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Author Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it  (Read 186053 times)
ccinet
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March 18, 2024, 08:10:34 PM
 #4641

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.

Anyone could do that with current any address with balance and known pubkey.

Not for any address,  that is only for low puzzles like 66, 67...  maybe up to 80 bit can be solved in some seconds or minutes depending of the program used.

This is only for the next specific addreses:

Code:
66		13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so
67 1BY8GQbnueYofwSuFAT3USAhGjPrkxDdW9
68 1MVDYgVaSN6iKKEsbzRUAYFrYJadLYZvvZ
69 19vkiEajfhuZ8bs8Zu2jgmC6oqZbWqhxhG
71 1PWo3JeB9jrGwfHDNpdGK54CRas7fsVzXU
72 1JTK7s9YVYywfm5XUH7RNhHJH1LshCaRFR
73 12VVRNPi4SJqUTsp6FmqDqY5sGosDtysn4
74 1FWGcVDK3JGzCC3WtkYetULPszMaK2Jksv
76 1DJh2eHFYQfACPmrvpyWc8MSTYKh7w9eRF
77 1Bxk4CQdqL9p22JEtDfdXMsng1XacifUtE
78 15qF6X51huDjqTmF9BJgxXdt1xcj46Jmhb
79 1ARk8HWJMn8js8tQmGUJeQHjSE7KRkn2t8

Other Regular wallets are safe
Then we should stop trying to solve the low puzzles and program a bot to catch the low transactions. Undecided
We will go from being hunters to leeches!!! Grin
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mcdouglasx
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March 18, 2024, 08:24:28 PM
 #4642


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.


And if the miner includes the transaction and his block is mined by others? Would the transaction be free for the sharks?

I'm not dead, long story... BTC bc1qxs47ttydl8tmdv8vtygp7dy76lvayz3r6rdahu
albert0bsd
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March 18, 2024, 08:44:16 PM
Last edit: March 18, 2024, 11:17:13 PM by albert0bsd
Merited by mcdouglasx (1)
 #4643

And if the miner includes the transaction and his block is mined by others? Would the transaction be free for the sharks?

Yest that can happend, it is bad luck, in that case there is nothing to do, but:

Code:
An orphan block is a block that has been solved within the blockchain network but was not accepted by the network. There can be two miners who solve valid blocks simultaneously. The network uses both blocks until one chain has more verified blocks.

So all depends of how much miner power do you have, to make your mined block the one that is accepted by the network, maybe only broadcast the block if you mined two simultaneos block in a row (the network must accept this as block winner)

Right now is not profitable wait for two simultaneus blocks to be mined, current block reward is 6.25 BTC, but in some weeks it is going to be 3.125 BTC, maybe then it can be useful or maybe in the near future when block reward is under 1 BTC

Maybe futures puzzles like 67, 68 etc.. may fit in that criteria, but all this is just speculation.

Regards

kTimesG
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March 18, 2024, 09:11:31 PM
 #4644

Friendly reminder that to solve #130 in 2**65.5 average steps you need to store 2**65 kangaroo jumps (e.g. thousands of exabytes with constant time random access).
If we go with distinguished points space goes down, expected time goes up. There's no free lunch.
Currently testing ~ 500 billion tame kangaroos footprints against wilds, I don't really expect a collision but who knows. I'd need like millions of times more storage.

I don't think #120 or #125 were solved by existing (public) software, it just seems unrealistic from a resources / cost perspective. There's something else going on there.
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March 18, 2024, 10:03:03 PM
Merited by albert0bsd (1)
 #4645

Friendly reminder that to solve #130 in 2**65.5 average steps you need to store 2**65 kangaroo jumps (e.g. thousands of exabytes with constant time random access).
If we go with distinguished points space goes down, expected time goes up. There's no free lunch.
Currently testing ~ 500 billion tame kangaroos footprints against wilds, I don't really expect a collision but who knows. I'd need like millions of times more storage.

I don't think #120 or #125 were solved by existing (public) software, it just seems unrealistic from a resources / cost perspective. There's something else going on there.

Lol, why?

Current speed for a sample of cards using Kangaroo:

RTX 2080Ti = 3,790 MKey/s
RTX 3070   = 3,085 MKey/s
RTX 4070   = 3,900 MKey/s
RTX 4080   = 5,260 MKey/s
RTX 4090   = 7,500 MKey/s
H100 SXM   = 13,600 MKey/s

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.

The software and hardware is out there.

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?
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.
ccinet
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March 18, 2024, 10:50:44 PM
 #4646

Friendly reminder that to solve #130 in 2**65.5 average steps you need to store 2**65 kangaroo jumps (e.g. thousands of exabytes with constant time random access).
If we go with distinguished points space goes down, expected time goes up. There's no free lunch.
Currently testing ~ 500 billion tame kangaroos footprints against wilds, I don't really expect a collision but who knows. I'd need like millions of times more storage.

