2019全国大学生信息安全竞赛初赛writeup
part_des
- 题目描述
Round n part_encode-> 0x92d915250119e12b
Key map -> 0xe0be661032d5f0b676f82095e4d67623628fe6d376363183aed373a60167af537b46abc2af53d97485591f5bd94b944a3f49d94897ea1f699d1cdc291f2d9d4a5c705f2cad89e938dbacaca15e10d8aeaed90236f0be2e954a8cf0bea6112e84
按题意keymap就是des加密过程中生成的子密钥Kn
n应该是des加密过程16轮循环内的某一个临时值
des加密过程
有一个pyDes库可以进行常规的des加解密，我们在此基础上修改比较容易
先对n和keymap进行处理，转换为二进制
print(bin(n)[2:])
print(bin(keymap)[2:])
然后看到一篇文章，可以将子密钥转化为原始密钥
由子密钥反推deskey
代码如下：
import libnum
import binascii
key1 = [1,1,1,0,0,0,0,0,1,0,1,1,1,1,1,0,0,1,1,0,0,1,1,0,0,0,0,1,0,0,0,0,0,0,1,1,0,0,1,0,1,1,0,1,0,1,0,1,]
__pc2 = [
13, 16, 10, 23, 0, 4,
2, 27, 14, 5, 20, 9,
22, 18, 11, 3, 25, 7,
15, 6, 26, 19, 12, 1,
40, 51, 30, 36, 46, 54,
29, 39, 50, 44, 32, 47,
43, 48, 38, 55, 33, 52,
45, 41, 49, 35, 28, 31
]
C1D1 = ['*']*56
for i in range(0,len(key1)):
C1D1[__pc2[i]] = key1[i]
print C1D1
C0='000000001*11111111*111*10*00'
D0='0000111*11*1001*0000100001*0'
__pc1 = [56, 48, 40, 32, 24, 16, 8,
0, 57, 49, 41, 33, 25, 17,
9, 1, 58, 50, 42, 34, 26,
18, 10, 2, 59, 51, 43, 35,
62, 54, 46, 38, 30, 22, 14,
6, 61, 53, 45, 37, 29, 21,
13, 5, 60, 52, 44, 36, 28,
20, 12, 4, 27, 19, 11, 3
]
C0D0 = C0+D0
res = ['*']*64
deskey = ""
for i in range(0,len(__pc1)):
res[__pc1[i]] = C0D0[i]
for i in res:
deskey += i
print deskey
def zuoyiwei(str,num):
my = str[num:len(str)]
my = my+str[0:num]
return my
def key_change_1(str):
key1_list = [57,49,41,33,25,17,9,1,58,50,42,34,26,18,10,2,59,51,43,35,27,19,11,3,60,52,44,36,63,55,47,39,31,23,15,7,62,54,46,38,30,22,14,6,61,53,45,37,29,21,13,5,28,20,12,4]
res = ""
for i in key1_list:
res+=str[i-1]
return res
def key_change_2(str):
key2_list = [14,17,11,24,1,5,3,28,15,6,21,10,23,19,12,4,26,8,16,7,27,20,13,2,41,52,31,37,47,55,30,40,51,45,33,48,44,49,39,56,34,53,46,42,50,36,29,32]
res = ""
for i in key2_list:
res+=str[i-1]
return res
def key_gen(str):
key_list = []
key_change_res = key_change_1(str)
key_c = key_change_res[0:28]
key_d = key_change_res[28:]
for i in range(1,17):
if (i==1) or (i==2) or (i==9) or (i==16):
key_c = zuoyiwei(key_c,1)
key_d = zuoyiwei(key_d,1)
else:
key_c = zuoyiwei(key_c,2)
key_d = zuoyiwei(key_d,2)
key_yiwei = key_c+key_d
key_res = key_change_2(key_yiwei)
key_list.append(key_res)
return key_list
#01100***01**011*0111001*0110101*0110010*01*00*0*0*1*010*0110010*
deskey = '01100abc01de011f0111001g0110101h0110010i01j00k0l0m1n010o0110010p'
print key_gen(deskey)
deskey = '0110011c0110011f0111001g0110101h0110010i0110010L0111010o0110010p'
def bintostr(str):
res = ""
for i in range(0,len(str),8):
res += chr(int(str[i:i+8],2))
return res
for c in "01":
for f in "01":
for g in "01":
for h in "01":
for i in "01":
for L in "01":
for o in "01":
for p in "01":
str = '0110011'+c+'0110011'+f+'0111001'+g+'0110101'+h+'0110010'+i+'0110010'+L+'0111010'+o+'0110010'+p
str = bintostr(str)
print str
输出如下：
ffrjddtd
ffrjddte
ffrjddud
ffrjddue
ffrjdetd
ffrjdete
ffrjdeud
...
