DASCTF2022.07赋能赛Crypto_复现

easyNTRU

keywords: 格密码私钥爆破

$Problem$

from Crypto.Hash import SHA3_256
from Crypto.Cipher import AES
from Crypto.Util.Padding import pad
from secret import flag# parameters
N = 10
p = 3
q = 512
d = 3
assert q>(6*d+1)*pR.<x> = ZZ[]#d1 1s and #d2 -1s
def T(d1, d2):assert N >= d1+d2s = [1]*d1 + [-1]*d2 + [0]*(N-d1-d2)shuffle(s)return R(s)def invertModPrime(f, p):Rp = R.change_ring(Integers(p)).quotient(x^N-1)return R(lift(1 / Rp(f)))def convolution(f, g):return (f*g) % (x^N-1)def liftMod(f, q):g = list(((f[i] + q//2) % q) - q//2 for i in range(N))return R(g)def polyMod(f, q):g = [f[i]%q for i in range(N)]return R(g)def invertModPow2(f, q):assert q.is_power_of(2)g = invertModPrime(f,2)while True:r = liftMod(convolution(g,f),q)if r == 1: return gg = liftMod(convolution(g,2 - r),q)def genMessage():result = list(randrange(p) - 1 for j in range(N))return R(result)def genKey():while True:try:f = T(d+1, d)g = T(d, d)Fp = polyMod(invertModPrime(f, p), p)Fq = polyMod(invertModPow2(f, q), q)breakexcept:continueh = polyMod(convolution(Fq, g), q)return h, (f, g)def encrypt(m, h):e = liftMod(p*convolution(h, T(d, d)) + m, q)return e# Step 1
h, secret = genKey()
m = genMessage()
e = encrypt(m, h)print('h = %s' % h)
print('e = %s' % e)# Step 2
sha3 = SHA3_256.new()
sha3.update(bytes(str(m).encode('utf-8')))
key = sha3.digest()cypher = AES.new(key, AES.MODE_ECB)
c = cypher.encrypt(pad(flag, 32))
print('c = %s' % c)"""
h = 39*x^9 + 60*x^8 + 349*x^7 + 268*x^6 + 144*x^5 + 469*x^4 + 449*x^3 + 165*x^2 + 248*x + 369
e = -144*x^9 - 200*x^8 - 8*x^7 + 248*x^6 + 85*x^5 + 102*x^4 + 167*x^3 + 30*x^2 - 203*x - 78
c = b'\xb9W\x8c\x8b\x0cG\xde\x7fl\xf7\x03\xbb9m\x0c\xc4L\xfe\xe9Q\xad\xfd\xda!\x1a\xea@}U\x9ay4\x8a\xe3y\xdf\xd5BV\xa7\x06\xf9\x08\x96="f\xc1\x1b\xd7\xdb\xc1j\x82F\x0b\x16\x06\xbcJMB\xc8\x80'
"""

$Analysis$

要想得到flag，需要知道step1中的m，而m进行了NTRU加密，已知NTRU的h，e，求m

简单回顾一下NTRU加密过程

首先在给定域内选择私钥$f$，求不同模意义下的逆
$$\begin{gathered} F_p\ast f\equiv 1(\mathrm{mod}p) \\ {F}_q\ast f\equiv 1(\mathrm{mod}q) \end{gathered}$$
计算公钥
$$h\equiv r\ast h+m(\mathrm{mod}q)$$
加密过程
$$e\equiv r\ast h+m(\mathrm{mod}q)$$
给出$h,e$

解密过程

第一步
$$\begin{array}{lcc}a& \equiv & e\ast f(\mathrm{mod}q)\\ & \equiv & r\ast p\ast g\ast F_q\ast f+m\ast f(\mathrm{mod}q)\\ & \equiv & p\ast r\ast g+m\ast f(\mathrm{mod}q)\end{array}$$

