hgame2025-Crypto小记
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hgame2025-Crypto小记
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hgame2025-Crypto小記
發現積壓在文件夾有一段時間了。整理一下發出來。
suprimeRSA
task.py
from Crypto.Util.number import *
import random
from sympy import prime
FLAG=b'hgame{xxxxxxxxxxxxxxxxxx}'
e=0x10001
def primorial(num):
print(num)
result = 1
for i in range(1, num + 1):
result *= prime(i)
return result
M=primorial(random.choice([39,71,126]))
def gen_key():
while True:
k = getPrime(random.randint(20,40))
a = getPrime(random.randint(20,60))
p = k * M + pow(e, a, M)
if isPrime(p):
return p
p,q=gen_key(),gen_key()
n=p*q
m=bytes_to_long(FLAG)
enc=pow(m,e,n)
print(n.bit_length())
print(f'{n=}')
print(f'{enc=}')
"""
n=787190064146025392337631797277972559696758830083248285626115725258876808514690830730702705056550628756290183000265129340257928314614351263713241
enc=365164788284364079752299551355267634718233656769290285760796137651769990253028664857272749598268110892426683253579840758552222893644373690398408
"""
其實是CVE-2017-15361
介紹這個攻擊的友聯：ROCA攻擊——CVE-2017-15361 | crumbling's secret room
GKCTF2020_Crypto_復現_仿射密碼在線解密器-CSDN博客之前就出現過，網上有exp
exp
?
from sage.all import *
from tqdm import tqdm
?
def solve(M, n, a, m):
# I need to import it in the function otherwise multiprocessing doesn't find it in its context
from sage_functions import coppersmith_howgrave_univariate
?
base = int(65537)
# the known part of p: 65537^a * M^-1 (mod N)
known = int(pow(base, a, M) * inverse_mod(M, n))
# Create the polynom f(x)
F = PolynomialRing(Zmod(n), implementation='NTL', names=('x',))
(x,) = F._first_ngens(1)
pol = x + known
beta = 0.1
t = m+1
# Upper bound for the small root x0
XX = floor(2 * n**0.5 / M)
# Find a small root (x0 = k) using Coppersmith's algorithm
roots = coppersmith_howgrave_univariate(pol, n, beta, m, t, XX)
# There will be no roots for an incorrect guess of a.
for k in roots:
# reconstruct p from the recovered k
p = int(k*M + pow(base, a, M))
if n%p == 0:
return p, n//p
?
def roca(n):
?
keySize = n.bit_length()
?
if keySize <= 960:
M_prime = 0x1b3e6c9433a7735fa5fc479ffe4027e13bea
m = 5
?
elif 992 <= keySize <= 1952:
M_prime = 0x24683144f41188c2b1d6a217f81f12888e4e6513c43f3f60e72af8bd9728807483425d1e
m = 4
print("Have you several days/months to spend on this ?")
?
elif 1984 <= keySize <= 3936:
M_prime = 0x16928dc3e47b44daf289a60e80e1fc6bd7648d7ef60d1890f3e0a9455efe0abdb7a748131413cebd2e36a76a355c1b664be462e115ac330f9c13344f8f3d1034a02c23396e6
m = 7
print("You'll change computer before this scripts ends...")
?
elif 3968 <= keySize <= 4096:
print("Just no.")
return None
?
else:
print("Invalid key size: {}".format(keySize))
return None
?
a3 = Zmod(M_prime)(n).log(65537)
order = Zmod(M_prime)(65537).multiplicative_order()
inf = a3 // 2
sup = (a3 + order) // 2
?
# Search 10 000 values at a time, using multiprocess
# too big chunks is slower, too small chunks also
chunk_size = 10000
for inf_a in tqdm(range(inf, sup, chunk_size)):
# create an array with the parameter for the solve function
inputs = [((M_prime, n, a, m), {}) for a in range(inf_a, inf_a+chunk_size)]
# the sage builtin multiprocessing stuff
from sage.parallel.multiprocessing_sage import parallel_iter
from multiprocessing import cpu_count
?
for k, val in parallel_iter(cpu_count(), solve, inputs):
if val:
p = val[0]
q = val[1]
print("found factorization:\np={}\nq={}".format(p, q))
return val
?
from sage.all_cmdline import *
?
def coppersmith_howgrave_univariate(pol, modulus, beta, mm, tt, XX):
"""
Taken from https://github.com/mimoo/RSA-and-LLL-attacks/blob/master/coppersmith.sage
Coppersmith revisited by Howgrave-Graham
?
finds a solution if:
* b|modulus, b >= modulus^beta , 0 < beta <= 1
* |x| < XX
More tunable than sage's builtin coppersmith method, pol.small_roots()
"""
#
# init
#
dd = pol.degree()
nn = dd * mm + tt
?
