Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 12.

JDScreesh

jr. member

Activity: 39

Merit: 12

Quote from: saatoshi_falling on April 06, 2024, 03:14:23 AM

Quote from: iceland2k14 on April 05, 2024, 01:47:52 PM

Quote from: JDScreesh on April 05, 2024, 08:57:34 AM

Then I just got the fork coins (only BCH, BSV, eCash, BTG, BCD and CDY) Wink

Good to know that even after 12 minutes the fork coins were intact. So probably the situation is not that bad as we are anticipating

During that time the puzzles had like 0.0064 bitcoin it and on top of that BTC was very low. Why would it create a world wide bot race and why would someone waste hundreds or thousands of dollars in fees to withdraw it?
It only became an issue once the creator increased the puzzle prize and since then, no lower end puzzle have been solved.

I suggest to take whatever time/scripts/motivation you have and try to solve puzzle #80 on wards instead where you will be guaranteed the prize.
Or, ask someone with knowledge about timings which is the lowest feasible puzzle to be solved that will leave enough time to withdraw without a bot snatchingup. I don't know this answer, could be even puzzle #70?

If someone can answer that, that would be great.

Hello again Cheesy

Puzzle #64 was solved on 09/09/2022. It's prize was 0.64 BTC and that day BTC price was US$21.268, so the prize in BTC was like US$13.611

Any puzzle below #80 or #90, I think, can be snatching up by a bot.

Quote from: Woz2000 on April 05, 2024, 06:31:57 PM

64 was solved before the 10x so the values of the fork coins were not as tempting as they are now.

Quote from: iceland2k14 on April 05, 2024, 01:47:52 PM

Quote from: JDScreesh on April 05, 2024, 08:57:34 AM

Then I just got the fork coins (only BCH, BSV, eCash, BTG, BCD and CDY) Wink

Good to know that even after 12 minutes the fork coins were intact. So probably the situation is not that bad as we are anticipating

That day I got like US$165 between BCH, BSV, XEC (eCash) and BTG. It was cool to get that here in a latinamerican country Grin

And, even today, fork coins weren't updated with 10x, so the prize is still the same for the fork coins (ie 0.66 BCH for puzzle #66 and so on)

saatoshi_falling

newbie

Activity: 6

Merit: 0

Quote from: iceland2k14 on April 05, 2024, 01:47:52 PM

Quote from: JDScreesh on April 05, 2024, 08:57:34 AM

Then I just got the fork coins (only BCH, BSV, eCash, BTG, BCD and CDY) Wink

Good to know that even after 12 minutes the fork coins were intact. So probably the situation is not that bad as we are anticipating

During that time the puzzles had like 0.0064 bitcoin it and on top of that BTC was very low. Why would it create a world wide bot race and why would someone waste hundreds or thousands of dollars in fees to withdraw it?
It only became an issue once the creator increased the puzzle prize and since then, no lower end puzzle have been solved.

I suggest to take whatever time/scripts/motivation you have and try to solve puzzle #80 on wards instead where you will be guaranteed the prize.
Or, ask someone with knowledge about timings which is the lowest feasible puzzle to be solved that will leave enough time to withdraw without a bot snatchingup. I don't know this answer, could be even puzzle #70?

If someone can answer that, that would be great.

CY4NiDE

newbie

Activity: 19

Merit: 5

Quote from: BitcoinOtter on April 05, 2024, 05:39:52 PM

I’m seeing so many variations of software throughout this thread and the other one. If I wanted to randomly let my pc waste electricity on weekends doing a random search for #66, what would be my best bet?

I have a 3070 and can switch to my Ubuntu partition if that’s better. I know my chances are probably like 0.001%, but I just want to feel like I’m joining the big boys on the hunt.

Your best option would be some implementation of KeyHunt-CUDA / Rotor-CUDA. Or BitCrack if you want to mess with the stride option.

In terms of raw performance BitCrack lacks tho.

