2022西湖论剑WriteUp

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2022西湖论剑WriteUp

MyErrorLearn

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def XennyOracle():
    r = getPrime(512)
    d = invert(secret+r, p) - getPrime(246)
    print('> r =', r)
    print('> d =', d)

def task():
    for _ in range(3):
        op = int(input())
        if op == 1:
            XennyOracle()
        elif op == 2:
            ss = int(input())

            if ss == secret:
                print('flag: ', flag)

与题目交互3次，前两次拿数据，最后一次验证secret拿flag

\[(s+r1)^{-1}-x \equiv d1 \pmod{p}\]

\[(s+r2)^{-1}-y \equiv d2 \pmod{p}\]

\(x,y\)分别表示两次数据中的\(Prime(246)\)

\[(d1+x)^{-1}-r1 \equiv s \pmod{p}\]

\[(d2+y)^{-1}-r2 \equiv s \pmod{p}\]

两式相减 \[ (d1+x)^{-1}-r1-(d2+y)^{-1}+r2 \equiv 0 \pmod{p} \] 上式乘\((d1+x)(d2+y)\) \[ (d2+y) - (d1+x) + (r2-r1)*(d1+x)*(d2+y) \equiv 0 \pmod{p} \] 一个二元的coppersmith 构造等式\(f = (r1-r2)*(d1+x)*(d2+y)-(d2+y-d1-x)\)， 找个多元copper的板子套上

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#sagemath
def small_roots(f, bounds, m=1, d=None):
    if not d:
        d = f.degree()

    R = f.base_ring()
    N = R.cardinality()

    f /= f.coefficients().pop(0)
    f = f.change_ring(ZZ)
 
    G = Sequence([], f.parent())
    for i in range(m + 1):
        base = N ^ (m - i) * f ^ i
        for shifts in itertools.product(range(d), repeat=f.nvariables()):
            g = base * prod(map(power, f.variables(), shifts))
            G.append(g)
 
    B, monomials = G.coefficient_matrix()
    monomials = vector(monomials)
 
    factors = [monomial(*bounds) for monomial in monomials]
    for i, factor in enumerate(factors):
        B.rescale_col(i, factor)
 
    B = B.dense_matrix().LLL()
 
    B = B.change_ring(QQ)
    for i, factor in enumerate(factors):
        B.rescale_col(i, 1 / factor)
 
    H = Sequence([], f.parent().change_ring(QQ))
    for h in filter(None, B * monomials):
        H.append(h)
        I = H.ideal()
        if I.dimension() == -1:
            H.pop()
        elif I.dimension() == 0:
            roots = []
            for root in I.variety(ring=ZZ):
                root = tuple(R(root[var]) for var in f.variables())
                roots.append(root)
            return roots
    return []


r1 = 10416632501801974694227453996514471230510822844141537215435310761282236706010374287476167228551720127564766961486909719901188052844926828718157982353509009
r2 = 10945694515092921935765033073472221225431374672659751536595840465276073402439042169265480667452817278318006398182876136457632746343409326182238053908029751
d1 = 56888336937500044415702919773416114217028770468080853282740190270108239476041614796007252946794331087237297216052651102131387831331715788882702697436356478481466698200484763202280365937244685692432494727290180152495648130890129145398986742781650456719538128804602194272879452329225793378627305015814235006449
d2 = 39564785910816285169732671501865632983688744846819855674275371020624933955200211958646008885482995237194222616193536005648361431696101118452470392982186472493910970565702639588941011814416028774758786985291250896861107704070339750852212760711413327701632586870290470689206960878069267347698863426921542358105
p = 117772635282729418285879412631216935490806524671311266185546790504745904196135114727754422648999093257461858612228556910365982066740540367638744157494110844029055193511998422999596748975390734628418184477450856691002269702533884505178634105672240426552752649855510423531462598366536500750787631689349504137087

import itertools
R.<x,y> = PolynomialRing(Zmod(p))
f = (r1-r2)*(d1+x)*(d2+y)-(d2+y-d1-x)
ans = small_roots(f, bounds=(2^246,2^246), m=2)
x, y = ans[0]
secret = 1/(d1+x)-r1
print(secret)

