Bitcoin puzzle transaction ~32 BTC prize to who solves it

[jonematt]

The solution is available only in periods 6-7. It saves time and effort and eliminates space 4-5


BIP39 Mnemonic : 17 +

 abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon about

Seven words remain to solve the puzzle 17+7 = 24 word

[Akito S. M. Hosana]

Seven words remain to solve the puzzle 17+7 = 24 word

That's approximately 4.79 billion years.

[benjaniah]

Roughly how long would it take to solve puzzle 135 with 8 x RTX-5090 using RCKangaroo, or something similar?

[snes22]

What to do if I found the Public Key and Private Key?

[Geshma]

use mara slipstream

[mahmood1356]

What to do if I found the Public Key and Private Key?

I give you one piece of advice: If you manage to solve the puzzle, do not trust intermediary websites for transferring your funds under any circumstances—this is the only way to keep your assets safe. The best option is to use the Electrum wallet, but make sure to download it from the official website. Then, send the funds like you normally do using the maximum allowed network fee. Don’t pay attention to the nonsense from users who are only trying to mislead others. Good luck!

[dextronomous]

yeah right,

obviously dont use electrum at all only if you want to be playing around with your funds safety, and yes tramsfer trough mara,
many non trustworthy links for electrum wallets, and goodluck.

[Wanderingaran]

What to do if I found the Public Key and Private Key?

Let the bots and thieves take everything. 

[teguh54321]

What to do if I found the Public Key and Private Key?

Let the bots and thieves take everything. 

Good idea 🤪😅

[teguh54321]

I have a question. Each address theoritically can be generated by how many different private key ?

Since h160 is less than the private key hex

[Denevron]

What to do if I found the Public Key and Private Key?

Be happy and transfer the funds to yourself, how to do it correctly - this thread has already been written more than once 


https://bitcointalk.org/index.php?topic=1306983.msg64379149#msg64379149 - Please, it's written here step by step

[kTimesG]

I have a question. Each address theoritically can be generated by how many different private key ?

Since h160 is less than the private key hex

Number of private keys: ~2**256
Number of SHA256 hashes: 2**256
Number of H160 hashes: 2**160

Naive answer: on average around 2**96 private keys for every address (one key in every other 2**160 tries, on average).

Better answer: since SHA256 is uniform, only around 63.4% of all possible SHA hashes will come up after 2**256 hash operations (any given hash has only a 63.4% probability to show up at least once or more). 36% of all possible SHAs won't make it. 28% will show up at least twice, etc.

So, around 0.63 * 2**256 unique SHA hashes go into RIPEMD-160.

Since RIPEMD-160 is also uniform, but the number of hash ops is much larger than 2**160, almost every possible H160 hash is likely to be seen at least once (but the average goes down a tiny little bit from the naive 2**96).

So, final answer: an address may be generated by no private key at all, or by a ton more of private keys than is expected, but the sweet spot of 99.9999% confidence is some small interval around the <2**96 average.

If you plan to find an address collision (this is what your idea sounds like, right?) best of luck. You'll likely need much more than 2**160 hashing ops before you can reach some 99.999% chances to get a collision. That's equivalent to solving Puzzle 71 several billion septillion times, instead of just once. If you take a billion RTX 4090 and start today, it would still take 10**21 years before you get a collision. That's 500 billion times more time than the age of our Universe.

[saeedxxx]

I have a question. Each address theoritically can be generated by how many different private key ?

Since h160 is less than the private key hex

~~ 2^97

[teguh54321]

I have a question. Each address theoritically can be generated by how many different private key ?

Since h160 is less than the private key hex

Number of private keys: ~2**256
Number of SHA256 hashes: 2**256
Number of H160 hashes: 2**160

Naive answer: on average around 2**96 private keys for every address (one key in every other 2**160 tries, on average).

Better answer: since SHA256 is uniform, only around 63.4% of all possible SHA hashes will come up after 2**256 hash operations (any given hash has only a 63.4% probability to show up at least once or more). 36% of all possible SHAs won't make it. 28% will show up at least twice, etc.

So, around 0.63 * 2**256 unique SHA hashes go into RIPEMD-160.

Since RIPEMD-160 is also uniform, but the number of hash ops is much larger than 2**160, almost every possible H160 hash is likely to be seen at least once (but the average goes down a tiny little bit from the naive 2**96).

So, final answer: an address may be generated by no private key at all, or by a ton more of private keys than is expected, but the sweet spot of 99.9999% confidence is some small interval around the <2**96 average.

If you plan to find an address collision (this is what your idea sounds like, right?) best of luck. You'll likely need much more than 2**160 hashing ops before you can reach some 99.999% chances to get a collision. That's equivalent to solving Puzzle 71 several billion septillion times, instead of just once. If you take a billion RTX 4090 and start today, it would still take 10**21 years before you get a collision. That's 500 billion times more time than the age of our Universe.

Haha just imaginary adress collision for experiment 😅.

Btw you always say h160 is always uniform.

But im scanning over 256 trilions keyspace in h160 prefix .  several private key position / range ,  and the result is vary , some even differ bout +10% 🤔.
Or it just a part of a bigger prefix cycle ? 🤔, anyone go on this path ?

[saeedxxx]

I have a question. Each address theoritically can be generated by how many different private key ?

Since h160 is less than the private key hex

Number of private keys: ~2**256
Number of SHA256 hashes: 2**256
Number of H160 hashes: 2**160

Naive answer: on average around 2**96 private keys for every address (one key in every other 2**160 tries, on average).