第二步

$$a\ast F_p\equiv m\ast f\ast F_p\equiv m(\mathrm{mod}p)$$

---

本题关键看N的大小，得知生成的私钥所在域太小了，可以爆破

d = 3
N = 10
R.<x> = ZZ[]
def T(d1, d2):assert N >= d1+d2s = [1]*d1 + [-1]*d2 + [0]*(N-d1-d2)shuffle(s)return R(s)f = T(d+1, d)

从这里可见私钥 $f$ 其实就是总共10个-1,0,1作为系数组成的多项式

那么爆破 $f$ ，给定判断条件为最后用key解出c的内容是否包含DASCTF即可

$Solution$

# type:ignore
import itertools
from Crypto.Hash import SHA3_256
from Crypto.Cipher import AESN = 10
p = 3
q = 512
d = 3
c = b'\xb9W\x8c\x8b\x0cG\xde\x7fl\xf7\x03\xbb9m\x0c\xc4L\xfe\xe9Q\xad\xfd\xda!\x1a\xea@}U\x9ay4\x8a\xe3y\xdf\xd5BV\xa7\x06\xf9\x08\x96="f\xc1\x1b\xd7\xdb\xc1j\x82F\x0b\x16\x06\xbcJMB\xc8\x80'
e = [-78, -203, 30, 167, 102, 85, 248, -8, -200, -144] # 提取多项式的系数
h = [369, 248, 165, 449, 469, 144, 268, 349, 60, 39]
ls = [1, 0, -1]
R.<x> = ZZ[]e = R(e)
h = R(h)#  这里要用到题目给出的多项式运算函数，所以导入
def liftMod(f, q):g = list(((f[i] + q//2) % q) - q//2 for i in range(N))return R(g)def polyMod(f, q):g = [f[i]%q for i in range(N)]return R(g)4def invertModPrime(f, p):Rp = R.change_ring(Integers(p)).quotient(x^N-1)return R(lift(1 / Rp(f)))def convolution(f, g):return (f*g) % (x^N-1)for i in itertools.product(ls, repeat=N):f = list(i)f = R(f)a = liftMod(convolution(e, f), q) try:Fp = polyMod(invertModPrime(f, p), p) except:continuem = liftMod(convolution(a, Fp), p)sha3 = SHA3_256.new()sha3 = sha3.update(bytes(str(m).encode('utf-8')))key = sha3.digest()aes = AES.new(key, AES.MODE_ECB)flag = aes.decrypt(c)if b"DASCTF" in flag:print(flag)break

