Bitcoin puzzle transaction ~32 BTC prize to who solves it

[Feron]

thousands of bots are drooling waiting for your transaction, your hard earned bitcoins will evaporate like steam over a pot
now no one will scan 66 address ddd

[1714631273]

1714631273

[1714631273]

1714631273

[1714631273]

1714631273

[1714631273]

1714631273

[WanderingPhilospher]

For #120, that is roughly 58 days with 64 RTX 4090s, to solve
For #125, with 128 RTX 4090s, that would be around 163 days, to solve.

And those are running on some zero-point module free energy?

So you basically stored 500 billion DP 0, tames, basically just printing pubs and privs to a file, and now are offsetting 130s pub by random amounts, and looking for a collision?

No (to all of the questions). Have you looked at the 2**65 keyspace? It's 36893488147419103232.
BTW it only takes around 16 bytes / key to store hundreds of billions of tame kangaroos for 129 bit case.
Ofcourse I'm not simply "printing pubs and privs" to a text file, that's an over-simplification.
If you want some hints: the more keys a hash table has, the less space/key is required.

For the traditional Kangaroo algo, for 130, with DP 32, I need to find only 9 billion tames and 9 billion wilds to solve. So it sounds like you just stored random pubs and privs, because 500 billion tames, with a decent DP, would take a loooooong time.

Also, you need to perform roughly 2^66.05 "steps" for #130, that would be the average.

I think you are kidding with your 9 billion kangaroos. You are missing something critical about the underlying theory.
Otherwise, you can solve all puzzles with 2 kangaroos, if you wait a trillion years.

If you are so sure #120 / #125 were solved with existing software, did you also do the math about how many kangaroos would have been needed? At DP 0 /1 / 2 etc? Your times have no meaning without space complexity attached to them.

Lol, yeah, I do not think you understand the Kangaroo algo.

It's all laid out for you in various readings/papers.

I never said 9 billion kangaroos. Do you understand the algo? When I say "find" x amount of tames and wilds, it is referring to the points/distances found by each type of kangaroo. You store tame and wild points (Based on DP used) and distances, that are generated from the tame and wild kangaroos, hopping all around.
2^66.05 - 2^32 (DP size that I stated) = 2^34.05 stored DPs. 2^33.05 tames and 2^33.05 wilds. 2^33.05 = 8,892,857,981; so roughly 9 billion points and distances stored (tames and wilds) to solve, on average. Could be a little higher, could be a little lower. So no, I was not "kidding". And yes, my times are based on math, and the space complexity is what I said, roughly 9 billion points & distances per tame and wild, to solve. I can't give you exact amount of GBs required because each Kangaroo program stores points differently, different amount of bytes and different formats, binary vs plain text. One would need to calculate it based on their DP and how the points/distances are stored.

But yes, you could set out 2 kangaroos, 1 tame, and 1 wild, and eventually solve, in many many years, or you could get lucky and solve within minutes, hours, days.

I doubt whoever solved 120/125, if they used the kangaroo algo, set a DP of less than 28. They would have an enormous amount of DP overhead, that JLP explains well in his github:

Code:

DP overhead according to the range size (N), DP mask size (dpBit) and number of kangaroos running in paralell (nbKangaroo).

110 and 115 were both solved with DP 25. I know that during the 115 run, the grid sizes for the GPUs were choked down and another part of the code was reduced, to prevent a massive DP overhead. And when finally solved, I do believe total DPs stored (points w distances) was a smidge over the expected total of 2^33.55

[Vvpgoat]

thousands of bots are drooling waiting for your transaction, your hard earned bitcoins will evaporate like steam over a pot
now no one will scan 66 address ddd

How can bots get ahead of me? Let's say I solved the puzzle without using the site and only I know the private key at the time of sending the transaction. please explain how this will happen because I apparently don’t understand something

[zahid888]

It is enough to mark the tx as without RBF and the winner gets it all!

Stop spreading bs.

This is not bullshit, as some others users already mentioning, Doesn't matter if you mark it with or without RBF, the transaction always can be repleaced and that only depends of Node configuracions to accept it or reject it.