#
# checks
#
if not 0 < beta <= 1:
raise ValueError("beta should belongs in [0, 1]")
?
if not pol.is_monic():
raise ArithmeticError("Polynomial must be monic.")
?
#
# calculate bounds and display them
#
"""
* we want to find g(x) such that ||g(xX)|| <= b^m / sqrt(n)
?
* we know LLL will give us a short vector v such that:
||v|| <= 2^((n - 1)/4) * det(L)^(1/n)
?
* we will use that vector as a coefficient vector for our g(x)
?
* so we want to satisfy:
2^((n - 1)/4) * det(L)^(1/n) < N^(beta*m) / sqrt(n)
?
so we can obtain ||v|| < N^(beta*m) / sqrt(n) <= b^m / sqrt(n)
(it's important to use N because we might not know b)
"""
#
# Coppersmith revisited algo for univariate
#
?
# change ring of pol and x
polZ = pol.change_ring(ZZ)
x = polZ.parent().gen()
?
# compute polynomials
gg = []
for ii in range(mm):
for jj in range(dd):
gg.append((x * XX) ** jj * modulus ** (mm - ii) * polZ(x * XX) ** ii)
for ii in range(tt):
gg.append((x * XX) ** ii * polZ(x * XX) ** mm)
?
# construct lattice B
BB = Matrix(ZZ, nn)
?
for ii in range(nn):
for jj in range(ii + 1):
BB[ii, jj] = gg[ii][jj]
?
BB = BB.LLL()
?
# transform shortest vector in polynomial
new_pol = 0
for ii in range(nn):
new_pol += x ** ii * BB[0, ii] / XX ** ii
# factor polynomial
potential_roots = new_pol.roots()
# test roots
roots = []
for root in potential_roots:
if root[0].is_integer():
result = polZ(ZZ(root[0]))
if gcd(modulus, result) >= modulus ** beta:
roots.append(ZZ(root[0]))
return roots
?
if __name__ == "__main__":
n = 787190064146025392337631797277972559696758830083248285626115725258876808514690830730702705056550628756290183000265129340257928314614351263713241
print("Starting factorization...")
#p,q = roca(n)
p=954455861490902893457047257515590051179337979243488068132318878264162627
q=824752716083066619280674937934149242011126804999047155998788143116757683
enc=365164788284364079752299551355267634718233656769290285760796137651769990253028664857272749598268110892426683253579840758552222893644373690398408
phi=(p-1)*(q-1)
e = 65537
d = gmpy2.invert(e,phi)
m = pow(enc,d,n)
print(libnum.n2s(int(m)))
?
Starting factorization...
