Bitcoin puzzle transaction ~32 BTC prize to who solves it

[Evillo]

Let's do it one more time. With that much money at stake, we should grow a huge community.

Everyone thinks he will be the lucky one and prefers to avoid having to share the X10 is made to encourage people to pool their power to demonstrate that even with a collective effort the 66 67 68 and 69 will be difficult to obtain otherwise it won't find away
Also give a good boost to this puzzle which is now 8 years old

With collective effort, we can go as far as 90 bits. Then from there, Quantum code should step in. But that's another level.

[1714644772]

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[Evillo]

There's the math comes again,

66 Bit puzzle  


the total number of keys in this range can be calculated by taking 16 to the power of 17 (i.e., 16^17), which is equal to 295,147,905,179,352,825,856.

So, there are a total of 295,147,905,179,352,825,856 keys in the given range.

If you have a speed of 343 million keys per second (Mkeys/s), the result's,

Total number of keys = 295,147,905,179,352,825,856
Time = Total number of keys / Speed

Substituting the values, we get:

Time = 295,147,905,179,352,825,856 / 343,000,000
Time = 859,488,566,568 seconds

Therefore, it would take approximately 859,488,566,568 seconds (or about 27,247 years) to generate all the possible keys in the given range at a speed of 343 million keys per second. This is a very long time and may not be practically feasible.

what if i have 5000 mkeys to scan those number of key ?
it would take approximately 59,029,581,036 seconds (or about 1,872 years) to generate all the possible keys in the given range at a speed of 5000 million keys per second, still lame haha  


and there's other scenario to win the ticket haha

For a 66-bit puzzle, the key range would be from 0 to 2^66 - 1, which is:

0 to 7,922,816,251,426,433

The number of keys in this range is:

2^66 = 73,786,976,294,838,206,464

At a scanning speed of 343 Mkey/s, the time it would take to scan this entire range would be:

Time = Number of Keys / Keys per Second

Time = (2^66) / 343000000 = 215625224 seconds
= 3593754 minutes
= 59896 hours
= 2495 days
≈ 6.84 years

So, it would take approximately 6.84 years to scan the entire key range of a 66-bit puzzle at the given scanning speed.

If we were to randomly hit the private key in the middle of the range, then we would only need to scan half of the key range, which is:

2^65 = 36,893,488,147,419,103,232

Using the same formula as above, the time it would take to scan half of the key range would be:

Time = (2^65) / 343000000 = 107812612 seconds
= 1796877 minutes
= 29948 hours
= 1248 days
≈ 3.42 years

So, if we randomly hit the private key in the middle of the key range, it would take approximately 3.42 years to scan half of the key range at the given speed of 343 Mkey/s.


btw, my close friend is neat, he's building something on ASICs L3+ to give a proof to scan the puzzle for mean time haha.
let's see..

Wrong, the entire range of puzz #66 contains 36.8 million trillion keys, half that range is obviously 18.44 million trillion keys.

Also the calc is misleading. With 1billion keys/sec, you need a little over 1000 years to scan the entire range sequentially.

[ineedhelpman]

"Pollard's rho algorithm" someone sent me this as a way to go at 66 is a way.

this is what ive kinda read up on so far about it. for anyone wondering if they don't know read below.


In the paper "A monte carlo method for factorization", Pollard introduces a method for factoring composite numbers that is better than trial division. The main idea is to generate random numbers and test if the difference between any pair has a non-1
 divisor in common with the number to be factored, which exploits the birthday paradox to find factors faster than trial division.

In his paper, Pollard uses the formula: xi+1≡x2i−1modn
, to generate random numbers, where n
 is the number that we want to factor and x0=2
. He notes that any polynomial of degree ≥2
 can be used.

It intuitively makes sense to me that linear functions can have additional properties that cause the generated sequence to not be random enough. However, I see no clear reason why the algorithm would only work for polynomials. Using different functions, for example with bit 2^64 manipulations and storing intermediate results modulo 264
 could potentially make the algorithm faster on real computers.


edit: how possibly hard could 66 be with a decent group.