跟文章不一样，密钥全是可以显示的字符
将字符再转换成Kn，所有Kn都是一样的，有些奇怪
后来搜索了一下，发现密钥的第8位，16位，24位好像是不影响加密的，所以这么多个密钥都可以用来加解密，结果完全一样
接下来把pyDes.py的全部代码复制进来，再加上如下代码
这里是一个标准的des加解密流程，先随便加密8字节的明文12345678
key='ffskeeue'
result = des(key)
y = result.encrypt('12345678')
print(y)
z = result.decrypt(y)
print(z)
然后修改des类中的__des_crypt函数，也就是加密实际执行的函数，插入题目所给的n
i = 0
while i < 16:
############
if(i==13 and crypt_type == des.ENCRYPT):
self.L=[1,0,0,1,0,0,1,0,1,1,0,1,1,0,0,1,0,0,0,1,0,1,0,1,0,0,1,0,0,1,0,1]
self.R=[0,0,0,0,0,0,0,1,0,0,0,1,1,0,0,1,1,1,1,0,0,0,0,1,0,0,1,0,1,0,1,1]
print(i,self.L,self.R)
############
tempR = self.R[:]
这里的i==13是从1开始一个一个试出来的
在等于13时，解密结果为y0ur9Ood，看起来很像flag
包上flag{y0ur9Ood}，交上去就对了
其实Kn反解出密钥应该不是必要步骤，在pyDes里面覆盖掉Kn应该也可以
测试确实可以
warmup
- 题目描述
from Crypto.Cipher import AES
from Crypto.Util import Counter
from Crypto import Random
import binascii
import SocketServer
pad = lambda s: s + (16 - len(s) % 16) * chr(16 - len(s) % 16)
flag = "******************************************"
key = Random.get_random_bytes(16)
print binascii.b2a_hex(key)
prefix = Random.get_random_bytes(4)
suffix = Random.get_random_bytes(4)
def enc(plaintext):
count = Counter.new(64, prefix=prefix, suffix=suffix)
cipher = AES.new(key, AES.MODE_CTR, counter=count)
print(binascii.hexlify(pad(plaintext)))
return cipher.encrypt(pad(plaintext + flag))
class ThreadedTCPServer(SocketServer.ThreadingMixIn, SocketServer.TCPServer):
pass
class EncHandler(SocketServer.BaseRequestHandler):
def handle(self):
self.request.sendall("Welcome to flag getting system\n")
while 1:
self.request.sendall("plaintext>")
plaintext = self.request.recv(1024).strip()
ciphertext = binascii.hexlify(enc(plaintext))
self.request.sendall("result>" + ciphertext + '\n')
if __name__ == "__main__":
HOST, PORT = "0.0.0.0", 7777
server = ThreadedTCPServer((HOST, PORT), EncHandler)
server.serve_forever()
就是aes加密，输入明文，服务器返回明文+flag加密后的密文
加密模式是CTR，8字节一组
AES-CTR
每次连接后，密钥流是不变的，如果我们一开始先输入空明文
那么加密的就是AES(''+flag)，记录为str1
然后再输入比flag长度长的fffffffffffffffffffffffffffffffffff
加密的就是AES(“fffffffffffffffffffffffffffffffffff”+flag)，记录为str2
将str2与fffffffffffffffffffffffffffffffffff异或，就得到密钥流keystream = xor(str2, “fffffffffffffffffffffffffffffffffff”)
再将密钥流与str1异或，就得到flagflag = xor(keystream, str1)
题目不难，但是对bytes型，str型，二进制，各种类型转换不熟练，卡了好一会
脚本如下：
#python3
import binascii
import itertools
ffff = 'fffffffffffffffffffffffffffffffffffffffffffffffff'
print(ffff)
miwenf = binascii.unhexlify('0e3ecd7afef871...')