WanderingPhilospher

full member

Activity: 1050

Merit: 219

Shooters Shoot...

Quote from: VinIVaderr on April 05, 2024, 06:55:33 PM

Total number of addresses that start with the same 11 characters = 2^66 / 58^(11 * log2(58))

First, calculate 58^(11 * log2(58)):

58^(11 * log2(58)) ≈ 58^42.614 ≈ 1.2107 × 10^74 (approximately)

Then, divide 2^66 by this result:

2^66 / 1.2107 × 10^74 ≈ 460708775.52

So, the approximate number of addresses that start with the same 11 characters from a key length of 2^66 bits is about 460,708,775.

This made me laugh out loud.

Where did 10^74 come from?

You realize 10^74 is larger than 2^66, correct? Not by a little but a lot.

Thanks for the laugh though! I appreciate it.

VinIVaderr

newbie

Activity: 12

Merit: 0

Total number of addresses that start with the same 11 characters = 2^66 / 58^(11 * log2(58))

First, calculate 58^(11 * log2(58)):

58^(11 * log2(58)) ≈ 58^42.614 ≈ 1.2107 × 10^74 (approximately)

Then, divide 2^66 by this result:

2^66 / 1.2107 × 10^74 ≈ 460708775.52

So, the approximate number of addresses that start with the same 11 characters from a key length of 2^66 bits is about 460,708,775.

Woz2000

jr. member

Activity: 74

Merit: 2

64 was solved before the 10x so the values of the fork coins were not as tempting as they are now.

Quote from: iceland2k14 on April 05, 2024, 01:47:52 PM

Quote from: JDScreesh on April 05, 2024, 08:57:34 AM

Then I just got the fork coins (only BCH, BSV, eCash, BTG, BCD and CDY) Wink

Good to know that even after 12 minutes the fork coins were intact. So probably the situation is not that bad as we are anticipating

BitcoinOtter

newbie

Activity: 4

Merit: 0

I’m seeing so many variations of software throughout this thread and the other one. If I wanted to randomly let my pc waste electricity on weekends doing a random search for #66, what would be my best bet?

I have a 3070 and can switch to my Ubuntu partition if that’s better. I know my chances are probably like 0.001%, but I just want to feel like I’m joining the big boys on the hunt.

iceland2k14

jr. member

Activity: 31

Merit: 52

Quote from: JDScreesh on April 05, 2024, 08:57:34 AM

Then I just got the fork coins (only BCH, BSV, eCash, BTG, BCD and CDY) Wink

Good to know that even after 12 minutes the fork coins were intact. So probably the situation is not that bad as we are anticipating

albert0bsd

hero member

Activity: 828

Merit: 657

Quote from: averagetoaster on April 04, 2024, 07:25:36 PM

i assume any wallet program will just refuse to send a TX due to invalid signature.

Yes that is correct, even if you broadcast it manually the network nodes will not accept your transaction.

JDScreesh

jr. member

Activity: 39

Merit: 12

Quote from: iceland2k14 on April 04, 2024, 09:13:02 AM

How was the Funds from Puzzle 64 transferred. There mist been a tug of war between the bots fighting each other during that time also. So how was it managed? Does someone know the story of that time. Same must have been for the Forks coin transfer of that puzzle as well. Another Tx another bot war ?

Hello there Cheesy

I don't remember if there was a bot war for BTC in puzzle #64, but I got the privkey like 12 minutes after the pubkey was exposed (but only with CPU, I don't have GPU Grin

)

Then I just got the fork coins (only BCH, BSV, eCash, BTG, BCD and CDY) Wink

WanderingPhilospher

full member

Activity: 1050

Merit: 219

Shooters Shoot...