NTRURSA

keywords: 多项式RSA，构造格子形成SVP问题

$Problem$

from Crypto.Util.number import *
from gmpy2 import *
from secret import flagdef gen():p1 = getPrime(256)while True:f = getRandomRange(1, iroot(p1 // 2, 2)[0])g = getRandomRange(iroot(p1 // 4, 2)[0], iroot(p1 // 2, 2)[0])if gcd(f, p1) == 1 and gcd(f, g) == 1 and isPrime(g) == 1:breakrand = getRandomRange(0, 2 ^ 20)g1 = g ^^ randh = (inverse(f, p1) * g1) % p1return h, p1, g, f, g1def gen_irreducable_poly(deg):while True:out = R.random_element(degree=deg)if out.is_irreducible():return outh, p1, g, f, g1 = gen()
q = getPrime(1024)
n = g * q 
e = 0x10001
c1 = pow(bytes_to_long(flag), e, n)
hint = list(str(h))
length = len(hint)
bits = 16
p2 = random_prime(2 ^ bits - 1, False, 2 ^ (bits - 1))
R.<x> = PolynomialRing(GF(p2))
P = gen_irreducable_poly(ZZ.random_element(length, 2 * length))
Q = gen_irreducable_poly(ZZ.random_element(length, 2 * length))
N = P * Q
S.<x> = R.quotient(N)
m = S(hint)
c2 = m ^ e
print("p1 =", p1)
print("c1 =", c1)
print("p2 =", p2)
print("c2 =", c2)
print("n =", n)
print("N =", N)'''
p1 = 106472061241112922861460644342336453303928202010237284715354717630502168520267
c1 = 20920247107738496784071050239422540936224577122721266141057957551603705972966457203177812404896852110975768315464852962210648535130235298413611598658659777108920014929632531307409885868941842921815735008981335582297975794108016151210394446009890312043259167806981442425505200141283138318269058818777636637375101005540308736021976559495266332357714
p2 = 64621
c2 = 19921*x^174 + 49192*x^173 + 18894*x^172 + 61121*x^171 + 50271*x^170 + 11860*x^169 + 53128*x^168 + 38658*x^167 + 14191*x^166 + 9671*x^165 + 40879*x^164 + 15187*x^163 + 33523*x^162 + 62270*x^161 + 64211*x^160 + 54518*x^159 + 50446*x^158 + 2597*x^157 + 32216*x^156 + 10500*x^155 + 63276*x^154 + 27916*x^153 + 55316*x^152 + 30898*x^151 + 43706*x^150 + 5734*x^149 + 35616*x^148 + 14288*x^147 + 18282*x^146 + 22788*x^145 + 48188*x^144 + 34176*x^143 + 55952*x^142 + 9578*x^141 + 9177*x^140 + 22083*x^139 + 14586*x^138 + 9748*x^137 + 21118*x^136 + 155*x^135 + 64224*x^134 + 18193*x^133 + 33732*x^132 + 38135*x^131 + 51992*x^130 + 8203*x^129 + 8538*x^128 + 55203*x^127 + 5003*x^126 + 2009*x^125 + 45023*x^124 + 12311*x^123 + 21428*x^122 + 24110*x^121 + 43537*x^120 + 21885*x^119 + 50212*x^118 + 40445*x^117 + 17768*x^116 + 46616*x^115 + 4771*x^114 + 20903*x^113 + 47764*x^112 + 13056*x^111 + 50837*x^110 + 22313*x^109 + 39698*x^108 + 60377*x^107 + 59357*x^106 + 24051*x^105 + 5888*x^104 + 29414*x^103 + 31726*x^102 + 4906*x^101 + 23968*x^100 + 52360*x^99 + 58063*x^98 + 706*x^97 + 31420*x^96 + 62468*x^95 + 18557*x^94 + 1498*x^93 + 17590*x^92 + 62990*x^91 + 27200*x^90 + 7052*x^89 + 39117*x^88 + 46944*x^87 + 45535*x^86 + 28092*x^85 + 1981*x^84 + 4377*x^83 + 34419*x^82 + 33754*x^81 + 2640*x^80 + 44427*x^79 + 32179*x^78 + 57721*x^77 + 9444*x^76 + 49374*x^75 + 21288*x^74 + 44098*x^73 + 57744*x^72 + 63457*x^71 + 43300*x^70 + 1508*x^69 + 13775*x^68 + 23197*x^67 + 43070*x^66 + 20751*x^65 + 47479*x^64 + 18496*x^63 + 53392*x^62 + 10387*x^61 + 2317*x^60 + 57492*x^59 + 25441*x^58 + 52532*x^57 + 27150*x^56 + 33788*x^55 + 43371*x^54 + 30972*x^53 + 39583*x^52 + 36407*x^51 + 35564*x^50 + 44564*x^49 + 1505*x^48 + 47519*x^47 + 38695*x^46 + 