2022DASCTF7月赋能赛CryptoWriteup

2022-07-25

字数统计: 2.4k字 | 阅读时长≈ 13分

Crypto方向四道题的全部题解：babysign、easyNTRU、NTRURSA、LWE？。

babysign

简单的ECDSA签名，给定nonce，根据签名(R，S)的关系即可得到私钥： $x=R^{-1}*(nonce*S- hash(m))$ 。

import hashlib
import ecdsa
from Crypto.Util.number import *
R = 0x7b35712a50d463ac5acf7af1675b4b63ba0da23b6452023afddd58d4891ef6e5
S = 0xa452fc44cc36fa6964d1b4f47392ff0a91350cfd58f11a4645c084d56e387e5c
k = 57872441580840888721108499129165088876046881204464784483281653404168342111855
hm = int(hashlib.sha256(b'welcome to ecdsa').hexdigest(), 16)
# r^-1*(k*s- hm)
gen = ecdsa.NIST256p.generator
n = int(gen.order())
x=inverse(R,n)*(k*S- hm)%n
print(long_to_bytes(int(x)))
# 11b7311d4f0137074a7256d3eb82f368

easyNTRU

NTRU格密码的标准实现，但是参数过小，可以用格的方法求解出私钥。核心思想如下： $h=p*f_q*g \mod q$ 。其中 $f_q$ 也就是f模q的逆。转换一下也就是： $g = h*f*p^{-1} \mod q$ 。其中g、f的系数都是1，0，-1，且一般来说1，-1的个数很少，也就是f，g系数组成的向量模长小，因此我们可以通过对 $h*x^i,i \in [0,n-1]$ 的系数构成的格进行格基规约，就可以得到f，g，因为 $[g.coefficients(),f.coefficients()]$ 在矩阵M行向量构成的格空间内，并且模长很小。M构造如下：

$M= \begin{bmatrix} qI & 0 \\ H & I \end{bmatrix} \quad where\ H[i]=(h*x^i).coefficients()$

一个完整的NTRU密码体系的分析和格攻击算法参考：LatticeHacks。

import random
from Crypto.Util.number import *

Zx.<x> = ZZ[]
# R.<x> = ZZ[]
def balancedmod(f,q):
    g = list( ((f[i] + q//2) % q) - q//2 for i in range(n) )
    return Zx(g)

def cyclicconvolution(f, g):
    return (f*g) % (x^n-1)

def invertmodprime(f,p):
    T = Zx.change_ring(Integers(p)).quotient(x^n-1)
    return Zx(lift(1 / T(f)))

def invertmodpowerof2(f,q):
    assert q.is_power_of(2)
    g = invertmodprime(f,2)
    while True:
        r = balancedmod(cyclicconvolution(g,f),q)
        if r == 1: return g
        g = balancedmod(cyclicconvolution(g,2 - r),q)

def encrypt(message, publickey):
    r = rpoly()
    return balancedmod(cyclicconvolution(publickey, r) + message, q)

def decrypt(cipher,f,fp):
    # cipher=Zx(cipher)
    a=balancedmod(cyclicconvolution(f, cipher), q)
    m=balancedmod(cyclicconvolution(fp, a),p)
    return m

def attack(publickey):
    recip3 = lift(1/Integers(q)(3))
    publickeyover3 = balancedmod(recip3 * publickey,q)
    M = matrix(2 * n)
    for i in range(n):
        M[i,i] = q
    for i in range(n):
        M[i+n,i+n] = 1
        c = cyclicconvolution(x^i,publickeyover3)
        for j in range(n):
            M[i+n,j] = c[j]
    M = M.LLL()
    for j in range(2 * n):
        try:
            f = Zx(list(M[j][n:]))
            f3 = invertmodprime(f,3)
            return (f,f3)
        except:pass
    return (f,f)

n = 10
p = 3
q = 512
d = 3
assert q>(6*d+1)*p

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'