[zahid888]

WOW!
Somebody (maybe the owner) increased the unsolved puzzles prizes again by x10 😱
Now the puzzle #66 prize is 6.6 BTC, #67 is 6.7 BTC... and so on .... puzzle # 160 prize is 16 BTC now
👍🏼🥳

OMG !!! 😱😱😱😱😱 Yesterday, I was thinking that I should give up now, but this has turned into a great motivation. need only one message from the puzzle creator in this situation, it would be the icing on the cake 👍🏼 👍🏼

[ineedhelpman]

WOW!
Somebody (maybe the owner) increased the unsolved puzzles prizes again by x10 😱
Now the puzzle #66 prize is 6.6 BTC, #67 is 6.7 BTC... and so on .... puzzle # 160 prize is 16 BTC now
👍🏼🥳

OMG !!! 😱😱😱😱😱 Yesterday, I was thinking that I should give up now, but this has turned into a great motivation. need only one message from the puzzle creator in this situation, it would be the icing on the cake 👍🏼 👍🏼

RIGHT!!!!!!! the man the myth the legend "the creator" would you kindly please give us a hint or maybe not a tool but a path to go for the ones who want to chase mr 125 -160 but your indeed right the cracking is to us everyone gave up a while back i feel like he brought the attention maybe this time to do as he said in the beginning test the ' community.'

[Evillo]

WOW!
Somebody (maybe the owner) increased the unsolved puzzles prizes again by x10 😱
Now the puzzle #66 prize is 6.6 BTC, #67 is 6.7 BTC... and so on .... puzzle # 160 prize is 16 BTC now
👍🏼🥳

OMG !!! 😱😱😱😱😱 Yesterday, I was thinking that I should give up now, but this has turned into a great motivation. need only one message from the puzzle creator in this situation, it would be the icing on the cake 👍🏼 👍🏼

RIGHT!!!!!!! the man the myth the legend "the creator" would you kindly please give us a hint of some sort a tool tool to look into something anything

Lol he did more than enough. The cracking method is on us to figure out.

[ineedhelpman]

uncomprewssed key for 66 1Pir2jKv35rpYgUh8dUxNL6JTTm6d7HX6U


Therefore, the uncompressed public key for the P2PKH address "1Pir2jKv35rpYgUh8dUxNL6JTTm6d7HX6U" is:  04ce42244bfa7731d7088ee2f20a72b0dd834a723d2b8f563bc58cfa5d5dc4911a2f8aae5c5b




could be wrong could be wright do what you want with the info im just tossing ideas.



does this look like itd do anything am i in the right path?

Code:

import hashlib
import base58

# uncompressed public key
public_key = "04ce42244bfa7731d7088ee2f20a72b0dd834a723d2b8f563bc58cfa5d5dc4911a2f8aae5c5b"

# step 1: compute SHA-256 hash of public key
sha256 = hashlib.sha256(bytes.fromhex(public_key)).digest()

# step 2: compute RIPEMD-160 hash of SHA-256 result
ripemd160 = hashlib.new("ripemd160")
ripemd160.update(sha256)
hash160 = ripemd160.digest()

# step 3: add version byte (0x00 for MainNet or 0x6f for TestNet) to hash160
version = "00"
hash160_version = version + hash160.hex()

# step 4: compute double-SHA-256 checksum of hash160_version
checksum = hashlib.sha256(hashlib.sha256(bytes.fromhex(hash160_version)).digest()).digest()

# step 5: append the first 4 bytes of the checksum to hash160_version
btc_address_hex = hash160_version + checksum.hex()[:8]

# step 6: encode the result using Base58Check encoding
btc_address = base58.b58encode(bytes.fromhex(btc_address_hex)).decode()

print("BTC public address: ", btc_address)

[NotATether]

Just converting a public key to an address will not help here, as there are already specialized programs in C++ and other faster, multithreaded languages, that can do that.

You don't need to compute base58 or bech32 in order to compare the addresses. Comparing the address hashes directly also works and saves an extra (expensive) computation, such that most of the work is pooled into making an efficient SHA256 algorithm OR optimizing the process of making private keys from public keys.

