Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
Guys with brute force we won't go anywhere.
Let us assume we would be able somehow to generate a brute force of 5 TH/s (The max ASIC Miner speed today that I found out there).
That's like 5000000000000 Hashes per second. Of course with CPU/GPU we won't get anywhere near.
But let us assume for a second that this performance would be possible somehow, we would need ~303150594339750000000000000000000000000000000000000000000 years to breakuntil the address 256 already counting with starting each address generation from 2^X-1. (EDIT: Not even counting with all the addresses before 
)
So it's kind of useless to brute force this.
x3 = λ2+a1λ−a2−x1−x2
y3 = −a1x3−a3−λx3+λx1−y1
λ = (y2−y1) / (x2−x1)
From: https://crypto.stanford.edu/pbc/notes/elliptic/explicit.html
How precise is this formula?
Let us assume we would be able somehow to generate a brute force of 5 TH/s (The max ASIC Miner speed today that I found out there).
That's like 5000000000000 Hashes per second. Of course with CPU/GPU we won't get anywhere near.
But let us assume for a second that this performance would be possible somehow, we would need ~303150594339750000000000000000000000000000000000000000000 years to break
)So it's kind of useless to brute force this.
Unlike hash operations, elliptic curve operations have unpredictable machine cycle count.
I would expect a single point addition operation (P3 = P1 + P2) to have a very predictable machine cycle count. It should be something like:x3 = λ2+a1λ−a2−x1−x2
y3 = −a1x3−a3−λx3+λx1−y1
λ = (y2−y1) / (x2−x1)
From: https://crypto.stanford.edu/pbc/notes/elliptic/explicit.html
How precise is this formula?