43107*x^45 + 1676*x^44 + 42057*x^43 + 49879*x^42 + 29083*x^41 + 42241*x^40 + 8853*x^39 + 33546*x^38 + 48954*x^37 + 30352*x^36 + 62020*x^35 + 39864*x^34 + 9519*x^33 + 24828*x^32 + 34696*x^31 + 2387*x^30 + 27413*x^29 + 55829*x^28 + 40217*x^27 + 30205*x^26 + 42328*x^25 + 6210*x^24 + 52442*x^23 + 58495*x^22 + 2014*x^21 + 26452*x^20 + 33547*x^19 + 19840*x^18 + 5995*x^17 + 16850*x^16 + 37855*x^15 + 7221*x^14 + 32200*x^13 + 8121*x^12 + 23767*x^11 + 46563*x^10 + 51673*x^9 + 19372*x^8 + 4157*x^7 + 48421*x^6 + 41096*x^5 + 45735*x^4 + 53022*x^3 + 35475*x^2 + 47521*x + 27544
n = 31398174203566229210665534094126601315683074641013205440476552584312112883638278390105806127975406224783128340041129316782549009811196493319665336016690985557862367551545487842904828051293613836275987595871004601968935866634955528775536847402581734910742403788941725304146192149165731194199024154454952157531068881114411265538547462017207361362857
N = 25081*x^175 + 8744*x^174 + 9823*x^173 + 9037*x^172 + 6343*x^171 + 42205*x^170 + 28573*x^169 + 55714*x^168 + 17287*x^167 + 11229*x^166 + 42630*x^165 + 64363*x^164 + 50759*x^163 + 3368*x^162 + 20900*x^161 + 55947*x^160 + 7082*x^159 + 23171*x^158 + 48510*x^157 + 20013*x^156 + 16798*x^155 + 60438*x^154 + 58779*x^153 + 9289*x^152 + 10623*x^151 + 1085*x^150 + 23473*x^149 + 13795*x^148 + 2071*x^147 + 31515*x^146 + 42832*x^145 + 38152*x^144 + 37559*x^143 + 47653*x^142 + 37371*x^141 + 39128*x^140 + 48750*x^139 + 16638*x^138 + 60320*x^137 + 56224*x^136 + 41870*x^135 + 63961*x^134 + 47574*x^133 + 63954*x^132 + 9668*x^131 + 62360*x^130 + 15244*x^129 + 20599*x^128 + 28704*x^127 + 26857*x^126 + 34885*x^125 + 33107*x^124 + 17693*x^123 + 52753*x^122 + 60744*x^121 + 21305*x^120 + 63785*x^119 + 54400*x^118 + 17812*x^117 + 64549*x^116 + 20035*x^115 + 37567*x^114 + 38607*x^113 + 32783*x^112 + 24385*x^111 + 5387*x^110 + 5134*x^109 + 45893*x^108 + 58307*x^107 + 33821*x^106 + 54902*x^105 + 14236*x^104 + 58044*x^103 + 41257*x^102 + 46881*x^101 + 42834*x^100 + 1693*x^99 + 46058*x^98 + 15636*x^97 + 27111*x^96 + 3158*x^95 + 41012*x^94 + 26028*x^93 + 3576*x^92 + 37958*x^91 + 33273*x^90 + 60228*x^89 + 41229*x^88 + 11232*x^87 + 12635*x^86 + 17942*x^85 + 4*x^84 + 25397*x^83 + 63526*x^82 + 54872*x^81 + 40318*x^80 + 37498*x^79 + 52182*x^78 + 48817*x^77 + 10763*x^76 + 46542*x^75 + 36060*x^74 + 49972*x^73 + 63603*x^72 + 46506*x^71 + 44788*x^70 + 44905*x^69 + 46112*x^68 + 5297*x^67 + 26440*x^66 + 28470*x^65 + 15525*x^64 + 11566*x^63 + 15781*x^62 + 36098*x^61 + 44402*x^60 + 55331*x^59 + 61583*x^58 + 16406*x^57 + 59089*x^56 + 53161*x^55 + 43695*x^54 + 49580*x^53 + 62685*x^52 + 31447*x^51 + 26755*x^50 + 14810*x^49 + 3281*x^48 + 27371*x^47 + 53392*x^46 + 2648*x^45 + 10095*x^44 + 25977*x^43 + 22912*x^42 + 41278*x^41 + 33236*x^40 + 57792*x^39 + 7169*x^38 + 29250*x^37 + 16906*x^36 + 4436*x^35 + 2729*x^34 + 29736*x^33 + 19383*x^32 + 11921*x^31 + 26075*x^30 + 54616*x^29 + 739*x^28 + 38509*x^27 + 19118*x^26 + 20062*x^25 + 21280*x^24 + 12594*x^23 + 14974*x^22 + 27795*x^21 + 54107*x^20 + 1890*x^19 + 13410*x^18 + 5381*x^17 + 19500*x^16 + 47481*x^15 + 58488*x^14 + 26433*x^13 + 37803*x^12 + 60232*x^11 + 34772*x^10 + 1505*x^9 + 63760*x^8 + 20890*x^7 + 41533*x^6 + 16130*x^5 + 29769*x^4 + 49142*x^3 + 64184*x^2 + 55443*x + 45925
'''