Quote from: JavaSandcrawler on April 04, 2024, 06:02:06 PM

Quote from: albert0bsd on April 04, 2024, 05:34:24 PM

Quote from: jacky19790729 on April 04, 2024, 02:11:44 PM

Code:

● 12-char prefix
20d45a6a7625355df5e5fc41b358e6f8581991e2
13zb1hQbWVsc1111111111111111111111 => 20d45a6a76253570014f1a30288bd88dc3ff6ffb
13zb1hQbWVsczzzzzzzzzzzzzzzzzzzzzz => 20d45a6a76253571d7012502fabee39e6bbaf2b4
1d5b20ad2d2330b10a7bb82b9

FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF / 1D5B20AD2D2330B10A7BB82B9 = 0x8B8758F7C8AD0286
(every 0x8B8758F7C8AD0286 will find 1 address 12-char prefix)

Did you realize that 0x8B8758F7C8AD0286 is near 1/3 of the 66 bit space?

Code:

>>> 2**65 // 0x8B8758F7C8AD0286
3
>>> 2**65 / 0x8B8758F7C8AD0286
3.6694959191703664

It just interesting

It seems to me that these calculations are very abstract, we are missing several steps for ripemd160. If this were really the case, then any addresses could be taken away without problems. I know you Alberto, we corresponded several times in tg. I am grateful to you in many ways, most likely thanks to you I started this activity. I have no goal of emptying other people’s wallets, but if one is found ownerless or from a puzzle, I will definitely send you a significant reward, rly. But, over time, I realized that this is more of a study, most likely this will never happen, maybe. Once again I apologize, I'm using a translator.

If you are talking addresses or better yet, address prefixes (which both are a reflection of their H160), you don't really need to know all of the steps, really any of them, the H160 or the SHA256, etc. You just have to know what comprises an address, in this case base58. And from there you can get a rough idea of how many specific prefixes will be in a specific, certain sized range.

First, the leading 1, isn't part of the prefix, it is a gimme because all of these type addresses start with 1.

Someone correct me if I am wrong, but couldn't we just take range size / 58^number of leading prefixes (minus the starting "1")
If everything is uniformly spread out, over the range/curve (as the experts like to say, but it's not proven, but is a good "guesstimator" of sorts), an 11-char prefix (1 + 11 characters = total of 12 characters), in a 2^65 bit range, there should be around 1.4 of them. 2^65 / 58^11
2^65 / 58^11 = 1.4
2^65 / 58^10 = 85
2^65 / 58^9 = 4,967
2^65 / 58^8 = 288,088

and I guess you could apply a similar logic to H160s as well, but use 16^number of leading H160 prefixes, instead of 58^number of leading prefixes (minus the starting "1")

Anyone?

averagetoaster

newbie

Activity: 2

Merit: 0

these last pages read like schizoposting.

anyway
what stops someone from sending invalid / badly signed TXs to the network so the bots are bogged down cracking the #66 with bad pubkeys? how does the network react in this case? does it just drop the bad transaction? or is it just not possible?
i could not find much on the subject of crafting invalid TXs, as i assume any wallet program will just refuse to send a TX due to invalid signature.

JavaSandcrawler

newbie

Activity: 0

Merit: 0

Quote from: albert0bsd on April 04, 2024, 05:34:24 PM

Quote from: jacky19790729 on April 04, 2024, 02:11:44 PM

Code:

● 12-char prefix
20d45a6a7625355df5e5fc41b358e6f8581991e2
13zb1hQbWVsc1111111111111111111111 => 20d45a6a76253570014f1a30288bd88dc3ff6ffb
13zb1hQbWVsczzzzzzzzzzzzzzzzzzzzzz => 20d45a6a76253571d7012502fabee39e6bbaf2b4
1d5b20ad2d2330b10a7bb82b9

FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF / 1D5B20AD2D2330B10A7BB82B9 = 0x8B8758F7C8AD0286
(every 0x8B8758F7C8AD0286 will find 1 address 12-char prefix)

Did you realize that 0x8B8758F7C8AD0286 is near 1/3 of the 66 bit space?

