hgame2025-Crypto Note
I found that I had been in the folder for a while. Organize and post it.
suprimeRSA
from 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(([39,71,126]))
def gen_key():
while True:
k = getPrime((20,40))
a = getPrime((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
"""
Actually it's CVE-2017-15361
A friendly link to introduce this attack:ROCA attack——CVE-2017-15361 | crumbling's secret room
GKCTF2020_Crypto_Reproduction_Affine Password Online Decryptor-CSDN BlogIt has appeared before, there is exp on the Internet
exp
from 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 .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 /mimoo/RSA-and-LLL-attacks/blob/master/
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 = ()
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 = ().gen()
# compute polynomials
gg = []
for ii in range(mm):
for jj in range(dd):
((x * XX) ** jj * modulus ** (mm - ii) * polZ(x * XX) ** ii)
for ii in range(tt):
((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 = ()
# 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:
(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 = (e,phi)
m = pow(enc,d,n)
print(libnum.n2s(int(m)))
Starting factorization...
b'hgame{ROCA_ROCK_and_ROll!}'
Sieve
#sage
from 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
It essentially calculates the first n term prime numbers and the first n term Euler function and
The prime numbers of the first n terms can be used with the sagemath built-in function prime_pi() function. As for finding Euler functions, I will not be able to find deepseek, so I will copy the official WP.
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 continue
for multiple in range(p, n+1, p):
phi[multiple] -= phi[multiple] // p
# Calculate the prefix and
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):
= m
, self.pre_s = compute_phi_and_prefix_sum(m) # Call new function
= {}
def S(self, n):
if n <= :
return self.pre_s[n]
if n in :
Return [n]
res = n * (n + 1) // 2
v = int(n**0.5)
sum1 = 0
for i in range(2, v + 1):
sum1 += (n // i)
u = n // (v + 1)
sum2 = 0
for k in range(1, u + 1):
sum2 += (k) * (n // k - n // (k + 1))
res -= (sum1 + sum2)
[n] = res
Return res
solver = EulerSumSolver(m=10**6)
print((65537**2 // 6)) # Note to correct the spelling of variable names slover->solver
#155763335410704472According to the formula
\[\phi(n) = n \prod_{i} \left(1 - \frac{1}{p_i}\right)
\]
So we only need it every time\(\phi(multiple) =\phi(multiple) -\frac{\phi(multiple)}{p}\)
Then we accumulate the Euler function values and get the prefix sum.
Then its recursive formula is
\[\begin{equation}
S(n) = \frac{n(n+1)}{2} - \sum_{i=2}^{v} S\left(\left\lfloor \frac{n}{i} \right\rfloor\right) - \sum_{k=1}^{u} S(k) \left(\left\lfloor \frac{n}{k}\right\rfloor - \left\lfloor \frac{n}{k+1}\right\rfloor\right)
\end{equation}
\]
HGame CTF 2025 week1 wpThis master's version is simpler
ezBag
from import *
import random
from import AES
import hashlib
from import pad
from secrets import flag
list = []
bag = []
p=(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
(a)
(b)
print(f'list={list}')
print(f'bag={bag}')
key = hashlib.sha256(str(p).encode()).digest()
cipher = (key, AES.MODE_ECB)
flag = pad(flag,16)
ciphertext = (flag)
print(f"ciphertext={ciphertext}")
There is a problem with backpack password, but the regular script is not allowed to be used.
\[\begin{pmatrix}
2 & 0 & \dots & 0 & a[0][63] & a[1][63] & a[2][63] & a[3][63] \\
0 & 2 & \dots & 0 & a[0][62] & a[1][62] & a[2][62] & a[3][62] \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots & \vdots \\
0 & 0 & \dots & 2 & a[0][0] & a[1][0] & a[2][0] & a[3][0] \\
1 & 1 & \dots & 1 & b[0] & b[1] & b[2] & b[3]
\end{pmatrix}
\]
import hashlib
from import *
from 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=()
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 = (key, AES.MODE_ECB)
flag = (ciphertext)
print(flag)
Why is this structure constructed in this way? Is there any master who understands the principle? However, on the surface, list and bag are both two-dimensional arrays, so it should be guessed that this is the structure.
