DASCTF七月赋能赛2022

​ AK了密码，写一写wp。

babysign

简单ecdsa，给了公钥和随机参数nounce求私钥，理一下加密公式即可：

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from gmpy2 import *
import hashlib
import ecdsa
from Crypto.Util.number import *

gen = ecdsa.NIST256p.generator
order = gen.order()

message = b'welcome to ecdsa'
hash_message = int(hashlib.sha256(message).hexdigest(), 16)
r,s=0x7b35712a50d463ac5acf7af1675b4b63ba0da23b6452023afddd58d4891ef6e5,0xa452fc44cc36fa6964d1b4f47392ff0a91350cfd58f11a4645c084d56e387e5c

nounce = 57872441580840888721108499129165088876046881204464784483281653404168342111855
q = order
f = (s * nounce -hash_message)* invert(r,q) % q
print(long_to_bytes(f))

easyNTRU

NTRU学习资料：
https://4xwi11.github.io/posts/a2b6ecd3/ 注意他这里加密的时候少乘了个p
https://blog.csdn.net/m0_46204256/article/details/122376414

这题非常的水啊，NTRU的N取太小了，仅仅是10，那么AES的私钥可以直接通过爆破获取。

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from Crypto.Hash import SHA3_256
from Crypto.Cipher import AES
import sys
from Crypto.Util.Padding import unpad

N = 10
p = 3
q = 512
d = 3

R.<x> = ZZ[]

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'

table = [-1, 0, 1]
for i1 in table:
    for i2 in table:
        for i3 in table:
            for i4 in table:
                for i5 in table:
                    for i6 in table:
                        for i7 in table:
                            for i8 in table:
                                for i9 in table:
                                    for i10 in table:
                                        result = [i1, i2, i3, i4, i5, i6, i7, i8, i9, i10]
                                        m = R(result)
                                        sha3 = SHA3_256.new()
                                        key = sha3.update(bytes(str(m).encode('utf-8'))).digest()
                                        dypher = AES.new(key, AES.MODE_ECB)
                                        try:
                                            flag = unpad(dypher.decrypt(c), 32)
                                            if flag.startswith(b'flag') or flag.startswith(b'DASCTF'):
                                                print(flag)
                                                sys.exit(0)
                                        except:
                                            pass

LWE?

构造矩阵为：

这里A、B、C都为矩阵，b为题目输出的向量；显然这样构造的矩阵行向量构成格；由于b=xA+yB+zC+e，所以e=b-(xA+yB+zC)，e是行向量组成的格上的格点(向量)，由于e系数很小，所以用LLL规约的最短向量应该就是e。
求出e以后再解下矩阵方程即可：

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import re

f=open('out','r')
def s2row(s):
    return list(map(int, re.split(r'\s+', s[1:-1].strip()))) #\s+ 意思是至少有一个空白字符存在

f.readline()
matA=[]
for _ in range(66):
    s = f.readline().strip()
    row = s2row(s)
    matA.append(row)

f.readline()
for _ in range(66):
    s = f.readline().strip()
    row = s2row(s)
    matA.append(row)

f.readline()
for _ in range(66):
    s = f.readline().strip()
    row = s2row(s)
    matA.append(row)

row= [-19786291, -713104590, 79700973, 23261288, 203038164, 430352288, 147848301, 633183638, 188651439, 243206160, -654830271, 335642059, -100511588, 180023362, 130607831, 227597861, 188424473, 175518170, -246987997, 180879649, 421934976, -227575274, -628937118, 5466646, -254939474, -438417079, 150434624, 327054986, 163561829, 816959939, -265298657, 82651050, 176899880, 174020455, -419656325, -101606182, 300413909, 237169571, -589213744, 121803611, -38080334, -255712509, -133782964, 106220001, 195767251, -397096116, -583305587, -182462561, -271478737, -32014717, 114385188, 437506115, -1165732, 179349265, -77761751, -233976783, 410153356, 476453640, 91892631, -242168750, 506769243, -384438362, 131852532, 586202810, 376719791, 578215353, 874304742, 163584566, 434260863, 98013671, 213627784, 59622886, -84912852, 156744856, 169652328, 178143615, 400046730, 408163110, -357990863, -269552089, -199410809, 187503858, -853206157, 134901027, 313984185, -162544217, -69722073, 43817388, -47389463, 210346729, -46516961, 72002967, 327714191, 45052266, 1010509210, 110937225, 448179404, 341448936, 446550865, 221914340, -804918424, -12007071, 151215468, 440279795, -73408566, -112121988, 40294376, 283179449, -193812410, -30061804, 20326854, 65412625, -260020045, -570090340, 1546454, 548030557, 618148316, 290333796, 665474379, 301709165, -104726821, -503111899, 480689642, -331192606, -518345784, -314602459, 25354403, 410995568, 179675848, -207010027, 400838662, 125916880, 501112567, 578261227, 24802586, 493171331, 383306766, -390093502, -389822626, -303615722, 20813851, -399678371, -566907567, -432647113, -280465568, 1002042393, -510901339, 316603766, -139701243, 211217523, 108545545, -12948109, -569199543, 37065919, -150542603, 417851006, -470173530, -628557669, -128339015, -427978763, 381402990, 205835334, -30976552, -357466556, -104985580, -115366372, 296031071, -8036087, 79340491, 650365147, 295521125, 885900267, 133049758, 217970062, 237420894, 358760095, -2684469, 475711698, 316770575, -25024622, -193442003, 200260606, 89183826, 567491985, 726371428, 222116554, 87397506, -29529094, 125968479, -50793004, 218035181, -210376687, 1025673749, -262390458, 467412984, -71097225, 259125517, -337232810, 143359550, 27115363]
matA.append(row)
L=Matrix(ZZ,matA)
e=vector(ZZ,L.LLL()[0])
print(e)
b=vector(ZZ,row)
s=b-e
m=L.solve_left(s)
print(bytes(list(m)))

