Here is the ELI5 version in 18 lines of SageMath / Python code. I use Sage because it provides elliptic curves as first-class citizens (`FiniteField` and `EllipticCurve`) and we can take multiplication operation for granted.
1 p = 103 2 F = FiniteField(p) 3 E = EllipticCurve(F, [0, 7]) 4 G = E([97, 10]) 5 n = G.order() 6 k = randint(2, n) 7 P = k * G 8 m = 6 9 t = randint(2, n) 10 R = t * G 11 r = mod(R, n) 12 s = mod(inverse_mod(t, n) * (m + r * k), n) 13 inv_s = inverse_mod(int(s), n) 14 u = mod(m * inv_s, n) 15 v = mod(r * inv_s, n) 16 F = int(u) * G + int(v) * P 17 f = mod(F, n) 18 print "YOU ARE A CRYPTOSTAR!" if r == f else "YOU SUCK!" YOU ARE A CRYPTOSTAR!
Besides lines #7, #10 and #16 that involves multiplication between a scalar and a point on elliptic curve, everything else is just modular arithmetic and two multiplicative inverse calculations (line #12 and #13)
Please have a look at Elliptic curves post for an introduction to elliptic curves then keep reading if you want line by line explanations and a few elliptic curve graphs.
Elliptic curve multiplication trapdoor
First thing first, let’s define the elliptic curve parameters and notations:
1 p = 103 2 F = FiniteField(p) 3 E = EllipticCurve(F, [0, 7]) 4 G = E([97, 10]) 5 n = G.order()
- uppercase letter - elements of the elliptic curve like finite fields (F), elliptic curve itself (E) or point on elliptic curve (G)
- lowercase letter - scalar value like the secret key (k)
- p - prime number
- F - finite field
- 0,7 - coefficients for elliptic curve
- E - elliptic curve
- G - point generator
- n - order (the number of points) of the cyclic subgroup generated by `G`
Now, it’s time to generate the secret key `k` and calculate the public key `P` using the trapdoor function called Elliptic Curve Discrete Logarithm Problem (ECDLP).
This is the bread and the butter of elliptic curve cryptography, given the generator `G` and public key `P’ it is infeasible to calculate the secret key `k`, hence the trapdoor
6 k = randint(2, n) 7 P = k * G
If you are curious how elliptic curve point `P` looks like then it is nothing than a Cartesian (x,y) coordinate:
P (65 : 31 : 1)
Let’s sign the message `m` and for this we need to generate a random integer `t` of order `n` and calculate the `R` point on elliptic curve.
8 m = 6 9 t = randint(2, n) 10 R = t * G 11 r = mod(R, n) 12 s = mod(inverse_mod(t, n) * (m + r * k), n)
The signature itself has two numbers `r` and `s` that are sent to third-party for verification:
- r - the `x` coordinate of point `R`
- s - is calculated with `s = (m + r*k) / t` but we use multiplication with inverse instead of division.
And this is how our signature looks like:
[r, s] [66, 51]
Given the signature `r, s` above and `G, P, r and n` that are public information we need to calculate the equation `F = u*G + v*P` where `u = m/s` and `v = r/s`.
The signature is valid if `R.x == F.x` is true, in other words the `x` coordinates are the same.
13 inv_s = inverse_mod(int(s), n) 14 u = mod(m * inv_s, n) 15 v = mod(r * inv_s, n) 16 F = int(u) * G + int(v) * P 17 f = mod(F, n) 18 print "YOU ARE A CRYPTOSTAR!" if r == f else "YOU SUCK!" YOU ARE A CRYPTOSTAR!
Remember that uppercase letters are points on elliptic curve, lowercase are scalars:
1 R == u*G + v*P 2 R == u*G + v*k*P 3 R == m/s * G + r/s * k*G 4 R == m/s * G + r*k/s * G 5 R == (m + r*k)/s * G 6 R == (m + r*k)/((m + r*k)/t) * G
As always we will do the math backwards:
- start with the equation of signature verification and substitute public key `P`
- substitute `u` and `v`
- multiplication is commutative and we can put `r` and `k` together
- extract `G` and `s` as factors
- substitute `s`
- after reduction we are left with `t*G`
And the intuition is valid if the `x` coordinates on both sides of the equation are equal.
print "MAGIC" if R == (int((m + r*k)*t/(m + r*k)) * G) else "ERROR" MAGIC