$Analysis$

flag经过了常数的RSA加密，其中n = g * q；而关于g给出了提示

def gen():p1 = getPrime(256)while True:f = getRandomRange(1, iroot(p1 // 2, 2)[0])g = getRandomRange(iroot(p1 // 4, 2)[0], iroot(p1 // 2, 2)[0])if gcd(f, p1) == 1 and gcd(f, g) == 1 and isPrime(g) == 1:breakrand = getRandomRange(0, 2 ^ 20)g1 = g ^^ randh = (inverse(f, p1) * g1) % p1return h, p1, g, f, g1h, p1, g, f, g1 = gen()
hint = list(str(h))
m = S(hint)

与g相关的是g1, h，之间的关联式子是
$$\begin{array}{rc}{g}_1=& g\oplus rand\\ h\ast f\equiv & g_1(\mathrm{mod}{p}_1)\end{array}$$
可以通过第二个关系式子，构造格子形成SVP问题求解g1，进而通过爆破rand来得到g
$$\begin{array}{rc}h\ast f\equiv & g_1(\mathrm{mod}{p}_1)\\ h\ast f-k\ast p_1=& g_1\end{array}$$
格子构造为
$$(f,-k)\left(\begin{array}{cc}1& h\\ 0& p_1\end{array}\right)=(f,g_1)$$
而构造格子需要的h又作为新的m进行了多项式版本的RSA加密

p2 = 64621
length = len(hint)
R.<x> = PolynomialRing(GF(p2))
P = gen_irreducable_poly(ZZ.random_element(length, 2 * length))
Q = gen_irreducable_poly(ZZ.random_element(length, 2 * length))
N = P * Q
S.<x> = R.quotient(N)
m = S(hint)
c2 = m ^ e

相当于在模p2的意义下对多项式进行了RSA加密，由于p2比较小，并且生成的P, Q的大小是以h的位数来生成的，考虑直接对N进行分解

N = ...
P, Q = factor(N)[0][0], factor(N)[1][0]
phi = (p2^P.degree() - 1) * (p2^Q.degree() - 1)

对于多项式来说，求多项式的欧拉函数，按照欧拉函数的定义应该意味着求次数小于等于该多项式的不可约多项式的个数

官方给出的欧拉函数公式是
$$\phi =(p_2^{P.degree()}-1)\ast (p_2^{Q.degree()}-1)$$
紧接着求逆元d即可，之后按照思路构造格子先后求解h，g1，g，flag