# publickey,secretkey = keypair()
donald = attack(h.coefficients(sparse=False))
m = decrypt(e,donald[0],donald[1])

from Crypto.Hash import SHA3_256
from Crypto.Cipher import AES
sha3 = SHA3_256.new()
sha3.update(bytes(str(Zx(m)).encode('utf-8')))
key = sha3.digest()

cipher = AES.new(key, AES.MODE_ECB)
flag = cipher.decrypt(c)
print('c = %s' % flag)

NTRURSA

首先是在模素数域上一个类似NTRU加密的过程，然后把公钥h进行多项式转换，再进行一个多项式RSA的加密。多项式RSA的解法很简单，这里多项式的次数和模数p都很小，可以分解得到两个本原多项式，模本原多项式的群的阶很好计算，就是对多项式每个位置系数进行穷举，除去零元，因此是 $p^{f.degree()}-1$ ，再利用积性即可得到阶。h通过RSA多项式群的阶就可以恢复。

NTRU过程的恢复和这篇博客是几乎相同的：NTRU，与上题多项式形式的NTRU过程的格是一致的。

$M= \begin{bmatrix} q & 0 \\ h & 1 \end{bmatrix}$

恢复了g1之后，对低20位进行穷举即可，另外一种方法是已知g高位的分解。

from Crypto.Util.number import *
p1 = 106472061241112922861460644342336453303928202010237284715354717630502168520267
c1 = 20920247107738496784071050239422540936224577122721266141057957551603705972966457203177812404896852110975768315464852962210648535130235298413611598658659777108920014929632531307409885868941842921815735008981335582297975794108016151210394446009890312043259167806981442425505200141283138318269058818777636637375101005540308736021976559495266332357714
p2 = 64621
R.<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 = 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
# S.<x> = R.quotient(N)
# factor(N)
q1, q2 = N.factor()
q1, q2 = q1[0], q2[0]

phi = (p2**q1.degree() - 1) * (p2**q2.degree() - 1)

e = 0x10001
d = inverse_mod(e, phi)
m = pow(c2,d,N)
h1 = list(m.coefficients(sparse=False))
h =int("".join([str(i) for i in h1]))
assert h<p1

M=matrix(ZZ,[
    [1,h],
    [0,p1]])
L=M.LLL()

from tqdm import tqdm
g_=abs(L[0,1])

for rand in tqdm(range(2^20)):
    g = g_^^rand
    if is_prime(g):
        if gcd(n,g)!=1:
            g,p=g,n//g
            print(f"[+] find {g,n//g}")
            break
phi=(p-1)*(g-1)
print(long_to_bytes(int(pow(c1,inverse(e,phi),p*g))))

LWE？

根据题目，显然与容错学习有关，其实就是最初始的LWE模式，A，B，C都是66×200的矩阵，值都定义在 $[-2^{19},2^{19}]$ 之间，x，y，z分别是flag一部分组成的1×66的向量： $b = x*A+y*B+z*C+e$ ，其实简单来写就是： $b=(x,y,z)*\begin{bmatrix} A \\ B \\ C \end{bmatrix} + e$ 。这就是标准的LWE问题。构造格：

$M =\begin{bmatrix} A \\ B \\ C \\ b \end{bmatrix} \quad size : \ 199*200$

显然误差向量e在M行向量构成的格空间内，因为： $[x,y,z,-1]*M=e$ 。通过LLL算法就可以得到e，然后解多元一次方程组即可得到flag。

import re
s2n=lambda x: [int(x) for x in re.findall(r"\-?\d+\.?\d*",x)]
f=open("./out","r").readlines()

m = 66
n = 200
p = 3
q = 2^20

A = [s2n(f[i+1]) for i in range(66)]
B = [s2n(f[i+68]) for i in range(66)]
C = [s2n(f[i+135]) for i in range(66)]
b= list(matrix(ZZ,s2n(f[-1])))

m=A+B+C+b
M = matrix(ZZ,m)
print("LLL start")
L = M.LLL()
print("LLL done")

print(L[0])
res=M.solve_left(L[0])
for i in res[:-1]:
    print(chr(abs(i)),end="")