b'hgame{ROCA_ROCK_and_ROll!}'
Sieve
#sage
from Crypto.Util.number import bytes_to_long
from sympy import nextprime
FLAG = b'hgame{xxxxxxxxxxxxxxxxxxxxxx}'
m = bytes_to_long(FLAG)
def trick(k):
if k > 1:
mul = prod(range(1,k))
if k - mul % k - 1 == 0:
return euler_phi(k) + trick(k-1) + 1
else:
return euler_phi(k) + trick(k-1)
else:
return 1
e = 65537
p = q = nextprime(trick(e^2//6)<<128)
n = p * q
enc = pow(m,e,n)
print(f'{enc=}')
#enc=2449294097474714136530140099784592732766444481665278038069484466665506153967851063209402336025065476172617376546
它本質上是求前n項素數和還有前n項歐拉函數和
前n項素數可以用sagemath內置函數prime_pi()函數。至于求歐拉函數我調教deepseek不出來，抄一下官方WP的。
e = 65537
limit = e**2 // 6
e = 65537
x=prime_pi(e^2//6)
#print(x)
#37030583
def sieve_of_eratosthenes(limit):
is_prime = [True] * (limit + 1)
p = 2
while p * p <= limit:
if is_prime[p]:
for i in range(p*p, limit+1, p):
is_prime[i] = False
p += 1
primes = [p for p in range(2, limit+1) if is_prime[p]]
return primes
def compute_phi_and_prefix_sum(n):
primes = sieve_of_eratosthenes(n)
phi = list(range(n + 1)) # phi[0]=0, phi[1]=1,...phi[i]=i
for p in primes:
if p < 2:
continue
for multiple in range(p, n+1, p):
phi[multiple] -= phi[multiple] // p
# 計算前綴和
pre_s = [0] * (n + 1)
current_sum = 0
for i in range(n + 1):
current_sum += phi[i]
pre_s[i] = current_sum
return phi, pre_s
class EulerSumSolver:
def __init__(self, m=10**6):
self.m = m
self.phi, self.pre_s = compute_phi_and_prefix_sum(m) # 調用新函數
self.cache = {}
def S(self, n):
if n <= self.m:
return self.pre_s[n]
if n in self.cache:
return self.cache[n]
res = n * (n + 1) // 2
v = int(n**0.5)
sum1 = 0
for i in range(2, v + 1):
sum1 += self.S(n // i)
u = n // (v + 1)
sum2 = 0
for k in range(1, u + 1):
sum2 += self.S(k) * (n // k - n // (k + 1))
res -= (sum1 + sum2)
self.cache[n] = res
return res
solver = EulerSumSolver(m=10**6)
print(solver.S(65537**2 // 6)) # 注意修正變量名拼寫slover->solver
#155763335410704472
根據公式
\[\phi(n) = n \prod_{i} \left(1 - \frac{1}{p_i}\right)
\]
\]
所以我們每次只需要\(\phi(multiple) =\phi(multiple) -\frac{\phi(multiple)}{p}\)
然后我們累加歐拉函數值，得到前綴和。
接著它的遞歸公式是
\[\begin{equation}
S(n) = \frac{n(n+1)}{2} - \sum_{i=2}^{v} S\left(\left\lfloor \frac{n}{i} \right\rfloor\right) - \sum_{k=1}^{u} S(k) \left(\left\lfloor \frac{n}{k}\right\rfloor - \left\lfloor \frac{n}{k+1}\right\rfloor\right)
\end{equation}
\]
S(n) = \frac{n(n+1)}{2} - \sum_{i=2}^{v} S\left(\left\lfloor \frac{n}{i} \right\rfloor\right) - \sum_{k=1}^{u} S(k) \left(\left\lfloor \frac{n}{k}\right\rfloor - \left\lfloor \frac{n}{k+1}\right\rfloor\right)
\end{equation}
\]
HGame CTF 2025 week1 wp這個師傅的版本簡單些
ezBag
task.py
from Crypto.Util.number import *
import random
from Crypto.Cipher import AES
import hashlib
from Crypto.Util.Padding import pad
from secrets import flag
list = []
bag = []
p=random.getrandbits(64)
assert len(bin(p)[2:])==64
for i in range(4):
t = p
a=[getPrime(32) for _ in range(64)]
b=0
for i in a:
temp=t%2
b+=temp*i
t=t>>1
list.append(a)
bag.append(b)
print(f'list={list}')
print(f'bag={bag}')
key = hashlib.sha256(str(p).encode()).digest()
cipher = AES.new(key, AES.MODE_ECB)
flag = pad(flag,16)
ciphertext = cipher.encrypt(flag)
print(f"ciphertext={ciphertext}")
