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Author Topic: == Bitcoin challenge Puzzle : Seed bruteforce attemps  (Read 303 times)
iceland2k14 (OP)
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March 17, 2024, 12:06:27 PM
Merited by ABCbits (2)
 #1

There has been several attempts to solve this bit range Puzzle https://bitcointalk.org/index.php?topic=5218972.0 through bruteforce of PrivateKey either the Address itself or the PubKeys through Kangaroo and BSGS algos.

This thread is another approach (a very slow one) which goes through the exact same procedure by which these puzzles were generated at the first hand. Creating the Wallet -> Mnemonics -> Seed -> List of Privatekeys -> Masked Privatekeys -> Check for known privatekeys collision.

Code : https://github.com/iceland2k14/btc_words/blob/main/seed_puzzle.py

This is more of an alternative exercise where if successful then all the remaining puzzles of ~ 1000 BTC will be solved all together. I know the probability of success is nearly zero. But still wanted to try to see how fast we can do it and what are the hurdles in this way. Any existing tool with what speed?

There are some assumptions which i think maybe correct/incorrect. Every opinion with logic could help in understanding more.
1. Wallet was new and puzzles 1-256 should corresponds to the Address index 0 to 255 from the wallet. Creator is very careful in making it. So this assumption maybe correct.
2. These Keys are sequentially generated by masking deterministic wallet privatekeys. So if any collision can be found with 10 masked Keys that is most likely the Correct seed.
3. The Puzzles are from year 2015 so the entropy bits could be either 128 or at max 192. Very little chance for 256.
4. Mnemonics used might have been English words, most probabilistic scenario.
5. Keys are generated using the standard BIP44 path with No Hardened Address route.

Assumption 2, 3, 4, 5 are dynamic and can be changed in the code easily. But if assumption 1 is incorrect then it is more difficult.

BTW, Just in case if the collision happens then you would see a result something like this image.
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March 17, 2024, 09:26:02 PM
 #2

What evidence do you have besides just making an assumption that the puzzle challenge's private keys could be from a hierarchical deterministic wallet?

I'm not a wallet historian and thus can't recall easily at what period of time what kind of deterministic wallets were available.

To me this brute-force approach seems unfeasible for various reasons:
  • too many variables and assumptions creating an even vaster search space
  • slow derivation due to PBKDF2 rounds involved in entropy-->seed-->extended private key-->further private key derivation
  • inability to reduce variables and assumptions from already found existing private keys

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c.h.
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iceland2k14 (OP)
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March 18, 2024, 06:35:31 AM
 #3

What evidence do you have besides just making an assumption that the puzzle challenge's private keys could be from a hierarchical deterministic wallet?

This is described very clearly by the Creator itself ....
Quote

Regarding your other concerns i am aware that
  • Yes this is a very very slow approach. How fast can it be done. Kwords/s or Mwords/s  or more ?
  • Assumptions, think in terms of probabilities. How strong the assumptions are.
  • The ability to use existing known privatekeys, any idea for any shortcut during the chain code generation process ?

The ability to exactly follow the same original process and ability to solve all the puzzles together is definitely an advantage, even if this can be called impractical approach.
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March 18, 2024, 07:07:41 AM
 #4

So how did the person who claims to be a constituent managed to create the HEX address 0000000000000000000000000000000000000000000000000000000000000000000000000000000 000001?
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March 18, 2024, 11:04:28 AM
 #5

Quote
So how did the person who claims to be a constituent managed to create the HEX address 0000000000000000000000000000000000000000000000000000000000000000000000000000000 000001?
The real private key was just bit-masked with appropriate bits. Garlo Nicon described it here: https://bitcointalk.org/index.php?topic=4453897.msg61848712#msg61848712

Of course, the real formula for making it, could be different than he described. For example: you can take the upper half, or the lower half of the key. You can use big endian, or little endian. There are many options, and even if you assume that everything would be guessed perfectly, then still, getting it right could be quite hard. However, there are 65 first private keys, which are revealed (and some more private and public keys above that), and getting the exact master private key could be possible in theory, because it seems there are enough revealed bits, to get only a single solution.

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satashi_nokamato
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March 19, 2024, 11:42:42 PM
 #6

Would it work if we place offsets of all puzzles converted to addresses?  That way even if 1 key is found then at least 1 puzzle could be solved.
iceland2k14 (OP)
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March 24, 2024, 01:59:55 PM
Merited by vapourminer (1)
 #7

Would it work if we place offsets of all puzzles converted to addresses?  That way even if 1 key is found then at least 1 puzzle could be solved.
There will be 2 Problems if we do this.
  • Decrease in performance. Speed<1/10. As we are not stopping till privatekey but instead reaching till Address. Also we have to derive all unsolved 85 puzzles derivation path. currently we are checking collision with 11 privatkeys only
  • We have to rely on a partial collision of masked Key of some other seed with same address but exactly at the same derivation path level. First is still possible but 2nd is very improbable
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March 26, 2024, 05:51:46 AM
Last edit: March 26, 2024, 06:33:44 AM by nomachine
 #8

So how did the person who claims to be a constituent managed to create the HEX address 0000000000000000000000000000000000000000000000000000000000000000000000000000000 000001?

It is simple script Grin

If you know what is last key - e.g. '910ed3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b'

Python script as an example

Code:
key = '910ed3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b'
last_index = 0
while last_index <= 255:
    mask = 1 << last_index
    mask_hex = mask + int(key, 16) % mask
    last_index += 1
    hexa = "%064x" % mask_hex
    print (hexa)

result

Code:
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0000006b5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
000000ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
000001ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
000003ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
000007ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
00000bab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
000013ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
000033ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
000053ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
0000d3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
0001d3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
0002d3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
0006d3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
000ed3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
001ed3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
002ed3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
004ed3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
008ed3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
010ed3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
030ed3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
050ed3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
090ed3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
110ed3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
310ed3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
510ed3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
910ed3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b

But no one knows the exact formula from the last private key to 1.
I even think he used logarithms and polynomials as steps between private keys..
So it ends up looking like there is no pattern.
But it's not random either according to the polynomial analysis.
https://bitcointalk.org/index.php?topic=1306983.msg63730770#msg63730770
There is an exact math formula for making this puzzle with some script, errors = ZERO.  
With high decimal precision (mp.dps = 20 at least)
And the formula is in the creator's mind.
iceland2k14 (OP)
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March 26, 2024, 04:31:54 PM
 #9

Simple Math is, Please Never ever mix the Integer modulo calculation with fraction, decimal, float, double or whatever 20 precision. They will never go along. Treat them as modulo Maths only.

log2 is just giving you the bit representation. Everone knows the Puzzle 66 is between the range 2**65 to -1+2**66 which is exactly between the bit 65 to 66. Same is for all the Puzzles. So off course there will be a straight regression line if you fit the log2 of known puzzle key values with puzzle number.
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