Code:

>>> 2**65 // 0x8B8758F7C8AD0286
3
>>> 2**65 / 0x8B8758F7C8AD0286
3.6694959191703664

It just interesting

It seems to me that these calculations are very abstract, we are missing several steps for ripemd160. If this were really the case, then any addresses could be taken away without problems. I know you Alberto, we corresponded several times in tg. I am grateful to you in many ways, most likely thanks to you I started this activity. I have no goal of emptying other people’s wallets, but if one is found ownerless or from a puzzle, I will definitely send you a significant reward, rly. But, over time, I realized that this is more of a study, most likely this will never happen, maybe. Once again I apologize, I'm using a translator.

albert0bsd

hero member

Activity: 828

Merit: 657

Quote from: jacky19790729 on April 04, 2024, 02:11:44 PM

Code:

● 12-char prefix
20d45a6a7625355df5e5fc41b358e6f8581991e2
13zb1hQbWVsc1111111111111111111111 => 20d45a6a76253570014f1a30288bd88dc3ff6ffb
13zb1hQbWVsczzzzzzzzzzzzzzzzzzzzzz => 20d45a6a76253571d7012502fabee39e6bbaf2b4
1d5b20ad2d2330b10a7bb82b9

FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF / 1D5B20AD2D2330B10A7BB82B9 = 0x8B8758F7C8AD0286
(every 0x8B8758F7C8AD0286 will find 1 address 12-char prefix)

Did you realize that 0x8B8758F7C8AD0286 is near 1/3 of the 66 bit space?

Code:

>>> 2**65 // 0x8B8758F7C8AD0286
3
>>> 2**65 / 0x8B8758F7C8AD0286
3.6694959191703664

It just interesting

jacky19790729

jr. member

Activity: 56

Merit: 8

Quote from: zahid888 on April 03, 2024, 02:08:04 PM

Are you sure about this? Where did you get this information from? Will you share the source of this information?

Just simple addition, subtraction, multiplication and division math

Code:

● 7-char prefix
13zb1hQ111111111111111111111111111 => 20d45a6a4b704db750a2c2bd3f07159698d80d6e
13zb1hQzzzzzzzzzzzzzzzzzzzzzzzzzzz => 20d45a6a9337b05029b466e310ca53586922d363
   47c76298d911a425d1c33dc1d04ac5f5

FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF / 47C76298D911A425D1C33DC1D04AC5F5 = 0x39106D151
(every 0x39106D151 will find 1 address 7-char prefix)

● 8-char prefix
13zb1hQb11111111111111111111111111 => 20d45a6a758414c1779b72697c32ea57a903e293
13zb1hQbzzzzzzzzzzzzzzzzzzzzzzzzzz => 20d45a6a76c0e5e769b1d1ee8d0e8eb7b1052c9c
   13cd125f2165f8510dba46008014a09

FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF / 13CD125F2165F8510DBA46008014A09 = 0xCEDB8B6C6A
(every 0xCEDB8B6C6A will find 1 address 8-char prefix)

● 9-char prefix
13zb1hQbW1111111111111111111111111 => 20d45a6a76227d5470a6a22c04a0bc87ad048798
13zb1hQbWzzzzzzzzzzzzzzzzzzzzzzzzz => 20d45a6a7627f3b1c8afd8c8cb8a0a66056e7ba1
   5765d5809369cc6e94dde5869f409

FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF / 5765D5809369CC6E94DDE5869F409 = 0x2EDDBD96902D
(every 0x2EDDBD96902D will find 1 address 9-char prefix)

● 10-char prefix
13zb1hQbWV111111111111111111111111 => 20d45a6a76252067051503284157fc9abd37adf4
13zb1hQbWVzzzzzzzzzzzzzzzzzzzzzzzz => 20d45a6a762538831cab3d7a68156376d939819c
   181c17963a5226bd66dc1c01d3a8

FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF / 181C17963A5226BD66DC1C01D3A8 = 0xA9E3CF41CAA3C
(every 0xA9E3CF41CAA3C will find 1 address 10-char prefix)

