简介
2021 AntCTF x D^3CTF 中共有四道Crypto方向的题目，题目难度适中，本文对这四道题目及本人的解题思路进行介绍，如有错误还请各位师傅指教。
babyLattice
题目分析
这道题的题目如下
from collections import namedtuple
PublicKey = namedtuple('PublicKey', ['n', 'b'])
SecretKey = namedtuple('SecretKey', ['p', 'q', 'A'])
def gen_key():
p = random_prime(2^512, lbound=2^511)
q = random_prime(2^512, lbound=2^511)
n = p * q
a11, a12, a21 = [random_prime(2^100) for _ in range(3)]
a22 = random_prime(2^100)
while a11 * a22 == a12 * a21:
a22 = random_prime(2^100)
A = Matrix(ZZ, [[a11, a12], [a21, a22]])
a1 = crt([a11, a21], [p, q])
a2 = crt([a12, a22], [p, q])
b = a1 * inverse_mod(a2, n) % n
PK = PublicKey(n, b)
SK = SecretKey(p, q, A)
return (PK, SK)
def encrypt(m, pk):
assert 0 < m < 2^400
r = randint(0, 2^400-1)
c = (pk.b*m + r) % pk.n
return c
def decrypt(c, sk):
a2 = crt([sk.A[0,1], sk.A[1,1]], [sk.p, sk.q])
s1 = a2 * c % sk.p
s2 = a2 * c % sk.q
m, r = sk.A.solve_right(vector([s1, s2]))
return m
def test(pk, sk, num=3):
for _ in range(num):
m = randint(0, 2^400-1)
c = encrypt(m, pk)
mm = decrypt(c, sk)
assert m == mm
if __name__ == '__main__':
from hashlib import sha256
from secret import m, FLAG
assert FLAG == 'd3ctf{%s}' % sha256(int(m).to_bytes(50, 'big')).hexdigest()
PK, SK = gen_key()
test(PK, SK)
c = encrypt(m, PK)
print(f"PK = {PK}")
print(f"c = {c}")
我们重点看看加密函数
也就是
这样就可以通过LLL算法还原出m了
EXP
from hashlib import sha256
n = 69804507328197961654128697510310109608046244030437362639637009184945533884294737870524186521509776189989451383438084507903660182212556466321058025788319193059894825570785105388123718921480698851551024108844782091117408753782599961943040695892323702361910107399806150571836786642746371968124465646209366215361
b = 65473938578022920848984901484624361251869406821863616908777213906525858437236185832214198627510663632409869363143982594947164139220013904654196960829350642413348771918422220404777505345053202159200378935309593802916875681436442734667249049535670986673774487031873808527230023029662915806344014429627710399196
c = 64666354938466194052720591810783769030566504653409465121173331362654665231573809234913985758725048071311571549777481776826624728742086174609897160897118750243192791021577348181130302572185911750797457793921069473730039225991755755340927506766395262125949939309337338656431876690470938261261164556850871338570
A = Matrix(ZZ,[[1,0,b],[0,2^400,c],[0,0,n]])
A = A.LLL()
m = int(A[0][0])
flag = 'd3ctf{%s}' % sha256(int(m).to_bytes(50, 'big')).hexdigest()
print(flag)
simpleGroup
题目分析
这道题的题目如下
from random import randint
from secret import FLAG
# A gift for key recovery in challenge [babyLattice]
n = 69804507328197961654128697510310109608046244030437362639637009184945533884294737870524186521509776189989451383438084507903660182212556466321058025788319193059894825570785105388123718921480698851551024108844782091117408753782599961943040695892323702361910107399806150571836786642746371968124465646209366215361
y = 12064801545723347322936991186738560311049061235541031580807549422258814170771738262264930441670708308259694588963224372530498305648578520552038029773849342206125074212912788823834152785756697757209804475031974445963691941515756901268376267360542656449669715367587909233618109269372332127072063171947435639328
e = 1928983487
M = int.from_bytes(FLAG, 'big')
C = []
while M != 0:
m = M % e
M //= e
r = randint(0, n-1)
c = power_mod(y, m, n) * power_mod(r, e, n)
C.append(c % n)
print(f"C = {C}")
通过注释我们可以大概猜测babyLattice本来是需要分解n的，但是因为被非预期了所以又出了这道题目
那么我们回到babyLattice题目里面，我们知道的参数实际上只有b,c,n，分解n应该和b有关，通过阅读b的生成代码我们可以得到
我们展开后两个式子
也就是
两边相乘得到
展开并变形得到
也就是
由于
所以我们同样可以用LLL还原出目标向量，然后使用factor进行分解（a11,a12,a21,a22它们都是素数）
当分解完毕后，通过猜测它们对应的值来分解n，即
得到p,q后，我们回来看题目里面的加密，其会对FLAG进行取模并分段加密余数，其中c的生成公式如下
r是随机生成的数字，而e可以被分解为e1和e2两个素数，这两个素数又分别是p-1和q-1的因子
那么我们可以得到
通过遍历j并判断得到的c'是不是为模p的e1次剩余，我们就可以得到m模e1的值
同样我们也可以用q得到m模e2的值，然后使用中国剩余定理即可还原m并最终得到flag
EXP
n = 69804507328197961654128697510310109608046244030437362639637009184945533884294737870524186521509776189989451383438084507903660182212556466321058025788319193059894825570785105388123718921480698851551024108844782091117408753782599961943040695892323702361910107399806150571836786642746371968124465646209366215361
