diceCTF 2021 | plagiarism

#diceCTF_2021 #good_challenges_2021

Two agents, Blex and Kane, have simultaneously known very secret message and transmitted it to Center. You know following:
1) They used RSA with this public key
2) They sent exactly the same messages except the signatures (name appended, eg. "[message]Blex")
3) They did encryption this way:

m = int(message.encode("hex"), 16)
c = pow(m, e, N)

4) And here are cryptograms you have intercepted:

N = 25898966400928827905718377946331123070958718286581765316565582158865227877882475404853218079499084099440419144196215764927720893687968939899067275095801562867742359933997487928281899714724738097735994026225339488710478292473051567851786254924548138570069406420407124627567648479424564834446192417334669768477661434992797176428220265984651288944265998446714590797833756720922745187467388408600309665467669255896919554072379878017822219455974525233467767926938557154083882126002952139561283708342676308894318951822068027821029295524097544028901807902120777407151278396388621981625398417573347316888458337381776303199529

e = 1048577

ciphertext_Blex = 11140520553087800834883326476247582685177207584737264356946559762068509060522907835540767944557089926814767920501376431871780404000550271362410228709616559148950928004959648199391157781102695421411667843970881959939688515679415870087007797819271601359811630724878746762862603629420061133824605384527474682526549557804674160851967543475275374840169790764048711047622418045734436512050742433282306694490346907876574514077395835974083376649624559301087384766644865104383786285302561584731767419571603248493060257358632833957327996996960955767927114473513709882904104552609194519132931270741118197821776138632855021619178

ciphertext_Kane = 2922817623733019475805146570530296261205732600738503605503192845278422660686627490817081424885152809772315629265930072636690234953045955503581182067349322827011065359648958225896393305011175960879100475146203207801533980643261035226402857047007061320653920746872424363923515091038846823007819033456503365649022294092944985887626605207259444051959239244136999684366533551627508385114998024232490369665950339127904350803268889205463047713233591604324960184727360413931125906144631968128488876241314939855024305076160092193380013725939761970042406866169417457376487954247442308318888399299295082898238584625937490546472



Now tell me that secret message! (The answer for this task starts from 'dice{') 

どうみても Franklin-Reiter Related Message Attackなのでそれをやればいいんだけど、が大きいという問題がある。一変数多項式のgcd の方法では ちょっと遅すぎる

N = 25898966400928827905718377946331123070958718286581765316565582158865227877882475404853218079499084099440419144196215764927720893687968939899067275095801562867742359933997487928281899714724738097735994026225339488710478292473051567851786254924548138570069406420407124627567648479424564834446192417334669768477661434992797176428220265984651288944265998446714590797833756720922745187467388408600309665467669255896919554072379878017822219455974525233467767926938557154083882126002952139561283708342676308894318951822068027821029295524097544028901807902120777407151278396388621981625398417573347316888458337381776303199529
e = 1048577


ciphertext_Blex = 11140520553087800834883326476247582685177207584737264356946559762068509060522907835540767944557089926814767920501376431871780404000550271362410228709616559148950928004959648199391157781102695421411667843970881959939688515679415870087007797819271601359811630724878746762862603629420061133824605384527474682526549557804674160851967543475275374840169790764048711047622418045734436512050742433282306694490346907876574514077395835974083376649624559301087384766644865104383786285302561584731767419571603248493060257358632833957327996996960955767927114473513709882904104552609194519132931270741118197821776138632855021619178
ciphertext_Kane = 2922817623733019475805146570530296261205732600738503605503192845278422660686627490817081424885152809772315629265930072636690234953045955503581182067349322827011065359648958225896393305011175960879100475146203207801533980643261035226402857047007061320653920746872424363923515091038846823007819033456503365649022294092944985887626605207259444051959239244136999684366533551627508385114998024232490369665950339127904350803268889205463047713233591604324960184727360413931125906144631968128488876241314939855024305076160092193380013725939761970042406866169417457376487954247442308318888399299295082898238584625937490546472

mb = b"Blex"
mk = b"Kane"
mb = int(mb.hex(), 16)
mk = int(mk.hex(), 16)

P.<x> = PolynomialRing(Zmod(N))
# x = m1 -> ([message]Blex)
g1 = x^e - ciphertext_Blex
g2 = (x - mb + mk)^e - ciphertext_Kane

gcd(g1, g2)

