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Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 19. (Read 245815 times)

hero member
Activity: 714
Merit: 1010
Crypto Swap Exchange
The amount of loudmouths and stupidity that has surfaced here recently is quite alarming. I don't know what some want to achieve to drive off users who actually achieved some success. But hey, you do yours and I'm quite confident that all the loudmouths simply won't achieve any progress or success here.

Just put the bullshitters on your ignore list. It's highly unlikely they'll contribute anything positive to this thread. Cleans the mess here to an acceptable level and makes following real and good ideas here far less annoying.

Don't waste your time to attack me, you're likely on my ignore list or earn a spot on it anyway.
newbie
Activity: 19
Merit: 0
That's correct. The more GPUs you have, the higher the chances of collisions due to the Birthday Paradox and the known key range. But I still don't believe 130 bits in 2 months  Grin
member
Activity: 503
Merit: 38
Even with multiple optimizations, doubling the speed of the Pollard's Kangaroo algorithm (e.g., achieving a 2x or 3x speedup) is achievable, but it's unlikely to result in a 600x increase in speed.

Theoretical limitations exist due to the inherent nature of the discrete logarithm problem and the algorithm's structure, which do not allow for drastic speed improvements without significant breakthroughs in mathematics or cryptography.

Using 600 GPUs, while providing a large parallel processing advantage, doesn’t change the size of the search space or the inherent difficulty of the problem itself.

No one will ever surpass Elon Musk in the number of GPUs he uses; he is a world record holder. Grin
newbie
Activity: 19
Merit: 0
Lol... RetiredCoder just came here to beg for 100 merit points and show off his money. That’s all he accomplished. We don’t need anything from him, nor do we need to waste time learning his slow codes from part 1, 2, or 3.

Anyone using the slowest code with 100+ GPUs will obviously become fast, and if that’s what he calls an "OPTIMIZED VERSION," then he’s living in a delusion. There are plenty of developers who could easily solve these puzzles if they had access to 100-500+ GPUs—nothing surprising about that.

As we’ve already said, if Elon Musk wakes up tomorrow with a whim to solve 135 or 140, he will. The real challenge isn’t throwing money at GPUs but finding a cheap and efficient algorithm capable of solving it. That's where the true magic lies.

The reason we call it a puzzle-solving community is simple: it's not a place for people to beg for merit points, flaunt their wealth, or brag about "optimized" codes without backing it up. It’s for individuals who genuinely help each other and contribute to collective progress.

Take JLP as an example—he openly shared one of the most effective codes, helping developers and pushing the community forward. That’s the kind of spirit this space is meant for, not empty boasting or selfish motives. If you're here for anything else, maybe you're in the wrong place.

?
Activity: -
Merit: -
Question for everyone.

If you find a effective method to complete 135 are you going to share your method/code ?
We have known for years that if you have money you have a head start in these puzzles if you bothered to read from page 70 odd if not before that.

No one here is going to hand you that.

snippets yes.

Me personally i wouldn't.
I would tip a few people in this forum for the intel posted if that's what helped me other than that why am i putting others in a million dollar race when i have a head start.Huh

Its a dog eat dog world full of delusional people.


Isn't this a super easy task, to test? Give me the same program, and I will run it with a GPU and then with a CPU, and let's see which solves the key first. Let's make it an 80 bit range. 1 GPU versus a single core, or do you want to use as many cores as the CPU has? Any bets on which one finds the key first?

80bit yes, because you can use low DP. For 120-130 and higher - you have to use high DP and you will get big overhead with JLP. And it's not easy to test 120-130bit range to confirm Smiley

Also, RetiredCoder, make mods to the program, to create less "kangs" when using a GPU, if it's to crazy for you...it's super easy to do. And another question, how does the speed of "kangs", impact the finding of High DP bits. Does a CPU (which the individual kangs are faster) find high DP bits, faster? Or does the GPU's slow, but many, find more, DP bits, faster?
And the last question, which "high puzzles" have you solved and what did you use to solve (CPU, GPU, DP, etc)

Yes I started with JLP code a long time ago, but after some time I got enough experience and I created my own software from scratch - much faster and efficient. So it's not a mod.