背包密碼問題，但是套常規腳本不彳亍。
\[\begin{pmatrix}
2 & 0 & \dots & 0 & a[0][63] & a[1][63] & a[2][63] & a[3][63] \\
0 & 2 & \dots & 0 & a[0][62] & a[1][62] & a[2][62] & a[3][62] \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots & \vdots \\
0 & 0 & \dots & 2 & a[0][0] & a[1][0] & a[2][0] & a[3][0] \\
1 & 1 & \dots & 1 & b[0] & b[1] & b[2] & b[3]
\end{pmatrix}
\]
2 & 0 & \dots & 0 & a[0][63] & a[1][63] & a[2][63] & a[3][63] \\
0 & 2 & \dots & 0 & a[0][62] & a[1][62] & a[2][62] & a[3][62] \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots & \vdots \\
0 & 0 & \dots & 2 & a[0][0] & a[1][0] & a[2][0] & a[3][0] \\
1 & 1 & \dots & 1 & b[0] & b[1] & b[2] & b[3]
\end{pmatrix}
\]
import hashlib
from Crypto.Util.number import *
from Crypto.Cipher import AES
list=[[2826962231, 3385780583, 3492076631, 3387360133, 2955228863, 2289302839, 2243420737, 4129435549, 4249730059, 3553886213, 3506411549, 3658342997, 3701237861, 4279828309, 2791229339, 4234587439, 3870221273, 2989000187, 2638446521, 3589355327, 3480013811, 3581260537, 2347978027, 3160283047, 2416622491, 2349924443, 3505689469, 2641360481, 3832581799, 2977968451, 4014818999, 3989322037, 4129732829, 2339590901, 2342044303, 3001936603, 2280479471, 3957883273, 3883572877, 3337404269, 2665725899, 3705443933, 2588458577, 4003429009, 2251498177, 2781146657, 2654566039, 2426941147, 2266273523, 3210546259, 4225393481, 2304357101, 2707182253, 2552285221, 2337482071, 3096745679, 2391352387, 2437693507, 3004289807, 3857153537, 3278380013, 3953239151, 3486836107, 4053147071], [2241199309, 3658417261, 3032816659, 3069112363, 4279647403, 3244237531, 2683855087, 2980525657, 3519354793, 3290544091, 2939387147, 3669562427, 2985644621, 2961261073, 2403815549, 3737348917, 2672190887, 2363609431, 3342906361, 3298900981, 3874372373, 4287595129, 2154181787, 3475235893, 2223142793, 2871366073, 3443274743, 3162062369, 2260958543, 3814269959, 2429223151, 3363270901, 2623150861, 2424081661, 2533866931, 4087230569, 2937330469, 3846105271, 3805499729, 4188683131, 2804029297, 2707569353, 4099160981, 3491097719, 3917272979, 2888646377, 3277908071, 2892072971, 2817846821, 2453222423, 3023690689, 3533440091, 3737441353, 3941979749, 2903000761, 3845768239, 2986446259, 3630291517, 3494430073, 2199813137, 2199875113, 3794307871, 2249222681, 2797072793], [4263404657, 3176466407, 3364259291, 4201329877, 3092993861, 2771210963, 3662055773, 3124386037, 2719229677, 3049601453, 2441740487, 3404893109, 3327463897, 3742132553, 2833749769, 2661740833, 3676735241, 2612560213, 3863890813, 3792138377, 3317100499, 2967600989, 2256580343, 2471417173, 2855972923, 2335151887, 3942865523, 2521523309, 3183574087, 2956241693, 2969535607, 2867142053, 2792698229, 3058509043, 3359416111, 3375802039, 2859136043, 3453019013, 3817650721, 2357302273, 3522135839, 2997389687, 3344465713, 2223415097, 2327459153, 3383532121, 3960285331, 3287780827, 4227379109, 3679756219, 2501304959, 4184540251, 3918238627, 3253307467, 3543627671, 3975361669, 3910013423, 3283337633, 2796578957, 2724872291, 2876476727, 4095420767, 3011805113, 2620098961], [2844773681, 3852689429, 4187117513, 3608448149, 2782221329, 4100198897, 