b = 65473938578022920848984901484624361251869406821863616908777213906525858437236185832214198627510663632409869363143982594947164139220013904654196960829350642413348771918422220404777505345053202159200378935309593802916875681436442734667249049535670986673774487031873808527230023029662915806344014429627710399196
c = 64666354938466194052720591810783769030566504653409465121173331362654665231573809234913985758725048071311571549777481776826624728742086174609897160897118750243192791021577348181130302572185911750797457793921069473730039225991755755340927506766395262125949939309337338656431876690470938261261164556850871338570
A = Matrix(ZZ,[[1,0,b^2],[0,1,b],[0,0,n]])
A = A.LLL()
x1 = -A[0][0]
x3 = A[0][2]
print(factor(x1))
print(factor(x3))
a12 = 1018979931854255696816714991181
a22 = 1151291153120610849180830073509
a11 = 1017199123798810531137951821909
a21 = 207806651167586080788016046729
print(gcd(b * a12 - a11,n))
print(gcd(b * a22 - a21,n))
p = 7669036313101194621345265255994200133006921565596653797956940811601516306410232120057637504305295677357422367597831138570269733177579895994340511712373309
q = 9102122415165681824420871673621781250311822805618731863628192549895693024247220594372897360668046264863189831887100676431439200352775676467952192600956629
assert(p * q == n)
#!/usr/bin/env python
from Crypto.Util.number import *
import gmpy2
p = 7669036313101194621345265255994200133006921565596653797956940811601516306410232120057637504305295677357422367597831138570269733177579895994340511712373309
q = 9102122415165681824420871673621781250311822805618731863628192549895693024247220594372897360668046264863189831887100676431439200352775676467952192600956629
n = 69804507328197961654128697510310109608046244030437362639637009184945533884294737870524186521509776189989451383438084507903660182212556466321058025788319193059894825570785105388123718921480698851551024108844782091117408753782599961943040695892323702361910107399806150571836786642746371968124465646209366215361
y = 12064801545723347322936991186738560311049061235541031580807549422258814170771738262264930441670708308259694588963224372530498305648578520552038029773849342206125074212912788823834152785756697757209804475031974445963691941515756901268376267360542656449669715367587909233618109269372332127072063171947435639328
e = 1928983487
e1 = 36493
e2 = 52859
def GCRT(mi, ai):
assert (isinstance(mi, list) and isinstance(ai, list))
curm, cura = mi[0], ai[0]
for (m, a) in zip(mi[1:], ai[1:]):
d = gmpy2.gcd(curm, m)
c = a - cura
assert (c % d == 0)
K = c // d * gmpy2.invert(curm // d, m // d)
cura += curm * K
curm = curm * m // d
cura %= curm
return (cura % curm, curm)
def check(d,p,n):
if((p - 1) % n == 0):
return pow(d,(p - 1) // n,p) == 1
else:
k = gmpy2.gcd(n, p - 1)
return pow(d,(p - 1) // k,p) == 1
def getM(c,e,p):
for i in range(2,e):
tmpc = (c * gmpy2.invert(pow(y,i,p),p)) % p
if check(tmpc,p,e):
return i
exit(0)
C = [63173987757788284988620600191109581820396865828379773315280703314093571300861961873159324234626635582246705378908610341772657840682572386153960342976445563045427986000105931341168525422286612417662391801508953619857648844420751306271090777865836201978470895906780036112804110135446130976275516908136806153488, 9763526786754236516067080717710975805995955013877681492195771779269768465872108434027813610978940562101906769209984501196515248675767910499405415921162131390513502065270491854965819776080041506584540996447044249409209699608342257964093713589580983775580171905489797513718769578177025063630080394722500351718, 37602000735227732258462226884765737048138920479521815995321941033382094711120810035265327876995207117707635304728511052367297062940325564085193593024741832905771507189762521426736369667607865137900432117426385504101413622851293642219573920971637080154905579082646915297543490131171875075081464735374022745371, 1072671768043618032698040622345664216689606325179075270470875647188092538287671951027561894188700732117175202207361845034630743422559130952899064461493359903596018309221581071025635286144053941851624510600383725195476917014535032481197737938329722082022363122585603600777143850326268988298415885565240343957, 27796821408982345007197248748277202310092789604135169328103109167649193262824176309353412519763498156841477483757818317945381469765077400076181689745139555466187324921460327576193198145058918081061285618767976454153221256648341316332169223400180283361166887912012807743326710962143011946929516083281306203120, 