Intergalactic Bound
from import *
from import AES
from 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 = (key, AES.MODE_ECB)
flag = pad(flag,16)
ciphertext = (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$\x00:\xf5"
"""
It looks very familiar at first glance because it happens to be studying curve-related issues. This is actually adapted from the combination of two curve topics from Yangcheng Cup 2024.
Attached link:/XiongSiqi_blog/article/details/141638136
I used the method of the blog above, without using the idiomatic method of distorted curves - mapped toWeierstrass
This question is to use two points to calculate a and d. Then when using the Pohlig_Hellman method later, the blog script you applied needs to be modified, that is, whether the last bit of the prime is removed. In the question of Yangcheng Cup, it is removed:
primes = [factors[i] ^ exponents[i] for i in range(len(factors))][:-1]
But this question does not need to be removed, I need to explore the specific reasons. I discovered it by chance
exp
from import *
from import AES
from import unpad
import hashlib
ciphertext=b"k\xe8\xbe\x94\x9e\xfc\xe2\x9e\x97\xe5\xf3\x04'\x8f\xb2\x01T\x06\x88\x04\xeb3Jl\xdd Pk$\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 = ()
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(()) // 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 = (key, AES.MODE_ECB)
flag = unpad((ciphertext), 16).decode()
print(f"Flag: {flag}")
#Flag: hgame{N0th1ng_bu7_up_Up_UP!}
Ancient Recall
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 = (tarot)
if card not in cards:
(card)
if card in Major_Arcana:
k = (reversals)
((card)^k)
if k == -1:
YOUR_initial_FATE.append("re-"+card)
else:
YOUR_initial_FATE.append(card)
else:
((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,^^
"""
Use deepseek R1 to kill directly?
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)
flag = "hgame{" + "&".join(cards) + "}"
print(flag)
#hgame{re-The_Moon&re-The_Sun&Judgement&re-Temperance&Six_of_Cups}
SPiCa
from import getPrime, long_to_bytes,bytes_to_long
from secrets import flag
from 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("", "w") as file:
(str(m) + "\n")
(str(p) + "\n")
for item in h:
(str(item) + "\n")
HSSP question, this question has actually been tested.
Exactly the same, this question is x\(1*n\)The random matrix of A is\(n*m\)random matrix, h=x*A
ctf-writeups/2022/zer0ptsctf/karen/solutions/ at main · roadicing/ctf-writeups · GitHubThe decryption script is here
principle:HSSP and Orthogonal grid study notes - 0xFFFF
use\(h\)Construct the grid basis, LLL finds m−n short vectors ui
Use ui to construct the grid\(L_x^{⊥}\),use\(Lx^{⊥}\)try to find\(L_x^{⊥}\)Orthogonal complement of\(L_x^{⊥}\)ˉ (can be regarded as\(L_x\)The same space, but the base is not xi\pmb{x}_ixxi)
right\(L_x^{⊥}\)ˉUse LLL recovery\(x_{i}\)
exp
#sage
from import long_to_bytes
n = 70
m = 247
p = 24727704801291912268835129736340977567569865784366882566681759917843647658060231409536848349518003784121914409876944135933654762801696486121844572452922377222301017649192408619831637530961997845860817966791811403512683444831050730277
h = (...)
# /2020/
# /raw/ZFk1qjfP
def orthoLattice(b,x0):
m = ()
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 = ()
for v in lv:
for i in range(n):
nv = allones(M5[i] - v)
if nv and nv not in lv:
(nv)
nv = allones(M5[i] + v)
if nv and nv not in lv:
(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 = ()
if m < 2 * n:
return M.right_kernel().matrix()
K = 2^(m // 2) * ()
MB = Matrix(ZZ, m + n, m)
MB[:n] = K * M
MB[n:] = identity_matrix(m)
MB2 = ().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 = ()
MOrtho = M2[:m - n]
ke = kernelLLL(MOrtho)
if n > 170:
return
beta = 2
while beta < n:
if beta == 2:
M5 = ()
else:
M5 = (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, ())), 2))
if (b"hgame{"):
FLAG = res
print(FLAG)