3705084667, 2753126641, 3477472717, 3202664393, 3422548799, 3078632299, 3685474021, 3707208223, 2626532549, 3444664807, 4207188437, 3422586733, 2573008943, 2992551343, 3465105079, 4260210347, 3108329821, 3488033819, 4092543859, 4184505881, 3742701763, 3957436129, 4275123371, 3307261673, 2871806527, 3307283633, 2813167853, 2319911773, 3454612333, 4199830417, 3309047869, 2506520867, 3260706133, 2969837513, 4056392609, 3819612583, 3520501211, 2949984967, 4234928149, 2690359687, 3052841873, 4196264491, 3493099081, 3774594497, 4283835373, 2753384371, 2215041107, 4054564757, 4074850229, 2936529709, 2399732833, 3078232933, 2922467927, 3832061581, 3871240591, 3526620683, 2304071411, 3679560821]]
bag=[123342809734, 118191282440, 119799979406, 128273451872]
ciphertext=b'\x1d6\xcc}\x07\xfa7G\xbd\x01\xf0P4^Q"\x85\x9f\xac\x98\x8f#\xb2\x12\xf4+\x05`\x80\x1a\xfa !\x9b\xa5\xc7g\xa8b\x89\x93\x1e\xedz\xd2M;\xa2'
L=matrix(ZZ,65,68)
for i in range(64):
L[i,i]=2
L[i,-1]=list[3][-i-1]
L[i,-2]=list[2][-i-1]
L[i,-3]=list[1][-i-1]
L[i,-4]=list[0][-i-1]
L[-1,:]=1
L[-1,-1]=bag[3]
L[-1,-2]=bag[2]
L[-1,-3]=bag[1]
L[-1,-4]=bag[0]
x=L.BKZ()
print(x[0])
p=''
for i in x[0][:64]:
if i==x[0][0]:
p+='1'
else:
p+='0'
p=int(p,2)
key = hashlib.sha256(str(p).encode()).digest()
cipher = AES.new(key, AES.MODE_ECB)
flag = cipher.decrypt(ciphertext)
print(flag)
這個格為什么這樣構造的，有沒有師傅懂原理。不過從表面上看，list和bag都是二維數組，應該可以猜測是這樣構造的吧。
Intergalactic Bound
task.py
from Crypto.Util.number import *
from Crypto.Cipher import AES
from Crypto.Util.Padding import pad
from random import randint
import hashlib
from secrets import flag
def add_THCurve(P, Q):
if P == (0, 0):
return Q
if Q == (0, 0):
return P
x1, y1 = P
x2, y2 = Q
x3 = (x1 - y1 ** 2 * x2 * y2) * pow(a * x1 * y1 * x2 ** 2 - y2, -1, p) % p
y3 = (y1 * y2 ** 2 - a * x1 ** 2 * x2) * pow(a * x1 * y1 * x2 ** 2 - y2, -1, p) % p
return x3, y3
def mul_THCurve(n, P):
R = (0, 0)
while n > 0:
if n % 2 == 1:
R = add_THCurve(R, P)
P = add_THCurve(P, P)
n = n // 2
return R
p = getPrime(96)
a = randint(1, p)
G = (randint(1,p), randint(1,p))
d = (a*G[0]^3+G[1]^3+1)%p*inverse(G[0]*G[1],p)%p
x = randint(1, p)
Q = mul_THCurve(x, G)
print(f"p = {p}")
print(f"G = {G}")
print(f"Q = {Q}")
key = hashlib.sha256(str(x).encode()).digest()
cipher = AES.new(key, AES.MODE_ECB)
flag = pad(flag,16)
ciphertext = cipher.encrypt(flag)
print(f"ciphertext={ciphertext}")
"""
p = 55099055368053948610276786301
G = (19663446762962927633037926740, 35074412430915656071777015320)
Q = (26805137673536635825884330180, 26376833112609309475951186883)
ciphertext=b"k\xe8\xbe\x94\x9e\xfc\xe2\x9e\x97\xe5\xf3\x04'\x8f\xb2\x01T\x06\x88\x04\xeb3Jl\xdd Pk$\x00:\xf5"
"""
一眼非常眼熟，因為剛好在研究curve相關問題。這其實是羊城杯2024的兩道curve題目結合改編而來。
附上鏈接：https://blog.csdn.net/XiongSiqi_blog/article/details/141638136
我采用上面博客的方法，沒有使用扭曲曲線慣用方法——映射到Weierstrass
本題就是先用兩個點算出a和d。然后后面使用Pohlig_Hellman方法的時候，套用的博客腳本要修改一下，就是它這個prime最后一位有沒有去掉的問題，羊城杯那題中,它是去掉的：
primes = [factors[i] ^ exponents[i] for i in range(len(factors))][:-1]
但是這題不用去掉，具體原因我還要探究一下。我是偶然間把它去了發現才可以解
exp
from Crypto.Util.number import *
from Crypto.Cipher import AES
from Crypto.Util.Padding import unpad
import hashlib
