CryptoCTF 2023 (part2)
这部分主要是复现赛时没做的medium类型的题目，虽说是medium难度实际上有的难度大于hard...
Medium
ASIv1
实际上就是\(\small R.seed=S\)，给定R和S求seed。计算过程在模3下进行。但是R为\(\small l^2\)行，\(\small l\)列，如果直接解矩阵方程复杂度太大不可行，考虑只取前\(\small 3l\)行进行计算，矩阵满秩且复杂度可接受。
f1 = open('ASlv_output.txt','r')
out = f1.read().splitlines()
R = eval(out[0][4:])
S = eval(out[1][4:])
l = 3 * len(R[0])
f = GF(3)
R = matrix(f,R[:l])
S= vector(f,S[:l])
x = R.solve_right(S)
print(x)
msg = ''.join(str(xi) for xi in x)
msg = int(msg, 3)
from Crypto.Util.number import *
print(long_to_bytes(msg))
# 3Xpl0i7eD_bY_AtT4ck3r!
Resuction
Suction的plus版本，p q都为1024比特，但d只有64比特，那么维纳一定可解。问题在于直接遍历n e的低位进行维纳攻击会很慢，根据连分数的生成原理可知，n的高位/e的高位展开连分数前若干项和对n/e直接展开是一致的，因此就可以直接用已知的高位展开，找到比特为64的质数，大概率就是d了，进一步爆破n e的低位结合刚才的d进行验证得到最终的n e d，最后爆破enc的低位解密即可。
PKEY =
enc =
b_PKEY = f'{PKEY:b}'
nh = ZZ(int(b_PKEY[:2040],2) << 8)
eh = ZZ(int(b_PKEY[2040:],2) << 8)
def get_d(N, e):
cf = continued_fraction (N / e)
i = 0
while True:
denom = cf . denominator ( i )
numer = cf . numerator ( i )
d = numer
if d.nbits() == 64 and is_prime(d):
return d
i += 1
d = get_d(nh,eh)
print(d)
for i in range(2^8):
for j in range(2^8):
n = nh + i
e = eh + j
if pow(2, e*d, n) == g:
print(f'n = {n}')
print(f'e = {e}')
Barak
参考春哥博客复现，利用sagemath的三次齐次方程的椭圆曲线进行构造，求阶，要注意参数morphism为True的时候为同构的Weierstrass elliptic curve才能用log求dlp，为False的时候才能求order。最后还需要枚举增加阶的倍数获得flag。
p = 73997272456239171124655017039956026551127725934222347
c = 1
d = 68212800478915688445169020404812347140341674954375635
# z=1的特殊情况
R.<x,y,z> = Zmod(p)[]
cubic = x^3 + y^3 + c * z^3 - d * x * y * z
EC = EllipticCurve_from_cubic(cubic, morphism=False)
mf = EllipticCurve_from_cubic(cubic, morphism=True)
print(EC.order())
P = (71451574057642615329217496196104648829170714086074852, 69505051165402823276701818777271117086632959198597714)
Q = (40867727924496334272422180051448163594354522440089644, 56052452825146620306694006054673427761687498088402245)
# PP = mf(P)
# QQ = mf(Q)
# dlp = PP.discrete_log(QQ)
# print(dlp)
# 73997272456239171124655016995459084401465136460086688
# 1780694557271320552511299360138314441283923223949197
n = 1780694557271320552511299360138314441283923223949197
p = 73997272456239171124655017039956026551127725934222347
o = 3083219685676632130193959041477461850061047352503612
from Crypto.Util.number import *
for i in range(100):
m = long_to_bytes(n + i * o)
try:
print(m.decode())
except:
continue
risk
若，m不小于3，则\(\small N>(ab)^m\)且\(\small N<(ab+1)^m\)，那么如果能找到m通过开根就能找到求出ab，并且知道r和s的比特为14，枚举一下验证得r和s仅一种可能取值，利用r、s，ab以及N的表达式可以联立方程用gb基求解出y，即\(b^m\)，那么rsa的q就拿到了。
pkey = (150953688, 373824666550208932851344358703053061405262438259996622188837935528607451817812480600479188884096072016823491996056842120586016323642383543231913508464973502962030059403443181467594283936168384790367731793997013711256520780338341018619858240958105689126133812557401122953030695509876185592917323138313818881164334490044163316692588337720342646339764174333821950301279825316497898035760524780198230860089280791887363472060435443944632303774987556026740232641561706904946900169858650106994947597642013168757868017723456208796677559254390940651802333382820063200360490892131573853635471525711894510477078934343423255983)