$Solution$

# type:ignore
from Crypto.Util.number import *
import gmpy2p1 = 106472061241112922861460644342336453303928202010237284715354717630502168520267
c1 = 20920247107738496784071050239422540936224577122721266141057957551603705972966457203177812404896852110975768315464852962210648535130235298413611598658659777108920014929632531307409885868941842921815735008981335582297975794108016151210394446009890312043259167806981442425505200141283138318269058818777636637375101005540308736021976559495266332357714
p2 = 64621
n = 31398174203566229210665534094126601315683074641013205440476552584312112883638278390105806127975406224783128340041129316782549009811196493319665336016690985557862367551545487842904828051293613836275987595871004601968935866634955528775536847402581734910742403788941725304146192149165731194199024154454952157531068881114411265538547462017207361362857
e = 0x10001R.<x> = PolynomialRing(GF(p2))c2 = 19921*x^174 + 49192*x^173 + 18894*x^172 + 61121*x^171 + 50271*x^170 + 11860*x^169 + 53128*x^168 + 38658*x^167 + 14191*x^166 + 9671*x^165 + 40879*x^164 + 15187*x^163 + 33523*x^162 + 62270*x^161 + 64211*x^160 + 54518*x^159 + 50446*x^158 + 2597*x^157 + 32216*x^156 + 10500*x^155 + 63276*x^154 + 27916*x^153 + 55316*x^152 + 30898*x^151 + 43706*x^150 + 5734*x^149 + 35616*x^148 + 14288*x^147 + 18282*x^146 + 22788*x^145 + 48188*x^144 + 34176*x^143 + 55952*x^142 + 9578*x^141 + 9177*x^140 + 22083*x^139 + 14586*x^138 + 9748*x^137 + 21118*x^136 + 155*x^135 + 64224*x^134 + 18193*x^133 + 33732*x^132 + 38135*x^131 + 51992*x^130 + 8203*x^129 + 8538*x^128 + 55203*x^127 + 5003*x^126 + 2009*x^125 + 45023*x^124 + 12311*x^123 + 21428*x^122 + 24110*x^121 + 43537*x^120 + 21885*x^119 + 50212*x^118 + 40445*x^117 + 17768*x^116 + 46616*x^115 + 4771*x^114 + 20903*x^113 + 47764*x^112 + 13056*x^111 + 50837*x^110 + 22313*x^109 + 39698*x^108 + 60377*x^107 + 59357*x^106 + 24051*x^105 + 5888*x^104 + 29414*x^103 + 31726*x^102 + 4906*x^101 + 23968*x^100 + 52360*x^99 + 58063*x^98 + 706*x^97 + 31420*x^96 + 62468*x^95 + 18557*x^94 + 1498*x^93 + 17590*x^92 + 62990*x^91 + 27200*x^90 + 7052*x^89 + 39117*x^88 + 46944*x^87 + 45535*x^86 + 28092*x^85 + 1981*x^84 + 4377*x^83 + 34419*x^82 + 33754*x^81 + 2640*x^80 + 44427*x^79 + 32179*x^78 + 57721*x^77 + 9444*x^76 + 49374*x^75 + 21288*x^74 + 44098*x^73 + 57744*x^72 + 63457*x^71 + 43300*x^70 + 1508*x^69 + 13775*x^68 + 23197*x^67 + 43070*x^66 + 20751*x^65 + 47479*x^64 + 18496*x^63 + 53392*x^62 + 10387*x^61 + 2317*x^60 + 57492*x^59 + 25441*x^58 + 52532*x^57 + 27150*x^56 + 33788*x^55 + 43371*x^54 + 30972*x^53 + 39583*x^52 + 36407*x^51 + 35564*x^50 + 44564*x^49 + 1505*x^48 + 47519*x^47 + 38695*x^46 + 43107*x^45 + 1676*x^44 + 42057*x^43 + 49879*x^42 + 29083*x^41 + 42241*x^40 + 8853*x^39 + 33546*x^38 + 48954*x^37 + 30352*x^36 + 62020*x^35 + 39864*x^34 + 9519*x^33 + 24828*x^32 + 34696*x^31 + 2387*x^30 + 27413*x^29 + 55829*x^28 + 40217*x^27 + 30205*x^26 + 42328*x^25 + 6210*x^24 + 52442*x^23 + 58495*x^22 + 2014*x^21 + 26452*x^20 + 33547*x^19 + 19840*x^18 + 5995*x^17 + 16850*x^16 + 37855*x^15 + 7221*x^14 + 32200*x^13 + 8121*x^12 + 23767*x^11 + 46563*x^10 + 51673*x^9 + 19372*x^8 + 4157*x^7 + 48421*x^6 + 41096*x^5 + 45735*x^4 + 53022*x^3 + 35475*x^2 + 47521*x + 27544
