Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 85.

nomachine

member

Activity: 462

Merit: 24

Quote from: iceland2k14 on March 19, 2024, 05:53:49 AM

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

AndrewWeb

jr. member

Activity: 46

Merit: 1

Quote from: albert0bsd on March 18, 2024, 03:22:17 PM

Quote from: pbies on March 18, 2024, 11:18:25 AM

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 ?

AliBah

newbie

Activity: 35

Merit: 0

Quote from: iceland2k14 on March 19, 2024, 05:53:49 AM

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 ?

iceland2k14

jr. member

Activity: 36

Merit: 68

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.

pbies

full member

Activity: 290

Merit: 133

Quote from: WanderingPhilospher on March 18, 2024, 02:43:07 PM

...

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.

Ovixx

newbie

Activity: 30

Merit: 0

Quote from: AliBah on March 19, 2024, 04:14:30 AM

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.

kTimesG

member

Activity: 165

Merit: 26

Quote from: WanderingPhilospher on March 18, 2024, 06:03:03 PM

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?

Quote from: WanderingPhilospher on March 18, 2024, 06:03:03 PM

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.

Quote from: WanderingPhilospher on March 18, 2024, 06:03:03 PM

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.

AliBah

newbie

Activity: 35

Merit: 0

may please someone tell me the correct command of Kangaroo or Keyhunt for bsgs mode?
I have 2 3070 cards.

Woz2000

jr. member

Activity: 82

Merit: 2

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.

albert0bsd

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Activity: 862

Merit: 662

Quote from: ccinet on March 18, 2024, 06:50:44 PM

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:

Quote from: WanderingPhilospher on March 18, 2024, 06:03:03 PM

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

ccinet

newbie

Activity: 41

Merit: 0

Quote from: WanderingPhilospher on March 18, 2024, 06:03:03 PM

Quote from: kTimesG on March 18, 2024, 05:11:31 PM

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...

WanderingPhilospher

full member

Activity: 1162

Merit: 237

Shooters Shoot...

Quote from: kTimesG on March 18, 2024, 05:11:31 PM

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.

kTimesG

member

Activity: 165

Merit: 26

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.

albert0bsd

hero member

Activity: 862

Merit: 662

Quote from: mcdouglasx on March 18, 2024, 04:24:28 PM

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

mcdouglasx

member

Activity: 239

Merit: 53

New ideas will be criticized and then admired.

Quote from: albert0bsd on March 18, 2024, 03:22:17 PM

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?

ccinet

newbie

Activity: 41

Merit: 0

Quote from: albert0bsd on March 18, 2024, 03:22:17 PM

Quote from: pbies on March 18, 2024, 11:18:25 AM

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.

Quote from: pbies on March 18, 2024, 02:37:34 PM

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

albert0bsd

hero member

Activity: 862

Merit: 662

Quote from: pbies on March 18, 2024, 11:18:25 AM

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.

Quote from: pbies on March 18, 2024, 02:37:34 PM

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

WanderingPhilospher

full member

Activity: 1162

Merit: 237

Shooters Shoot...

Quote from: Baskentliia on March 18, 2024, 02:52:54 PM

Quote from: WanderingPhilospher on March 18, 2024, 02:43:07 PM

Quote from: pbies on March 18, 2024, 02:37:34 PM

Quote from: citb0in on March 18, 2024, 01:44:30 PM

Quote from: pbies on March 18, 2024, 11:18:25 AM

...

Stop spreading bs.

better you stop spreading bullshit. You may mark the TX with RBF or not, it doesn't change anything. The transaction can always be replaced independently if the TX has been signaled RBF or not. Believe it or not, saatoshi_falling is right about what he said.

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

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.

What do you think, brother? Is such a thing possible? Do you agree with what is said?
If so, there will be no point in searching for low-end puzzles.

Everything I’ve read, says it’s possible.
It will be interesting to see how many sharks are in the water.

With the low end challenges/puzzle, their time is almost up. Can search for 130, 135, 140 with same or lower ops as 67, 68, 69. For me, I stopped putting a lot of hash on #66 and more on 130; but that’s just me.

ccinet

newbie

Activity: 41

Merit: 0

Quote from: Baskentliia on March 18, 2024, 02:52:54 PM

Quote from: WanderingPhilospher on March 18, 2024, 02:43:07 PM

Quote from: pbies on March 18, 2024, 02:37:34 PM

Quote from: citb0in on March 18, 2024, 01:44:30 PM

Quote from: pbies on March 18, 2024, 11:18:25 AM

...

Stop spreading bs.

better you stop spreading bullshit. You may mark the TX with RBF or not, it doesn't change anything. The transaction can always be replaced independently if the TX has been signaled RBF or not. Believe it or not, saatoshi_falling is right about what he said.

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

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.

What do you think, brother? Is such a thing possible? Do you agree with what is said?
If so, there will be no point in searching for low-end puzzles.

The problem comes in the way the bitcoin protocol solves the double spending problem.
If two transactions from the same private address simultaneously without confirmations bid, the one with the most confirmations wins... then there is a possibility that the thief will keep your coins, and this is due to the bitcoin protocol itself.

nomachine

member

Activity: 462

Merit: 24

This is like throwing one fish into a pond of sharks. Either it will be torn apart and there will be nothing left for anyone, or one will be the fastest.

Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 85. (Read 215611 times)