# gcd(g1, g2) -> (x - m1) # tooo slow

そこで half-GCDを使う。これまでのGCDがていどの計算量だったのに対してこちらはと非常に高速

非常に高速とはいえがでかいので解くのには5000秒以上かかった。おいおい

def pdivmod(u, v):
    """
    polynomial version of divmod
    """
    q = u // v
    r = u - q*v
    return (q, r)

def hgcd(u, v, min_degree=10):
    """
    Calculate Half-GCD of (u, v)
    f and g are univariate polynomial
    http://web.cs.iastate.edu/~cs577/handouts/polydivide.pdf
    """
    x = u.parent().gen()

    if u.degree() < v.degree():
        u, v = v, u

    if 2*v.degree() < u.degree() or u.degree() < min_degree:
        q = u // v
        return matrix([[1, -q], [0, 1]])

    m = u.degree() // 2
    b0, c0 = pdivmod(u, x^m)
    b1, c1 = pdivmod(v, x^m)

    R = hgcd(b0, b1)
    DE = R * matrix([[u], [v]])
    d, e = DE[0,0], DE[1,0]
    q, f = pdivmod(d, e)

    g0 = e // x^(m//2)
    g1 = f // x^(m//2)

    S = hgcd(g0, g1)
    return S * matrix([[0, 1], [1, -q]]) * R

def pgcd(u, v):
    """
    fast implementation of polynomial GCD
    using hgcd
    """
    if u.degree() < v.degree():
        u, v = v, u

    if v == 0:
        return u

    if u % v == 0:
        return v

    if u.degree() < 10:
        while v != 0:
            u, v = v, u % v
        return u

    R = hgcd(u, v)
    B = R * matrix([[u], [v]])
    b0, b1 = B[0,0], B[1,0]
    r = b0 % b1
    if r == 0:
        return b1

    return pgcd(b1, r)


import time

N = 25898966400928827905718377946331123070958718286581765316565582158865227877882475404853218079499084099440419144196215764927720893687968939899067275095801562867742359933997487928281899714724738097735994026225339488710478292473051567851786254924548138570069406420407124627567648479424564834446192417334669768477661434992797176428220265984651288944265998446714590797833756720922745187467388408600309665467669255896919554072379878017822219455974525233467767926938557154083882126002952139561283708342676308894318951822068027821029295524097544028901807902120777407151278396388621981625398417573347316888458337381776303199529
e = 1048577


ciphertext_Blex = 11140520553087800834883326476247582685177207584737264356946559762068509060522907835540767944557089926814767920501376431871780404000550271362410228709616559148950928004959648199391157781102695421411667843970881959939688515679415870087007797819271601359811630724878746762862603629420061133824605384527474682526549557804674160851967543475275374840169790764048711047622418045734436512050742433282306694490346907876574514077395835974083376649624559301087384766644865104383786285302561584731767419571603248493060257358632833957327996996960955767927114473513709882904104552609194519132931270741118197821776138632855021619178
ciphertext_Kane = 2922817623733019475805146570530296261205732600738503605503192845278422660686627490817081424885152809772315629265930072636690234953045955503581182067349322827011065359648958225896393305011175960879100475146203207801533980643261035226402857047007061320653920746872424363923515091038846823007819033456503365649022294092944985887626605207259444051959239244136999684366533551627508385114998024232490369665950339127904350803268889205463047713233591604324960184727360413931125906144631968128488876241314939855024305076160092193380013725939761970042406866169417457376487954247442308318888399299295082898238584625937490546472

mb = b"Blex"
mk = b"Kane"
mb = int(mb.hex(), 16)
mk = int(mk.hex(), 16)

P.<x> = PolynomialRing(Zmod(N))
# x = m1 -> ([message]Blex)
g1 = x^e - ciphertext_Blex
g2 = (x - mb + mk)^e - ciphertext_Kane


t1 = time.time()
r = pgcd(g1, g2)
t2 = time.time()
print(t2-t1)
print(r)
print(r, g1 % r, g2 % r)
print(bytes.fromhex(hex(inverse_mod(int(-r[1]), N) * r[0] % N)[2:]))
# gcd(g1, g2) -> (x - m1)