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Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 7. (Read 187143 times)

member
Activity: 235
Merit: 12
i have access to a quantum comp

Imagine if an immensely advanced quantum computer suddenly start the mining generating 2,016  blocks within a minute, and then vanishes.This would skyrocket the Bitcoin mining difficulty by a factor of 20,160. Conventional miners worldwide would need to labor for 140 days instead of the usual 10 minutes to uncover the hash value of a solitary block. Consequently, the blockchain would come to a standstill, halting all Bitcoin transactions during this period.  Grin
hero member
Activity: 828
Merit: 657
Is this puzzle really worth trying to solve? It's been close to a decade. Can someone give me a TL;DR?

There are several puzzles with different bits of diffculty it is exponential, puzzle 66 is 6.6 BTC BTC puzzle 67 is 6.7 BTC.

Puzzles prizes have been incremented in a x10 Factor two times since its creation.

Check the list of all unsolved puzzles.
https://docs.google.com/spreadsheets/d/1vZyRZ67iNl7aIwYNB55OY8LvZLAaoBxSanxqnGMZlWA/edit?usp=sharing

There are some puzzles addres with publickey avaiable those have another method to be solved current unsolved with publickey is puzle 130 13 BTC

I have another idea,

The puzzle creator said the only thing all the private keys have in common, is that all of them were created from a HD wallet, and then masked from the left with zeros.

Is it any weakness on the HD method of creating wallets? I mean, with the number of private keys already revealed, we could just try to target the searches for the the master private key that was used in the HD wallet.

What do you think?

Regards.

I think that author or this puzzle may used a custom method to generate those keys

This puzzle is very strange. If it's for measuring the world's brute forcing capacity, 161-256 are just a waste (RIPEMD160 entropy is filled by 160, and by all of P2PKH Bitcoin). The puzzle creator could improve the puzzle's utility without bringing in any extra funds from outside - just spend 161-256 across to the unsolved portion 51-160, and roughly treble the puzzle's content density.

If on the other hand there's a pattern to find... well... that's awfully open-ended... can we have a hint or two? Cheesy

I am the creator.

You are quite right, 161-256 are silly.  I honestly just did not think of this.  What is especially embarrassing, is this did not occur to me once, in two years.  By way of excuse, I was not really thinking much about the puzzle at all.

I will make up for two years of stupidity.  I will spend from 161-256 to the unsolved parts, as you suggest.  In addition, I intend to add further funds.  My aim is to boost the density by a factor of 10, from 0.001*length(key) to 0.01*length(key).  Probably in the next few weeks.  At any rate, when I next have an extended period of quiet and calm, to construct the new transaction carefully.

A few words about the puzzle.  There is no pattern.  It is just consecutive keys from a deterministic wallet (masked with leading 000...0001 to set difficulty).  It is simply a crude measuring instrument, of the cracking strength of the community.

Finally, I wish to express appreciation of the efforts of all developers of new cracking tools and technology.  The "large bitcoin collider" is especially innovative and interesting!

He never mention HD wallet it only mentioned some deterministic wallet
full member
Activity: 427
Merit: 105
I have another idea,

The puzzle creator said the only thing all the private keys have in common, is that all of them were created from a HD wallet, and then masked from the left with zeros.

Is it any weakness on the HD method of creating wallets? I mean, with the number of private keys already revealed, we could just try to target the searches for the the master private key that was used in the HD wallet.

What do you think?

Regards.

nice catch, go go go from me,
when update ready.
newbie
Activity: 55
Merit: 0
I have another idea,

The puzzle creator said the only thing all the private keys have in common, is that all of them were created from a HD wallet, and then masked from the left with zeros.

Is it any weakness on the HD method of creating wallets? I mean, with the number of private keys already revealed, we could just try to target the searches for the the master private key that was used in the HD wallet.

What do you think?

Regards.
newbie
Activity: 0
Merit: 0
i have access to a quantum comp give the script hahah
jr. member
Activity: 90
Merit: 7
Is this puzzle really worth trying to solve? It's been close to a decade. Can someone give me a TL;DR?

Of course not ...

Think for a moment: why someone keeps so much money locked in btc puzzle instead of cashing it out ?

Because he has a much more and this puzzle is used for monitoring current cracking possibilities. It's not some kind of competition someone made for fun. It's made on purpose.

Nobody will crack 66 or more bit puzzle using python scripts on home computer. It's a total waste of time.

The puzzle is there to meter if there is anybody who can use enough processing power to crack it.

