Bitcoin puzzle transaction ~32 BTC prize to who solves it

[JavaSandcrawler]

Deepseek how do you know bc1qgp48hjxp9uctzysq458dtlhk7ewtf9k4xpjpjj is the creator, their reason for sending 184USD TO #66 is not clear, but you are assuming it is a clue to #67?

Also if it was a clue why ignore the zeros, 0.00189717 = 0.007C553B1ADE27BE0A11   and 0.00010392  =  0.0006CF7D005BC5789A9B

I think you stretching cause as far as I know nobody ever correctly guessed #66 started with 283 , how could they guess #67 but not #66

https://www.talkimg.com/images/2025/02/17/qMGlW.png

I was about 8 hours short of opening 66, I was rummaging through this range that day. It's a shame.
I started with the logarithm 19.666 and didn't get there. Which was found through triangles in AutoCAD.
Sry i use translater.

[karrask]

I have an idea.  the question arises - is it possible to change the bsgs keyhant, namely the random mode, to suit my mathematics? so that the values are not random, but are calculated using a formula.

BSGS requires a public key (e.g. puzzles 135, 140, 145, 150, 155, 160, or a signed message or TX for all other puzzles).

IDK what you're doing but BSGS is fully deterministic, unless you run multiple smaller BSGS over (random) subintervals, or you tradeoff the 100% chances to find the key with a smaller memory footprint (potentially missing the solution). But before all, you should answer yourself why you want to use BSGS in the first place. For 135 there might not be enough storage on our planet to even finish off iterating through the baby steps, let alone enough fast RAM to store all of that, to run through the giant steps. At each giant step, a lookup must be performed in the table, but if you can't have the table (or it is not a fast table, which means that all and any of it needs to be directly accessible instantly, not using any network, cloud, cables, compression, etc.) you're at a dead end.

then can you explain - random values are generated, and each of them is checked for a match using the public key?

maybe you have a version of kangaroo (or brute force) for GPU and CPU (for comparing addresses and public keys), which would iterate in a row with a given step (example 1+1+2+3+4..etc) ?
to make it easier for me to figure out how to change the step logic. I want to test my theory.

[deep_seek]

He was talking about the 66th puzzle. after it was solved.

Okay, my bad. I thought he was talking about predicting the next puzzle, but he was actually referring to Puzzle 66 with some BS. However, I still have my doubts that retired_coder and satoshi_rising are the same person.
This thread spans 369 pages filled with debates, groundbreaking ideas, successes, and failures. It holds a wealth of knowledge—some theories dismissed, others leading to real breakthroughs. Many thanks to those who contributed to this field by developing BitCrack, VanitySearch, keyhunt CUDA, JLP Kangaroo, Sota Kangaroo, Rotor CUDA, wifcudasolver, and many more.

Now, with all this information, it’s time to build something new.

During my reading of more threads, I came across another person's idea that seems very close to the actual method used for creating these puzzles. Here is the post:

https://bitcointalk.org/index.php?topic=4453897.msg61848712#msg61848712

In Python, it's quite straightforward to implement multiple derivation methods for private keys that are masked with leading zeros. I'm curious—has anyone developed a CUDA version of these methods? If so, could you share a reference link?

Thanks in advance...

[brainless]

Nikola Tesla lucky number 369
Page 369 appear before and my post was 369 too and about this lucky number 369
After my post 369 Late night, mod shuffle and maybe removed unused post and page appear 367

[deep_seek]

Nikola Tesla lucky number 369
Page 369 appear before and my post was 369 too and about this lucky number 369
After my post 369 Late night, mod shuffle and maybe removed unused post and page appear 367

So sad Bro... Now my 3rd post becomes 7369 in page no. 369
What does that mean we have to go for above idea ? one more thing when i try to post i got a warning - as a newbie you have to wait 360 seconds for your new post. Lol 

[cctv5go]

At most, I will solve puzzle 67 by September this year. Everyone is waiting for my good news.