Probably time for me to stop by Cornell university's internet arxiv for some more finite field research papers.

uncomprewssed key for 66 1Pir2jKv35rpYgUh8dUxNL6JTTm6d7HX6U

Therefore, the uncompressed public key for the P2PKH address "1Pir2jKv35rpYgUh8dUxNL6JTTm6d7HX6U" is:  04ce42244bfa7731d7088ee2f20a72b0dd834a723d2b8f563bc58cfa5d5dc4911a2f8aae5c5b

could be wrong could be wright do what you want with the info im just tossing ideas.



does this look like itd do anything am i in the right path?

Code:

import hashlib
import base58

# uncompressed public key
public_key = "04ce42244bfa7731d7088ee2f20a72b0dd834a723d2b8f563bc58cfa5d5dc4911a2f8aae5c5b"

# step 1: compute SHA-256 hash of public key
sha256 = hashlib.sha256(bytes.fromhex(public_key)).digest()

# step 2: compute RIPEMD-160 hash of SHA-256 result
ripemd160 = hashlib.new("ripemd160")
ripemd160.update(sha256)
hash160 = ripemd160.digest()

# step 3: add version byte (0x00 for MainNet or 0x6f for TestNet) to hash160
version = "00"
hash160_version = version + hash160.hex()

# step 4: compute double-SHA-256 checksum of hash160_version
checksum = hashlib.sha256(hashlib.sha256(bytes.fromhex(hash160_version)).digest()).digest()

# step 5: append the first 4 bytes of the checksum to hash160_version
btc_address_hex = hash160_version + checksum.hex()[:8]

# step 6: encode the result using Base58Check encoding
btc_address = base58.b58encode(bytes.fromhex(btc_address_hex)).decode()

print("BTC public address: ", btc_address)

[digaran]

Using tools will take much more time, this needs to be solved by math and by hands.😉

[Gianluca95]

Using tools will take much more time, this needs to be solved by math and by hands.😉

Agree with you, but how can you explain maths behind this puzzle? It is really hard and this will cost to you a lot of headache while you find a solution (if it exists). As I can see, sequence seems to be casual, there isn't any

math behind it. The only thing to do if you don't want to use tools, is to try with bit 0 and 1 and be lucky.

[ProjectSe7en]

id love to help in any way i have 3 3070ti gpus i would constintly run 24/7 for someone if you can push me in the right direction if youd want to split the price if the chance came one of hit. in conclusion what im saying is im a script kiddie who cant code worth shit but this is too apealing not to learn how or help atleast. under the correct cicomstances id be willing to basically give 24\7 acess to my computing power "probably not much" to help in this if anyone wants to help let me know.P.s ive tried kangaroo its too confusing i cant even get it to work thats how "not hip " i am lol... best of luck to all though ive been reading this thread and multipule others about it for a few months now with the right teacher i feel like i could be somewhat of help.

Hello, I've been following this thread for a year. I stocked all of them, including many archives that were deleted from the internet, along with their source code.

I just came up with a new idea and I need a little help with it.

Do you have fast point extraction algorithm for python ?

Thanks..

Not: sorry for my bad english, i don't know english i use google translate

Guys! Add me to discord, we can work together.
I'm working with python, C++ and Rust.
Need help with some coding functions, and more power will really help.

 

DaMonsterAJ#4341


Alberto igual podrías escribirme de favor! hay funciones y experiencia que tú tienes que nos pueden beneficiar.


 

[albert0bsd]

Alberto igual podrías escribirme de favor! hay funciones y experiencia que tú tienes que nos pueden beneficiar.

 

Hi, there is no need to write in Spanish i can read English without need to translate it, so if you want to speak in Spanish you can do it in PM or telegram.