Why are your feelings getting hurt? Are you the puzzle solver? Because to me, it sure looks like you’re RetiredCoder. A guy who solves his own mini-puzzle for attention—what more can we expect from him? A guy who supposedly holds a world record but shares code with the public that’s slower than what is already available—what can we expect from him? A millionaire who begs for 100 merit points despite having millions of dollars—what can we expect from him?

If he doesn’t want to share the code, he should just be a man and say, 'I don’t want to share it,' instead of teasing us with 'I’ll share the GPU version in Part 3.' What are we supposed to do with his Part 1, 2, and 3 if he never actually shares his method?

People like JLP have earned respect because they openly shared code that was effective at the time and helped many developers. Meanwhile, RetiredCoder himself said he started with JLP’s code. But where’s the credit to JLP? Instead, we get empty promises and slower code. What can we expect from someone like that?




You think I'm retiredcoder  Cheesy Cheesy Cheesy Cheesy Cheesy Cheesy
I can tell you 100 fact i am not the person.
I'm a broke ass working a crappy 40 hour week i got lucky with the mini puzzle that's all simple luck and saw the post early enough and using my brain.
I'm certainly not sitting on millions i wish i was life would be great but sitting on my ass getting hand out code to solve 500k plus puzzles is just not going to happen.

Maybe the whole point retiredcoder is doing with part 1 part 2 part 3 is making people learn ? Ever think that.
You come across as just wanting a handout for doing nothing.

My feelings are hurt ? probably the fact that the guy/women that's solved the most difficult puzzles to date actually came forward (not that they owe any explanation to anyone on this forum) and shared info.
Real technical info and then is meet would a bunch of babies thinking that he must give up his code or he owes them a explanation.

Now after the backlash you think anyone else who solves any puzzles will bother to share methods or details ?

Anyways thanks for the laugh in thinking im a multi millionaire if you want we can put a 1btc bet on ? i mean i dont even have 0.000000000000000000000000000000001 btc but hey it will be a win win bet with a delusional person  Wink

newbie
Activity: 68
Merit: 0
The issue I need to explain is that the reason I wrote the first 2 and then changed it to 4 is because there are 2 more in the hex codes starting with 5.... I didn't want to write them at first because I knew they were out of range.

Okay, has anyone else found "1BY8GQbnueY"?

I'm wrong in my calculations. (Maybe it's true.)
If there's something different than your base58 or Ripemd calculations?

Standard written calculations, space void, etc. are all created by mathematical theorems.

Yes, sometimes silence is the best.

cctv5go brother, I promise I have a gift for you when I find the wallet.

If anyone else finds "1BY8GQbnueY", please let me know, I'll have a gift for them too.

Then I'll share the Theorem with a few people like Alberto and Zielar who are serious about it.

why you always bring Alberto and Zielar??? look like you need legitimates or you not confidence.
I know someone from telegram group that declare he is a math expert from turky and he has method same with you. maybe you're same person, btw goodluck.

The reason I said Ziealer is because of the articles he wrote during the 52nd wallet period in 2017, the topics he opened and the importance he gave to togetherness. (Except for Chris' TTD sales system.) Someone who gave importance to the togetherness system was betrayed by stupid thieves who thought they were smart.

Alberto has many reasons for why I said it.
Again, like many people, I learned about the Pubkey incident thanks to Alberto at that time. Like the kangaroo system etc. (But I didn't look into it in detail. Because I was just doing research at that time.)


By the way, I don't use Telegram.
newbie
Activity: 68
Merit: 0
Standard written calculations, space void, etc. are all created by mathematical theorems.

Yes, sometimes silence is the best.

cctv5go brother, I promise I have a gift for you when I find the wallet.

If anyone else finds "1BY8GQbnueY", please let me know, I'll have a gift for them too.

Yeah, let's all stop whatever we're doing and find the hundreds (thousands?) prefixes for you, shall we.

I don't think you understand that they can only be found if you actually scan the entire range, which is what address-only puzzles are about. Which makes your entire point/idea/whatever useless.

You're the typical case that happens here whenever BTC is rising:

- step 1: you found some glitch / bug / hidden secret / magic theorem in some ECC / hash function / solved puzzles / whatever
- step 2: you "think/believe/guess/attempt to prove" that what you found is valid, though it is not
- step 2.1: add some random people names into the mix even though they most likely don't give a shit about your idea
- step 3: brag about how you're gonna solve it tomorrow / next month / whenever
- step 4: disappear after a while since everyone ignores you; puzzle never gets solved.