I don't think #120 or #125 were solved by existing (public) software, it just seems unrealistic from a resources / cost perspective. There's something else going on there.

Lol, why?

Current speed for a sample of cards using Kangaroo:

RTX 2080Ti = 3,790 MKey/s
RTX 3070   = 3,085 MKey/s
RTX 4070   = 3,900 MKey/s
RTX 4080   = 5,260 MKey/s
RTX 4090   = 7,500 MKey/s
H100 SXM   = 13,600 MKey/s

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.

The software and hardware is out there.

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?
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.

But then it's easy! A pair of RTX 4090 and in six months I solve 120 and 125 and become a millionaire???
Unfortunately, I don't think that's the case, for some reason thousands of people have been trying for years...
albert0bsd
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March 18, 2024, 11:16:04 PM
 #4647

But then it's easy! A pair of RTX 4090 and in six months I solve 120 and 125 and become a millionaire???
Unfortunately, I don't think that's the case, for some reason thousands of people have been trying for years...

It is not just "A pair" it is some tens of it please read it again:


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.

First example are 64 cards ( 32 pairs)
Second example are 128 cards ( 64 pairs )

Now those puzzles 120 and 125 were already solved by someone

For puzzle 130 it will  more cards and more time, also they need to know what are they doing, most people don't even know what their commands do


Woz2000
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March 19, 2024, 04:29:46 AM
Last edit: March 19, 2024, 05:15:18 AM by Woz2000
 #4648

I am re-visiting JLP's Kangaroo program and have a few questions.

The docs mention that the work files should be merged for server/client mode when there are multiple (disconnected) servers. If I am running only 1 server with 2 clients, I do not need to merge the 2 client work files, is that correct (the server will collect the work from both clients and check for collisions)?

The docs also mention the following "To build such an architecture, the total number of kangaroo running in parallel must be know at the starting time to estimate the DP overhead. It is not recommended to add or remove clients during running time, the number of kangaroo must be constant.". How important is this and how does it apply (adding clients to server or adding GPU to client)? Simply - do I need to wait until all equipment is ready or can I add GPUs slowly over time? Some how I think this has more to do with slower/older computers.
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March 19, 2024, 08:14:30 AM
 #4649

may please someone tell me the correct command of Kangaroo or Keyhunt for bsgs mode?
I have 2 3070 cards.
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March 19, 2024, 08:48:29 AM
 #4650

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? Smiley

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.
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March 19, 2024, 08:51:04 AM
 #4651

may please someone tell me the correct command of Kangaroo or Keyhunt for bsgs mode?
I have 2 3070 cards.
If you want to use keyhunt you don't need graphics cards, but you do need a good CPU and as much RAM as possible.
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March 19, 2024, 08:58:56 AM
 #4652


...

Really? That’s your reply?
No it can’t be done to any address. #66 will be solved within seconds of public key being broadcast, because its range is known. That’s why this is different versus just any old address.

Ah, ok, so when space is 1,4615016373309029182036848327163e+48 it can't be done, but when the space is 36893488147419103232 these are seconds.

Nice joke.

BTC: bc1qmrexlspd24kevspp42uvjg7sjwm8xcf9w86h5k
I have 9900K and 1080 Ti, gathering funds for new desktop PC for Bitcoin operations - 14900K and RTX 4090
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March 19, 2024, 09:53:49 AM
Merited by albert0bsd (2)
 #4653

Why Speculate.... Let's do a simulation....

Creating a Random Key in the Range of Puzzle 66
Code:
import secp256k1 as ice
import random
p66_key = random.randint(2**65, -1+2**66)
P66 = ice.scalar_multiplication(p66_key)

The Values obtained are
Code:
hex(p66_key) = 0x318c1cdee7973e9a7
P66.hex() = '0426aef3f353caaf022cc7dd0fd5e0a76cf5a06f274aae0457bf759171ebfc5ac8300749efa4eae951dbfab6008e45935cc18bf0c5a390e8c0643b4985b43fc085'

Running Kangaroo algo on this Pubkey
Code:
Kangrand.exe -gpu -t 2 -st 20000000000000000 -en 3ffffffffffffffff t66.txt
Kangaroo v2.1 : Added Start End Options
Start:20000000000000000
Stop :3FFFFFFFFFFFFFFFF
Keys :1
Number of CPU thread: 2
Range width: 2^65
Jump Avg distance: 2^31.99
Number of kangaroos: 2^18.18
Suggested DP: 11
Expected operations: 2^33.60
Expected RAM: 254.5MB
DP size: 11 [0xFFE0000000000000]
SolveKeyCPU Thread 0: 1024 kangaroos
SolveKeyCPU Thread 1: 1024 kangaroos
GPU: GPU #0 Quadro K2100M (3x192 cores) Grid(6x384) (29.5 MB used)
SolveKeyGPU Thread GPU#0: creating kangaroos...
SolveKeyGPU Thread GPU#0: 2^18.17 kangaroos [1.5s]
[2241.49 TK/s][GPU 1459.68 TK/s][Count 2^30.85][Dead 0][45s (Avg 04:35)][31.0/64.1MB]
Key# 0 [1S]Pub:  0x0326AEF3F353CAAF022CC7DD0FD5E0A76CF5A06F274AAE0457BF759171EBFC5AC8
       Priv: 0x318C1CDEE7973E9A7

Done: Total time 50s

If you look, this has been run on Laptop with 9 year old GPU, still < 1 minute.
What will happen if someone run it on Recent Workstation GPU. It will be few seconds only.

