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.
Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 32. (Read 192699 times)
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...
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
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.
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...
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.
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.
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.
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.
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
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?
It is enough to mark the tx as without RBF and the winner gets it all!
Stop spreading bs.
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
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
We will go from being hunters to leeches!!!
It is enough to mark the tx as without RBF and the winner gets it all!
Stop spreading bs.
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
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
...
Stop spreading bs.
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.
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.
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.
...
Stop spreading bs.
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.
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.
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.
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.
...
Stop spreading bs.
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.
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.
...
Stop spreading bs.
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.
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.
Maybe we won't find the puzzle, but the person who finds it will be the victim. People with bitcoin technical background and any kind of knowledge on this subject should share what they know.
Don't get too upset. I think there are at least 1000 bots already set up and waiting. There will be chaos the moment puzzle 66 public key appears. If there is any money left in that bidding. 
...
Stop spreading bs.
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.
If we find puzzle 66, then someone else will steal it, right?
obviously yes
what is the solution ?
either
a) you know someone who has a monstrous mining rig in the EH/s (exahash per second) range and you could ask him to include your transaction which you manually baked into a block
or
b) you just try extending your luck by risking to make the transaction with the vague hope that it will take place within a few seconds because a block containing your transaction will be mined within the next second.
Note: Please avoid consecutive posts.
Brother, your information is very valuable, this is a very important issue, for this reason, I wrote messages frequently to ensure that this situation is not left behind.
Are you absolutely sure about this? So I'm in shock right now. I think there is definitely another way. Maybe we won't find the puzzle, but the person who finds it will be the victim. People with bitcoin technical background and any kind of knowledge on this subject should share what they know.
If we find puzzle 66, then someone else will steal it, right?
obviously yes
what is the solution ?
either
a) you know someone who has a monstrous mining rig in the EH/s (exahash per second) range and you could ask him to include your transaction which you manually baked into a block
or
b) you just try extending your luck by risking to make the transaction with the vague hope that it will take place within a few seconds because a block containing your transaction will be mined within the next second.
Note: Please avoid consecutive posts.
It is enough to mark the tx as without RBF and the winner gets it all!
Stop spreading bs.
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.
If we find puzzle 66, then someone else will steal it, right?
If someone found #66 and withdrew with an extremely high transaction fee say 10k USD and then within 1 minute the scammers find the private key with kangaroo and submit their withdrawal with lets say 100k USD fee! Would the scammers still get it? Cause this means all the lower bit range puzzles are useless to even attempt.
Correct, but I won't call it a "scammer", perhaps just a leech. And it won't be found within 1 minute. It will be found within a few seconds. The leech will still get 6 bitcoins, so even 200k RBF won't be too far fetched, as he'll get double his money back.
I feel sorry for people who enthusiastically tried puzzle 66 for many months or even a year or two, without realizing even if they found the private key there's no way in hell they will be able to withdraw it because unless you have 20/30/40k sitting around to gamble on the RBF lottery, you're not getting those funds.
The reason this did not happen before is because puzzle 65 and lower had very little to no reward money in it, so it was not worth the bidding war, but this all changes when the puzzle creator decided to put 6+ BTC in each puzzle. Now there's more people sitting and waiting with bots than people trying to solve.
what is the solution ?
It is enough to mark the tx as without RBF and the winner gets it all!
Stop spreading bs.
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.
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