ciphertext=b"k\xe8\xbe\x94\x9e\xfc\xe2\x9e\x97\xe5\xf3\x04'\x8f\xb2\x01T\x06\x88\x04\xeb3Jl\xdd Pk$\x00:\xf5"
a=39081810733380615260725035189
p = 55099055368053948610276786301
P = (19663446762962927633037926740, 35074412430915656071777015320)
Q = (26805137673536635825884330180, 26376833112609309475951186883)
d = (a * P[0] ** 3 + P[1] ** 3 + 1) * inverse(P[0] * P[1], p) % p
# construct ECC to get a solution of 2X^3+Y^3+Z^3=dXYZ
R.<x,y,z> = Zmod(p)[]
cubic = a* x^3 + y^3 + z^3 - d*x*y*z
E = EllipticCurve_from_cubic(cubic,morphism=True)
P = E(P)
Q = E(Q)
P_ord = P.order()
def Pohlig_Hellman(n,P,Q):
factors, exponents = zip(*factor(n))
primes = [factors[i] ^ exponents[i] for i in range(len(factors))]
print(primes)
dlogs = []
for fac in primes:
t = int(int(P.order()) // int(fac))
dlog = discrete_log(t*Q,t*P,operation="+")
dlogs += [dlog]
print("factor: "+str(fac)+", Discrete Log: "+str(dlog)) #calculates discrete logarithm for each prime order
num2 = crt(dlogs,primes)
return num2
num2 = Pohlig_Hellman(P_ord,P,Q)
print(num2)
key = hashlib.sha256(str(num2).encode()).digest()
cipher = AES.new(key, AES.MODE_ECB)
flag = unpad(cipher.decrypt(ciphertext), 16).decode()
print(f"Flag: {flag}")
#Flag: hgame{N0th1ng_bu7_up_Up_UP!}
Ancient Recall
task.py
import random
Major_Arcana = ["The Fool", "The Magician", "The High Priestess","The Empress", "The Emperor", "The Hierophant","The Lovers", "The Chariot", "Strength","The Hermit", "Wheel of Fortune", "Justice","The Hanged Man", "Death", "Temperance","The Devil", "The Tower", "The Star","The Moon", "The Sun", "Judgement","The World"]
wands = ["Ace of Wands", "Two of Wands", "Three of Wands", "Four of Wands", "Five of Wands", "Six of Wands", "Seven of Wands", "Eight of Wands", "Nine of Wands", "Ten of Wands", "Page of Wands", "Knight of Wands", "Queen of Wands", "King of Wands"]
cups = ["Ace of Cups", "Two of Cups", "Three of Cups", "Four of Cups", "Five of Cups", "Six of Cups", "Seven of Cups", "Eight of Cups", "Nine of Cups", "Ten of Cups", "Page of Cups", "Knight of Cups", "Queen of Cups", "King of Cups"]
swords = ["Ace of Swords", "Two of Swords", "Three of Swords", "Four of Swords", "Five of Swords", "Six of Swords", "Seven of Swords", "Eight of Swords", "Nine of Swords", "Ten of Swords", "Page of Swords", "Knight of Swords", "Queen of Swords", "King of Swords"]
pentacles = ["Ace of Pentacles", "Two of Pentacles", "Three of Pentacles", "Four of Pentacles", "Five of Pentacles", "Six of Pentacles", "Seven of Pentacles", "Eight of Pentacles", "Nine of Pentacles", "Ten of Pentacles", "Page of Pentacles", "Knight of Pentacles", "Queen of Pentacles", "King of Pentacles"]
Minor_Arcana = wands + cups + swords + pentacles
tarot = Major_Arcana + Minor_Arcana
reversals = [0,-1]
Value = []
cards = []
YOUR_initial_FATE = []
while len(YOUR_initial_FATE)<5:
card = random.choice(tarot)
if card not in cards:
cards.append(card)
if card in Major_Arcana:
k = random.choice(reversals)
Value.append(tarot.index(card)^k)
if k == -1:
YOUR_initial_FATE.append("re-"+card)
else:
YOUR_initial_FATE.append(card)
else:
Value.append(tarot.index(card))
YOUR_initial_FATE.append(card)
else:
continue
print("Oops!lets reverse 1T!")
