Kanav

Aspiring Cryptology Researcher

@kanav99

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Sage Math notes for CTF Crypto

Me and my friends at SDSLabs participated in BSides Ahmedabad 2021 CTF, and the outcome of the CTF was not something I liked, I found the crypto side a bit weak. The two problems I couldn’t solve (ecc-rsa and there-were-eleven) were not that hard, but my lack of Sagemath experience was a bit of reason why I couldn’t do it.

So here it is, my notes on Sage math.

Choose correct field #

For RSA and any modular math, use Zmod(p).

F = Zmod(p)

Polynomial Fields #

We want to make a field of polynomials where the arithmetic is happening over the F field and has variables x, y, z

PR.<x, y, z> = PolynomialRing(F)

or

PR = PolynomialRing(F, 3, 'x,y,z')
x, y, z = PR.gens()

Solving multivariate polynomials over Zmod(p) #

One might be tempted to use solve_mod function to solve a system of equations over a modulo. But if you have a special case when -

  1. Functions are multivariate polynomials.
  2. p is huge

In this case, solve_mod sometimes fails with Overflow Error. Instead, use Groebner Basis of the polynomials.

f1 = x*y + x^2 - 88
f2 = y^2 - z^2 + 51
f3 = z^2 - z + x*y - 102
I = (f1, f2, f3)*P
print(I.groebner_basis())

(This is a random example and doesn’t print anything good, but when equations are well formed the basis literally spits out the solution).

What is Groebner Basis? #

TODO

Chinese remainder theorem #

CRT is a useful function that given moduli n1, n2, n3... and respective remainders r1, r2, r3..., it finds a number m modulo n1*n2*n3... such that m = ri mod ni for all i.

m = crt([r1, r2, r3], [n1, n2, n3])

Lagrange Interpolation #

Lagrange interpolation finds a polynomial f such that given m pairs of (xi, yi), f(xi) = yi. Degree of f is at max m-1

shares = [(1, 1651293975450381579706844999808202297670211173037061827272908790114230592434748044848097133563469251678879059156225205298834971071359017469397331605782920), (2, 49656064002974834481096383104316375265711545391722811288216446968986145936494966876404160910407919885451814058823146922107458035910700220495010462147112), (3, 1481214561214496310917942246038921499126047497749957535731608952096552856013930232284898279007009260107597472601959627310496773682697020898442717240484400), (4, 1950790377868548708758723604473108315857898618124646291056275632619091046294238343215502355242288776617394025418770078552886012721353626716473759644786481)]

shares = [(F(x), F(y)) for x,y in shares ]
reconstructed_polynomial = PR.lagrange_polynomial(shares)

Always remember to convert the number to the correct field element!

Polynomial GCD #

When two polynomials are known to have a common root, the best way to get that root is by calculating the polynomial gcd. Passing two polynomial objects in gcd could work but is not fast. Use half-GCD algorithm.

(Source: https://github.com/death-of-rats/CTF/tree/master/Dice2021/plagiarism)

def hgcd(a0,a1):
    if a1.degree() <= (a0.degree()//2):
        return np.array([[1,0],[0,1]])
    m = a0.degree()//2
    X = a0.variables()[0]
    b0 = a0 // X**m
    b1 = a1 // X**m
    R = hgcd(b0,b1)
    [d,e] = (R.dot(np.array([a0,a1]).transpose())).transpose()
    ff = d % e
    m = m // 2
    g0 = e // X**m
    g1 = ff // X**m
    S = hgcd(g0,g1)
    q = d // e
    return S.dot(np.array([[0,1],[1,-q]])).dot(R)

def gcd(a0,a1):
    while True:
        print(a0.degree(), end=", ", flush=True)
        if a0 % a1 == 0:
            return a1
        if a0.degree() == a1.degree():
            a1 = a0%a1
        #print(a0.degree())
        R = hgcd(a0,a1)
        [b0,b1] = R.dot(np.array([a0,a1]).transpose()).transpose()
        if b0%b1==0:
            return b1
        c = b0 % b1
        a0 = b1
        a1 = c

Lattices #

TODO

Coppersmith short-pad attack #

TODO