# 首先构造矩阵为:
#			  A
#             B
#             C
#             b
# LLL一下得到误差e
# 解一下矩阵方程就得到xyz:
# s=b-e
# m=L.solve_left(s)

NTRURSA

多项式环上的RSA，第一步先把模多项式分解，然后类似于普通RSA把hint解出来：

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R.<x> = PolynomialRing(GF(64621))
n = R('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')
p,q = n.factor()
p,q = p[0],q[0]
print(p)
print(q)
phi = (64621**p.degree()-1)*(64621**q.degree()-1)
e = 65537
d = inverse_mod(e,phi)
print(d)
c = R('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')
m = pow(c,d,n)
print(m)
print("".join([str(c) for c in m.list()]))
# 88520242910362871448352317137540300262448941340486475602003226117035863930302

另一部分主要是svp问题，首先我们已知：

也就是：

我们可以构造一个由下面这个矩阵M中的两个行向量(1,h), (0,p)所张成的lattice：

可以证明向量(f,g1)可以由两组基向量M的某种整系数线性组合(f, -u)来表示，因此向量(f,g1)就在这个lattice上。相对于两个基底向量(1, h), (0, p)来说，向量(f, g1)的长度要小得多得多，所以可以规约，求出f和g1：

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def GaussLatticeReduction(v1, v2):
    while True:
        if v2.norm() < v1.norm():
            v1, v2 = v2, v1
        m = round( v1*v2 / v1.norm()^2 )
        if m == 0:
            return (v1, v2)
        v2 = v2 - m*v1
h=88520242910362871448352317137540300262448941340486475602003226117035863930302
p=106472061241112922861460644342336453303928202010237284715354717630502168520267
v1 = vector(ZZ, [1, h])
v2 = vector(ZZ, [0, p])
print(GaussLatticeReduction(v1, v2)[0])

求得g1以后可以还原g，分解对flag加密的rsa的n，最后解密：

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from gmpy2 import *
from Crypto.Util.number import *

h = 88520242910362871448352317137540300262448941340486475602003226117035863930302
p1 = 106472061241112922861460644342336453303928202010237284715354717630502168520267
c1 = 20920247107738496784071050239422540936224577122721266141057957551603705972966457203177812404896852110975768315464852962210648535130235298413611598658659777108920014929632531307409885868941842921815735008981335582297975794108016151210394446009890312043259167806981442425505200141283138318269058818777636637375101005540308736021976559495266332357714
n = 31398174203566229210665534094126601315683074641013205440476552584312112883638278390105806127975406224783128340041129316782549009811196493319665336016690985557862367551545487842904828051293613836275987595871004601968935866634955528775536847402581734910742403788941725304146192149165731194199024154454952157531068881114411265538547462017207361362857

f,g1 = (183610829622016944154542682943585488074, 228679177303871981036829786447405151037)

for i in range(1,2**20):
    g = i^g1
    q = n//g
    if g*q == n:
        print(q)

q = 137302287745437841210048169960228468385713294771962040184193847476345777625160375267870201668501051400029025050935236430696922380184245300369661070711315995173977120623620591621664587405983128326834510282043350556688416400426843142968182300454634262900982709191955329829399906351225536027386080596329406517693
g = n//q
phi = (g-1)*(q-1)
d = invert(65537,phi)
print(long_to_bytes(pow(c1,d,n)))
# b'DASCTF{P01yn0m141RS4_W17h_NTRU}'

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