In our country, everything is given to those who deserve it, and no value is given to those who don't.

There are many people here who claim to know how to code and program.
Those who see themselves in space because they have coded the articles they read. Like (kTimesG) example.

kTimesG, nothing is asked of you.

Step 1 - read some ECCs/hash functions.
Step 2 - you think what you know is valid.
Step 3 - I probably don't care about your opinion or comment.
Step 4 - brag that you will solve it tomorrow/next month/whenever. (You didn't have to explain yourself.)
Step 5 - disappear after a while because I and a few others ignore you.
Step 6 - Keep being jealous and thinking highly of yourself. Now keep reading the articles.
member
Activity: 165
Merit: 26
Standard written calculations, space void, etc. are all created by mathematical theorems.

Yes, sometimes silence is the best.

cctv5go brother, I promise I have a gift for you when I find the wallet.

If anyone else finds "1BY8GQbnueY", please let me know, I'll have a gift for them too.

Yeah, let's all stop whatever we're doing and find the hundreds (thousands?) prefixes for you, shall we.

I don't think you understand that they can only be found if you actually scan the entire range, which is what address-only puzzles are about. Which makes your entire point/idea/whatever useless.

You're the typical case that happens here whenever BTC is rising:

- step 1: you found some glitch / bug / hidden secret / magic theorem in some ECC / hash function / solved puzzles / whatever
- step 2: you "think/believe/guess/attempt to prove" that what you found is valid, though it is not
- step 2.1: add some random people names into the mix even though they most likely don't give a shit about your idea
- step 3: brag about how you're gonna solve it tomorrow / next month / whenever
- step 4: disappear after a while since everyone ignores you; puzzle never gets solved.
newbie
Activity: 15
Merit: 0
The issue I need to explain is that the reason I wrote the first 2 and then changed it to 4 is because there are 2 more in the hex codes starting with 5.... I didn't want to write them at first because I knew they were out of range.

Okay, has anyone else found "1BY8GQbnueY"?

I'm wrong in my calculations. (Maybe it's true.)
If there's something different than your base58 or Ripemd calculations?

Standard written calculations, space void, etc. are all created by mathematical theorems.

Yes, sometimes silence is the best.

cctv5go brother, I promise I have a gift for you when I find the wallet.

If anyone else finds "1BY8GQbnueY", please let me know, I'll have a gift for them too.

Then I'll share the Theorem with a few people like Alberto and Zielar who are serious about it.

why you always bring Alberto and Zielar??? look like you need legitimates or you not confidence.
I know someone from telegram group that declare he is a math expert from turky and he has method same with you. maybe you're same person, btw goodluck.
newbie
Activity: 68
Merit: 0
The issue I need to explain is that the reason I wrote the first 2 and then changed it to 4 is because there are 2 more in the hex codes starting with 5.... I didn't want to write them at first because I knew they were out of range.

Okay, has anyone else found "1BY8GQbnueY"?

I'm wrong in my calculations. (Maybe it's true.)
If there's something different than your base58 or Ripemd calculations?

Standard written calculations, space void, etc. are all created by mathematical theorems.

Yes, sometimes silence is the best.

cctv5go brother, I promise I have a gift for you when I find the wallet.

If anyone else finds "1BY8GQbnueY", please let me know, I'll have a gift for them too.

Then I'll share the Theorem with a few people like Alberto and Zielar who are serious about it.
member
Activity: 165
Merit: 26
Quote
I disagree with you. Firstly, puzzle #67 has a range width of 66 bits.
The second is that 1BY8GQbnueY has a length of 56 bits and accordingly in the entire range of 66 bits there are approximately 1024 such combinations.

1BY8GQbnueY How did you know it was 56 bit?

You lack proper understanding of things, every char in base58 encodes, well, one of 58 bits values.

So 10 base58 chars can encode at most 58**10 values. This is a rough estimate without even needing to bother going deeper to binary or even hex.

2**66 / 58**10 is around 171, and Etar was even more precise and correct with his calculus, whether you like it or not. Nowhere near your "correct answer is 4", which is an illusion and correct for you only.

Sorry to smash your nice dreams of finding a hash collision 250x faster than mathematically possible.
newbie
Activity: 7
Merit: 1
Hey there, how can the "unfortunate event" have place? There's something I'm not getting about it. Could you please explain to me how could the sniper get the private key and snatch all the money on the address?