print(miwenf)
miwen = binascii.unhexlify('0e34ca7be3a7272b37da7245bbbf9.......')
print(miwen)
def xor(s, key):
key = key * (len(s) / len(key) + 1)
return ''.join(chr(ord(x) ^ ord(y)) for (x,y) in itertools.izip(s, key))
keyliu = xor(miwenf,ffff)
print(keyliu)
flag = xor(miwen,keyliu)
print(flag)
Asymmetric
- 题目描述
import gmpy2
import random
from Crypto.Util.number import *
from flag import flag
def generate_key(nbit):
p = getPrime(nbit)
r = random.randint(2, 10)
s = random.randint(r, nbit)
while True:
e = random.randint(3, p**r*(p-1))
if gmpy2.gcd(e, p**s*(p-1)) == 1:
break
pubkey = (long(e), long(p**r))
return pubkey
def crypt(msg, pkey):
e, n = pkey
m = bytes_to_long(msg)
assert m < n - 1
enc = pow(m, e, n)
return long_to_bytes(enc)
nbit = 1024
pubkey = generate_key(nbit)
print 'pubkey =', pubkey
msg = flag
enc = crypt(msg, pubkey)
print 'enc =\n', enc.encode('base64')
pubkey = (58134567416061346246424950552806959952164141873988197038339318172373514096258823300468791726051378264715940131129676561677588167620420173326653609778206847514019727947838555201787320799426605222230914672691109516799571428125187628867529996213312357571123877040878478311539048041218856094075106182505973331343540958942283689866478426396304208219428741602335233702611371265705949787097256178588070830596507292566654989658768800621743910199053418976671932555647943277486556407963532026611905155927444039372549162858720397597240249353233285982136361681173207583516599418613398071006829129512801831381836656333723750840780538831405624097443916290334296178873601780814920445215584052641885068719189673672829046322594471259980936592601952663772403134088200800288081609498310963150240614179242069838645027877593821748402909503021034768609296854733774416318828225610461884703369969948788082261611019699410587591866516317251057371710851269512597271573573054094547368524415495010346641070440768673619729280827372954003276250541274122907588219152496998450489865181536173702554116251973661212376735405818115479880334020160352217975358655472929210184877839964775337545502851880977049299029101466287659419446724781305689536816523774995178046989696610897508786776845460908137698543091418571263630383061605011820139755322231913029643701770497299157169690586232187419462594477116374977216427311975598620616618808494138669546120288334682865354702356192972496556372279363023366842805886601834278434406709218165445335977049796015123909789363819484954615665668979L, 754600786340927688096652328072061561501667781193760284816393637647032362908189628005150802929636396969230958922073774180726205402897453096041624408154494621307262657492560975357997726055874834308239749992507552325614973631556754707427580134609221878324704469965450463088892083264951442562525825243127575048386573246756312509362222667015490013299327398464802116909245529065994770788125182846841016932803939806558559335886481214931253578226314057242462834149031625361286317307273138514126289052003214703248070256059405676891634792175775697355408418965738663732479622148276007308404691800186837579126431484536836513358124181380166971922188839934522356902295160649189850427580493328509329115798694580347461641487270793993129066433242544366683131231903590153844590595882428219010673818765995719694470668924781499987923250883546686344997580959954960334567874040563037167422839228466141912000421309282727363913908613116739074234989825489075148091144771967111113068647060175231126374070143480727000247378471525286907200601035581143391602569836131345909055708005758380081303860198696570649330092070410465978479841469533490522594827330661914537170063053059393550673731195548189192109328158876774080143171304333338291909598353550442855717204721L)
enc = YXmuOsaD1W4poL...