FLAG=("hgame{"+"&".join(YOUR_initial_FATE)+"}").replace(" ","_")
YOUR_final_Value = Value
def Fortune_wheel(FATE):
FATEd = [FATE[i]+FATE[(i+1)%5] for i in range(len(FATE))]
return FATEd
for i in range(250):
YOUR_final_Value = Fortune_wheel(YOUR_final_Value)
print(YOUR_final_Value)
YOUR_final_FATE = []
for i in YOUR_final_Value:
YOUR_final_FATE.append(tarot[i%78])
print("Your destiny changed!\n",",".join(YOUR_final_FATE))
print("oh,now you GET th3 GOOd lU>k,^^")
"""
Oops!lets reverse 1T!
[2532951952066291774890498369114195917240794704918210520571067085311474675019, 2532951952066291774890327666074100357898023013105443178881294700381509795270, 2532951952066291774890554459287276604903130315859258544173068376967072335730, 2532951952066291774890865328241532885391510162611534514014409174284299139015, 2532951952066291774890830662608134156017946376309989934175833913921142609334]
Your destiny changed!
Eight of Cups,Ace of Cups,Strength,The Chariot,Five of Swords
oh,now you GET th3 GOOd lU>k,^^
"""
直接使用deepseek R1秒殺？
exp
Major_Arcana = ["The Fool", "The Magician", "The High Priestess","The Empress", "The Emperor", "The Hierophant","The Lovers", "The Chariot", "Strength","The Hermit", "Wheel of Fortune", "Justice","The Hanged Man", "Death", "Temperance","The Devil", "The Tower", "The Star","The Moon", "The Sun", "Judgement","The World"]
wands = ["Ace of Wands", "Two of Wands", "Three of Wands", "Four of Wands", "Five of Wands", "Six of Wands", "Seven of Wands", "Eight of Wands", "Nine of Wands", "Ten of Wands", "Page of Wands", "Knight of Wands", "Queen of Wands", "King of Wands"]
cups = ["Ace of Cups", "Two of Cups", "Three of Cups", "Four of Cups", "Five of Cups", "Six of Cups", "Seven of Cups", "Eight of Cups", "Nine of Cups", "Ten of Cups", "Page of Cups", "Knight of Cups", "Queen of Cups", "King of Cups"]
swords = ["Ace of Swords", "Two of Swords", "Three of Swords", "Four of Swords", "Five of Swords", "Six of Swords", "Seven of Swords", "Eight of Swords", "Nine of Swords", "Ten of Swords", "Page of Swords", "Knight of Swords", "Queen of Swords", "King of Swords"]
pentacles = ["Ace of Pentacles", "Two of Pentacles", "Three of Pentacles", "Four of Pentacles", "Five of Pentacles", "Six of Pentacles", "Seven of Pentacles", "Eight of Pentacles", "Nine of Pentacles", "Ten of Pentacles", "Page of Pentacles", "Knight of Pentacles", "Queen of Pentacles", "King of Pentacles"]
Minor_Arcana = wands + cups + swords + pentacles
tarot = Major_Arcana + Minor_Arcana
final_values = [
2532951952066291774890498369114195917240794704918210520571067085311474675019,
2532951952066291774890327666074100357898023013105443178881294700381509795270,
2532951952066291774890554459287276604903130315859258544173068376967072335730,
2532951952066291774890865328241532885391510162611534514014409174284299139015,
2532951952066291774890830662608134156017946376309989934175833913921142609334
]
def reverse_fortune_wheel(current):
v0 = (current[0] + current[4] - current[1] - current[3] + current[2]) // 2
v1 = current[0] - v0
v2 = current[1] - v1
v3 = current[2] - v2
v4 = current[3] - v3
assert v4 + v0 == current[4], "Reverse step failed"
return [v0, v1, v2, v3, v4]
current = final_values.copy()
for _ in range(250):
current = reverse_fortune_wheel(current)
initial_values = current
def get_card_name(value):
k_reversed = value ^ -1
if 0 <= k_reversed < len(Major_Arcana):
