N1CTF baby_N1ES writeup
原文链接 http://findneo.tech/180310N1CTFWP/
注:以下为加速网络访问所做的原文缓存,经过重新格式化,可能存在格式方面的问题,或偶有遗漏信息,请以原文为准。
baby_N1ES
题目提供两个文件,challenge.py
和N1ES.py
。
虽然似乎在模仿AES,但是实际上明文和密文是一一映射的,复杂度不是恶心的100^48
而只是100*48
,穷举是很快的。一个小障碍是N1ES.py
第71行的L, R = R, L
,这导致了明文的[0:8]
、[8:16]
、[16:24]
、[24:32]
、[32:40]
、[40:48]
分别对应的是密文的[8:16]
、[0:8]
、[24:32]
、[16:24]
、[40:48]
、 [32:40]
,写穷举脚本时需要注意。
crack.py
import base64,string,N1ES
key = "wxy191iss00000000000cute"
c = base64.b64decode("HRlgC2ReHW1/WRk2DikfNBo1dl1XZBJrRR9qECMNOjNHDktBJSxcI1hZIz07YjVx")
n1es = N1ES.N1ES(key)
f=""
for i in xrange(3):
for j in xrange(16):
for k in string.printable:
s="x"*i*16+"x"*j+k+"x"*(48-i*16-j-1)
e=n1es.encrypt(s)
check=c[i*16+j+8]==e[i*16+j+8] if j<8 else c[i*16+j-8]==e[i*16+j-8]
if check:
f+=k
break
print f
# N1CTF{F3istel_n3tw0rk_c4n_b3_ea5i1y_s0lv3d_/--/}
challenge.py
from N1ES import N1ES
import base64
key = "wxy191iss00000000000cute"
n1es = N1ES(key)
flag = "N1CTF{*****************************************}"
cipher = n1es.encrypt(flag)
print base64.b64encode(cipher) # HRlgC2ReHW1/WRk2DikfNBo1dl1XZBJrRR9qECMNOjNHDktBJSxcI1hZIz07YjVx
N1ES.py
# -*- coding: utf-8 -*-
def round_add(a, b):
f = lambda x, y: x + y - 2 * (x & y)
res = ''
for i in range(len(a)):
res += chr(f(ord(a[i]), ord(b[i])))
return res
def permutate(table, block):
return list(map(lambda x: block[x], table))
def string_to_bits(data):
data = [ord(c) for c in data]
l = len(data) * 8
result = [0] * l
pos = 0
for ch in data:
for i in range(0,8):
result[(pos<<3)+i] = (ch>>i) & 1
pos += 1
return result
s_box = [54, 132, 138, 83, 16, 73, 187, 84, 146, 30, 95, 21, 148, 63, 65, 189, 188, 151, 72, 161, 116, 63, 161, 91, 37, 24, 126, 107, 87, 30, 117, 185, 98, 90, 0, 42, 140, 70, 86, 0, 42, 150, 54, 22, 144, 153, 36, 90, 149, 54, 156, 8, 59, 40, 110, 56,1, 84, 103, 22, 65, 17, 190, 41, 99, 151, 119, 124, 68, 17, 166, 125, 95, 65, 105, 133, 49, 19, 138, 29, 110, 7, 81, 134, 70, 87, 180, 78, 175, 108, 26, 121, 74, 29, 68, 162, 142, 177, 143, 86, 129, 101, 117, 41, 57, 34, 177, 103, 61, 135, 191, 74, 69, 147, 90, 49, 135, 124, 106, 19, 89, 38, 21, 41, 17, 155, 83, 38, 159, 179, 19, 157, 68, 105, 151, 166, 171, 122, 179, 114, 52, 183, 89, 107, 113, 65, 161, 141, 18, 121, 95, 4, 95, 101, 81, 156, 17, 190, 38, 84, 9, 171, 180, 59, 45, 15, 34, 89, 75, 164, 190, 140, 6, 41, 188, 77, 165, 105, 5, 107, 31, 183, 107, 141, 66, 63, 10, 9, 125, 50, 2, 153, 156, 162, 186, 76, 158, 153, 117, 9, 77, 156, 11, 145, 12, 169, 52, 57, 161, 7, 158, 110, 191, 43, 82, 186, 49, 102, 166, 31, 41, 5, 189, 27]
def generate(o):
k = permutate(s_box,o)
b = []
for i in range(0, len(k), 7):
b.append(k[i:i+7] + [1])
c = []
for i in range(32):
pos = 0
x = 0
for j in b[i]:
x += (j<<pos)
pos += 1
c.append((0x10001**x) % (0x7f))
return c
class N1ES:
def __init__(self, key):
if (len(key) != 24 or isinstance(key, bytes) == False ):
raise Exception("key must be 24 bytes long")
self.key = key
self.gen_subkey()
def gen_subkey(self):
o = string_to_bits(self.key)
k = []
for i in range(8):
o = generate(o)
k.extend(o)
o = string_to_bits([chr(c) for c in o[0:24]])
self.Kn = []
for i in range(32):
self.Kn.append(map(chr, k[i * 8: i * 8 + 8]))
return
def encrypt(self, plaintext):
if (len(plaintext) % 16 != 0 or isinstance(plaintext, bytes) == False):
raise Exception("plaintext must be a multiple of 16 in length")
res = ''
for i in range(len(plaintext) / 16):
block = plaintext[i * 16:(i + 1) * 16]
L = block[:8]
R = block[8:]
for round_cnt in range(32):
L, R = R, (round_add(L, self.Kn[round_cnt]))
L, R = R, L
res += L + R
return res
签到
IRC频道公告有字符串TjFDVEZ7V2VsYzBtZV90b19OMUNURl8yMDE4fQ
,base64解码后即flag。