Look at TX (the real solver)
{
  https://www.blockchain.com/explorer/transactions/btc/8c8ec6b3511c62500ea9b3a1c30ca937e15d251b55d30290a2a6da2f1124f3fb
}

It is safe to assume once he found the key, he accessed his wallet using electrum (or similar) and swept the #66 address (13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so)
{
  You can sweep a key with the WIF -> p2pkh:KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZfFoWMiwBt943V7CQeX
}

Now, that is a massive mistake, since there were 34 UTXO and only 2 actually worth taking.
{
 1) 0.066 BTC
 4) 5.940 BTC
}

About a minute later this TX shows up.
{
  https://www.blockchain.com/explorer/transactions/btc/57a88f47e4c047740b782a5562fca143ce85de0373cbff3a7d406e9ae7fc2f5f
}

Now, this one looks scripted.
{
 A) Only the highest UTXO was selected (5.940 BTC)

 B) The fee offered was 76,395 sats (+50K sats than first TX)
}

So miners dropped the first TX and replaced with the second TX, since it gave them more sats per byte.

The idea here is once you broadcast a TX, the pubkey shows up in the mempool.

Once the bot had the pubkey, it found the secret number in seconds.

And the problem with #66 is the secret number is very very small [46346217550346335726]

And this can possibly happen to keys all the way to #129, assuming the bot has the required precomputed DPs.

This is a theory, you have to do a lot of research on the topic if you want the real answer to #66.
newbie
Activity: 23
Merit: 0
Quote

So I guess everyone learned the answer to the question I wrote before,
How many 1BY8GQbnueY starts are there in our 67 bit range? The correct answer is 4. Wink
where does 4 come from if there were 2? Grin
it will take another couple of days and it turns out that it is no longer 2 or 4, or even 16, but 21, or maybe more))))

Yes, you are very funny, really. Smiley

Now go put on your glasses. Cheesy Then dream. Then wake up in the morning and see if they are still there.

Quote
I disagree with you. Firstly, puzzle #67 has a range width of 66 bits.
The second is that 1BY8GQbnueY has a length of 56 bits and accordingly in the entire range of 66 bits there are approximately 1024 such combinations.

1BY8GQbnueY How did you know it was 56 bit?


[/quote]
I remember that it was originally written 2, after you corrected it to 4. I will monitor the increase)))
sr. member
Activity: 652
Merit: 316
1BY8GQbnueY How did you know it was 56 bit?
If you look at the RMD hash (739437bb3dd6d1 9 83e66629c5f08c70e52769371), you will see that for your combination 1BY8GQbnueY 14 hex numbers are used, plus one more digit from 4 to a. In total, this gives 2^56 bits.
And how did you come to the conclusion that in the 66-bit range there are only 4 combinations?
member
Activity: 348
Merit: 34
What will happen if
#135 #140 #145 #150 #155 #160 appear tx made
Listed addresses will remain , or next broken wif will appear as addon ?
newbie
Activity: 68
Merit: 0
Quote
where does 4 come from if there were 2? Grin
it will take another couple of days and it turns out that it is no longer 2 or 4, or even 16, but 21, or maybe more))))

Yes, you are very funny, really. Smiley

Now go put on your glasses. Cheesy Then dream. Then wake up in the morning and see if they are still there.

Quote
I disagree with you. Firstly, puzzle #67 has a range width of 66 bits.
The second is that 1BY8GQbnueY has a length of 56 bits and accordingly in the entire range of 66 bits there are approximately 1024 such combinations.

1BY8GQbnueY How did you know it was 56 bit?

sr. member
Activity: 652
Merit: 316
So I guess everyone learned the answer to the question I wrote before,
How many 1BY8GQbnueY starts are there in our 67 bit range? The correct answer is 4. Wink
I disagree with you. Firstly, puzzle #67 has a range width of 66 bits.
The second is that 1BY8GQbnueY has a length of 56 bits and accordingly in the entire range of 66 bits there are approximately 1024 such combinations.
newbie
Activity: 23
Merit: 0
1BY8GQbnueYofwSuFAT3USAhGjPrkxDdW9
1BY8GQbnueYebq5d6CE1wDfbdAWWy33ZyW
HEX:7545bf10859946eca
WIF:KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qbRyCkJQmfJhcqQ1bT4q

Thank you very much for this information. cctv5go

Thank you for the 1BY8GQbnueY that I want to include in my own system.


