Quote from: kTimesG on Today at 04:18:43 PM

Quote from: nomachine on Today at 04:02:29 PM

Quote from: kTimesG on Today at 03:41:48 PM

Quote from: nomachine on Today at 02:07:37 PM

[Hops: 229969 h/s] [00:00:01]
total time: 1.47 sec
KANGAROO: Sun Jul 14 16:01:53 2024 PUZZLE SOLVED
Private key (hex) : 000000000000000000000000000000000000000000000000000000e9ae4933d6
Hops: 335150
Average time to solve: 1.47 sec

I can bet that you can make a .so file that works 20 times faster point_addition & point_doubling than this. Grin

A single point addition according to your program requires:

m = (y1 - y2) / (x1 - x2)
x3 = m**2 - x1 - x2
y3 = m(x2 - x3) - y2

result_bytes = ice.point_addition(P_bytes, Q_bytes)

You have to ask the iceland how it works. Since the source is closed.

But this is irrelevant. I was pointing out the flaw in the high-level strategy:

while (condition)
   add_point(P, Q)

which is a serial one-by-one element addition. If you really aim for higher speed you need to rethink what you are doing, not pretend that some black-box "addPoints(a, b)" can do wonders for you. The black-box does whatever the formula to add 2 points does, and has no knowledge of your context (adding many points).

Here's a clue:

z = p[0].x - jp[0].x
for (i = 1 .. n) {
  d = p[i ].x - jp[i ].x
  z *= d
  q[i ] = z
}

t = inv(z)
for (i = n - 1 ... 0) {
   xd = i > 0 ? t * q[i-1] : t
   t *= q[i ]
  // finish addition..
}

This trades 3(N-1) multiplications with a single inversion
Depending on size of N you can get a speedup up to the factor of around cost(1 inversion) / cost(3 multiplications).

You are going off topic and pretending to be too smart without need. Just like Digaran... This is my answer @WanderingPhilospher what would happen if we switch same script from GMP to iceland. And I don't intend to talk with with you at all. You just raise the pressure without any script ever being shown..