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Author Topic: Is it possible to generate private key for public key with vanitygen ?  (Read 113 times)
COBRAS (OP)
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July 29, 2021, 04:34:49 AM
 #1

Hello

Is it possible to generate private key for public key with vanitygen or vanitysearch ?

for example:
04791892936f9026ec674f5a620b58e4e4347a9fc50414c90e8fc63d332618e0f65d8e9d9b84f41 a46f8bc36b24bfbd4d892fc6d940165d32eb8cdbf73fdcb7830

12CNHLuvx6JME5kCKkiBfoGrWM2MAQaXHo

or

04308cb32e35947e6b960d3d182f90def3f65309d2b6d70ad69806ae71e19c0b3a6f150cbc21b21 849573b43c3e710871625137b826eb1cc9671b815174f98ecc8

1G3Y5xa4AJRo2tPYvUwRUhwsEBek7znfHs

?

Regards !!!


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July 29, 2021, 07:05:46 AM
Last edit: July 29, 2021, 07:21:52 AM by Pmalek
 #2

So you want to extract a private key from its public key? If that was possible, Bitcoin wouldn't make sense and wouldn't be safe to use. I doubt you would have success with that even if you ran a bruteforce software your whole life and that of your next of kin.

Edit: A quantum computer might one day have success with re-used addresses where the public keys are revealed on the blockchain. But that's an if and when for now. 

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COBRAS (OP)
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July 29, 2021, 09:05:45 AM
Last edit: July 29, 2021, 09:23:55 AM by COBRAS
 #3

So you want to extract a private key from its public key? If that was possible, Bitcoin wouldn't make sense and wouldn't be safe to use. I doubt you would have success with that even if you ran a bruteforce software your whole life and that of your next of kin.

Edit: A quantum computer might one day have success with re-used addresses where the public keys are revealed on the blockchain. But that's an if and when for now.  

Hello.

But I know parT(no partial with all parts unfortunately (( ) of private key, How to reconstruct full privkey Huh

Regard

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July 29, 2021, 10:12:29 AM
 #4

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.
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July 29, 2021, 10:15:00 AM
 #5

--snip--

Hello.

But I know parT(no partial with all parts unfortunately (( ) of private key, How to reconstruct full privkey Huh

Regard

doesn't matter if you know part of the private key... Unless you're just missing a handfull of characters and you have the rest of the private key at hand... In that case, you could probably bruteforce the handfull of missing characters.

The other way around doesn't work... It's not because you know half of the private key that it would be easyer to find the other half.

Like Pmalek already said: bitcoin would not have existed if this was possible.

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COBRAS (OP)
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July 29, 2021, 10:38:59 AM
Last edit: July 29, 2021, 11:28:56 AM by COBRAS
 #6

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:

Quote
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/VanitySearch


Any idea ?

Let's come in, guys, to the discussion. Why I need to do everything alone....

For someone who  know answer - please HELP.

regards.






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July 29, 2021, 01:31:29 PM
 #7

Let's come in, guys, to the discussion. Why I need to do everything alone....
Why don't you ask it in the official thread of VanitySearch? You may also not need to do that either, just search it and find out if someone else had the same query you're having. You should check it on ninjastic.space that gives a more user-friendly experience during your search. It also loads faster than bitcointalk in terms of bandwidth.

Think resourcefully; the users here aren't working for the ones who question. They may get paid for doing it, but they aren't forced to behave you as their customer. You also don't seem to explain your problem very properly, due to lack of language ability (?).

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