[WanderingPhilospher]

Nikola Tesla lucky number 369
Page 369 appear before and my post was 369 too and about this lucky number 369
After my post 369 Late night, mod shuffle and maybe removed unused post and page appear 367

3-6-9, damn she fine
Hoping she can sock it to me one more time
Get low, get low (get low), get low (get low), get low (get low)
To the window (to the window), to the wall (to the wall)
'Til the sweat drop down my balls (my balls)
'Til all these females crawl
'Til all skeet-skeet, mosukcer (mosukcer)
'Til all skeet-skeet, gotdang (gotdang)
'Til all skeet-skeet, mosukcer (mosukcer)
'Til all skeet-skeet, gotdang (gotdang)

[jdx009]

Nikola Tesla lucky number 369
Page 369 appear before and my post was 369 too and about this lucky number 369
After my post 369 Late night, mod shuffle and maybe removed unused post and page appear 367

I'm just leaving a comment here on page 369 - as I was waiting for a long time to see what comes at 369 - I have been so obsessed with this number for a long time.

[frozenen]

Deepseek how do you know bc1qgp48hjxp9uctzysq458dtlhk7ewtf9k4xpjpjj is the creator, their reason for sending 184USD TO #66 is not clear, but you are assuming it is a clue to #67?

Also if it was a clue why ignore the zeros, 0.00189717 = 0.007C553B1ADE27BE0A11   and 0.00010392  =  0.0006CF7D005BC5789A9B

I think you stretching cause as far as I know nobody ever correctly guessed #66 started with 283 , how could they guess #67 but not #66

https://www.talkimg.com/images/2025/02/17/qMGlW.png

I was about 8 hours short of opening 66, I was rummaging through this range that day. It's a shame.
I started with the logarithm 19.666 and didn't get there. Which was found through triangles in AutoCAD.
Sry i use translater.

Could you explain more how you were doing this?

[kTimesG]

Guys, while flipping randomly through my 3 notebooks full of OCD-induced EC investigations, I found on one corner the proof on how ECDSA is broken. Unluckily, my cat made confetti out of the rest of the page, and I don't remember squat about how I anded up at that result. All I could recover is that it has to do with inter-dimensional mathematics, specifically fractal theory.

For those unaware, fractals are a cutting edge area of mathematics, since they behave outside of the normal framework of dimensionality (such as points on a line, 2D or 3D shapes). They were developed in the 80s, so they are actually newer than elliptic curves. Since they were impossible to analyze analiticaly due to their immense complexity and recurrent relations, they only made sense once the first computers started to became mainstream in the science community.

In short, there is a way to map Weierstrass curves over to some fractal (of course, not the same one). The fractal can then be looked up and zoomed-in to analyze a particular range. Depending on the intensity of the particular XY (which is simply the fractal recurrent formula landing there), we can derive back private keys. Unfortunately, while my proof seemed to work out OK, the part of the paper that contained the formula for the mapping was lost (but at least the cat had an happy hour).

Note: this doesn't work with addresses.

[JavaSandcrawler]

Could you explain more how you were doing this?

It will be pointless. 99.9% of ideas are criticized and easily destroyed.
Therefore, I see no point in showing my results, because they are rather based on chance and the dark side. lol.

But for 67 my approach is still the same. I also keep a log of prefixes so I can understand what ranges I've gone through. It makes absolutely no sense, but I just like to see thats prefixes.

I have some good work on random libraries, but not perfect. For example, for 130 bits, my library guesses from 90 to 95 bits in 15-25 seconds.
But for example for 66 from 50 to 58 bits.
Likewise for 40 from 30 to 39 bits.
https://www.talkimg.com/images/2025/02/18/qPIol.png
https://www.talkimg.com/images/2025/02/18/qPsx1.png
https://www.talkimg.com/images/2025/02/18/qPNuo.png

For any prefix 67bits these are the results.
https://www.talkimg.com/images/2025/02/18/qPyCC.png
67, I have absolutely no idea where he might be. I’m probably doing this for the sake of an idea, since bots will instantly change tx. This is of course very disappointing.
Good luck everyone.

[karrask]

Could you explain more how you were doing this?

It will be pointless. 99.9% of ideas are criticized and easily destroyed.
Therefore, I see no point in showing my results, because they are rather based on chance and the dark side. lol.

But for 67 my approach is still the same. I also keep a log of prefixes so I can understand what ranges I've gone through. It makes absolutely no sense, but I just like to see thats prefixes.

