ex n=947 = 2^9+2^8+2^7+2^5+2^4+2^1+2^0
which means that 947P = P+2P+16P+32P+128P+256P+512P. Thats 9 doublings and 6 additions, much better than 946 additions....
if we allow subtraction as well (ternary expansion) its even better.
want to calculate nP?
algorithm:
1. input point P from E(Fp) and n>=1
2. set Q=P and R=Ø (point at infinity)
3. loop while n>0{
if n = 1 (mod2) set R= R+Q
set Q=2Q and n=n/2 (floor)}
4.return R (which equals nP)
I've now posted working sourcecode available at http://christelbach.com
(source code contained in master thesis)
for instance this:
//double and add algorithm
private BigInteger[] calculatenP(BigInteger x, BigInteger y, BigInteger n, BigInteger A, BigInteger modp)
{
BigInteger[] Q = { x, y };
BigInteger[] R = { 0, 0 };
while (n > 0)
{
BigInteger t = BigInteger.ModPow(n, 1, 2);
if ((t) == 1)
{
R = addPoints(Q[0], Q[1], R[0], R[1], modp);
}
Q = doublePoints(Q[0], Q[1], A, modp);
n = n / 2;
}
BigInteger[] resPoint = { R[0], R[1] };
return resPoint;
}
Look in the pdf file for the methods addPoints() and doublePoints()
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