刚开始给的题目里没有flag，随后更新了题目。
一开始以为只是添加了flag，做了半天发现解不出来，才发现p和e都更新了，以后做题细心点。
这题的p在factordb.com可解(1657407551...67<309>)^4，则r=4
有点类似RSA，猜测欧拉函数就是p**s*(p-1)，则d = gmpy2.invert(e,p**s*(p-1))，测试可解，代码如下：
import gmpy2
import random
from Crypto.Util.number import *
import binascii
p=165740755190793304655854506052794072378181046252118367693457385632818329041540419488625472007710062128632942664366383551452498541560538744582922713808611320176770401587674618121885719953831122487280978418110380597358747915420928053860076414097300832349400288770613227105348835005596365488460445438176193451867L
r=4
e=58134567416061346246424950552806959952164141873988197038339318172373514096258823300468791726051378264715940131129676561677588167620420173326653609778206847514019727947838555201787320799426605222230914672691109516799571428125187628867529996213312357571123877040878478311539048041218856094075106182505973331343540958942283689866478426396304208219428741602335233702611371265705949787097256178588070830596507292566654989658768800621743910199053418976671932555647943277486556407963532026611905155927444039372549162858720397597240249353233285982136361681173207583516599418613398071006829129512801831381836656333723750840780538831405624097443916290334296178873601780814920445215584052641885068719189673672829046322594471259980936592601952663772403134088200800288081609498310963150240614179242069838645027877593821748402909503021034768609296854733774416318828225610461884703369969948788082261611019699410587591866516317251057371710851269512597271573573054094547368524415495010346641070440768673619729280827372954003276250541274122907588219152496998450489865181536173702554116251973661212376735405818115479880334020160352217975358655472929210184877839964775337545502851880977049299029101466287659419446724781305689536816523774995178046989696610897508786776845460908137698543091418571263630383061605011820139755322231913029643701770497299157169690586232187419462594477116374977216427311975598620616618808494138669546120288334682865354702356192972496556372279363023366842805886601834278434406709218165445335977049796015123909789363819484954615665668979L
n=p**r
for s in xrange(r,1024):
if gmpy2.gcd(e, p**s*(p-1)) == 1:
break
d = gmpy2.invert(e,p**s*(p-1))
enc ='YXmuOsaD1...'
enc = enc.decode('base64')
def decrypt(msg, d):
m = bytes_to_long(msg)
#assert m < n - 1
dec = pow(m, d, n)
return long_to_bytes(dec)
flag = decrypt(enc,d)
print flag
deep_leaning
- 题目描述
from PIL import Image
import sys
import os
import numpy as np
import random
import time
import base64
import inception
import string
import hashlib
SALT_LEN = 10
HEX_LEN = 4
std_image_name="/home/ctf/images/image.jpg"
input_image_name=''
os.environ['TF_CPP_MIN_LOG_LEVEL']='2'
def base_str():
return "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
def random_string(length):
string = [random.choice(base_str()) for i in range(length)]
return ("".join(string))
def tofile(data):
try:
data=base64.b64decode(data)
except:
return ""
filename="/backup/"+str(time.time())+".jpg"
fd=open(filename,"wb")
fd.write(data)
fd.close()
filename="/home/ctf/images/"+str(time.time())+".jpg"
fd=open(filename,"wb")
fd.write(data)
fd.close()
return filename
def check():
global input_image_name
try:
input_image=Image.open(input_image_name)
std_image=Image.open(std_image_name)
except:
print("[-]give me a real image!!")