Regards

[WanderingPhilospher]

WOW!
Somebody (maybe the owner) increased the unsolved puzzles prizes again by x10 😱
Now the puzzle #66 prize is 6.6 BTC, #67 is 6.7 BTC... and so on .... puzzle # 160 prize is 16 BTC now
👍🏼🥳

Well, all the remaining addresses received a total of 872.2 BTC, so this challenge is getting damn huge. 

https://blockchair.com/bitcoin/transaction/12f34b58b04dfb0233ce889f674781c0e0c7ba95482cca469125af41a78d13b3

I guess time to fire up the GPUs and make an AMD / Xilinx build of these solvers

Have you tried messing with a stride function with vanitysearch? If I had that, I think I could solve 125 faster than kangaroo...

I need to implement GPU stride function for vanitysearch/keyhunt-cuda

[ineedhelpman]

has anyone tried this website or pulling data looks like they use the stride method as long as a data saver

561,301,131,286 so far keys scanned on 66 on this maybe its not trustworthy but i think they are heading in the right direction. it also allows u to change your stride if ur computing power can handle it.

https://privatekeys.pw/scanner?mode=combined&startKey=0000000000000000000000000000000000000000000000020000000000000000&stopKey=000000000000000000000000000000000000000000000003ffffffffffffffff&addrType=compressed&bitcoin

[s.john]

I think the puzzle creator should reveal the public keys for all the keys bigger than 120bit

[algorithm32]

WOW!
Somebody (maybe the owner) increased the unsolved puzzles prizes again by x10 😱
Now the puzzle #66 prize is 6.6 BTC, #67 is 6.7 BTC... and so on .... puzzle # 160 prize is 16 BTC now
👍🏼🥳

new methods are needed

[Evillo]

has anyone tried this website or pulling data looks like they use the stride method as long as a data saver

561,301,131,286 so far keys scanned on 66 on this maybe its not trustworthy but i think they are heading in the right direction. it also allows u to change your stride if ur computing power can handle it.

https://privatekeys.pw/scanner?mode=combined&startKey=0000000000000000000000000000000000000000000000020000000000000000&stopKey=000000000000000000000000000000000000000000000003ffffffffffffffff&addrType=compressed&bitcoin

Too slow, an average gpu can finish scanning these keys in 30 minutes. But the website is popular and helpful enough in other areas

I think the puzzle creator should reveal the public keys for all the keys bigger than 120bit

Agreed, conventional private key cracking won't work on these high bits. Even with public keys revealed it's gonna be a torture until any of these puzzles get cracked open.

[algorithm32]

Agreed, conventional private key cracking won't work on these high bits. Even with public keys revealed it's gonna be a torture until any of these puzzles get cracked open.

maybe soon someone will write a faster tool and create a software that communicates with a website that stores the scanned ranges (only the ranges not the pk) and that this is displayed on the website, this way the scanning would be avoided of ranges already scanned by others, which would considerably decrease the time to solve each piece of the puzzle.
obviously with the code on github for assembly and transparency.

[Evillo]

Agreed, conventional private key cracking won't work on these high bits. Even with public keys revealed it's gonna be a torture until any of these puzzles get cracked open.

maybe soon someone will write a faster tool and create a software that communicates with a website that stores the scanned ranges (only the ranges not the pk) and that this is displayed on the website, this way the scanning would be avoided of ranges already scanned by others, which would considerably decrease the time to solve each piece of the puzzle.
obviously with the code on github for assembly and transparency.

If we construct the pool, this will be done. I've seen it in older pools.

[digaran]

So what was the cost of sequential search for #66?, was it more than 6.6 BTC? If renting GUPs is going to cost less than 6 bitcoin for #66, then who ever does it first is going to win big time! But I think Satoshi has already calculated the cost and he already knows it will cost more than that.

Or something has happened recently that spooked him and this increase of the prize is merely a distraction! Lol.

I'm only good at making up conspiracy theories. Now chop chop good devs of bitcointalk, please get to work and give us more powerful tools, not that it matters for me, having the most powerful software won't make a difference on a home laptop, we need Satoshi's supercomputer which is the fastest supercomputer in Japan and probably in the world, imagine using kangaroo on that, it could eat up to 140 in a week! 😉