● 11-char prefix
13zb1hQbWVs11111111111111111111111 => 20d45a6a7625352fc9f79f5d6b915546d55c908e
13zb1hQbWVszzzzzzzzzzzzzzzzzzzzzzz => 20d45a6a7625359a344e13210b21d70cd5d82eb0
   6a6a5673c39f9081c6007b9e22

FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF / 6A6A5673C39F9081C6007B9E22 = 0x267D9CF4E7E91B2
(every 0x267D9CF4E7E91B2 will find 1 address 11-char prefix)

● 12-char prefix
13zb1hQbWVsc1111111111111111111111 => 20d45a6a76253570014f1a30288bd88dc3ff6ffb
13zb1hQbWVsczzzzzzzzzzzzzzzzzzzzzz => 20d45a6a76253571d7012502fabee39e6bbaf2b4
   1d5b20ad2d2330b10a7bb82b9

FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF / 1D5B20AD2D2330B10A7BB82B9 = 0x8B8758F7C8AD0286
(every 0x8B8758F7C8AD0286 will find 1 address 12-char prefix)

========= use My GPU - RTX 3090 ====== EXE: VanitySearch-1.15.4_bitcrack

>> Search 7-char prefix
   VanitySearch.exe -t 0 -gpu -o 2D3C0C.txt --keyspace 2D3C0C00000000000:+39106D151FF 13zb1hQ
   255 keys were supposed to be found, 267 keys were found

   VanitySearch.exe -t 0 -gpu -o 2E2B98.txt --keyspace 2E2B9800000000000:+39106D151FF 13zb1hQ
   255 keys were supposed to be found, 282 keys were found

   VanitySearch.exe -t 0 -gpu -o 3A82BC.txt --keyspace 3A82BC00000000000:+39106D151FF 13zb1hQ
   255 keys were supposed to be found, 289 keys were found

   VanitySearch.exe -t 0 -gpu -o 354BAC.txt --keyspace 354BAC00000000000:+39106D15100 13zb1hQ
   255 keys were supposed to be found, 256 keys were found

   VanitySearch.exe -t 0 -gpu -o 222222.txt --keyspace 22222200000000000:+EFFFFFFFFFF 13zb1hQ
   1076 keys were supposed to be found, 1012 keys were found

>> Search 8-char prefix
   VanitySearch.exe -t 0 -gpu -o 2A82BC.txt --keyspace 2A82BC00000000000:+CEDB8B6C6AF 13zb1hQb
   15 keys were supposed to be found, 10 keys were found

   VanitySearch.exe -t 0 -gpu -o 2D13BA.txt --keyspace 2D13BA00000000000:+CEDB8B6C6AF 13zb1hQb
   15 keys were supposed to be found, 10 keys were found

so I say.... ( puzzle 66 range 0x40000000000000000 - 0x20000000000000000 = 0x20000000000000000 )

13zb1hQbWV 10-char prefix match in ripemd160 hash will be little reference significance
0x20000000000000000 / 0xA9E3CF41CAA3C = 0x3038 ( 10000+ private key has 10 char prefix match compressed address )

13zb1hQbWVs 11-char prefix match
0x20000000000000000 / 0x267D9CF4E7E91B2 = 0xd4 ( 200~250 private key has 11 char prefix match compressed address )

13zb1hQbWVsc 12-char prefix match
0x20000000000000000 / 0x8B8758F7C8AD0286 = 0x3 ( 3~4 private key has 12 char prefix match compressed address )

I have two 11-char prefix match address and private in 66 range

13zb1hQbWVsoqtKv6BRRpKZ81caz7KWea5
13zb1hQbWVsSAaNvpKfmc43NsuAj33LyRM

albert0bsd

hero member

Activity: 828

Merit: 657

Quote from: iceland2k14 on April 04, 2024, 09:13:02 AM

How was the Funds from Puzzle 64 transferred. There mist been a tug of war between the bots fighting each other during that time also. So how was it managed? Does someone know the story of that time. Same must have been for the Forks coin transfer of that puzzle as well. Another Tx another bot war ?

