Post
Topic
Board Development & Technical Discussion
Re: 5-7 kangaroo method
by
kTimesG
on 17/10/2024, 12:38:11 UTC
#120 was pretty easy, just 20pcs of 4090 for six weeks.

Let's say you needed 1.36 sqrt(n) ops, that would mean

Code:
1.36*sqrt(2**119)/(14*86400)/20

op/s per each GPU, which is close to 46 Gk/s

Now, a 4090 can do at max 82 Tflop/s

That's 1782 float32 ops for a single jump.

For a jump you'd need around 6 256-bit muls and 6 additions (if inversions amortized to negligible).

To do a 256-bit mod mul you'd need around 8 int32 * 8 int32 = 64 32-bit multiplications and 56 32-bit additions = 119 operations just for the first step and around another 18 mul + a few adds or so to reduce mod N.
For a single 256-bit addition you'd need around 8 operations (16 to do mod N).

That adds up to around 170 or so ops per 256-bit mul and 16 per add.

Estimated cost: 170 * 6 + 16 * 6 = 1116 ops.

Theoretically, feasible since 1116 < 1782, but are you some kind of a genius or WTF? I didn't even put in the rest of the overhead work required.