N = 25081*x^175 + 8744*x^174 + 9823*x^173 + 9037*x^172 + 6343*x^171 + 42205*x^170 + 28573*x^169 + 55714*x^168 + 17287*x^167 + 11229*x^166 + 42630*x^165 + 64363*x^164 + 50759*x^163 + 3368*x^162 + 20900*x^161 + 55947*x^160 + 7082*x^159 + 23171*x^158 + 48510*x^157 + 20013*x^156 + 16798*x^155 + 60438*x^154 + 58779*x^153 + 9289*x^152 + 10623*x^151 + 1085*x^150 + 23473*x^149 + 13795*x^148 + 2071*x^147 + 31515*x^146 + 42832*x^145 + 38152*x^144 + 37559*x^143 + 47653*x^142 + 37371*x^141 + 39128*x^140 + 48750*x^139 + 16638*x^138 + 60320*x^137 + 56224*x^136 + 41870*x^135 + 63961*x^134 + 47574*x^133 + 63954*x^132 + 9668*x^131 + 62360*x^130 + 15244*x^129 + 20599*x^128 + 28704*x^127 + 26857*x^126 + 34885*x^125 + 33107*x^124 + 17693*x^123 + 52753*x^122 + 60744*x^121 + 21305*x^120 + 63785*x^119 + 54400*x^118 + 17812*x^117 + 64549*x^116 + 20035*x^115 + 37567*x^114 + 38607*x^113 + 32783*x^112 + 24385*x^111 + 5387*x^110 + 5134*x^109 + 45893*x^108 + 58307*x^107 + 33821*x^106 + 54902*x^105 + 14236*x^104 + 58044*x^103 + 41257*x^102 + 46881*x^101 + 42834*x^100 + 1693*x^99 + 46058*x^98 + 15636*x^97 + 27111*x^96 + 3158*x^95 + 41012*x^94 + 26028*x^93 + 3576*x^92 + 37958*x^91 + 33273*x^90 + 60228*x^89 + 41229*x^88 + 11232*x^87 + 12635*x^86 + 17942*x^85 + 4*x^84 + 25397*x^83 + 63526*x^82 + 54872*x^81 + 40318*x^80 + 37498*x^79 + 52182*x^78 + 48817*x^77 + 10763*x^76 + 46542*x^75 + 36060*x^74 + 49972*x^73 + 63603*x^72 + 46506*x^71 + 44788*x^70 + 44905*x^69 + 46112*x^68 + 5297*x^67 + 26440*x^66 + 28470*x^65 + 15525*x^64 + 11566*x^63 + 15781*x^62 + 36098*x^61 + 44402*x^60 + 55331*x^59 + 61583*x^58 + 16406*x^57 + 59089*x^56 + 53161*x^55 + 43695*x^54 + 49580*x^53 + 62685*x^52 + 31447*x^51 + 26755*x^50 + 14810*x^49 + 3281*x^48 + 27371*x^47 + 53392*x^46 + 2648*x^45 + 10095*x^44 + 25977*x^43 + 22912*x^42 + 41278*x^41 + 33236*x^40 + 57792*x^39 + 7169*x^38 + 29250*x^37 + 16906*x^36 + 4436*x^35 + 2729*x^34 + 29736*x^33 + 19383*x^32 + 11921*x^31 + 26075*x^30 + 54616*x^29 + 739*x^28 + 38509*x^27 + 19118*x^26 + 20062*x^25 + 21280*x^24 + 12594*x^23 + 14974*x^22 + 27795*x^21 + 54107*x^20 + 1890*x^19 + 13410*x^18 + 5381*x^17 + 19500*x^16 + 47481*x^15 + 58488*x^14 + 26433*x^13 + 37803*x^12 + 60232*x^11 + 34772*x^10 + 1505*x^9 + 63760*x^8 + 20890*x^7 + 41533*x^6 + 16130*x^5 + 29769*x^4 + 49142*x^3 + 64184*x^2 + 55443*x + 45925P, Q = factor(N)[0][0], factor(N)[1][0]
phi = (p2^P.degree() - 1) * (p2^Q.degree() - 1)
d = inverse_mod(e, phi)m2 = pow(c2, d, N)
h = int(("".join(list(map(str, list(m2)[::-1])))))
M = matrix([[1, h],[0, p1]])
print(M.LLL())
g1 = M.LLL()[0][1]for i in range(2**20):tmp = i ^^ g1if n % tmp == 0:g = tmpq = n // tmpbreak
phi = (g - 1) * (q - 1)
d = inverse(e, phi)
m = pow(c1, d, n)
print(long_to_bytes(int(m)))