27578857139265869760149251280906035333246393024444009493717159606257881466594628022512140403127178174789296810502616834123420723261733024810610501421455454191654733275226507268803879479462533730695515454997186867769363797096196096976825300792616487723840475500246639213793315097434400920355043141319680299224, 29771574667682104634602808909981269404867338382394257360936831559517858873826664867201410081659799334286847985842898792091629138292008512383903137248343194156307703071975381090326280520578349920827357328925184297610245746674712939135025013001878893129144027068837197196517160934998930493581708256039240833145, 33576194603243117173665354646070700520263517823066685882273435337247665798346350495639466826097821472152582124503891668755684596123245873216775681469053052037610568862670212856073776960384038120245095140019195900547005026888186973915360493993404372991791346105083429461661784366706770467146420310246467262823, 5843375768465467361166168452576092245582688894123491517095586796557653258335684018047406320846455642101431751502161722135934408574660609773328141061123577914919960794180555848119813522996120885320995386856042271846703291295871836092712205058173403525430851695443361660808933971009396237274706384697230238104, 61258574367240969784057122450219123953816453759807167817741267194076389100252707986788076240792732730306129067314036402554937862139293741371969020708475839483175856346263848768229357814022084723576192520349994310793246498385086373753553311071932502861084141758640546428958475211765697766922596613007928849964, 13558124437758868592198924133563305430225927636261069774349770018130041045454468021737709434182703704611453555980636131119350668691330635012675418568518296882257236341035371057355328669188453984172750580977924222375208440790994249194313841200024395796760938258751149376135149958855550611392962977597279393428]
m = 0
for c in C[::-1]:
cp = c % p
cq = c % q
m1 = getM(cp,e1,p)
m2 = getM(cq,e2,q)
mm,lcm = GCRT([e1,e2],[m1,m2])
print("Get mm: " + hex(mm))
m *= e
m += mm
flag = long_to_bytes(m)
print(flag)
EasyCurve
题目分析
这道题目的主要部分如下
import socketserver
from Crypto.PublicKey import RSA
from Crypto.Util.number import getPrime , bytes_to_long
from Curve import MyCurve
from hashlib import sha256
import os
import string
import random
import signal
from secret import flag
BIT = 2048
p = 9688074905643914060390149833064012354277254244638141162997888145741631958242340092013958501673928921327767591959476890238698855704376126231923819603296257
class Task(socketserver.BaseRequestHandler):
def proof_of_work(self):
random.seed(os.urandom(8))
proof = ''.join([random.choice(string.ascii_letters+string.digits) for _ in range(20)])
_hexdigest = sha256(proof.encode()).hexdigest()
self.send(f"sha256(XXXX+{proof[4:]}) == {_hexdigest}".encode())
self.send(b'Give me XXXX: ')
x = self.recv()
if len(x) != 4 or sha256(x+proof[4:].encode()).hexdigest() != _hexdigest:
self.send('wrong')
return False
return True
def recv(self):
data = self.request.recv(1024)
return data.strip()
def send(self, msg, newline=True):
if isinstance(msg , bytes):
msg += b'\n'
else:
msg += '\n'
msg = msg.encode()
self.request.sendall(msg)
def key_gen(self , bit):
key = RSA.generate(bit)
return key
def ot(self , point):
x , y = point
random.seed(os.urandom(8))
key = self.key_gen(BIT)
self.send('n = ' + str(key.n))
self.send('e = ' + str(key.e))
x0 = random.randint(1 , key.n)
x1 = random.randint(1 , key.n)
self.send("x0 = " + str(x0))
self.send("x1 = " + str(x1))
self.send("v = ")
v = int(self.recv())
m0_ = (x + pow(v - x0, key.d, key.n)) % key.n
m1_ = (y + pow(v - x1, key.d, key.n)) % key.n
self.send("m0_ = " + str(m0_))
self.send("m1_ = " + str(m1_))
def handle(self):
signal.alarm(180)
if not self.proof_of_work():
return 0
e = bytes_to_long(os.urandom(32))
u = random.randint(1 , p)
D = random.randint(1 , p)
curve = MyCurve(p , D , u)
self.send('p = ' + str(p))
self.send('D = ' + str(D))
for i in range(3):
G = curve.getPoint()
self.ot(G)
P = curve.mul(e , G)
self.ot(P)
self.send("do you know my e?")
guess = int(self.recv())
if guess == e:
self.send("oh no!")
self.send(flag)
return 0
else:
self.send("Ha, I know you can't get it.")