LockbyLock

比赛的时候脑瘫代错数据了，半小时没跑出来

比赛结束一看改过来，三分钟就出了

第一次第二次分别发2和4过去，通过rsa同态，gcd拿到n

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

io = remote('tcp.cloud.dasctf.com', 21488)

io.recvuntil(b'msg1 = ')
m1 = int(io.recvline())
io.recvuntil(b'msg2 = ')
m2 = int(io.recvline())
io.recvuntil(b'msg3 = ')
m3 = int(io.recvline())

io.sendline(b'2')
io.recvuntil(b'msg1 = ')
m11 = int(io.recvline())
io.recvuntil(b'msg2 = ')
m12 = int(io.recvline())
io.recvuntil(b'msg3 = ')
m13 = int(io.recvline())

io.sendline(b'4')
io.recvuntil(b'msg1 = ')
m21 = int(io.recvline())
io.recvuntil(b'msg2 = ')
m22 = int(io.recvline())
io.recvuntil(b'msg3 = ')
m23 = int(io.recvline())

n = gcd(m11**2-m21, m12**2-m22, m13**2-m23)
print('n = ', n)
print('m1 = ', m1)
print('m2 = ', m2)
print('m3 = ', m3)
print('m11 = ', m11)
print('m12 = ', m12)
print('m13 = ', m13)
print('m21 = ', m21)
print('m22 = ', m22)
print('m23 = ', m23)

对发送2（或者4）时拿到的数据，使用Pollard’s kangaroo algorithm求解离散对数，时间复杂度大概在10^7

Ps：比赛的时候误拿了最开始给的第一组数据去跑了我有罪qwq

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#sagemath
n =  27950092418645918795271296476246539971431421044767995947796205802943478655421653573048132032794234798415984952652376438949021032706491508283019165573213412785457536560835247461258996700097675642961189941397147832121215798258718677899047430424242917316567794068767871489924536948337626278187554943376441787903310776418948278970828056614007480123321047526250014275143243282932429377977743696352171804147086603083279288062958658420062643379022044426846987241250137102448598685606712521741583607474213255315746187154198844169065935475113145789594643131816301778344794413391236892556628344738627222814821415207388109885721
m11 =  20376839897483150202591245279988785405360453405518229860314688355462074916580107129330682477462552663163126795528589023633648581007626813266919326759350958000841142246828917207676253853232169554216424361180522508797731315528852217133666113558436155410718944260169577145534697157368360316333345542201829151854115374449392740319226345217386316755786443394636434084405240052107608897042547234279512095541661926355330342347645389032183986346153775559868246848334225734203525123266991086247401891517240101646138518686721872066661769309263715716056740461960894538873962031902453537598108351613389507079566414498498479169599
print('e1 = ', discrete_log_lambda(m11,mod(2,n),(10**14,10**15)))
m13 =  25654168760396958029792298552570736468682480823084565803373285663407780762485503411904748880883813230583714779620465503738003648395595616814054655108314643554108891626126605868539398053448468263238189719286443578861455115451473982124850952368091893597737777395501987491639060745712330371967127518105827612033705562056542312813576000021698561975189704496509514959272625270304453384259337771998645760374945175685616354851961881768292039874827087997461444008148949745457037468733732431091509683231366382582177637282818474046552143645887960397145203641778786138850215567179584185583145503621183907129930648092263851567807
print('e2 = ', discrete_log_lambda(m13,mod(2,n),(10**14,10**15)))

#e1 = 710179922790353
#e2 = 871767685631399

最后一步，rsa共模攻击，直接套

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e1 = 710179922790353
e2 = 871767685631399
m1 =  18750007013174536423532792814200623641168684158537875023748157501043254834580359946005782798296094859389089373500698907014613039137711295832309784735974363506728965819640087175515796834513477450080858989016689349728236691024985081556114367071105810780697020942697446551986754568210757566280492351447198167578093986707571108589625817862709984743133957428410783190472709108509345889765505376018262071861959201900814249687461034976969943031661314448282010075462891703692709064089097420481900561158105185207229830775312377353386152497009445846705513245630211941745950480797566412389719032698035360150224973575890501234349
m3 =  7797790268283288534585277663447783093020314808581197168654721551908450973265373361401405475956787957261162271142906099769330725245515346716540341701573417181548009002140256280036084891830998092431760582730910699403915059015408500826152174941526987251255119134941854685885210496410310463675404149461859807594357045329568535924128371829778827795984952587210383466369410277156253887474458529307937131048870052343339798306152534342663301893903024266179991962280108122818029085323024355732887033339956485454178220409801766544479498601241502177418664898422266425022618582009534572842700671485483690331035082266707254176987
n =  27950092418645918795271296476246539971431421044767995947796205802943478655421653573048132032794234798415984952652376438949021032706491508283019165573213412785457536560835247461258996700097675642961189941397147832121215798258718677899047430424242917316567794068767871489924536948337626278187554943376441787903310776418948278970828056614007480123321047526250014275143243282932429377977743696352171804147086603083279288062958658420062643379022044426846987241250137102448598685606712521741583607474213255315746187154198844169065935475113145789594643131816301778344794413391236892556628344738627222814821415207388109885721

from gmpy2 import *
_, s1, s2 = gcdext(e1, e2)
m = pow(m1, s1, n) * pow(m3, s2, n)
m %= n
flag = long_to_bytes(m)
print(flag)