sys.stdout.flush()
return False
input_image_np=np.array(input_image)
std_image_np=np.array(std_image)
input_x=len(input_image_np)
input_y=len(input_image_np[0])
input_z=len(input_image_np[0][0])
std_x=len(std_image_np)
std_y=len(std_image_np[0])
std_z=len(std_image_np[0][0])
if std_x!=input_x or std_y!=input_y or std_z!=input_z:
return False
diff=0
for i in range(input_x):
for j in range(input_y):
for k in range(input_z):
if input_image_np[i][j][k]>std_image_np[i][j][k]:
diff+=input_image_np[i][j][k]-std_image_np[i][j][k]
else:
diff+=std_image_np[i][j][k]-input_image_np[i][j][k]
diff=diff/(input_x*input_y*input_z)
if diff>0.8:
return False
return True
def classify(image_path):
model = inception.Inception()
pred = model.classify(image_path=image_path)
return (pred.argmax(),model.name_lookup.cls_to_name(pred.argmax(),only_first_name=True))
def getflag():
fd=open("/home/ctf/flag")
flag=fd.readline()
fd.close()
return flag
def main():
salt=random_string(SALT_LEN)
tmpvalue=random_string(20)+salt
md5=hashlib.md5()
md5.update(tmpvalue.encode("utf-8"))
submd5=md5.hexdigest()[:4]
print ("[*]Proof of work:")
print ("\tMD5(key+\"%s\")[:4]==%s"%(salt,submd5))
print ("[+]Give me the key:")
sys.stdout.flush()
value=sys.stdin.readline()[:-1]
value=value+salt
md5=hashlib.md5()
md5.update(value.encode("utf-8"))
md5value=md5.hexdigest()
if(md5value[:HEX_LEN]!=submd5):
print ("[-]Access Failed")
return;
print ("[*]I am the world smartest CV system!")
print ("[+]Give me a wing to fly?")
sys.stdout.flush()
global input_image_name
image=sys.stdin.readline()[:-1]
if(len(image)>200000):
print("[-]input too long!")
return;
input_image_name=tofile(image)
if input_image_name=="":
print ("[-]base64 please!")
sys.stdout.flush()
return
if not check():
print ("[-]You cannot fool me!")
sys.stdout.flush()
return
(input_image_class,input_image_classname)=classify(input_image_name)
(std_image_class,std_image_classname)=classify(std_image_name)
if input_image_class!=std_image_class and input_image_class==503:
print("[*]Wow I get the wing")
print("[*]Give you the flag")
print(getflag())
sys.stdout.flush()
return
else:
print("[*]Give me the wing!")
sys.stdout.flush()
return
main()
题目原图
简单来说，就是生成对抗样本，扰动率低于0.8且神经网络识别为wing，属于白盒有目标攻击。
本地环境是Anaconda+python3+keras-gpu
在Anaconda里直接装tensorflow-gpu装不上，装keras-gpu就装上了，还带了tensorflow-gpu。
题目提示使用的神经网络是inception-2015-12-05.tgz。
然而题目只用了一句import inception就导入了，找了很久教程，最后找到一个TensorFlow-Tutorials，只要把压缩包里的解压放在一个目录下，就可以一句话导入，第一次使用会自动下载inception-2015-12-05.tgz，下载比较慢，可以自己下好解压放在对应目录下。
一开始是参考这篇文章的TensorFlow 教程 #11 - 对抗样本 - 知乎
使用的应该是FGSM方法，一步步下来确实生成了对抗样本。
对抗样本1
修改一下这个plt输出的图片，去掉白边黑框。上传上去没拿到flag。
本地查了一下扰动率，得到diff=1.349157168264337，尝试修改各种生成参数，都无法降低扰动率，直到比赛结束。
后续看到有foolbox advbox cleverhans，三个对抗样本库，以后再试一下。
方法综述
攻击方法对比
对抗攻击概念
工具