LWE?

keywords: LWE问题，Embedding Technique

$Problem$

from secret import secret
assert len(secret)==66*3
sec = [ord(x) for x in secret]DEBUG = False
m = 66
n = 200
p = 3
q = 2^20def errorV():return vector(ZZ, [1 - randrange(p) for _ in range(n)])def matrixMn():return matrix(ZZ, [[q//2 - randrange(q) for _ in range(n)] for _ in range(m)])A, B, C = matrixMn(), matrixMn(), matrixMn()
x = vector(ZZ, sec[0:m])
y = vector(ZZ, sec[m:2*m])
z = vector(ZZ, sec[2*m:3*m])
e = errorV()
b = x*A+y*B+z*C+eif DEBUG:print('x = %s' % x)print('y = %s' % y)print('z = %s' % z)print('e = %s' % e)
print('A = \n%s' % A)
print('B = \n%s' % B)
print('C = \n%s' % C)
print('b = %s' % b)

$Analysis$

将flag的各个字符按顺序作为向量的元素赋值，关键代码

b = x*A+y*B+z*C+e

进行了LWE问题的实现，LWE问题应该是
$$b=Ax+e$$
这里题目我们可以看作
$$\begin{array}{lc}X& =(x,y,z)\\ A& =\left(\begin{array}{c}A\\ B\\ C\end{array}\right)\end{array}$$
这样一来就构成了
$$b=AX+e$$
这样的LWE问题

应用Embedding Technique构造格子
$$M=\left[\begin{array}{c}Ab\\ 0\beta \end{array}\right]$$
其中 $\beta$ 是自己选取的值，为了尽可能构造出正确的格（可以通过求LLL算法得到正确结果），这里将 $\beta$ 选作1
$$\left[\begin{array}{c}Ab\\ 0\beta \end{array}\right]\left[\begin{array}{c}-X\\ 1\end{array}\right]=\left[\begin{array}{c}e\\ \beta \end{array}\right]$$
得到e之后与b向量相减，再将 $A$ 看作整体求矩阵乘法即可

PS：题目实际构造格子的时候因为 $b$ 加入矩阵会让矩阵不是方阵，所以考虑构造了$\left[\begin{array}{c}A0\\ b\beta \end{array}\right]$

$Solution$

# type:ignore
from Crypto.Util.number import *
import gmpy2
import rem = 66
n = 200
p = 3
q = 2^20with open ("C:\\Users\\Menglin\\Desktop\\out","r") as f:data = f.read().split('\n')f.close()A = matrix(m, n, [int(j) for i in data[1: m + 1] for j in re.findall(r'-?\d+', i)])
B = matrix(m, n, [int(j) for i in data[m + 1: 2 * m + 2] for j in re.findall(r'-?\d+', i)])
C = matrix(m, n, [int(j) for i in data[2 * m + 2: 3 * m + 3] for j in re.findall(r'-?\d+', i)])
b = vector([int(j) for i in data[3 * m + 3:] for j in re.findall(r'-?\d+', i)])A = block_matrix([[A], [B], [C]])
M = block_matrix([[A, matrix(3 * m, 1, [0] * 3 * m)], [matrix(b), matrix([1])]])
e = M.LLL()[0][:n]
x = A.solve_left(b - e)
for i in x:print(chr(i),end="")

$Peference$

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