class ForkedServer(socketserver.ForkingMixIn, socketserver.TCPServer):
pass
if __name__ == "__main__":
HOST, PORT = '0.0.0.0', 10000
server = ForkedServer((HOST, PORT), Task)
server.allow_reuse_address = True
server.serve_forever()
其使用了一个随机生成参数的MyCurve并生成了随机的e，给我们三次交互的机会，每次交互会随机生成点G和点P并使用OT将这两个点的信息传递给我们，点P是e倍的点G，当我们给服务器正确的e的时候我们可以得到flag
这其实就是一个离散对数问题，我们首先关注服务器的参数，MyCurve所使用的p是512比特的，而OT中RSA的n是2048比特的，这样生成的点的x和y乘起来也没有n的大。那么可以参考2020hackergame的不经意传输中的攻击方式来同时获取点的x和y坐标
之后便是如何通过点G和点P来获取e了，我们可以注意到p-1是光滑的
sage: factor(p-1)
2^21 * 3^10 * 7^4 * 11 * 13^2 * 17 * 19 * 29 * 31 * 37 * 43^3 * 47 * 71 * 83 * 89 * 97 * 223 * 293 * 587 * 631 * 709 * 761 * 1327 * 1433 * 1733 * 1889 * 2503 * 3121 * 6043 * 6301 * 49523 * 98429 * 140683 * 205589 * 1277369 * 1635649 * 5062909 * 45698189 * 67111151 * 226584089 * 342469397
那么我们可以通过Pohlig-Hellman algorithm来解决离散对数问题并最终得到flag
EXP
exp有概率成功，如果报错或者答案错误多跑几次即可
#!/usr/bin/env python
import string, gmpy2
from hashlib import sha256
from pwn import *
context.log_level = "debug"
dic = string.ascii_letters + string.digits
def solvePow(prefix,h):
for a1 in dic:
for a2 in dic:
for a3 in dic:
for a4 in dic:
x = a1 + a2 + a3 + a4
proof = x + prefix.decode("utf-8")
_hexdigest = sha256(proof.encode()).hexdigest()
if _hexdigest == h.decode("utf-8"):
return x
def getData():
r.recvuntil("n = ")
n = int(r.recvuntil("\n", drop = True))
r.recvuntil("e = ")
e = int(r.recvuntil("\n", drop = True))
r.recvuntil("x0 = ")
x0 = int(r.recvuntil("\n", drop = True))
r.recvuntil("x1 = ")
x1 = int(r.recvuntil("\n", drop = True))
offset = 2 << 1024
offset_e = int(pow(offset, e, n))
v = ((offset_e * x0 - x1) * gmpy2.invert(offset_e - 1, n)) % n
r.sendlineafter("v = ",str(v))
r.recvuntil("m0_ = ")
m0 = int(r.recvuntil("\n", drop = True))
r.recvuntil("m1_ = ")
m1 = int(r.recvuntil("\n", drop = True))
m = (m0 * offset - m1) % n
x = m // offset + 1
y = x * offset - m
return x,y
r = remote("47.100.50.252",10000)
r.recvuntil("sha256(XXXX+")
prefix = r.recvuntil(") == ", drop = True)
h = r.recvuntil("\n", drop = True)
result = solvePow(prefix,h)
r.sendlineafter("Give me XXXX: \n",result)
r.recvuntil("p = ")
r.recvuntil("\n", drop = True)
r.recvuntil("D = ")
D = int(r.recvuntil("\n", drop = True))
Gx,Gy = getData()
Px,Py = getData()
with open("data.txt","wb") as f:
f.write(str(D).encode() + b"\n")
f.write(str(Gx).encode() + b"\n")
f.write(str(Gy).encode() + b"\n")
f.write(str(Px).encode() + b"\n")
f.write(str(Py).encode() + b"\n")
s = process(argv=["sage", "exp.sage"])
e = int(s.recv())
s.close()
r.sendline(str(e))
r.interactive()
# exp.sage
load("Curve.sage")
p = 9688074905643914060390149833064012354277254244638141162997888145741631958242340092013958501673928921327767591959476890238698855704376126231923819603296257
F = GF(p)
fac = [2^21,3^10,7^4,11,13^2,17,19,29,31,37,43^3,47,71,83,89,97,223,293,587,631,709,761,1327,1433,1733,1889,2503,3121,6043,6301,49523,98429,140683,205589,1277369,1635649,5062909,45698189,67111151,226584089,342469397]
def bsgs(g, y, p):
m = int(ceil(sqrt(p - 1)))
S = {}
point = (u,0)
for i in range(m):
point = curve.add(point,g)
pointg = point[0] << 800 | point[1]
S[pointg] = i
gs = curve.mul(m,g)
for i in range(m):
pointy = y[0] << 800 | y[1]
if pointy in S:
return S[pointy] - i * m + 1
y = curve.add(y,gs)
return None
def Pohlig_Hellman(G,P):
ea = []
na = []
for i in range(len(fac)):
c = fac[i]
n = (p - 1) // c
gi = curve.mul(n, G)
yi = curve.mul(n, P)
ei = bsgs(gi,yi,c)
ea.append(ei%c)
na.append(c)
ee = crt(ea,na)
return ee
data = open("data.txt","rb").read().decode("utf-8")
data = data.split("\n")
D = int(data[0])
Gx = int(data[1])
Gy = int(data[2])
Px = int(data[3])
Py = int(data[4])
G = (F(Gx),F(Gy))
P = (F(Px),F(Py))
u2 = (Gx ^ 2 - D * Gy ^ 2)
u2 = F(u2)
u = int(u2.sqrt())
curve = MyCurve(p , D , u)
e = Pohlig_Hellman(G,P)
e %= p - 1
print(e)
AliceWantFlag
题目分析
这道题目分为server端和Alice端，其中server端的代码如下
from elgamal import elgamal
import socketserver
from prikey import server_prikey , AlicePasswd
from pubkey import Alice_pubkey
from secret import Alice_flag , ctfer_flag
import random
import signal
from os import urandom
from Crypto.Util.number import long_to_bytes , bytes_to_long
from Crypto.Cipher import AES
MENU = "1. signup 2.signin"
