XYCTF-crypto

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前言

第一次打比赛，被新生赛狠狠拷打了QAQ

Complex_dlp

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from Crypto.Util.number import *
from secrets import flag


class Complex:
    def __init__(self, re, im):
        self.re = re
        self.im = im

    def __mul__(self, c):
        re_ = self.re * c.re - self.im * c.im
        im_ = self.re * c.im + self.im * c.re
        return Complex(re_, im_)

    def __str__(self):
        if self.im == 0:
            return str(self.re)
        elif self.re == 0:
            if abs(self.im) == 1:
                return f"{'-' if self.im < 0 else ''}i"
            else:
                return f"{self.im}i"
        else:
            return f"{self.re} {'+' if self.im > 0 else '-'} {abs(self.im)}i"


def complex_pow(c, exp, n):
    result = Complex(1, 0)
    while exp > 0:
        if exp & 1:
            result = result * c
            result.re = result.re % n
            result.im = result.im % n
        c = c * c
        c.re = c.re % n
        c.im = c.im % n
        exp >>= 1
    return result


flag = flag.strip(b"XYCTF{").strip(b"}")
p = 1127236854942215744482170859284245684922507818478439319428888584898927520579579027
g = Complex(3, 7)
x = bytes_to_long(flag)
#complex_pow(g, x, p)==5699996596230726507553778181714315375600519769517892864468100565238657988087817 + 198037503897625840198829901785272602849546728822078622977599179234202360717671908i

做的时候一直卡在复数转实数上，想到了共轭复数但是没有去仔细研究。

赛后看wp知道了定义image-20240428191957162

这样题目中的g和c都可以通过实部和虚部平方相加的方式转为实数再进行普通的dlp求解

exp：

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from Cryptodome.Util.number import *
import gmpy2
from sympy.ntheory import discrete_log
p=1127236854942215744482170859284245684922507818478439319428888584898927520579579027
g=58
c=(5699996596230726507553778181714315375600519769517892864468100565238657988087817**2)+(198037503897625840198829901785272602849546728822078622977599179234202360717671908**2)
flag=discrete_log(p,c,g)
print(long_to_bytes(flag))

Complex_rsa

与上题类似，也是一个复数域上的问题

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from Crypto.Util.number import *
from secrets import flag


class Complex:
    def __init__(self, re, im):
        self.re = re
        self.im = im

    def __mul__(self, c):
        re_ = self.re * c.re - self.im * c.im
        im_ = self.re * c.im + self.im * c.re
        return Complex(re_, im_)

    def __str__(self):
        if self.im == 0:
            return str(self.re)
        elif self.re == 0:
            if abs(self.im) == 1:
                return f"{'-' if self.im < 0 else ''}i"
            else:
                return f"{self.im}i"
        else:
            return f"{self.re} {'+' if self.im > 0 else '-'} {abs(self.im)}i"


def complex_pow(c, exp, n):
    result = Complex(1, 0)
    while exp > 0:
        if exp & 1:
            result = result * c
            result.re = result.re % n
            result.im = result.im % n
        c = c * c
        c.re = c.re % n
        c.im = c.im % n
        exp >>= 1
    return result


m = bytes_to_long(flag)
key = getRandomNBitInteger(m.bit_length())
c = m ^ key
com = Complex(key, c)
p = getPrime(512)
q = getPrime(512)
e = 9
enc = complex_pow(com, e, p * q)
print(enc)
print(Complex(p, q) * Complex(p, q))
# 66350931528185981323649477263900844564494528747802437244889229343520648562164950914433406604580038018765783183569276743239888668912948977370163046257917321742455772852779551569446155827368453262479370103326286297164105599131090881306108546341785251895116423206455175290083968296281375908109039893280371271943 + 65266730684129269806656018828265187384002656633231286337130178390517924611751697965395744944541329793503617856896706439809241745206839328124348693486741656130890593895436661857688522977866438805549144904296596887218275440542852887148071837153436265722614658566275517205322945316112048487893204059562830581004i
# -28814875173103880290298835537218644402443395484370652510062722255203946330565951328874411874019897676900075613671629765922970689802650462822117767927082712245512492082864958877932682404829188622269636302484189627580600076246836248427780151681898051243380477561480415972565859837597822263289141887833338111120 + 235362412848885579543400940934854106052672292040465052424316433330114813432317923674803623227280862945857543620663672974955235166884830751834386990766053503640556408758413592161122945636548462064584183165189050320898315823173824074873376450569212651128285746330837777597290934043912373820690250920839961482862i

已知Complex(p, q) * Complex(p, q)，明显通过该式子求得p，q，同时在复数域中phi=（p^2^-1)（q^2^-1）,按照一般的思路求e关于phi的逆元即可得到结果，但对于这道题有

gcd(e，p^2^-1)=gcd(e,q^2^-1)=3,所以逆元不存在。

反方向的密码 相思

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from Crypto.Util.number import *
import hashlib
from secrets import flag


def hash(x):
    return hashlib.sha256(x.encode()).digest()


def pad(message):
    return message + hash(str(len(message)))


m = bytes_to_long(pad(flag))
p = getStrongPrime(512)
q = getStrongPrime(512)
n = p * q
e = 3
print(pow(m, e, n))
print(n)
# c=120440199294949712392334113337541924034371176306546446428347114627162894108760435789068328282135879182130546564535108930827440004987170619301799710272329673259390065147556073101312748104743572369383346039000998822862286001416166288971531241789864076857299162050026949096919395896174243383291126202796610039053
# n=143413213355903851638663645270518081058249439863120739973910994223793329606595495141951165221740599158773181585002460087410975579141155680671886930801733174300593785562287068287654547100320094291092508723488470015821072834947151827362715749438612812148855627557719115676595686347541785037035334177162406305243
# m=1252734178929444996144243699830218307941015280272407466795285950142407741311749512416560379489724065530428888294520405996969464281930538422811270188075397985

其实就是将sha256看做256进制，256^32^代表将m作为高位拼接到32字节的h之前，其实与二进制移位乘2^n^道理一样。最后copper求小根。

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from Crypto.Util.number import *
import hashlib
from tqdm import *


def hash(x):
    return hashlib.sha256(x.encode()).digest()

e = 3
c = 120440199294949712392334113337541924034371176306546446428347114627162894108760435789068328282135879182130546564535108930827440004987170619301799710272329673259390065147556073101312748104743572369383346039000998822862286001416166288971531241789864076857299162050026949096919395896174243383291126202796610039053
n = 143413213355903851638663645270518081058249439863120739973910994223793329606595495141951165221740599158773181585002460087410975579141155680671886930801733174300593785562287068287654547100320094291092508723488470015821072834947151827362715749438612812148855627557719115676595686347541785037035334177162406305243

PR.<x> = PolynomialRing(Zmod(n))
for length in trange(20,50):
    suffix = bytes_to_long(hash(str(length)))
    f = (256^32*x + suffix)^3 - c
    f = f.monic()
    res = f.small_roots(X=256^length,beta=1,epsilon=0.05)
    if(res != []):
        print(long_to_bytes(int(res[0])))
        break