The only safest way to move the founds is ask to a miner if they mined your transaction without broadcast it (previously to the block begin mined), also it is even better if you are a miner and you include the transaction without broadcast it before it is mined.

Thanks Alberto, then I have to sell the key to the miner who bids the highest.

Whats the best way for me to prove that I have the key, without disclosing the private key and public key to a miner ?

Address = 13zb1hQbWVuYdZoAkztVrNrm65aReL2pYD
Message = I own this address
Signature = HxSDzJj3BhS8lh7qoU1L9zJsymorFkJv8/roi8o6+269PEDPr8psAE1QEOerWAWKv/ypXss7hhXEe0FutUR4b0s=
https://bitaps.com/signature

I am still curious to know if someone is able to crack these signatures as well

Thanks Alberto, then I have to sell the key to the miner who bids the highest.

Whats the best way for me to prove that I have the key, without disclosing the private key and public key to a miner ?

There is no safe way to do that, even signature text messages include the public key.

Can you crack and post the PVK of that signed address??

[Feron]

Just kidding, but the risk here is if the hacker set a high transaction fee, the hackers transaction may be confirmed before yours
now there is a lot of money at stake, I think that there are a lot of bots attached to the 66 address puzzle

[albert0bsd]

Can you crack and post the PVK of that signed address??

Well, the puzzle 66 is expected to be in the 66 bit key space, is your key also in that range?

If that is true, then anyone would be able to crack it

[Feron]

the advantage is that no one knows exactly when the 66 puzzle will be solved, it could be a day, a month, a year or 100 years
I think that no one will monitor the 66 puzzle address for that long

[albert0bsd]

Can you crack and post the PVK of that signed address??

Code:

-----BEGIN BITCOIN SIGNED MESSAGE-----
I also own this address
-----BEGIN SIGNATURE-----
13zb1hQbWVuYdZoAkztVrNrm65aReL2pYD
H8SlTYJ7a/Mp5cra9VzqgDFMGRQUYfA5NLrSCb0GkwbeEDqx8vKWGYWX3YmiqIU8nl6THdprDK/k34y1GQrFFDk=
-----END BITCOIN SIGNED MESSAGE-----

It took 5 minutes in my laptop and less of 1 minute in main computer

the advantage is that no one knows exactly when the 66 puzzle will be solved, it could be a day, a month, a year or 100 years
I think that no one will monitor the 66 puzzle address for that long

That is cheap to do that, Actually I am doing it.

[kTimesG]

Lol, yeah, I do not think you understand the Kangaroo algo.

It's all laid out for you in various readings/papers.

I never said 9 billion kangaroos. Do you understand the algo? When I say "find" x amount of tames and wilds, it is referring to the points/distances found by each type of kangaroo. You store tame and wild points (Based on DP used) and distances, that are generated from the tame and wild kangaroos, hopping all around.
2^66.05 - 2^32 (DP size that I stated) = 2^34.05 stored DPs. 2^33.05 tames and 2^33.05 wilds. 2^33.05 = 8,892,857,981; so roughly 9 billion points and distances stored (tames and wilds) to solve, on average. Could be a little higher, could be a little lower. So no, I was not "kidding". And yes, my times are based on math, and the space complexity is what I said, roughly 9 billion points & distances per tame and wild, to solve. I can't give you exact amount of GBs required because each Kangaroo program stores points differently, different amount of bytes and different formats, binary vs plain text. One would need to calculate it based on their DP and how the points/distances are stored.

But yes, you could set out 2 kangaroos, 1 tame, and 1 wild, and eventually solve, in many many years, or you could get lucky and solve within minutes, hours, days.

I doubt whoever solved 120/125, if they used the kangaroo algo, set a DP of less than 28. They would have an enormous amount of DP overhead, that JLP explains well in his github:

Code:

DP overhead according to the range size (N), DP mask size (dpBit) and number of kangaroos running in paralell (nbKangaroo).