XOR = lambda s1,s2 :bytes([x1^x2 for x1 , x2 in zip(s1,s2)])
def pad(m):
m += bytes([16 - len(m) % 16] * (16 - len(m) % 16))
return m
def unpad(m):
padlen = m[-1]
for i in range(1 , padlen + 1):
if m[-i] != m[-1]:
return b''
return m[:-m[-1]]
class server(socketserver.BaseRequestHandler):
def setup(self):
self.pubkey = {}
self.passwd = {}
self.prikey = elgamal(server_prikey)
self.pubkey[b'Alice'] = elgamal(Alice_pubkey)
self.passwd[b'Alice'] = AlicePasswd
def _recv(self):
data = self.request.recv(1024)
return data.strip()
def _send(self, msg, newline=True):
if isinstance(msg , bytes):
msg += b'\n'
else:
msg += '\n'
msg = msg.encode()
self.request.sendall(msg)
def enc_send(self, msg , usrid , enc_key = b''):
if enc_key == b'':
pubenc = self.pubkey[usrid]
y1 , y2 = pubenc.encrypt(bytes_to_long(msg))
self._send(str(y1) + ', ' + str(y2))
else:
assert len(enc_key) == 16
aes = AES.new(enc_key , AES.MODE_ECB)
self._send(aes.encrypt(pad(msg)))
def dec_recv(self, enc_key = b''):
msg = self._recv()
if enc_key == b'':
c = [int(i) for i in msg.split(b', ')]
m = self.prikey.decrypt(c)
print(long_to_bytes(m))
return long_to_bytes(m)
else:
assert len(enc_key) == 16
aes = AES.new(enc_key , AES.MODE_ECB)
return unpad(aes.decrypt(msg))
def signup(self):
if len(self.passwd) > 5:
self._send('sorry, the number of users is out of limit')
return 0
self._send('please give me your name')
userid = self._recv()
if len(userid) > 20:
self._send('your id can\'t be too long')
return 0
elif userid in self.passwd:
self._send('the name has been used')
return 0
else:
self._send('please give me your passwd(encrypted)')
userpasswd = self.dec_recv()
if len(userpasswd) > 11:
self._send('your password can\'t be too long')
return 0
else:
self.passwd[userid] = userpasswd
self._send('please give me your publickey')
userpubkey = self._recv()
try:
userpubkey = [int(i) for i in userpubkey[1:-1].split(', ')]
except:
self._send('publickey format error')
self.passwd.pop(userid)
return 0
self.pubkey[userid] = elgamal(userpubkey)
self._send('sign up success')
return 1
def signin(self):
self._send('please give me your name')
userid = self._recv()
if userid not in self.passwd:
self._send('sorry the userid is not existed')
return 0
while 1:
random.seed(urandom(8))
r = random.getrandbits(8 * 11)
self._send('please give me your passwd(encrypted and xored by r)')
self._send(str(r))
userdata = self.dec_recv()
if bytes_to_long(userdata) == r ^ bytes_to_long(self.passwd[userid]):
self._send('signin success')
break
else:
self._send('password error')
endkey = urandom(5)
key = userdata + endkey
self._send('now let\'s communicate with this key')
self.enc_send(endkey , userid)
return userid , key
def handle(self):
signal.alarm(240)
key = b''
userid = ''
while 1:
self._send(MENU)
choice = self._recv()
if choice == b'1':
self.signup()
elif choice == b'2':
temp = self.signin()
if temp != 0:
userid , key = temp
break
else:
self._send('error')
msg = self.dec_recv(enc_key = key)
if msg == b'I am a ctfer.Please give me flag':
self.enc_send(b'ok, your flag is here ' + ctfer_flag , userid , enc_key= key)
elif msg == b'I am Alice, Please give me true flag' and userid == b'Alice':
self.enc_send(b'Hi Alice, your flag is ' + Alice_flag , userid , enc_key= key)
return 0
def finish(self):
self.request.close()
class ForkedServer(socketserver.ForkingMixIn, socketserver.TCPServer):
pass
if __name__ == "__main__":
HOST, PORT = '0.0.0.0', 10001
server = ForkedServer((HOST, PORT), server)
server.allow_reuse_address = True
server.serve_forever()
Alice端的代码如下
import socket
from elgamal import elgamal
from pubkey import server_pubkey
from prikey import Alice_prikey , AlicePasswd
from Crypto.Util.number import long_to_bytes , bytes_to_long
from Crypto.Cipher import AES
import socketserver , signal
def pad(m):
m += bytes([16 - len(m) % 16] * (16 - len(m) % 16))
return m
def unpad(m):
return m[:-m[-1]]
class Alice:
def __init__(self , ip , port):
self.pridec = elgamal(Alice_prikey)
self.pubenc = elgamal(server_pubkey)
self.s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
self.s.connect((ip, port))
def _recv(self):
data = self.s.recv(1024)
return data.strip()
def _send(self, msg):
if isinstance(msg , str):
msg = msg.encode()
self.s.send(msg)