return f"re-{Major_Arcana[k_reversed]}"
if 0 <= value < len(Major_Arcana):
return Major_Arcana[value]
index = value % len(tarot)
return tarot[index]
cards = []
for v in initial_values:
card = get_card_name(v)
card = card.replace(" ", "_")
cards.append(card)
flag = "hgame{" + "&".join(cards) + "}"
print(flag)
#hgame{re-The_Moon&re-The_Sun&Judgement&re-Temperance&Six_of_Cups}
SPiCa
task.py
from Crypto.Util.number import getPrime, long_to_bytes,bytes_to_long
from secrets import flag
from sage.all import *
def derive_M(n):
iota=0.035
Mbits=int(2 * iota * n^2 + n * log(n,2))
M = random_prime(2^Mbits, proof = False, lbound = 2^(Mbits - 1))
return Integer(M)
m = bytes_to_long(flag).bit_length()
n = 70
p = derive_M(n)
F = GF(p)
x = random_matrix(F, 1, n)
A = random_matrix(ZZ, n, m, x=0, y=2)
A[randint(0, n-1)] = vector(ZZ, list(bin(bytes_to_long(flag))[2:]))
h = x*A
with open("data.txt", "w") as file:
file.write(str(m) + "\n")
file.write(str(p) + "\n")
for item in h:
file.write(str(item) + "\n")
HSSP問題，這題其實考過，y011d4.log
一模一樣的，這題就是x是\(1*n\)的隨機矩陣，A是\(n*m\)的隨機矩陣，h=x*A
ctf-writeups/2022/zer0ptsctf/karen/solutions/solver.sage at main · roadicing/ctf-writeups · GitHub解密腳本在這
原理：HSSP與正交格學習筆記 - 0xFFFF
用\(h\)構造格基，LLL找到m?n個短向量ui
用ui構造格\(L_x^{⊥}\)，用\(Lx^{⊥}\)找\(L_x^{⊥}\)的正交補\(L_x^{⊥}\)ˉ（可以看作是和\(L_x\)同一個空間，但基不是xi\pmb{x}_ixxi）
對\(L_x^{⊥}\)ˉ使用LLL恢復\(x_{i}\)
exp
#sage
from Crypto.Util.number import long_to_bytes
n = 70
m = 247
p = 24727704801291912268835129736340977567569865784366882566681759917843647658060231409536848349518003784121914409876944135933654762801696486121844572452922377222301017649192408619831637530961997845860817966791811403512683444831050730277
h = (...)
# https://eprint.iacr.org/2020/461.pdf
# https://pastebin.com/raw/ZFk1qjfP
def orthoLattice(b,x0):
m = b.length()
M = Matrix(ZZ, m, m)
for i in range(1, m):
M[i, i] = 1
M[1:m, 0] = -b[1: m] * inverse_mod(b[0], x0)
M[0,0] = x0
for i in range(1, m):
M[i, 0] = mod(M[i, 0], x0)
return M
def allones(v):
if len([vj for vj in v if vj in [0, 1]]) == len(v):
return v
if len([vj for vj in v if vj in [0, -1]]) == len(v):
return -v
return None
def recoverBinary(M5):
lv = [allones(vi) for vi in M5 if allones(vi)]
n = M5.nrows()
for v in lv:
for i in range(n):
nv = allones(M5[i] - v)
if nv and nv not in lv:
lv.append(nv)
nv = allones(M5[i] + v)
if nv and nv not in lv:
lv.append(nv)
return Matrix(lv)
def allpmones(v):
return len([vj for vj in v if vj in [-1, 0, 1]]) == len(v)
def kernelLLL(M):
n = M.nrows()
m = M.ncols()
if m < 2 * n:
return M.right_kernel().matrix()
K = 2^(m // 2) * M.height()
MB = Matrix(ZZ, m + n, m)
MB[:n] = K * M
MB[n:] = identity_matrix(m)
MB2 = MB.T.LLL().T
assert MB2[:n, :m - n] == 0
Ke = MB2[n:, :m - n].T
return Ke
def NSattack(n, m, p, h):
iota = 0.035
nx0 = int(2 * iota * n^2 + n * log(n, 2))
x0 = p
b = vector(h)
M = orthoLattice(b, x0)
M2 = M.LLL()
MOrtho = M2[:m - n]
ke = kernelLLL(MOrtho)
if n > 170:
return
beta = 2
while beta < n:
if beta == 2:
M5 = ke.LLL()
else:
M5 = M5.BKZ(block_size = beta)
if len([True for v in M5 if allpmones(v)]) == n:
break
if beta == 2:
beta = 10
else:
beta += 10
MB = recoverBinary(M5)
return MB
MB = NSattack(n, m, p, h)
for r in MB:
res = long_to_bytes(int(''.join(map(str, r.list())), 2))
if res.startswith(b"hgame{"):
FLAG = res
print(FLAG)
總結
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