I have some good work on random libraries, but not perfect. For example, for 130 bits, my library guesses from 90 to 95 bits in 15-25 seconds.
But for example for 66 from 50 to 58 bits.
Likewise for 40 from 30 to 39 bits.
https://www.talkimg.com/images/2025/02/18/qPIol.png
https://www.talkimg.com/images/2025/02/18/qPsx1.png
https://www.talkimg.com/images/2025/02/18/qPNuo.png

For any prefix 67bits these are the results.
https://www.talkimg.com/images/2025/02/18/qPyCC.png
67, I have absolutely no idea where he might be. I’m probably doing this for the sake of an idea, since bots will instantly change tx. This is of course very disappointing.
Good luck everyone.

share the code))) бpo)

or write to me in private messages, I have a couple of questions.

[Baskentliia]

At most, I will solve puzzle 67 by September this year. Everyone is waiting for my good news.

İMPOSSİBLE BROOO

[Gtsg]

I've been analyzing the puzzle formation algorithm for two years. Just look at this information. After you figure this out, I'm ready for further dialogue.
https://i.postimg.cc/dVHh5k8h/XLS.jpg

It's part of the algorithm, there's another part, but what's the point of it all if the bots take the entire reward?

[singlethread1]

Wondering if anyone can help me with this.

Just curious - how long would it take for the average household desktop computer now in 2025 to get through 50% of the keys for puzzle 67?

I want to know to better explain this puzzle to my family/friends.

Thanks!!

[HABJo12]

Dear Sir first try to know your RAM processor unit quantity your device generation and graphics cards then you can check it by making random code like this 
import ecdsa
import time

def generate_key():
    sk = ecdsa.SigningKey.generate(curve=ecdsa.SECP256k1)
    return sk.to_string().hex()

start_time = time.time()
num_keys = 1000  # number of keys to generate for testing
for _ in range(num_keys):
    generate_key()

end_time = time.time()
speed = num_keys / (end_time - start_time)
print(f"Generated {num_keys} keys in {end_time - start_time:.2f} seconds.")
print(f"Speed: {speed:.2f} keys per second.")             

[frozenen]

Wondering if anyone can help me with this.

Just curious - how long would it take for the average household desktop computer now in 2025 to get through 50% of the keys for puzzle 67?

I want to know to better explain this puzzle to my family/friends.

Thanks!!

#67 range is  73.79 quintillion
#68 range is 147.57 quintillion
#69 range is 295.15 quintillion

73.79 quintillion = 73790000000000000000

your new houshold Pc can prob do 8 Mkeys/s without high end GPU!

so about 292,277 years or  148,639 years for 50% of the range in #67 to answer your question!

[nomachine]

The fact that there are only 2^77 grains of sand on all the beaches in the world helps put the scale of the puzzle into perspective. This is a lot. It's like trying to find a specific grain of sand in a massive desert that’s far bigger than all the beaches in the world combined.

[fecell]

You know, guys, I noticed that if you generate tame and wild correctly, then the kangaroo gets to the target very quickly.
with Python (3.11) on a 40-bit PK, at i5-9300H (2.4Ghz) with 32Gb (used only 20%)... it's only 3 seconds to solve. no any RND was used - only bit range and bit unique rules.
No CUDA, multithreading, and so on - just one-thread calculation.

Code:

[+] Precomputed jump points ready: 33
[+] Puzzle complexity: 40-bit
[+] Search range: 2^39 - 2^40
[+] Distinguished point rarity: 2^10
[+] Distinguished point rarity: 1024
[+] Expected hops: 2^21.09 (2233466)
[+] Tame and wild herds initialized. 00:00:00
[+] Tame and wild points initialized. 00:00:00

[+] PUZZLE SOLVED
[+] Private key (dec): 1003651412950

[+] Total hops: 1144040

[+] Search duration: 00:00:03
[+] Total execution time: 00:00:03

for 135-bit tame's and wild's values needs very long time (about 48h) to fill.

[Akito S. M. Hosana]

for 135-bit tame's and wild's values needs very long time (about 48h) to fill.

You would need approximately 2^68.50 (417402170410649059328) hops to complete this. It’s unrealistic to achieve this within 48 hours, even if you precompute 2^33 hops.

Could you provide us some code to verify these calculations?