0x00 题目源码
#!/usr/bin/python3
import sys
from random import *
#from secret import flag
flag = "a1234567890123456789"
HACK=False
NUMS=20
rd2=lambda:randint(0,1)
getBase=lambda lens=NUMS:[rd2() for i in range(lens)]
getBins=lambda strings:[list(map(int,list(bin(ord(i))[2:].rjust(8,'0')))) for i in strings]
def inits():
while True:
a_base,b_base=getBase(),getBase()
check=[1 if a_base[i] == b_base[i] else 0 for i in range(NUMS)]
if sum(check) > 8:
break
print('check ',check)
return a_base,b_base,check
def hack(hack_base,a_base,data):
res=[]
assert len(hack_base) == len(a_base) == len(data)
for i in range(len(a_base)):
if hack_base[i] != a_base[i]:
data[i]=rd2()
return data
def qb(x,y):
assert len(x)==len(y)
tmp=[['0','1'],['o','l']]
return [tmp[x[i]][y[i]] for i in range(len(x))]
def comm(a_base,b_base,check,hack_base=None):
bf,index=getBins(flag),0
for i in bf:
x=0
m=[]
for j in check:
if j and x<8:
m.append(i[x])
x+=1
else:
m.append(rd2())
enc=qb(a_base,m)
print("len of m:", len(m))
if HACK:
hack_m=hack(hack_base,a_base,m)
hack_enc=qb(hack_base,hack_m)
print ('hack->',''.join(hack_enc))
x=0
c=''
for j in range(NUMS):
if check[j] and x<8:
c+=str(m[j])
x+=1
if chr(int(c,2)) == flag[index]:
print("Received the %d char successfully."%index)
index+=1
else:
print("Received the %d char error, maybe have hack"%index)
break
if __name__ == '__main__':
#print(getBins(flag))
a_base,b_base,check=inits()
try:
print('Do you want to monitor communication? (y/n)',end=' ')
if input()[0] == 'y':
sys.stdout.flush()
hack_base=list(map(int,input("Please input your base(use ',' to split): ").split(',')))
assert len(hack_base)==NUMS
for i in hack_base:
assert i in [0,1]
HACK=True
else:
hack_base=None
print("Good boy~")
except:
print("Oh~~ bad boooooooooy")
sys.exit(0)
print('hackse',hack_base)
comm(a_base,b_base,check,hack_base)
0x01 源码分析
- 小函数
rd2：生成随机的0或者1
getBase：生成长度为NUMS的01字符串
getBins：将字符串中的每个字符，编码为8个比特的01串（ascii码值的二进制表示） - def inits()函数
随机生成01序列的a_base[20]和b_base[20]
check=[1 if a_base[i] == b_base[i] else 0 for i in range(NUMS)]
上一行可以优化为check[i]=a_base[i] ^ b_base[i] ^ 1
其中check中1的位数大于等于8 - def hack(hack_base,a_base,data)函数
hack_base[i] == a_base[i]时，data[i]不变，否则data[i]变成随机的01 - def qb(x,y)函数
就是一个编码函数，通过输出的01ol可以确定出来x[i]和y[i]的值 - def comm(a_base,b_base,check,hack_base=None)
编码函数，将flag的每个字符ch散列到长度为NUMS的01串中，相应于check中前8个1的位置放置ch[i]，其他位置为rand 0 1，生成m。其中，check的前8个1的位置为有效位。
例如，当chek=[0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1]
那么第2,5,9,10,11,13,14,16位为有效位。
在HACK模式下：
hack_m=hack(hack_base,a_base,m) //在有效位，如果hack_base[i] != a_base[i]，m会被修改，导致解密失败
hack_enc=qb(hack_base,hack_m)//只是为了输出，通过输出的01ol来反推hack_base和hack_m
print ('hack->',''.join(hack_enc))
0x02 结论
针对每一个有效位，如果hack_base与a_base的值相同，那么m就不会被修改，即可解密成功，根据输出即可解密flag的ch[i]。有效位为8位，因此，碰撞空间为2**8=256
a_base随机生成，本来考虑当a_base确定后，去穷举hack_base。但是，一旦a_base确定后，只能试一次，程序即运行结束，无法穷举hack_base。
那就固定hack_base，穷举a_base，也即通过多次调用nc xxxx xx，经过O(256)量级的穷举，总有一次会使得hack_base==a_base的有效位，输出hack_enc即可解密。
为了方便解密，令hack_base=[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
穷举a_base的脚本（半手工，因为pwn需要手动停），本人运行了几十次，成功获取了flag
from pwn import *
#sh = process('./pwn')
sh=remote("47.94.21.141","21325")
payload = "0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0"
print sh.readline()
sh.recvuntil(") ")
sh.sendline("y")
sh.sendline(payload)
sh.interactive()
exit(0)
优化后的批量脚本：
from pwn import *
payload = "0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0"
for i in range(0, 256):
print "trying[", i , "]"
#sh = process('./haha.py')
sh=remote("xxxx","4444")
print sh.readline()
sh.recvuntil(") ")
sh.sendline("y")
sh.sendline(payload)
#sh.interactive()
res = sh.recvall()
print res
if("Received the 19 char successful" in res):
print res
break
sh.close()
当碰撞成功后，获得了42个01字符串，解密脚本如下：
check = [1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0]
ss = ["00001110001100000110",
"01101010011010001010",
"01011011000001111110",
"01111011001101111100",
"01111011110100111010",
"01101010001011010101",
"01111011000000101100",
"00110111100011100100",
"01101011000010110011",
"00111111001010010110",
"01000010101010111111",
"00100110100100000011",
"00000111100100111000",
"01110110011010101110",
"01010111100011001000",
"01010011101011100101",
"01111010001110010111",
"00010111110010011000",
"00000011011000101110",
"01010010101011000001",
"01010111101001000111",
"00000011101101010000",
"01101110001110001110",
"00000110011010100101",
"00011011000010110000",
"01110010110010101011",
"01001010001100011010",
"01011110000010100101",
"00110010011000111001",
"00101110001010101011",
"01011011000010110000",
"01000110110010100000",
"01010010100101000001",
"00110110101000100001",
"01000010100110000001",
"01010110110000010011",
"01011110001011000100",
"01100011100001001100",
"00100011101101001011",
"01000111100000001001",
"01110011110010001111",
"01101111111000100111"]
NUMS = 20
res = ""
print len(ss)
for item in ss:
x = 0
c = ''
for j in range(NUMS):
if check[j] and x < 8:
c += str(item[j])
x += 1
res += chr(int(c, 2))
print res
获得flag
flag{da1ad523-05f8-446f-a9fa-ea92528d0608}