Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
It's not that I'm a pessimist but every time I think about the magnitude of the spaces we face I get chills 
For example, in puzzle 66 the number of possible positions is 36893488147419103231. To understand this it is necessary to resort to an analogy. If each of those numbers were a millimeter, that would be equivalent to 3901.95 light years!!!
Any attempt at sequential search is then ruled out outright. Only with methods like kangaroo or bsgs would there be a possibility if the public key was known.
By the way, I leave a py script to help find the public key from the public address.
https://github.com/Dedaloo/PublicKeyHunt
For example, in puzzle 66 the number of possible positions is 36893488147419103231. To understand this it is necessary to resort to an analogy. If each of those numbers were a millimeter, that would be equivalent to 3901.95 light years!!!
Any attempt at sequential search is then ruled out outright. Only with methods like kangaroo or bsgs would there be a possibility if the public key was known.
By the way, I leave a py script to help find the public key from the public address.
https://github.com/Dedaloo/PublicKeyHunt
and 2^130 which requires a search between 2^129 and 2^130 equals 680,564,733,841,876,926,926,749,214,863,536 km.
therefore 71,986,683,526 light years (almost 72 billion).
I agree with the great luck with random and many subkeys....but many.