        'd': 0.0349835, 'e': 0.1041442, 'f': 0.0197881,
        'g': 0.0158610, 'h': 0.0492888, 'i': 0.0558094,
        'j': 0.0009033, 'k': 0.0050529, 'l': 0.0331490,
        'm': 0.0202124, 'n': 0.0564513, 'o': 0.0596302,
        'p': 0.0137645, 'q': 0.0008606, 'r': 0.0497563,
        's': 0.0515760, 't': 0.0729357, 'u': 0.0225134,
        'v': 0.0082903, 'w': 0.0171272, 'x': 0.0013692,
        'y': 0.0145984, 'z': 0.0007836, ' ': 0.1918182}
    for x in sentence.lower():
        if x in freqs:
            score += freqs[x]
    return score

def hanming(x,y):
    num=0
    for i in range(0,len(x)):
        t=ord(x[i])^ord(y[i])
        while t:
           if t&1 : num+=1
           t>>=1
    return num

def thechar(st1):
    score = 0
    for i in range(0, 255):
        tmp = []
        for j in range(0,len(st1)):  # 任意两个字符的字符串
            tmp += chr(i ^ int(st1[j],16))
        tmpstr = "".join(tmp)

        #num=len(re.findall(r'[a-zA-Z ,.;?!:]',tmpstr))  #'[a-zA-Z ,.?!:;]'
        num=english_test(tmpstr)
        if num > score:
            score = num  # 用于更新用
            key = chr(i)
            #print key,score
    return key


ans = []
for i in range(1,41):
    str1=[]
    str2=[]
    str3=[]
    str4=[]
    for j in range(0,i): str1+=[wenben[j]]
    for j in range(i,2*i): str2+=[wenben[j]]
    for j in range(2*i,3*i): str3+=[wenben[j]]
    for j in range(3*i,4*i): str4+=[wenben[j]]
    str1="".join(str1)
    str2="".join(str2)
    str3="".join(str3)
    str4="".join(str4)
    x1=float(hanming(str1,str2))/i
    x2=float(hanming(str2,str3))/i
    x3=float(hanming(str3,str4))/i
    x4=float(hanming(str1,str4))/i
    x5=float(hanming(str1,str3))/i
    x6=float(hanming(str2,str4))/i
    aa=(x1+x2+x3+x4+x5+x6)/6
    ans+=[(i,aa)]
ans.sort(lambda x,y:cmp(x[1],y[1]))
for i in range(len(ans)):
    print ans[i][0],ans[i][1]
#print len(wenben)
#print len(wenben)%29
wenben=wenben.encode('hex')

block=[re.findall(r'(.{2})',z)  for z in re.findall(r'(.{58})',wenben)]



keyy = []
for i in range(0,29):
    tmp=[]
    for j in range(0,len(block)):
        tmp+=[block[j][i]]
    keyy+=[thechar(tmp)]
keyy="".join(keyy)

print keyy
keyy=keyy*10000

wenben=wenben.decode('hex')
an=[]
for i in range(0,len(wenben)):
    an+=[chr(ord(wenben[i])^ord(keyy[i]))]
an="".join(an)
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

#coding:utf-8
import re
import base64
with open("ex6.txt","r") as fp:
    wenben=[base64.b64decode(i) for i in fp.readlines()]
print len(wenben[2])
wenben="".join(wenben)

def english_test(sentence):
    score = 0
    freqs = {
        'a': 0.0651738, 'b': 0.0124248, 'c': 0.0217339,
        'd': 0.0349835, 'e': 0.1041442, 'f': 0.0197881,
        'g': 0.0158610, 'h': 0.0492888, 'i': 0.0558094,
        'j': 0.0009033, 'k': 0.0050529, 'l': 0.0331490,
        'm': 0.0202124, 'n': 0.0564513, 'o': 0.0596302,
        'p': 0.0137645, 'q': 0.0008606, 'r': 0.0497563,
        's': 0.0515760, 't': 0.0729357, 'u': 0.0225134,
        'v': 0.0082903, 'w': 0.0171272, 'x': 0.0013692,
        'y': 0.0145984, 'z': 0.0007836, ' ': 0.1918182}
    for x in sentence.lower():
        if x in freqs:
            score += freqs[x]
    return score