enc = 275574424285842306309073814038154403551700455145115884031072340378743712325975683329051874910297915882286569143815715537085387363420246497061870251520240399514896001311724695996978111559476733709139080970977190150345474341853495386364275702356438666152069791355990718058189043717952080875207858163490627801836274404446661613239167700736337269924479349700031535265765885064606399858172168036794462235707003475360358004643720927563261787867952228496769300443415094124132722170498229611285689671203272698693505808912907778910378274197503048226322090611405601517624884408718689404556983397217070272851442351897456769883
e, N = pkey
m = 4
# 需要枚举e的14位因子确定r和s
r = 10728
s = 14071
ab = floor(N^(1/m))
PR.<x, y> = PolynomialRing(ZZ)
f1 = x * y - ab^m
f2 = (x + r) * (y + s) - N
print(Ideal([f1, f2]).groebner_basis())
PR.<y> = PolynomialRing(ZZ)
f3 = 10728*y^2 - 511086150867555670991182215716996884365017094048011334498502864759458604412974525192545554902255616748785034586283311417267948637161905335789524940513213807388789747893057116518524986565352587286484213819873973314973576845522082944759362821020072842990178502502907835104629002180230922192278601476281549539304919*y + 5260086883027989894151266471310659627033447768756412470819138590823035454528439414529342666788115829348723355876515825478765835689971978836816254977610642160178724965865849006430519169265825342385264354073331979931090503900140796473000025308521505151693828876495191201072094916519467807477939653879213738285370121752035575512256453146199781476491291347483720934255498391686419228296642359873666276793532997868425427822887434754439647641247870006446823852067931010198965229083394533490439946570165059178219301430708237320778256595039419216917256059584687394490563748927035618459815041808985900104721604927460617006077696
print(f3.roots())
q = 15040222622096320078383580808680733765955114958694997949647342925417877088612792495485641348591026281373930569798925789027166056695954731923306109646611840570310396750856642056018981080439916663195842593441587057719678555907050674529272376248049062724657792390788687452049496308886252188791975094655675924736 + s
p = pkey[1]//q
assert p * q == pkey[1]
print(p,q)
后面参考little case的做法：
from Crypto.Util.number import *
import itertools
pkey = (150953688, 373824666550208932851344358703053061405262438259996622188837935528607451817812480600479188884096072016823491996056842120586016323642383543231913508464973502962030059403443181467594283936168384790367731793997013711256520780338341018619858240958105689126133812557401122953030695509876185592917323138313818881164334490044163316692588337720342646339764174333821950301279825316497898035760524780198230860089280791887363472060435443944632303774987556026740232641561706904946900169858650106994947597642013168757868017723456208796677559254390940651802333382820063200360490892131573853635471525711894510477078934343423255983)
enc = 275574424285842306309073814038154403551700455145115884031072340378743712325975683329051874910297915882286569143815715537085387363420246497061870251520240399514896001311724695996978111559476733709139080970977190150345474341853495386364275702356438666152069791355990718058189043717952080875207858163490627801836274404446661613239167700736337269924479349700031535265765885064606399858172168036794462235707003475360358004643720927563261787867952228496769300443415094124132722170498229611285689671203272698693505808912907778910378274197503048226322090611405601517624884408718689404556983397217070272851442351897456769883