Quote

lets make me clear, did you mean after your script found prefix 1BY8GQbnueY then automatically jump to the next prefix 1BY8GQbnueY?

A similar example.

1BY8GQbnueY how many do you think are in our 67 bit range?
1BY8GQbnue how many do you think are in our 67 bit range?

I found the answers to these in part.

So I confirmed these with hardware that was not too high, but my range is still far but not as far as space. Smiley

So I guess everyone learned the answer to the question I wrote before,
How many 1BY8GQbnueY starts are there in our 67 bit range? The correct answer is 4. Wink
where does 4 come from if there were 2? Grin
it will take another couple of days and it turns out that it is no longer 2 or 4, or even 16, but 21, or maybe more))))
newbie
Activity: 0
Merit: 0
If I were the finder, I would've moved the three UTXOs worth 0.066, 0.594 and 5.94BTC in a private transaction via slipstream.mara.com ..
could someone show how to make a raw transaction for 67? it can be created without privatekey, are there any services for creating raw transactions, or using bitcoincore? to spend all outputs on one address with the required commission. so that I can then sign it and send it via MARA

ps: I haven't found the private key yet, but if I do I'll thank you for this tip))

If you find the private key to 67 and you want to send it through mara I think this is the way.  Someone please correct me if i am wrong.

bitcoin-cli createrawtransaction '[{"txid":"12f34b58b04dfb0233ce889f674781c0e0c7ba95482cca469125af41a78d13b3","vout":2},{"txid":"5d45587cfd1d5b0fb826805541da7d94c61fe432259e68ee26f4a04544384164","vout":15},{"txid":"08389f34c98c606322740c0be6a7125d9860bb8d5cb182c02f98461e5fa6cd15","vout":66}]' '{"bc1q0pju92kh0qyaygvnm074mpdhdg28qa357w7usd":6.69}'

The bitcoin address at the end should be replaced with your address otherwise you will send me 6.69BTC.

The output gives you this.

0200000003b3138da741af259146ca2c4895bac7e0c08147679f88ce3302fb4db0584bf31202000 00000fdffffff6441384445a0f426ee689e2532e41fc6947dda41558026b80f5b1dfd7c58455d0f 00000000fdffffff15cda65f1e46982fc082b15c8dbb60985d12a7e60b0c742263608cc9349f380 84200000000fdffffff014021e027000000001600147865c2aad77809d22193dbfd5d85b76a1470 763400000000

After you get the private key you need to sign it.

bitcoin-cli signrawtransactionwithkey "0200000003b3138da741af259146ca2c4895bac7e0c08147679f88ce3302fb4db0584bf31202000 00000fdffffff6441384445a0f426ee689e2532e41fc6947dda41558026b80f5b1dfd7c58455d0f 00000000fdffffff15cda65f1e46982fc082b15c8dbb60985d12a7e60b0c742263608cc9349f380 84200000000fdffffff014021e027000000001600147865c2aad77809d22193dbfd5d85b76a1470 763400000000" '["PRIVATE KEY WIF"]'

The result is what you put into mara slipstream. 





hello this is the first time for me posting or replying here , supposed some did this methode , from 1% to 100 % what is the propability of succes ??



newbie
Activity: 2
Merit: 0
Hey, new to Bitcoin but I have a question that one of you may be able to help me understand.

Given the unfortunate event of puzzle#66
{
  the real solver likely panic and tried to sweep the WIF using electrum (34 inputs and paying only 5.217/B)

  the sniper within a minute made a clean TX (1 input and paying 406.356/B)
}

If saatoshi_rising used a static padding for low entropy puzzles (like #66)

-> "123456789abcdef123456789abcdef123456789abcdef12" + [2832ed74f2b5e35ee]

Would a sniper still be able to find the private key from the public key being exposed in the mempool within 2 block confirmations?

From my limited understanding this would not be possible because the pubkey would be from a 256-bit key, but maybe I am missing something.

Thanks!

Hey there, how can the "unfortunate event" have place? There's something I'm not getting about it. Could you please explain to me how could the sniper get the private key and snatch all the money on the address?
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