def enc_send(self, msg , enc_key = b''):
if enc_key == b'':
y1 , y2 = self.pubenc.encrypt(bytes_to_long(msg))
self._send(str(y1) + ', ' + str(y2))
else:
assert len(enc_key) == 16
aes = AES.new(enc_key , AES.MODE_ECB)
self._send(aes.encrypt(pad(msg)))
def dec_recv(self, enc_key = b''):
msg = self._recv()
if enc_key == b'':
c = [int(i) for i in msg.split(b', ')]
m = self.pridec.decrypt(c)
return long_to_bytes(m)
else:
assert len(enc_key) == 16
aes = AES.new(enc_key , AES.MODE_ECB)
return unpad(aes.decrypt(msg))
def main(self):
firstmsg = self._recv()
if firstmsg != b'1. signup 2.signin':
return 0
self._send('2')
self._recv()
self._send('Alice')
self._recv()
r = int(self._recv())
userdata = long_to_bytes(bytes_to_long(AlicePasswd) ^ r)
self.enc_send(userdata)
self._recv()
self._recv()
endkey = self.dec_recv()
key = userdata + endkey
self.enc_send(b'I am a ctfer.Please give me flag' , enc_key = key)
return self.dec_recv(enc_key = key)
class Task(socketserver.BaseRequestHandler):
def _recv(self):
data = self.request.recv(1024)
return data.strip()
def _send(self, msg, newline=True):
if isinstance(msg , bytes):
msg += b'\n'
else:
msg += '\n'
msg = msg.encode()
self.request.sendall(msg)
def handle(self):
signal.alarm(60)
self._send('Hello, I am Alice, can you tell me the address of the server?\nIn return, I will give you the ctf_flag')
try:
addr = self._recv()
ip, port = [x.strip() for x in addr.split(b':')]
port = int(port)
except:
ip, port = '0.0.0.0', 10001
a = Alice(ip , port)
msg = a.main()
self._send(b'Thanks, here is your flag')
self._send(msg)
class ForkedServer(socketserver.ForkingMixIn, socketserver.TCPServer):
pass
if __name__ == "__main__":
HOST, PORT = '0.0.0.0', 10003
server = ForkedServer((HOST, PORT), Task)
server.allow_reuse_address = True
server.serve_forever()
server和Alice的密钥生成和使用都和elgamal算法一致，这里不再阐述
服务端的大概逻辑如下
U: User
S: Server
UserPasswd = UserPublicKey
Sign Up:
S -> U : 'please give me your name'
U -> S : userid
S : assert len(userid) <= 20 and userid not in passwd
S -> U : 'please give me your passwd(encrypted)'
U -> S : c = elgamal.enc(UserPasswd,ServerPublicKey)
S : userpasswd = elgamal.dec(c,ServerPrivateKey
S : assert len(userpasswd) <= 11
S : passwd[userid] = userpasswd
S -> U : 'sign up success'
Sign In:
S -> U : 'please give me your name'
U -> S : userid
S : assert userid in passwd
S -> U : 'please give me your passwd(encrypted and xored by r)'
S -> U : r = random.getrandbits(8 * 11)
U -> S : c = elgamal.enc(UserPasswd ^ r,ServerPublicKey)
S : assert elgamal.dec(c,ServerPrivateKey) == passwd[userid] ^ r
S : key = userdata + endkey, userdata = passwd[userid] ^ r, endkey = urandom(5)
S -> U : 'now let\'s communicate with this key'
S -> U : k = elgamal.enc(key,UserPasswd)
U -> S : m = AES.enc(key,msg)
S : dm = AES.dec(key,m)
S : if dm == 'I am a ctfer.Please give me flag':
S -> U : r1 = AES.enc(key,ctfer_flag)
S : if dm == 'I am Alice, Please give me true flag' and userid == 'Alice'
S -> U : r2 = AES.enc(key,Alice_flag)
大概理解下来就是一个利用elgamal进行密钥交换然后加密通信的逻辑
由于这里我们需要以Alice的身份登陆并使用交换的AES通信密钥进行密文的加密，所以我们需要知道AlicePasswd和key
这道题目中由于也有Alice端的服务，所以我们可以伪装成服务端来和Alice端进行通信，也就是进行中间人攻击
首先来看看如何获得AlicePasswd，观察Alice端的如下代码
def main(self):
firstmsg = self._recv()
if firstmsg != b'1. signup 2.signin':
return 0
self._send('2')
self._recv()
self._send('Alice')
self._recv()
r = int(self._recv())
userdata = long_to_bytes(bytes_to_long(AlicePasswd) ^ r)
self.enc_send(userdata)
self._recv()
self._recv()
endkey = self.dec_recv()
key = userdata + endkey
self.enc_send(b'I am a ctfer.Please give me flag' , enc_key = key)
return self.dec_recv(enc_key = key)
def dec_recv(self, enc_key = b''):
msg = self._recv()
if enc_key == b'':
c = [int(i) for i in msg.split(b', ')]
m = self.pridec.decrypt(c)
return long_to_bytes(m)
else:
assert len(enc_key) == 16
aes = AES.new(enc_key , AES.MODE_ECB)
return unpad(aes.decrypt(msg))
由于endkey长度为5，而key的长度是16，那么可以自然推断出userdata的长度为11
但是如果我们控制r使得userdata的第一个字节异或为了\x00，那么userdata的长度就变成了10，如果endkey的长度不变，再使用userdata + endkey作为AES的key，那么会通不过assert len(enc_key) == 16，即连接会断开
这样我们可以通过单字节爆破AlicePasswd的值使得key的长度从16变成15，这样就能获得AlicePasswd的第一个字节，然后重复该过程便可以获得AlicePasswd（每爆破出一个字节，在爆破下一个字节的时候将endkey的长度变长一位即可连续爆破）
PS：实际上的操作过程中AlicePasswd的最后一个字节爆破不出来，我们使用服务端的Sign In功能来爆破最后一个字节即可