e, N = pkey
m = 4
r = 10728
s = 14071
bm = 15040222622096320078383580808680733765955114958694997949647342925417877088612792495485641348591026281373930569798925789027166056695954731923306109646611840570310396750856642056018981080439916663195842593441587057719678555907050674529272376248049062724657792390788687452049496308886252188791975094655675924736
q = bm + s
p = N // q
def get_oneroot(p, e):
while True:
Zp = Zmod(p)
g = Zp.random_element()
g = g^((p-1) // e)
for mult in divisors(e):
if (mult != e):
g2 = g^mult
if (g2 == 1):
break
else:
return g
def decrypt(p, c, e):
w = gcd(e, p-1)
e1, p1 = e // w, (p-1) // w
d = inverse_mod(e1, p1)
c1 = pow(c, d, p)
g, a, b = xgcd(p1, w)
g = get_oneroot(p, w)
m = pow(c1, b, p)
return [ZZ(m * g^i) for i in range(w)]
mp_list = decrypt(p, enc, e)
print('Find root p OK')
mq_list = decrypt(q, enc, e)
print('Find root q OK')
for mp, mq in itertools.product(mp_list, mq_list):
m = crt([mp, mq], [p, q])
msg = long_to_bytes(int(m))
if (b'CCTF' in msg):
print(msg)
Insights（预期）
参考春哥，利用p和q的高位改改Boneh-Durfee的形式即可。
from __future__ import print_function
import time
############################################
# Config
##########################################
"""
Setting debug to true will display more informations
about the lattice, the bounds, the vectors...
"""
debug = True
"""
Setting strict to true will stop the algorithm (and
return (-1, -1)) if we don't have a correct
upperbound on the determinant. Note that this
doesn't necesseraly mean that no solutions
will be found since the theoretical upperbound is
usualy far away from actual results. That is why
you should probably use `strict = False`
"""
strict = False
"""
This is experimental, but has provided remarkable results
so far. It tries to reduce the lattice as much as it can
while keeping its efficiency. I see no reason not to use
this option, but if things don't work, you should try
disabling it
"""
helpful_only = True
dimension_min = 7 # stop removing if lattice reaches that dimension
############################################
# Functions
##########################################
# display stats on helpful vectors
def helpful_vectors(BB, modulus):
nothelpful = 0
for ii in range(BB.dimensions()[0]):
if BB[ii,ii] >= modulus:
nothelpful += 1
print(nothelpful, "/", BB.dimensions()[0], " vectors are not helpful")
# display matrix picture with 0 and X
def matrix_overview(BB, bound):
for ii in range(BB.dimensions()[0]):
a = ('%02d ' % ii)
for jj in range(BB.dimensions()[1]):
a += '0' if BB[ii,jj] == 0 else 'X'
if BB.dimensions()[0] < 60:
a += ' '
if BB[ii, ii] >= bound:
a += '~'
print(a)
# tries to remove unhelpful vectors
# we start at current = n-1 (last vector)
def remove_unhelpful(BB, monomials, bound, current):
# end of our recursive function
if current == -1 or BB.dimensions()[0] <= dimension_min:
return BB
# we start by checking from the end
for ii in range(current, -1, -1):
# if it is unhelpful:
if BB[ii, ii] >= bound:
affected_vectors = 0
affected_vector_index = 0
# let's check if it affects other vectors
for jj in range(ii + 1, BB.dimensions()[0]):
# if another vector is affected:
# we increase the count
if BB[jj, ii] != 0:
affected_vectors += 1
affected_vector_index = jj
# level:0
# if no other vectors end up affected
# we remove it
if affected_vectors == 0:
print("* removing unhelpful vector", ii)