You will have ample time to resubmit the transaction pending in current block by using your own Address having a little bit higher key.
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March 19, 2024, 10:12:14 AM
 #4654

Why Speculate.... Let's do a simulation....

Creating a Random Key in the Range of Puzzle 66
Code:
import secp256k1 as ice
import random
p66_key = random.randint(2**65, -1+2**66)
P66 = ice.scalar_multiplication(p66_key)

The Values obtained are
Code:
hex(p66_key) = 0x318c1cdee7973e9a7
P66.hex() = '0426aef3f353caaf022cc7dd0fd5e0a76cf5a06f274aae0457bf759171ebfc5ac8300749efa4eae951dbfab6008e45935cc18bf0c5a390e8c0643b4985b43fc085'

Running Kangaroo algo on this Pubkey
Code:
Kangrand.exe -gpu -t 2 -st 20000000000000000 -en 3ffffffffffffffff t66.txt
Kangaroo v2.1 : Added Start End Options
Start:20000000000000000
Stop :3FFFFFFFFFFFFFFFF
Keys :1
Number of CPU thread: 2
Range width: 2^65
Jump Avg distance: 2^31.99
Number of kangaroos: 2^18.18
Suggested DP: 11
Expected operations: 2^33.60
Expected RAM: 254.5MB
DP size: 11 [0xFFE0000000000000]
SolveKeyCPU Thread 0: 1024 kangaroos
SolveKeyCPU Thread 1: 1024 kangaroos
GPU: GPU #0 Quadro K2100M (3x192 cores) Grid(6x384) (29.5 MB used)
SolveKeyGPU Thread GPU#0: creating kangaroos...
SolveKeyGPU Thread GPU#0: 2^18.17 kangaroos [1.5s]
[2241.49 TK/s][GPU 1459.68 TK/s][Count 2^30.85][Dead 0][45s (Avg 04:35)][31.0/64.1MB]
Key# 0 [1S]Pub:  0x0326AEF3F353CAAF022CC7DD0FD5E0A76CF5A06F274AAE0457BF759171EBFC5AC8
       Priv: 0x318C1CDEE7973E9A7

Done: Total time 50s

If you look, this has been run on Laptop with 9 year old GPU, still < 1 minute.
What will happen if someone run it on Recent Workstation GPU. It will be few seconds only.

You will have ample time to resubmit the transaction pending in current block by using your own Address having a little bit higher key.


May please tell me the command of Kangaroo for 2 Gpus ?
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March 19, 2024, 10:20:11 AM
 #4655

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 ?
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March 19, 2024, 10:25:29 AM
Last edit: March 19, 2024, 10:47:34 AM by nomachine
 #4656


If you look, this has been run on Laptop with 9 year old GPU, still < 1 minute.
What will happen if someone run it on Recent Workstation GPU. It will be few seconds only.



You simply create a script that pings:

https://blockchain.info/q/pubkeyaddr/13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so.

If the pubkey returned is not 'error', extract the key and run Kangaroo.

Upon finding the private key, initiate bitcoin-cli to execute the transaction,

all within a 10-second timeframe.

You don't have to be at the computer and watch at all.  No manual supervision required.  Grin
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March 19, 2024, 10:50:05 AM
 #4657

Can a kangaroo program be run using only the processor? If there is such a kangaroo program, could you share it?
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March 19, 2024, 10:51:41 AM
Last edit: March 19, 2024, 11:36:19 AM by Vvpgoat
 #4658

Hi guys
I have a question. Let's say someone solved puzzle 66 using third-party software. How quickly will a HEX appear on privatekeyfinder.io? after the transaction is sent immediately? or over a certain period of time? how does it even work
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March 19, 2024, 11:22:23 AM
 #4659


...

Really? That’s your reply?
No it can’t be done to any address. #66 will be solved within seconds of public key being broadcast, because its range is known. That’s why this is different versus just any old address.

Ah, ok, so when space is 1,4615016373309029182036848327163e+48 it can't be done, but when the space is 36893488147419103232 these are seconds.

Nice joke.

It's no joke. After I found out the public key from #65, I scanned with keyhunt in bsgs mode and found the private key in 7 minutes, after searching for it in rmd mode for 3 years without results, and this with a computer without much performance.
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March 19, 2024, 12:36:31 PM
 #4660

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.

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

Here nobody mention the electric cost of those cards, but yes, you need to include it to your budget and it will be a lot.


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