110 and 115 were both solved with DP 25. I know that during the 115 run, the grid sizes for the GPUs were choked down and another part of the code was reduced, to prevent a massive DP overhead. And when finally solved, I do believe total DPs stored (points w distances) was a smidge over the expected total of 2^33.55

So your estimated times include the DP overhead?
Because I did a lot of simulations on lower puzzles to get min/avg/max jumps and stored footprints, and, as you can guess, when the DP criteria kicks in, storage goes down, number of jumps until a collision goes up. And there's a lot of ways to improve either space or time, but not both.

So it is not fair to still call the time complexity anywhere near O(sqrt(n)) when you're using 2*sqrt(sqrt(n)) stored items instead of the average 2*sqrt(n).

Steps until a collision can range between expected average / 10, and 3 * the expected average.

And JLP kangaroo is not some godly reference, I did not use it at all, for one I don't even agree with the way jumps and hashmap keys are used, but that's another story.

No one says you need to have 50/50 tames and kangs. And no one says you even need to store the entire traveled distance. There are various techniques.

[zahid888]

Can you crack and post the PVK of that signed address??

Code:

-----BEGIN BITCOIN SIGNED MESSAGE-----
I also own this address
-----BEGIN SIGNATURE-----
13zb1hQbWVuYdZoAkztVrNrm65aReL2pYD
H8SlTYJ7a/Mp5cra9VzqgDFMGRQUYfA5NLrSCb0GkwbeEDqx8vKWGYWX3YmiqIU8nl6THdprDK/k34y1GQrFFDk=
-----END BITCOIN SIGNED MESSAGE-----

It took 5 minutes in my laptop and less of 1 minute in main computer

WTF, I'm utterly astounded to realize that the smaller keys are not secure by any means, and searching for the larger ones is proving to be an insurmountable challenge for us. Given the magnitude of this predicament, it seems wise for us to disengage from this futile endeavor. The journey thus far has been both eye-opening and taxing.  

[nomachine]

"Do not disturb my kangaroos." --  Archimedes

 

[AliBah]

Can you crack and post the PVK of that signed address??

Code:

-----BEGIN BITCOIN SIGNED MESSAGE-----
I also own this address
-----BEGIN SIGNATURE-----
13zb1hQbWVuYdZoAkztVrNrm65aReL2pYD
H8SlTYJ7a/Mp5cra9VzqgDFMGRQUYfA5NLrSCb0GkwbeEDqx8vKWGYWX3YmiqIU8nl6THdprDK/k34y1GQrFFDk=
-----END BITCOIN SIGNED MESSAGE-----

It took 5 minutes in my laptop and less of 1 minute in main computer

the advantage is that no one knows exactly when the 66 puzzle will be solved, it could be a day, a month, a year or 100 years
I think that no one will monitor the 66 puzzle address for that long

That is cheap to do that, Actually I am doing it.

By which program you cracked that? and how to find thats in what range?

[Easteregg69]

"Do not disturb my kangaroos." --  Archimedes

 

"It's Vogon poetry to me" -- Me.

So it's here you keep all the bright heads. Stuck on a riddle.

[brainless]

Can you crack and post the PVK of that signed address??

Code:

-----BEGIN BITCOIN SIGNED MESSAGE-----
I also own this address
-----BEGIN SIGNATURE-----
13zb1hQbWVuYdZoAkztVrNrm65aReL2pYD
H8SlTYJ7a/Mp5cra9VzqgDFMGRQUYfA5NLrSCb0GkwbeEDqx8vKWGYWX3YmiqIU8nl6THdprDK/k34y1GQrFFDk=
-----END BITCOIN SIGNED MESSAGE-----

It took 5 minutes in my laptop and less of 1 minute in main computer

WTF, I'm utterly astounded to realize that the smaller keys are not secure by any means, and searching for the larger ones is proving to be an insurmountable challenge for us. Given the magnitude of this predicament, it seems wise for us to disengage from this futile endeavor. The journey thus far has been both eye-opening and taxing.  