def hanming(x,y):
    num=0
    for i in range(0,len(x)):
        t=ord(x[i])^ord(y[i])
        while t:
           if t&1 : num+=1
           t>>=1
    return num

def thechar(st1):
    score = 0
    for i in range(0, 255):
        tmp = []
        for j in range(0,len(st1)):  # 任意两个字符的字符串
            #print str1
            tmp += chr(i ^ ord(st1[j]))
        tmpstr = "".join(tmp)
        #num=len(re.findall(r'[a-zA-Z ,.;?!:]',tmpstr))  #'[a-zA-Z ,.?!:;]'
        num=english_test(tmpstr)
        if num > score:
            score = num  # 用于更新用
            key = chr(i)
            #print key,score
    return key

ans = []
for i in range(1,41):
    str1=[]
    str2=[]
    str3=[]
    str4=[]
    for j in range(0,i): str1+=[wenben[j]]
    for j in range(i,2*i): str2+=[wenben[j]]
    for j in range(2*i,3*i): str3+=[wenben[j]]
    for j in range(3*i,4*i): str4+=[wenben[j]]
    str1="".join(str1)
    str2="".join(str2)
    str3="".join(str3)
    str4="".join(str4)
    x1=float(hanming(str1,str2))/i
    x2=float(hanming(str2,str3))/i
    x3=float(hanming(str3,str4))/i
    x4=float(hanming(str1,str4))/i
    x5=float(hanming(str1,str3))/i
    x6=float(hanming(str2,str4))/i
    aa=(x1+x2+x3+x4+x5+x6)/6
    ans+=[(i,aa)]
ans.sort(lambda x,y:cmp(x[1],y[1]))
for i in range(len(ans)):
    print ans[i][0],ans[i][1]
#print len(wenben)
#print len(wenben)%29
#wenben=wenben.encode('hex')
#print wenben
#wenben=wenben.decode('utf-8')
block=re.findall(r'[sS]{29}',wenben)
#block=[wenben[i:i+29] for i in xrange(0,len(wenben),29)]
#for i in range(0,len(block)):
#    block[i]=block[i]
b1=[]
print block
#print block[0][4].encode('hex')
print block[0]
print len(block[0])
nn=0
for i in range(0,len(block)):
    for j in range(0,len(block[i])):
        nn+=1
print nn

print [z.encode('hex')  for z in block[3]]
print block[3][28].encode('hex')
print len(block[3])
print block[3][0].encode('hex')
print block[3]
#['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']

keyy = []
for i in range(0,29):
    tmp=[]
    for j in range(0,len(block)):
        tmp+=block[j][i]
    tmp="".join(tmp)
    keyy+=thechar(tmp)
keyy="".join(keyy)

print keyy
keyy=keyy*10000

#wenben=wenben.decode('hex')
an=[]
for i in range(0,len(wenben)):
    an+=[chr(ord(wenben[i])^ord(keyy[i]))]
an="".join(an)
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。

from Crypto.Cipher import AES
import base64,re
with open("ex7.txt","r") as fp:
    C=[base64.b64decode(i.replace("
","")) for i in fp.readlines()]
C="".join(C)

key = "YELLOW SUBMARINE"
cipher=AES.new(key,AES.MODE_ECB)
m= cipher.decrypt(C)
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

import re
with open("ex8.txt","r") as fp:
    wenben=[i.replace("
","")  for i in fp.readlines()]
for ecb in wenben:
    block =re.findall(".{16}",ecb)
    if len(block)-len(set(block)):
        print ecb