And you can only crack it  randomly guessing private keys if you don't have public key available. Those with public key available can be solved using kangaroo or bsgs but still it needs too much processing power for an average person for a random guy (that's why the higher that 66 bit ranges are already solved but still nobody can't beat 130bit - FYI  Every fifth puzzle here have intentionally released it's public key by the creator)

Everything here can be calculated - i mean you can calculate how much processing power you need to crack it, how many and what kind of hardware you need to achieve that. You must invest hundreds of thousands $$$ to even try to crack it.
newbie
Activity: 0
Merit: 0

In the context of the polynomial generated using the Lagrange interpolation method in Sage, the x value represents the variable of the polynomial. It's not directly related to a public key's x value in cryptography. Here, x is the independent variable in the polynomial.

In Sage, when you construct a polynomial, x represents the variable in that polynomial. In the given code:


Code:
R = PolynomialRing(QQ,'x')                                 
f = R.lagrange_polynomial([(0,1),(1,3),(2,7),(3,8),(4,21),(5,49),(6,76),(7,224),(8,467),(9,514),(10,1155),(11,2683),(12,5216),(13,10544),(14,26867),(15,51510)])
for i in range(16):
    print(f(i))
Here, x is the variable in the polynomial f. So, x in the polynomial formula is just a placeholder for the variable in the polynomial equation, and it doesn't have any direct relationship to a public key's x value in cryptography.






member
Activity: 316
Merit: 34
guys i found a method to get this number -673909/1307674368000*x^15 + 5004253/87178291200*x^14 - 151337/52254720*x^13 + 9320029/106444800*x^12 - 25409989753/14370048000*x^11 + 2192506957/87091200*x^10 - 19011117413/73156608*x^9 + 1200887962891/609638400*x^8 - 3585932821063/326592000*x^7 + 647416874047/14515200*x^6 - 18586394742863/143700480*x^5 + 30899291755337/119750400*x^4 - 274137631043849/825552000*x^3 + 36933161067083/151351200*x^2 -   anyone give me how i can use my method Huh i m realy didnt know how thst method help us to solve the puzzle ??
What is x value ?
Is it public key x value ?
newbie
Activity: 0
Merit: 0
guys i found a method to get this number -673909/1307674368000*x^15 + 5004253/87178291200*x^14 - 151337/52254720*x^13 + 9320029/106444800*x^12 - 25409989753/14370048000*x^11 + 2192506957/87091200*x^10 - 19011117413/73156608*x^9 + 1200887962891/609638400*x^8 - 3585932821063/326592000*x^7 + 647416874047/14515200*x^6 - 18586394742863/143700480*x^5 + 30899291755337/119750400*x^4 - 274137631043849/825552000*x^3 + 36933161067083/151351200*x^2 -   anyone give me how i can use my method Huh i m realy didnt know how thst method help us to solve the puzzle ??
legendary
Activity: 1512
Merit: 7340
Farewell, Leo
Is this puzzle really worth trying to solve? It's been close to a decade. Can someone give me a TL;DR?
newbie
Activity: 0
Merit: 0
guys any result puzzle solved Huh
full member
Activity: 1050
Merit: 219
Shooters Shoot...
I removed the space between the private keys.
See how the first 4 digits of the private key are known?
Now guess 66!



What is this sorcery?!
hero member
Activity: 828
Merit: 657
Alberto, of course it was BS. That's the point, just emphasizing it Wink  Grin
Nice to know, in this kind of comment I am bad getting the sarcarm of it
hero member
Activity: 630
Merit: 731
Bitcoin g33k
No, 05 because of its special compression level  Tongue
member
Activity: 235
Merit: 12
Nope. It's starting with 34

And the public key starts with 03?   Grin
hero member
Activity: 630
Merit: 731
Bitcoin g33k
Alberto, of course it was BS. That's the point, just emphasizing it Wink  Grin
hero member
Activity: 828
Merit: 657
Nope. It's starting with 34

TBH this is also BS, There is no way to know that unless you already solved it.
hero member
Activity: 630
Merit: 731
Bitcoin g33k
I removed the space between the private keys.
See how the first 4 digits of the private key are known?
Now guess 66!



So the private key is starting with 37 in hex?

Nope. It's starting with 34
newbie
Activity: 43
Merit: 0
I removed the space between the private keys.
See how the first 4 digits of the private key are known?
Now guess 66!

https://www.talkimg.com/images/2024/04/23/jWRcT.gif

So the private key is starting with 37 in hex?
hero member
Activity: 828
Merit: 657
I removed the space between the private keys.
See how the first 4 digits of the private key are known?
Now guess 66!

Bullshit
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