if someone found 66 PK, can share with me, for withdraw, i know secure way to transfer, other you have seen above test, and definatly, slowly slowly lot of people would have 66 PK, but no one take step for attempt withdraw

[Baskentliia]

thousands of bots are drooling waiting for your transaction, your hard earned bitcoins will evaporate like steam over a pot
now no one will scan 66 address ddd

How can bots get ahead of me? Let's say I solved the puzzle without using the site and only I know the private key at the time of sending the transaction. please explain how this will happen because I apparently don’t understand something

As soon as the person who finds puzzle 66 sends the transaction to the network, pubkey will appear without requiring network approval. Once Pubkey appears it will take 1 second to find the private key. Then, someone else will spend again because the network approval has not yet occurred. The person with the highest network approval will own the bitcoins. Even if you send with a high transaction fee, the person who receives a higher transaction fee and network approval than you will win. For this reason, Low puzzles are problematic.

[AliBah]

thousands of bots are drooling waiting for your transaction, your hard earned bitcoins will evaporate like steam over a pot
now no one will scan 66 address ddd

How can bots get ahead of me? Let's say I solved the puzzle without using the site and only I know the private key at the time of sending the transaction. please explain how this will happen because I apparently don’t understand something

As soon as the person who finds puzzle 66 sends the transaction to the network, pubkey will appear without requiring network approval. Once Pubkey appears it will take 1 second to find the private key. Then, someone else will spend again because the network approval has not yet occurred. The person with the highest network approval will own the bitcoins. Even if you send with a high transaction fee, the person who receives a higher transaction fee and network approval than you will win. For this reason, Low puzzles are problematic.

If find the Pkey by pubkey is so easy like this, then why puzzle 130 didnt solve yet?

[albert0bsd]

By which program you cracked that? and how to find thats in what range?

I use keyhunt the program that i develop:

https://bitcointalk.org/index.php?topic=5322040.0
https://github.com/albertobsd/keyhunt

But other programs can also be used like Kangaroo

https://github.com/JeanLucPons/Kangaroo

There is no way to know the range of an address.

But since we are talking here of puzzle 66 i did the test in the bit 66 just to check if that address was in that range, and actually the key is of that address is in that range!!

Other address besides of low bit puzzles aren't vulnerable to this.

As soon as the person who finds puzzle 66 sends the transaction to the network, pubkey will appear without requiring network approval. Once Pubkey appears it will take 1 second to find the private key. Then, someone else will spend again because the network approval has not yet occurred. The person with the highest network approval will own the bitcoins. Even if you send with a high transaction fee, the person who receives a higher transaction fee and network approval than you will win. For this reason, Low puzzles are problematic.

Exactly all the non-confirmed transacctions are public avaible in the mempool of each node.

Also there is sites to check them https://mempool.space/ also they offer some api to check for some values

https://mempool.space/docs/api/rest#get-address-transactions-mempool

Once that you get the TX id, you need to download the Raw transaction, decode it and extract the publickey and that is easy to do if you know what are you doing.

If find the Pkey by pubkey is so easy like this, then why puzzle 130 didnt solve yet?

Well for puzzle  66 is really easy (some minutes/seconds) but since the complexity is exponential it will take a lot of time do that for those bits ranges, months or years depending of hardware.

Here we are talking of low bit puzzles less than 80 bits are solvables by GPU almost in less than 10 minutes (in avarage the time needed to mine a block)

[smracer]

If there are thousands of bots watching the 66 address, which I think there are, they will all watch the mempool as well and keep increasing the fee against each other.  Maybe the answer is to send the transaction first with the entire amount as the fee minus 1 sat to your address and hope whichever miner that finds it will return it to you.  This way no one can increase the fee any higher and your timestamp is first.  Would that work?

[albert0bsd]

Would that work?

Yes, but if you are the minner you can mine your own trasactions without broadcast it publicly. The mined block is only broadcasted if you found the solution for the block, in this case if you are luck enough and no other miner mine the same block height at the same time you will be able to redeem it without problem.

