Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 225. (Read 244277 times)

That's just wishful thinking.
The most sane reason he/they did is because the "community" already discussed that puzzle 66+ is already hard enough to be worthy of that prize.

Also, the creator (for some reason) tends to include puzzle #1 whenever he decided to increase the prize.
Like when he spent 161~256 and sent it to the previously unclaimed ranges, he included #1 to the recipient: 5d45587cfd1d5b0fb826805541da7d94c61fe432259e68ee26f4a04544384164

I think when the creator increased the value of puzzle he was trying to give us some hint, by putting 0.09 bitcoin in puzzle1, maybe puzzle1's address or sha160 of the address or integer value of sha160 or its compressed or uncompressed Public's or public's integer, all this may be seed or there may be any relation with puzzle1, or there may be some conspiracy to mislead us.

what improvements - an example is possible? He is a brute force on the moon - a brute force, and in Antarctica)) it all depends on the power of the GPUs and how many of them there are! I'm not a laboratory rabbit Roger - to test meaningless ranges on me! There is a science of mathematics - it clearly shows - an unthinkable time!! And this is an unthinkable time - it does not depend on the size of the prize in any way!!!

What hint do you need more than having such amounts which are directly pointing to the exact bit range where the keys reside? It is a game and a test, we play this game and they use our feedback to analyze the system.

Have you missed the recent movements and the signs of improving the old tools to make them faster?
If it wasn't for the latest prize increase, we'd be scratching our heads and using the old stuff without any serious improvements being done.

Don't sit around and wait for others to come up with a solution, try 50 methods, if all of them failed try 50 more, we never achieve anything in life if we stop trying when we hit a dead end, finding a way around it is the challenge and that should be entertaining to find without frustration, don't worry no one is going to collect all the puzzles any time soon, we have time and we need to focus without stressing ourselves.😉

The creator said - these are deterministic addresses with keys that are truncated to the right bits! I think it's true! The keys have no dependency and no logic! Not only were they randomly generated, they were also cut off after that!!! No need to look for a black cat in a dark room - especially if it's not a cat there!!)) Without hints to the main key - only bruteforce!!! It's very sad!

Faith will move mountains, but knowledge will move them faster. The keys position within its range are randomly. You can proof yourself, just check the keys and calculate its position. That's no rocket science. Hint: look at puzzle 25, 31, 60, 63 to get a feeling of what a high position is and puzzles 38, 50, 85 to understand what a low position means There is no "narrowing the search range solution" for this. You wouldn't have solve the mentioned puzzle otherwise. The key positions for the known keys of today for this puzzle vary from 0.069 (puzzle 38) to 0.97 (puzzle 60), while disregarding puzzles 1-10.

Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense.

Did I forget to mention there is no possible way to determine any specific narrowed down range for an address, that is even if we know the exact start/ end range, what you can only do is to guess a small range and then trying to brute force it, doing that would be equal to searching half the previous puzzle's range, considering even if you could shrink the 66 range to 25%, you'd be searching half of 65's range.

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I have looked at solved puzzles for weeks to find a pattern or to just reduce the search range for unsolved ones, I was able to reduce the range of the next puzzle by 10%, lol.

What you should know, I believe Satoshi placed the keys at least in 50% + of each range, which makes solving them counterproductive financially.

https://privatekeyfinder.io/

Give me the same puzzle as the creator with provided keys until 65 and I will give you range narrowed down by 50-70% percent for #66.

period. There is no 'but...'

what makes you think that you have an advantage of being capable renting a 4090 card? Seriously, that's amusing. Most of us can rent cloud computing power, I can rent 128x4090 but that won't be helpful cause you either run out of money or time. i don't want to intimidate you, but separate yourself from the illusion that you or anyone else can anticipate a pattern, because there is no such pattern.

To help you understand, here is a simple example. In this example I play the puzzle creator and create the keys for the puzzle 50-55. Here are my made up numbers:

#50: 27F9B1013CAA4
#51: 7FEF1D8A4141A
#52: 94FBB3882E756
#53: 1FFF93C8AB39DF
#54: 2010A46AB8CC41
#55: 6D7DB29029A8F5

Now it's your turn. Show me the pattern. You can't? Ok, so show me the formula so you can exclude a certain search range that will help reducing calculation time. I'm curious to see your findings...
Hopefully you now understand what it's all about 

https://poe.com/ChatGPT


ask chatgpt it will share you all the code You keep your money and your algorithm becomes your resource for the next puzzle

There is not pattern, that's true. All next puzzles are just random numbers masked with leading zeros. But if you use math you can narrow down the range of searching by 50-70% for 90% of puzzles. I've formula what shows that I need to compare only 9 600 000 000 000 000 000 addresses with address for puzzle #66. I need someone who can help creating a script for that because I'm not into IT much. I can rent hundred 4090 for that and share with reward.

In my humble understanding, he -as many others- thinks that there is a pattern and he tries to predict it. But it's as simple as the puzzle creator said: THERE IS NO PATTERN. The puzzle creator would not be so foolish to use a predictable pattern, would you in his position?

Why are you wasting your time on addresses and their 160 bit binary values? Don't you know an address does not exist in bitcoin related equations? An address is just a name base58 encoding gives us, a rmd160 is just a name public key hash gives us, public key hash is just a name public key gives us, public key is not a "name" given by private key.