在拿到了AlicePasswd之后，我们便可以伪造成Alice登陆服务端，但是由于endkey是使用AlicePublicKey来进行加密的，所以我们还需要拿到endkey的值才能获得AES的key并进行任意文本加解密
前面我们提过，endkey长度为5，但是实际上elgamal的p是512比特的，也就是说endkey远比p小
如果我们将elgamal加密后的endkey的y2乘以一个倍数k，那么elgamal解密后的endkey就会变大k倍，这个值如果特别大，则userdata + endkey就也会变大，这样便会通不过assert len(enc_key) == 16，即连接会断开
那么我们就可以通过遍历k并查看连接是否断开来得到endkey的大致范围，然后通过控制r让userdata的长度变小来使得我们的k可以不断变大，进而将endkey的取值范围不断缩小来得到endkey
最后我们便可以进行任意文本加解密来获取flag
EXP
爆破AlicePasswd的脚本（除了最后一个字节）
#!/usr/bin/env python
from elgamal import elgamal
from os import urandom
from Crypto.Util.number import *
from pwn import *
from time import *
import random
#context.log_level = 'debug'
Alice_pubkey = (10701440058624032601015137538928332495339102166449611910023158626004456760436930147541475696463030881833656888220652983522600176918743749340172660134163173, 1564399668655593150166497641453625075939863931648697579307, 7485644640971189066076867813504769638089749022750276585841131549227880841063823940682209946365975810625990180843110530957715179877761206203179636693608929, 10399272689500457356753299445284422908920074489727610618928888372268024186959263604721857776550008093778901180936272708522371781846820901338928077050396521)
pubenc = elgamal(Alice_pubkey)
def enc(msg):
y1 , y2 = pubenc.encrypt(bytes_to_long(msg))
return [y1,y2]
def attackAlice(rr,m):
try:
middle_shell = listen(8888)
alice_shell = remote("47.100.0.15",10003)
alice_shell.recvuntil(b"Hello, I am Alice, can you tell me the address of the server?\nIn return, I will give you the ctf_flag\n")
alice_shell.sendline("xxxx:8888") # xxxx -> your vps's ip
middle_shell.sendline(b'1. signup 2.signin')
middle_shell.recv()
middle_shell.sendline(b'please give me your name')
middle_shell.recv()
middle_shell.sendline(b'please give me your passwd(encrypted and xored by r)')
middle_shell.sendline(str(rr))
middle_shell.recv()
middle_shell.sendline(b'signin success')
middle_shell.sendline(b'now let\'s communicate with this key')
middle_shell.sendline(str(m[0]) + ', ' + str(m[1]))
sleep(0.3)
result = middle_shell.recv()
if result != b"":
middle_shell.close()
alice_shell.close()
return True
except:
if middle_shell:
middle_shell.close()
if alice_shell:
alice_shell.close()
return False
known_pwd = b""
# 0x35343764643163636333xx
for i in range(11):
for r in range(0,256):
rr = ((bytes_to_long(known_pwd) << 8) + r) << ((11 - i - 1) * 8)
print("try:" + hex(rr))
msg = b"A" * (5 + i)
c = enc(msg)
if not attackAlice(rr,c):
known_pwd += long_to_bytes(r)
print(known_pwd.hex())
break
爆破AlicePasswd的最后一个字节的脚本
#!/usr/bin/env python
from elgamal import elgamal
from os import urandom
from Crypto.Util.number import *
from pwn import *
from time import *
import random
#context.log_level = 'debug'
dic = "0123456789abced"
server_pubkey = (8299337325013713958100496214277076548352330213422739951900206795659160881192662528217175848727001874097369338994314737585158671248737646741717255122339339, 1168114014665994438995759247944846107956060291607878556427, 6500863983405565947154848535503122330952083500341721347265599161478330537510643776384164499549064061675517930495094496645911948535824156417648599603482256, 1567838365897620258270310904624368598290758028096181970817619626094906443214320401208038763050717813632540079799097716376430981907783174222429828480377116)
pubenc = elgamal(server_pubkey)
known_pwd = "547dd1ccc3"
server_shell = remote("47.100.0.15",10001)
server_shell.sendlineafter("\n","2")
server_shell.recvuntil("please give me your name\n")
server_shell.sendline("Alice")
for c in dic:
server_shell.recvuntil("please give me your passwd(encrypted and xored by r)\n")
rr = int(server_shell.recvline())
pwd = bytes_to_long((known_pwd + c).encode())
prefix = long_to_bytes(pwd ^ rr)
assert(len(prefix) == 11)
y1 , y2 = pubenc.encrypt(pwd ^ rr)
server_shell.sendline(str(y1) + ', ' + str(y2))
result = server_shell.recv()
if b"success" in result:
print(known_pwd + c)
break
server_shell.interactive()
获取endkey并进行中间人攻击的脚本
#!/usr/bin/env python
from pwn import *
from Crypto.Util.number import *
from Crypto.Cipher import AES
from elgamal import elgamal
#context.log_level = "debug"
pwd = "547dd1ccc38".encode()
pwd = bytes_to_long(pwd)