[zahid888]

if someone found 66 PK, can share with me, for withdraw, i know secure way to transfer, other you have seen above test, and definatly, slowly slowly lot of people would have 66 PK, but no one take step for attempt withdraw

For a drowning person, even a straw is a great support. I will definitely contact you.

When everything is dependent on fate, it's best that we only utilize this code. There's no need for undue force with the computer, no burden on the machine, no excessive worry about electricity costs, and most importantly, without any concern for double spending.

2^129 - 2^130 | PUB : 0383af736b4eefaa2f697e63b4530129ff34ff945be353ee498b0473238ea91ef4 | PVK : 0x2264f44bd23a25b4a5b8390e650cf9cae
2^134 - 2^135 | PUB : 03a987392d57dd82a9ee630f99a75b3925ca5417869bbdfc76b6f59997e558ee73 | PVK : 0x7a5330f4174f496ca6c4364888e9aef5db
2^139 - 2^140 | PUB : 0389d6d9b49a09988a70238b4c79639ef2881c686134075a485843e4814e36320d | PVK : 0xdffa0bf08d3584c46a48904da5df25016af
2^144 - 2^145 | PUB : 0226d47bfd7e231e9c123683137bf0f34522c03a8604fee9a9f626f0334daf9c0b | PVK : 0x167b8d6bec4db640ca2560f120f0a072bf658
2^149 - 2^150 | PUB : 0363abdd6ae3bc62c6ccf3d34d22b44d90f710fa99931d566a6a8c0da68acca933 | PVK : 0x3e1954a5af437841b085ba454c8b9e1149268a
2^154 - 2^155 | PUB : 035bebccb841e7b812fcfbc5fe86d9f9549e68d3d3da1bffd613e8a3960ca26ddc | PVK : 0x7a7179d2027c342b336fb8fd92347d50d937d1e
2^159 - 2^160 | PUB : 022cc56adb8c5385aefbb3cd38dfc2294560c4f73ad4562f68916535d6ab8ebf66 | PVK : 0xc09735b95a670ac005324cb8ddcf3576df7b3f1e

Code:

from bit import *
import random

L = [
'03633cbe3ec02b9401c5effa144c5b4d22f87940259634858fc7e59b1c09937852',
'02145d2611c823a396ef6712ce0f712f09b9b4f3135e3e0aa3230fb9b6d08d1e16',
'031f6a332d3c5c4f2de2378c012f429cd109ba07d69690c6c701b6bb87860d6640',
'03afdda497369e219a2c1c369954a930e4d3740968e5e4352475bcffce3140dae5',
'03137807790ea7dc6e97901c2bc87411f45ed74a5629315c4e4b03a0a102250c49',
'035cd1854cae45391ca4ec428cc7e6c7d9984424b954209a8eea197b9e364c05f6',
'02e0a8b039282faf6fe0fd769cfbc4b6b4cf8758ba68220eac420e32b91ddfa673'
]

while True:
    seed = random.getrandbits(161)
    random.seed(seed)
    for exp in range(129, 164, 5):
        a = random.randrange(2**exp, 2**(exp+1))
        pk = Key.from_int(a)
        key = Key.from_int(a)
        pub = pk._pk.public_key.format(compressed=True)
        addrpub = pub.hex()
        if addrpub in L:
            print('================found=====================\n')
            f=open(u"Key_Found.txt","a")
            f.write("Seed : " + str(seed) + "\n")
            f.write("Prvk : " + hex(a) + "\n")
            f.close()
            input('================Exit=====================\n')
            exit()
        #if addrpub.startswith('03633cbe3') or addrpub.startswith('02145d26') or addrpub.startswith('031f6a33') or addrpub.startswith('03afdda4') or addrpub.startswith('03137807') or addrpub.startswith('035cd185') or addrpub.startswith('02e0a8b0'):
        print(f"2^{exp} - 2^{exp+1} | PUB : {addrpub} | PVK : {hex(a)}", end = '\n')
        

If fortune is on your side, a single core of a computer is sufficient for you..  Good...Luck...Luck...And....Only....Luck