That's the only mathematical relationship we should focus, leave the rest be, even if you manage to brute force 66, what about 67? so the main focus should be on public keys, because hash256, hash160, base58 etc are just distraction, they are the firewalls and you are trying to solve a 160 bit firewall to then reach a 256 bit firewall. Lol.

-p = "Public Key in hex format (compressed or uncompressed)
-keyspace "Keyspace Range ( hex ) to search from min:max. default=1:order of curve
-ncore= "Number of CPU to use. default = Total-1
-n = "Total range search in 1 loop. default=72057594037927935
-rand = "Start from a random value in the given range from min:max and search 0XFFFFFFFFFFFFFF values then again take a new r
-rand1= "First Start from a random value, then go fully sequential, in the given range from min:max

run the following code in the same folder as the files, write the range you want to search, set the settings, save it as a bat file and run it

kangaroo.py -p 02CEB6CBBCDBDF5EF7150682150F4CE2C6F4807B349827DCDBDD1F2EFA885A2630 -keyspace 800000000000000000000000000000:FFFFFFFFFFFFFFFFFFFFFFFFFFFFFF -n 720575940379279350 -rand -ncore 7

No here you must use one pubkey but in https://github.com/JeanLucPons/Kangaroo you can use CPU or GPU or both with multiple pubkeys.

The script is easy to run from cmd.
You should first install python package (For here https://www.python.org/downloads/ )
Then use this format in cmd(HEX = Hexadecimal)
python kangaroo.py -p PUBLICK_KEY_HEX -keyspace START_RANGE_HEX:END_RANGE_HEX

For example(search in range 80):
python kangaroo.py -p 0352b1af31d67e6a83ec7931c148f56b0755ce40c836f20c6fe2b6da612c89cf3e -keyspace 80000000000000000000:100000000000000000000

who is fast in programming, try to make such an analysis.

for example rmd160 puzzle 66 like this

13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so 20d45a6a762535700ce9e0b216e31994335db8a5
0010000011010100010110100110101001110110001001010011010101110000000011001110100 1111000001011001000010110111000110001100110010100001100110101110110111000101001 01 160 len

"1" 73, "0" 87

according to this criterion

160!/73!/87!
50039953558241343191231898620403129563706328000

50039953558241343191231898620403129563706328000/2^65 1356335658972975302954605575

2^160/50039953558241343191231898620403129563706328000 29
2^65/29 1272189246462727697

2^65/2^20 35184372088832

2^160/2^65 39614081257132168796771975168
50039953558241343191231898620403129563706328000/39614081257132168796771975168  1263186017957493013   \
                                                                                           35184372088832×35968 1265511495291109376     > 2^60-2^61  
                                                                                                                 2^65/29 1272189246462727697   /

for every 1048576 step of puzzle 66, will fall around ~36000 "1" 73, "0" 87 and if we add fishing on the first 20 bits (for example)
001000001101010001011010011010100111011000100101001101010111000000001100111010011110000010110010000 1011011100011000110011001010000110011010111011011100010100101

then, based on the probability of dropping 20 bits, you need 1048576 outcomes

1048576/36000 29

1048576 × 30 31457280
1048576 × 29 30408704  there will be only 1 00100000110101000101 "1" 73, "0" 87
1048576 × 28 29360128

what is a full turn for example by 3

001
100
010

010
100
001

100
001
010

there may be such

100
100
001

001
001
001

etc

but in theory, when hashing, the data is simply shuffled, that is, rotated

this means that 20 bits (1048576  steps) in the first 00100000110101000101 will simply move to another place in the second (1048576  steps), third (1048576  steps), etc.

1048576×1048576 = 1099511627776 1 twist

2^65/1048576 = 35184372088832

35184372088832/1099511627776 32 twists for all puzzle 66


1048576×32 = 33554432 (there will be only 1 00100000110101000101 "1" 73, "0" 87)

2^65/33554432 = 1099511627776

all puzzle be

33554432 steps by 1099511627776 len or

1099511627776 steps by 33554432 len  

during the analysis, 1-3 drops out on such steps

we can rotate this space as we like, even take a square

6074001000
6074001000

imagine that we fill with zeros those addresses that do not suit us according to the sorting criterion and mark 1 those that do

we will get a similar picture

000001000001000100000000000000000001000000100000000000000001
001000000000000010000000000110000000000000000100000000100000
etc...

if we take another piece of 20 bits from the address, it will behave similarly

11011011100010100101

so these pieces will jump around the whole puzzle according to the random distribution and in total, as I wrote above, there will be 32 full turns

the idea is to take and randomly generate all possible collisions

select statistics from the puzzle space and try to jump by sorting the template


2^10*2^10                                                           divide to state 2^10, abbreviated example  2^65/33554432 = 1099511627776 (33554432  can be divided into 1024 parts)

001 010                                                                      000 000    <  000000000000000000000000000000000000001000000000000100000000000000  33554432 step
100 100                                                                      000 000    <  000000000000000000000000000000000000010000000000000000000000000001  33554432 step
010 001                                                                      000 000    <  000000000000000000000001000000100000000000000000000000000000000000  33554432 step
  

Bro thanks for replying, can you pl tell me How to run it in cmd with my particular range & pub key, plus can it take multiple pubkeys at a time, all within same range of 2^80??

Check this https://github.com/iceland2k14/kangaroo

Is there any python script for BSGS or Kangaroo algorithm to search with bit range of 2^80???

This one is not working, I tested it. https://github.com/Telariust/pollard-kangaroo/blob/master/pollard-kangaroo.py