Well that is a good question, but unless someone recorded the mempool TX related to that address at that time there is no way to know for sure because after some time all those TX that didn't win the "bid" fight are promptly discarded from most of the servers that show TX details online.

One example of that is what i write in Is FullRBF allowing double spend? where i show some TX tha are related to one of my addresses and those TX no longer exists because they were FullRBF with new TX.

iceland2k14

jr. member

Activity: 31

Merit: 52

How was the Funds from Puzzle 64 transferred. There mist been a tug of war between the bots fighting each other during that time also. So how was it managed? Does someone know the story of that time. Same must have been for the Forks coin transfer of that puzzle as well. Another Tx another bot war ?

brainless

member

Activity: 316

Merit: 34

Quote from: ccinet on April 04, 2024, 08:49:44 AM

Quote from: saatoshi_falling on April 04, 2024, 07:24:50 AM

There's simply no feasible way to withdraw the funds on lower end puzzles like #66. It will be snatched up by bots. Not maybe, but it's 100% guaranteed. There will be hundreds of withdrawal transactions with varying fees all battling each other. You will simply be left in the dust.

It's a fundamental flaw in this puzzle that was originally aimed to bring the community together and try to solve the puzzle. The puzzle creator did not think much about the puzzle, and this is proof of it.

The only way the original solver will get the funds of puzzle #66 is if they can prove to be in possession of the private keys directly to the puzzle creator. The puzzle creator then withdraws the funds at a 6BTC fee (they can afford it) to your address.

The goal of the puzzles is to demonstrate the strength of bitcoin and so far it has more than proven that.
I don't think the creator cares about the method by which the private key is obtained, whether it's years of brute force, whether it's discovering a pattern or through social engineering, the owner of the private key is the owner of the coins. If there is more than one transaction with the same private key, the protocol designed to resolve double spending will be applied.
End of story

How much tx you have played, ?
Sure tx you had made is different subject.

ccinet

newbie

Activity: 33

Merit: 0

Quote from: saatoshi_falling on April 04, 2024, 07:24:50 AM

There's simply no feasible way to withdraw the funds on lower end puzzles like #66. It will be snatched up by bots. Not maybe, but it's 100% guaranteed. There will be hundreds of withdrawal transactions with varying fees all battling each other. You will simply be left in the dust.

It's a fundamental flaw in this puzzle that was originally aimed to bring the community together and try to solve the puzzle. The puzzle creator did not think much about the puzzle, and this is proof of it.

The only way the original solver will get the funds of puzzle #66 is if they can prove to be in possession of the private keys directly to the puzzle creator. The puzzle creator then withdraws the funds at a 6BTC fee (they can afford it) to your address.

The goal of the puzzles is to demonstrate the strength of bitcoin and so far it has more than proven that.
I don't think the creator cares about the method by which the private key is obtained, whether it's years of brute force, whether it's discovering a pattern or through social engineering, the owner of the private key is the owner of the coins. If there is more than one transaction with the same private key, the protocol designed to resolve double spending will be applied.
End of story

saatoshi_falling

newbie

Activity: 6

Merit: 0

There's simply no feasible way to withdraw the funds on lower end puzzles like #66. It will be snatched up by bots. Not maybe, but it's 100% guaranteed. There will be hundreds of withdrawal transactions with varying fees all battling each other. You will simply be left in the dust.

It's a fundamental flaw in this puzzle that was originally aimed to bring the community together and try to solve the puzzle. The puzzle creator did not think much about the puzzle, and this is proof of it.

The only way the original solver will get the funds of puzzle #66 is if they can prove to be in possession of the private keys directly to the puzzle creator. The puzzle creator then withdraws the funds at a 6BTC fee (they can afford it) to your address.

Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 12. (Read 187143 times)