BB = BB.delete_columns([ii])
BB = BB.delete_rows([ii])
monomials.pop(ii)
BB = remove_unhelpful(BB, monomials, bound, ii-1)
return BB
# level:1
# if just one was affected we check
# if it is affecting someone else
elif affected_vectors == 1:
affected_deeper = True
for kk in range(affected_vector_index + 1, BB.dimensions()[0]):
# if it is affecting even one vector
# we give up on this one
if BB[kk, affected_vector_index] != 0:
affected_deeper = False
# remove both it if no other vector was affected and
# this helpful vector is not helpful enough
# compared to our unhelpful one
if affected_deeper and abs(bound - BB[affected_vector_index, affected_vector_index]) < abs(bound - BB[ii, ii]):
print("* removing unhelpful vectors", ii, "and", affected_vector_index)
BB = BB.delete_columns([affected_vector_index, ii])
BB = BB.delete_rows([affected_vector_index, ii])
monomials.pop(affected_vector_index)
monomials.pop(ii)
BB = remove_unhelpful(BB, monomials, bound, ii-1)
return BB
# nothing happened
return BB
"""
Returns:
* 0,0 if it fails
* -1,-1 if `strict=true`, and determinant doesn't bound
* x0,y0 the solutions of `pol`
"""
def boneh_durfee(pol, modulus, mm, tt, XX, YY):
"""
Boneh and Durfee revisited by Herrmann and May
finds a solution if:
* d < N^delta
* |x| < e^delta
* |y| < e^0.5
whenever delta < 1 - sqrt(2)/2 ~ 0.292
"""
# substitution (Herrman and May)
PR.<u, x, y> = PolynomialRing(ZZ)
Q = PR.quotient(x*y + 1 - u) # u = xy + 1
polZ = Q(pol).lift()
UU = XX*YY + 1
# x-shifts
gg = []
for kk in range(mm + 1):
for ii in range(mm - kk + 1):
xshift = x^ii * modulus^(mm - kk) * polZ(u, x, y)^kk
gg.append(xshift)
gg.sort()
# x-shifts list of monomials
monomials = []
for polynomial in gg:
for monomial in polynomial.monomials():
if monomial not in monomials:
monomials.append(monomial)
monomials.sort()
# y-shifts (selected by Herrman and May)
for jj in range(1, tt + 1):
for kk in range(floor(mm/tt) * jj, mm + 1):
yshift = y^jj * polZ(u, x, y)^kk * modulus^(mm - kk)
yshift = Q(yshift).lift()
gg.append(yshift) # substitution
# y-shifts list of monomials
for jj in range(1, tt + 1):
for kk in range(floor(mm/tt) * jj, mm + 1):
monomials.append(u^kk * y^jj)
# construct lattice B
nn = len(monomials)
BB = Matrix(ZZ, nn)
for ii in range(nn):
BB[ii, 0] = gg[ii](0, 0, 0)
for jj in range(1, ii + 1):
if monomials[jj] in gg[ii].monomials():
BB[ii, jj] = gg[ii].monomial_coefficient(monomials[jj]) * monomials[jj](UU,XX,YY)
# Prototype to reduce the lattice
if helpful_only:
# automatically remove
BB = remove_unhelpful(BB, monomials, modulus^mm, nn-1)
# reset dimension
nn = BB.dimensions()[0]
if nn == 0:
print("failure")
return 0,0
# check if vectors are helpful
if debug:
helpful_vectors(BB, modulus^mm)
# check if determinant is correctly bounded
det = BB.det()
bound = modulus^(mm*nn)
if det >= bound:
print("We do not have det < bound. Solutions might not be found.")
print("Try with highers m and t.")
if debug:
diff = (log(det) - log(bound)) / log(2)
print("size det(L) - size e^(m*n) = ", floor(diff))
if strict:
return -1, -1
else:
print("det(L) < e^(m*n) (good! If a solution exists < N^delta, it will be found)")
# display the lattice basis
if debug:
matrix_overview(BB, modulus^mm)
# LLL
if debug:
print("optimizing basis of the lattice via LLL, this can take a long time")
BB = BB.LLL()
if debug:
print("LLL is done!")
