Bitcoin puzzle transaction ~32 BTC prize to who solves it

[nochkin]

Would a 5090 produce even better results than a 4090 in what we're trying to do? You can only go so fast. Thoughts?

5090 is about 50% faster compared to 4090.

[MrGPBit]

Would a 5090 produce even better results than a 4090 in what we're trying to do? You can only go so fast. Thoughts?

5090 is about 50% faster compared to 4090.

The RTX 5090 is about 29% faster than the RTX 4090. In searching for the private key with a perfect cuda programming

[nochkin]

5090 is about 50% faster compared to 4090.

The RTX 5090 is about 29% faster than the RTX 4090. In searching for the private key with a perfect cuda programming

I don't know what you mean by "perfect cuda programming". This is clearly an oxymoron like "perfect programming".
Anyway, in my own benchmarks I get about 50% difference. Vanity shows up to 5500MK/s using 4090 and up to 8000MK/s using 5090.
If you measure per Watt speed, both 4090 and 5090 perform the same.
It's all just estimates, so YMMV.

[Frequence]

i dont get the idea of range f6f543.
while all talk, range scanned to be empty

Lady, this has nothing to do with the prefix scan range. This is the HASH160 of a Bitcoin address.

And here is the correct sequence from private key to Bitcoin address:

Private Key


→ Public Key (via Elliptic Curve multiplication)

→ SHA-256 of the public key

→ RIPEMD-160 of the SHA-256 result (this gives the HASH160) ( this step )
they are talking about this as they want to get the public key for fast decryption and they will fail !

→ Add Network Byte (e.g., 0x00 for mainnet)

→ SHA-256 of that

→ SHA-256 again

→ Take first 4 bytes (checksum)

→ Append checksum to HASH160 + network byte

→ Base58Check encode

= Bitcoin Address

[analyticnomad]

Hey has the privatekeys.pw site been down all day for you guys too?

[nameavailable]

Hey has the privatekeys.pw site been down all day for you guys too?

Yes it has. Whoever runs the site recently updated some of the user interfaces. Now the page with the unsolved keys has tabs 'solved', 'unsolved' and 'all'. I assume that they might be doing some other updates. But then again those updates could have posed some vulnerabilities.

[FrozenThroneGuy]

A few Cyclone updates: threads and public key skipping if it starts with K leading zeroes, because 71 partial hash generates with 1 max 2 leading zeroes  public key X ccordinates.
https://github.com/Dookoo2/Cyclone

[Nodemath]

A few Cyclone updates: threads and public key skipping if it starts with K leading zeroes, because 71 partial hash generates with 1 max 2 leading zeroes  public key X ccordinates.
https://github.com/Dookoo2/Cyclone

Can u help me achieve my process with ur using modules because I am getting lot errors

My process was decimal to hash 160

[Benjade]

A few Cyclone updates: threads and public key skipping if it starts with K leading zeroes, because 71 partial hash generates with 1 max 2 leading zeroes  public key X ccordinates.
https://github.com/Dookoo2/Cyclone

You have an error in AVX512 version, the 10th key, index 9 will never receive its actual SHA-256 hash but a copy of the previous result (outPtr[8]), so the hash will be incorrect for this key. On line 258.

Code:

sha256_avx512_16B(
    inPtr[0], inPtr[1], inPtr[2], inPtr[3],
    inPtr[4], inPtr[5], inPtr[6], inPtr[7],
    inPtr[8], inPtr[9], inPtr[10], inPtr[11],
    inPtr[12], inPtr[13], inPtr[14], inPtr[15],
    outPtr[0], outPtr[1], outPtr[2], outPtr[3],
    outPtr[4], outPtr[5], outPtr[6], outPtr[7],
    outPtr[8], outPtr[8], outPtr[10], outPtr[11],
    outPtr[12], outPtr[13], outPtr[14], outPtr[15]);

[dextronomous]

the avx512 version give me the error,
./Cyclone -a 128z5d7nN7PkCuX5qoA4Ys6pmxUYnEy86k -r FAC875:6FAC3875 -p 6 -j 10000000 -s
Illegal instruction

[FrozenThroneGuy]

the avx512 version give me the error,
./Cyclone -a 128z5d7nN7PkCuX5qoA4Ys6pmxUYnEy86k -r FAC875:6FAC3875 -p 6 -j 10000000 -s
Illegal instruction

I won’t update it, may be later.

[kTimesG]

Is there a quick math trick to determine if two points share the same higher level endomorphic ring? Just by their X and Y?

