Can you search for an address with vanitygen with a public key? I was thinking would that be able to shorten the search space like kangaroo.
Not 100% sure with vanitygen but vanity search takes inputted address and converts it to its RIPEMD160, then searches for a match for the RIPEMD160. One could tweak code to search for a pubkey which would save one sha256 and the one RIPEMD160 function.
Priv key
Pub key
sha256
ripemd160
so you would save two functions but I am not sure on the speed gained since normally, the most time consuming part. whether its CPU or GPU. is doing the math from priv key to pub key.
I asked this a few days ago.
For the makin discussion more intrestig, I hope it will be. this a a example:
This a our target address and pubkey
1Kn5h2qpgw9mWE5jKpk8PP4qvvJ1QVy8su
0348e843dc5b1bd246e6309b4924b81543d02b16c8083df973a89ce2c7eb89a10d
this is a parts of of privkey:
0xdb09b0615ad40a0
and anower variant of part:
0xffffffffffffff01aeecd29b4137e0
How to find a secret key like in method what JeanLuc write:How it works
Basically the -sp (start public key) adds the specified starting public key (let's call it Q) to the starting keys of each threads. That means that when you search (using -sp), you do not search for addr(k.G) but for addr(kpart.G+Q) where k is the private key in the first case and kpart the "partial private key" in the second case. G is the SecpK1 generator point.
Then the requester can reconstruct the final private key by doing kpart+ksecret (mod n) where kpart is the partial private key found by the searcher and ksecret is the private key of Q (Q=ksecret.G). This is the purpose of the -rp option.
The searcher has found a match for addr(kpart.G+ksecret.G) without knowing ksecret so the requester has the wanted address addr(kpart.G+Q) and the corresponding private key kpart+ksecret (mod n). The searcher is not able to guess this final private key because he doesn't know ksecret (he knows only Q).
Note: This explanation is simplified, it does not take care of symmetry and endomorphism optimizations but the idea is the same.
https://github.com/JeanLucPons/VanitySearchAny idea ?
Let's come in, guys, to the discussion. Why I need to do everything alone....For someone who know answer - please HELP.
regards.