# transform vector i & j -> polynomials 1 & 2
if debug:
print("looking for independent vectors in the lattice")
found_polynomials = False
for pol1_idx in range(nn - 1):
for pol2_idx in range(pol1_idx + 1, nn):
# for i and j, create the two polynomials
PR.<w,z> = PolynomialRing(ZZ)
pol1 = pol2 = 0
for jj in range(nn):
pol1 += monomials[jj](w*z+1,w,z) * BB[pol1_idx, jj] / monomials[jj](UU,XX,YY)
pol2 += monomials[jj](w*z+1,w,z) * BB[pol2_idx, jj] / monomials[jj](UU,XX,YY)
# resultant
PR.<q> = PolynomialRing(ZZ)
rr = pol1.resultant(pol2)
# are these good polynomials?
if rr.is_zero() or rr.monomials() == [1]:
continue
else:
print("found them, using vectors", pol1_idx, "and", pol2_idx)
found_polynomials = True
break
if found_polynomials:
break
if not found_polynomials:
print("no independant vectors could be found. This should very rarely happen...")
return 0, 0
rr = rr(q, q)
# solutions
soly = rr.roots()
if len(soly) == 0:
print("Your prediction (delta) is too small")
return 0, 0
soly = soly[0][0]
ss = pol1(q, soly)
solx = ss.roots()[0][0]
#
return solx, soly
def example():
############################################
# How To Use This Script
##########################################
#
# The problem to solve (edit the following values)
#
# the modulus
N = 12765231982257032754070342601068819788671760506321816381988340379929052646067454855779362773785313297204165444163623633335057895252608396010414744222572161530653104640020689896882490979790275711854268113058363186249545193245142912930804650114934761299016468156185416083682476142929968501395899099376750415294540156026131156551291971922076435528869024742993840057342092865203064721826362149723366381892539617642364692012936270150691803063945919154346756726869466855557344213050973081755499746750276623648407677639812809665472258655462846021403503851719008687214848550916999977775070011121527941755954255781343103086789
# the public exponent
e = 459650454686946706615371845737527916539205656667844780634386049268800615782964920944229084502752167395446158290854047696006034750210758341744841762479191173017773034647739346927390580848998121830029134542880713409306092967282675122699586503684943407535067216738556403169403622104762516293879994387324370835718056251706150557820106296417750402984941838652433642298378976899556042987560946508887315484380807248331504458640857234708123277403252632993828101306072382329879857946191508782246793011691530554606521701055094223574951862129713872918021549814674387049788995785872980320871421550616327471735316980754238323013
L = 0b10100000111001001100001011000110111010001101001011000110110000101101100
sh = L << (1024 - 71)
# the hypothesis on the private exponent (the theoretical maximum is 0.292)
delta = .292 # this means that d < N^delta
#
# Lattice (tweak those values)
#
# you should tweak this (after a first run), (e.g. increment it until a solution is found)
m = 5 # size of the lattice (bigger the better/slower)
# you need to be a lattice master to tweak these
t = int((1-2*delta) * m) # optimization from Herrmann and May
X = 2*floor(N^delta) # this _might_ be too much
Y = floor(2^953) # correct if p, q are ~ same size
#
# Don't touch anything below
#
# Problem put in equation
P.<x,y> = PolynomialRing(ZZ)
A = int((N+1)/2 - sh)
pol = 1 + x * (A + y)
'''
e d = 1 + k (p-1) (q-1)
e d = 1 + k (N+1) + k s
e d = 1 + 2k ((N+1)/2 + s/2)
e d = 1 + 2k ((N+1)/2 + sh/2 + sl/2)
'''
#
# Find the solutions!
#
# Checking bounds
if debug:
print("=== checking values ===")
print("* delta:", delta)
print("* delta < 0.292", delta < 0.292)
print("* size of e:", int(log(e)/log(2)))
print("* size of N:", int(log(N)/log(2)))
print("* m:", m, ", t:", t)
# boneh_durfee
if debug:
print("=== running algorithm ===")
start_time = time.time()
solx, soly = boneh_durfee(pol, e, m, t, X, Y)
# found a solution?
if solx > 0:
print("=== solution found ===")
if False:
print("x:", solx)
print("y:", soly)
d = int(pol(solx, soly) / e)
print("private key found:", d)
else:
print("=== no solution was found ===")
if debug:
print(("=== %s seconds ===" % (time.time() - start_time)))
if __name__ == "__main__":
example()