By higher-level, I mean 149 or 631. 2 and 3 are obvious. I know that all of the points that share the same subgroup sum up to O. It's also possible to build up the subgroup entirely, but is this really required, in order to make such a check, that the other point is in the same subgroup?

[E36cat]

A few Cyclone updates: threads and public key skipping if it starts with K leading zeroes, because 71 partial hash generates with 1 max 2 leading zeroes  public key X ccordinates.
https://github.com/Dookoo2/Cyclone

Cyclone_avx2# g++ -std=c++17 -Ofast -funroll-loops -ftree-vectorize -fstrict-aliasing -fno-semantic-interposition -fvect-cost-model=unlimited -fno-trapping-math -fipa-ra -fipa-modref -flto -fassociative-math -fopenmp -mavx2 -mbmi2 -madx -o Cyclone Cyclone.cpp SECP256K1.cpp Int.cpp IntGroup.cpp IntMod.cpp Point.cpp ripemd160_avx2.cpp p2pkh_decoder.cpp sha256_avx2.cpp
g++: error: unrecognized command-line option ‘-fipa-modref’

[smracer]

If 100,000 people were independently running RCkangaroo searching for Puzzle 135 with a 4090 each, would theoretically 1 person get lucky and find the key earlier or does RCkangaroo not work that way?

[kTimesG]

If 100,000 people were independently running RCkangaroo searching for Puzzle 135 with a 4090 each, would theoretically 1 person get lucky and find the key earlier or does RCkangaroo not work that way?

Absolutely not. Kangaroo isn't a lottery contest (like a hash brute force search is). It's only reason of being more and more likely to find a solution after some number of operations is because it uses the output of those some number of operations. 100.000 separate guys will need to each compute for themselves those ops before ever reaching equal odds. Not a one-ticket lottery. at all

Some may say this is wrong and anyone can hit the key from the first try.

Totally agree. Just as feasible as having that one-legged unicorn riding a monocycle, on a tight rope, over a volcano, blind-folded, with meteors popping off.

[smracer]

If 100,000 people were independently running RCkangaroo searching for Puzzle 135 with a 4090 each, would theoretically 1 person get lucky and find the key earlier or does RCkangaroo not work that way?

Absolutely not. Kangaroo isn't a lottery contest (like a hash brute force search is). It's only reason of being more and more likely to find a solution after some number of operations is because it uses the output of those some number of operations. 100.000 separate guys will need to each compute for themselves those ops before ever reaching equal odds. Not a one-ticket lottery. at all

Some may say this is wrong and anyone can hit the key from the first try.

Totally agree. Just as feasible as having that one-legged unicorn riding a monocycle, on a tight rope, over a volcano, blind-folded, with meteors popping off.

Thank you.  That is what I thought. So they only way would be to have a central server that is running RCkangaroo periodically saving the progress (in case you have to restart it) and then you would have client GPU's that would connect to the server.

[fixedpaul]

Is there a quick math trick to determine if two points share the same higher level endomorphic ring? Just by their X and Y?

By higher-level, I mean 149 or 631. 2 and 3 are obvious. I know that all of the points that share the same subgroup sum up to O. It's also possible to build up the subgroup entirely, but is this really required, in order to make such a check, that the other point is in the same subgroup?

If I understood your question right, maybe one possible way to simplify the check is to first verify whether both points are n-torsion points (nP = O and nQ = O). If not, you can already rule out that they belong to the same n-order subgroup. If they both are, i think you have to build up the entire subgroup as you said, i don't think there is a direct way to check. But at least the n-torsion check can help "quickly" exclude cases that definitely don’t match, or at least use less computation. Could this make sense?

[analyticnomad]

Is there a quick math trick to determine if two points share the same higher level endomorphic ring? Just by their X and Y?

By higher-level, I mean 149 or 631. 2 and 3 are obvious. I know that all of the points that share the same subgroup sum up to O. It's also possible to build up the subgroup entirely, but is this really required, in order to make such a check, that the other point is in the same subgroup?

Every time one of you guys starts talking about n-torsion points or endomorphic rings I go down a completely new rabbit hole!

it never ends...

[fecell]

Hey has the privatekeys.pw site been down all day for you guys too?

use https://privatekeyfinder.io/bitcoin-puzzle/

[GonzAlex]

Hi everyone. Can anyone share the private key they found in range #71 starting with f6f5431d25bbf7... or less characters.
Thanks for your help!