Better answer: since SHA256 is uniform, only around 63.4% of all possible SHA hashes will come up after 2**256 hash operations (any given hash has only a 63.4% probability to show up at least once or more). 36% of all possible SHAs won't make it. 28% will show up at least twice, etc.

So, around 0.63 * 2**256 unique SHA hashes go into RIPEMD-160.

Since RIPEMD-160 is also uniform, but the number of hash ops is much larger than 2**160, almost every possible H160 hash is likely to be seen at least once (but the average goes down a tiny little bit from the naive 2**96).

So, final answer: an address may be generated by no private key at all, or by a ton more of private keys than is expected, but the sweet spot of 99.9999% confidence is some small interval around the <2**96 average.

If you plan to find an address collision (this is what your idea sounds like, right?) best of luck. You'll likely need much more than 2**160 hashing ops before you can reach some 99.999% chances to get a collision. That's equivalent to solving Puzzle 71 several billion septillion times, instead of just once. If you take a billion RTX 4090 and start today, it would still take 10**21 years before you get a collision. That's 500 billion times more time than the age of our Universe.

I think you forgot that each private key can generate 2 addresses (compressed and uncompressed in legacy version), so I believe that you have to take this fact into account as well.

[kTimesG]

I think you forgot that each private key can generate 2 addresses (compressed and uncompressed in legacy version), so I believe that you have to take this fact into account as well.

True. You're right, the average is 2**96.78 private keys / address.

Btw you always say h160 is always uniform.

But im scanning over 256 trilions keyspace in h160 prefix .  several private key position / range ,  and the result is vary , some even differ bout +10% 🤔.

You're analyzing a grain of sand and making assumptions about the entire beach.

[teguh54321]

I think you forgot that each private key can generate 2 addresses (compressed and uncompressed in legacy version), so I believe that you have to take this fact into account as well.

True. You're right, the average is 2**96.78 private keys / address.

Btw you always say h160 is always uniform.

But im scanning over 256 trilions keyspace in h160 prefix .  several private key position / range ,  and the result is vary , some even differ bout +10% 🤔.

You're analyzing a grain of sand and making assumptions about the entire beach.

So you suggest the entire beach should be uniform ? Howbout  a "beach" compare to the next "beach" eg 10 quantilion keyspace? 😅.  Or there might be something we can use 🤔

I believe there some kind of tiny bias , but still figure out how to use it 🙃

[kTimesG]

So you suggest the entire beach should be uniform ? Howbout  a "beach" compare to the next "beach" eg 10 quantilion keyspace? 😅.  Or there might be something we can use 🤔

I believe there some kind of tiny bias , but still figure out how to use it 🙃

No, the beach isn't more or less uniform than some other beach, or the island that contains it, or the planet that holds the island. Maybe a dog pissed over the sand you're looking at, it doesn't say anything about the whole picture. That is the whole point: randomness. If you flip a coin 2 quadrillion times, it won't get to exactly 1 quadrillion heads and 1 quadrillion tails, even after quadrillions of repeats. If it does end up like that more times than expected, it doesn't make it a fair coin, it makes it a rigged coin (since you can predict the results in your favor).

Even if H160 is rigged (the beach is "uniform", or prefixes are spread out predictably in other words), it would only mean you can rig the input (the SHA256), but then you have another problem: how to find the public key (if one even exists) that hashes to that particular SHA256. And assuming you do find it, you'll then need to find the corresponding private key, which is itself a 2**124 bits problem anyway.

[nomachine]

https://bitcointalk.org/index.php?topic=1306983.msg64379149#msg64379149 - Please, it's written here step by step

Private transaction services represent a potentially complementary or orthogonal migration
strategy that specifically addresses short-range attacks during the window of vulnerability
between transaction broadcast and confirmation. These services, such as existing private
mempools like MARA’s Slipstream [MDH24], allow users to submit transactions directly to
trusted miners rather than broadcasting them publicly, preventing quantum adversaries
from observing and hijacking transactions before confirmation. While this approach doesn’t
eliminate the need for protocol-level quantum-resistant signature schemes, it could reduce,
but not eliminate, risks during the transition period after such schemes are implemented.

from --> https://chaincode.com/bitcoin-post-quantum.pdf

From the report, private transaction services like MARA's Slipstream let users send transactions directly to trusted miners instead of broadcasting them publicly. This helps prevent quantum adversaries from seeing and hijacking transactions before they're confirmed. But it doesn't replace the need for quantum-resistant signatures at the protocol level. However, during the transition period after implementing those signatures, this approach can reduce risks. The text also mentions that these services are contentious now because they're used for non-financial data storage and create a two-tier system. This is supposed to help against quantum attacks because a quantum computer could potentially steal the transaction details (like the signature) and create a conflicting transaction if they can see it in the mempool before confirmation.

[Akito S. M. Hosana]

From the report, private transaction services like MARA's Slipstream let users send transactions directly to trusted miners instead of broadcasting them publicly. This helps prevent quantum adversaries from seeing and hijacking transactions before they're confirmed. But it doesn't replace the need for quantum-resistant signatures at the protocol level. However, during the transition period after implementing those signatures, this approach can reduce risks. The text also mentions that these services are contentious now because they're used for non-financial data storage and create a two-tier system. This is supposed to help against quantum attacks because a quantum computer could potentially steal the transaction details (like the signature) and create a conflicting transaction if they can see it in the mempool before confirmation.

Will all mempools become like MARA?