server_pubkey = (8299337325013713958100496214277076548352330213422739951900206795659160881192662528217175848727001874097369338994314737585158671248737646741717255122339339, 1168114014665994438995759247944846107956060291607878556427, 6500863983405565947154848535503122330952083500341721347265599161478330537510643776384164499549064061675517930495094496645911948535824156417648599603482256, 1567838365897620258270310904624368598290758028096181970817619626094906443214320401208038763050717813632540079799097716376430981907783174222429828480377116)
pubenc = elgamal(server_pubkey)
p = 10701440058624032601015137538928332495339102166449611910023158626004456760436930147541475696463030881833656888220652983522600176918743749340172660134163173
def pad(m):
m += bytes([16 - len(m) % 16] * (16 - len(m) % 16))
return m
def unpad(m):
padlen = m[-1]
for i in range(1 , padlen + 1):
if m[-i] != m[-1]:
return b''
return m[:-m[-1]]
def oracle(m,n,p):
y1 = m[0]
y2 = m[1]
y2 = (y2 * n) % p
return [y1,y2]
def combineKey(m,rr):
for i in range(3):
sleep(0.3)
try:
middle_shell = listen(8888)
alice_shell = remote("47.100.0.15",10003)
alice_shell.recvuntil(b"Hello, I am Alice, can you tell me the address of the server?\nIn return, I will give you the ctf_flag\n")
alice_shell.sendline("xxxx:8888") # xxxx -> your vps's ip
middle_shell.sendline(b'1. signup 2.signin')
middle_shell.recv()
middle_shell.sendline(b'please give me your name')
middle_shell.recv()
middle_shell.sendline(b'please give me your passwd(encrypted and xored by r)')
middle_shell.sendline(str(rr))
middle_shell.recv()
middle_shell.sendline(b'signin success')
middle_shell.sendline(b'now let\'s communicate with this key')
middle_shell.sendline(str(m[0]) + ', ' + str(m[1]))
sleep(0.5)
result = middle_shell.recv()
print(result)
if result != b"":
middle_shell.close()
alice_shell.close()
return True
except:
if middle_shell:
middle_shell.close()
if alice_shell:
alice_shell.close()
return False
server_shell = remote("47.100.0.15",10001)
server_shell.sendlineafter("\n","2")
server_shell.recvuntil("please give me your name\n")
server_shell.sendline("Alice")
server_shell.recvuntil("please give me your passwd(encrypted and xored by r)\n")
rr = int(server_shell.recvline())
prefix = long_to_bytes(pwd ^ rr)
assert(len(prefix) == 11)
y1 , y2 = pubenc.encrypt(pwd ^ rr)
server_shell.sendline(str(y1) + ', ' + str(y2))
server_shell.recvuntil("now let's communicate with this key\n")
y1,y2 = [int(i) for i in server_shell.recvuntil("\n", drop = True).decode("utf-8").split(", ")]
print(y1,y2)
success("Get communicate key:" + str(y1) + "," + str(y2))
l = 0
h = 2**40
idx = 0
prefix_length = 0
bound = 2**40
count = 0
flag = False
for _ in range(11):
if flag:
break
binary_ptr = 0x80
diff = binary_ptr // 2
assert_arr = [-1] * 256
for i in range(10):
count += 1
if binary_ptr != 0 and assert_arr[binary_ptr-1] ^ assert_arr[binary_ptr] == 1:
prefix_length += 1
l = bound // multiple
h = bound // (multiple - 1)
idx = multiple - 1
print(hex(l),hex(h),count)
bound *= 0x100
if abs(h - l) < 2:
flag = True
break
if binary_ptr != 255 and assert_arr[binary_ptr] ^ assert_arr[binary_ptr+1] == 1:
prefix_length += 1
l = bound // (multiple + 1)
h = bound // multiple
idx = multiple
print(hex(l),hex(h),count)
bound *= 0x100
if abs(h - l) < 2:
flag = True
break
rr = bytes_to_long(long_to_bytes(pwd)[:prefix_length]) * 2**(8*(11 - prefix_length))
multiple = idx * 0x100 + binary_ptr
m = oracle([y1,y2],multiple,p)
if combineKey(m,rr):
if binary_ptr == 255:
prefix_length += 1
h = bound // (multiple + 1)
idx = multiple + 1
print(hex(l),hex(h),count)
bound *= 0x100
if abs(h - l) < 2:
exit(0)
break
assert_arr[binary_ptr] = 0
binary_ptr += diff
diff //= 2
else:
assert_arr[binary_ptr] = 1
binary_ptr -= diff
diff //= 2
context.log_level = "debug"
endkey = long_to_bytes(h)
key = prefix + endkey
success("get key:" + key.hex())
data = b'I am Alice, Please give me true flag'
cipher = AES.new(key , AES.MODE_ECB)
data = cipher.encrypt(pad(data))
server_shell.sendline(data)
msg = server_shell.recvuntil("\n",drop = True)
print(cipher.decrypt(msg))
参考
https://github.com/USTC-Hackergame/hackergame2020-writeups/blob/master/official/不经意传输/README.md
https://ctf-wiki.org/crypto/asymmetric/discrete-log/discrete-log/#pohlig-hellman-algorithm
发表评论
您还未登录，请先登录。
登录