0
$\begingroup$

I'm trying to implement elliptic curve point addition in hardware. I've managed to create a working module using Affine coordinates however I've been reading about how points can be added by using Jacobian coordinates more efficently due to needing less modular inversions. I then found this repeated point doubling algorithm which I've attempted to implement in C.

enter image description here

I take the Jacobian coordinates returned from this algorithm and convert them into Affine coordinates using the modular inversions $x = X/Z^2$ and $y = Y/Z^3$. I get the correct point on the curve when calculating $2P$ with $m = 1$ however if I set $m > 1$ to calculate $4P$, $8P$ etc. I get the wrong points.

I'm wondering if I'm doing something obviously wrong? I've included my c code below. I'm fairly new to all this so I'm sorry if there is some obvious errors.

int extendedEuclid(int *c, int phi, int e) {

    int a[2] = {1,0};
    int b[2] = {0,1};
    int q;

    while(1) {
        q = phi / e;
        phi = phi % e;
        a[0] = a[0] - q*a[1];
        b[0] = b[0] - q*b[1];
        if(phi == 0) {
            c[0] = e; 
            c[1] = a[1];
            c[2] = b[1];
            return;
        };
        q = e / phi; 
        e = e % phi;
        a[1] = a[1] - q*a[0];
        b[1] = b[1] - q*b[0];
        if(e == 0) {
            c[0] = phi; 
            c[1] = a[0];
            c[2] = b[0];
            return;
        };
    };
}

int jacobianDouble(int* x, int* y, int* z)
{

    int a, b, w, m = 1;

    *y = 2*(*y);
    w = pow(*z, 4);

    while(m > 0)
    {
        a = 3*(pow(*x, 2) - w);
        b = *x*pow(*y, 2);
        *x = pow(a, 2) - 2*b;
        *z = *z*(*y);
        m = m - 1;
        if(m > 0)
        {
            w = w*pow(*y, 4);
        }
        *y = 2*a*(b - *x) - pow(*y, 4);
    }
    *y = *y/2;
}

int main() {

    int x = 5, y = 4, z = 1, p = 97;
    int ans[2];

    jacobianDouble(&x, &y, &z);

    extendedEuclid(ans, pow(z, 2), p);
    x = fmod(x*ans[1], p);
    if(x < 0)
    {
        x = p + x;
    }

    extendedEuclid(ans, pow(z, 3), p);
    y = fmod(y*ans[1], p);
    if(y < 0)
    {
        y = p + y;
    }

    printf("x = %d\n", x);
    printf("y = %d\n", y);

    return 0;
}
$\endgroup$
3
$\begingroup$

All arithmetic operations in jacobianDouble should be in the finite field $\mathbb{F}_p$, i.e. all results should mod $p$, not plain integer arithmetics. Note division Y/2 is defined as $Y\cdot 2^{-1}$ where $2^{-1}$ is the multiplicative inverse of 2 modulo $p$.

$\endgroup$
  • 1
    $\begingroup$ @Andrew: an easy way to implement $Y \cdot 2^{-1} \bmod p$ is "if $Y$ is even, then it's $Y/2$, otherwise, it's $(Y+p)/2$, where the $/$ is integer divide. $\endgroup$ – poncho Sep 5 '18 at 1:12
  • $\begingroup$ Thanks, much appreciated. It's working correctly now, I assume the trick for $Y*2^{-1}\ mod\ p$ only works for power of 2 numbers? $\endgroup$ – Andrew Sep 5 '18 at 8:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.