|
albert0bsd
|
 |
April 22, 2024, 04:27:55 AM |
|
Please reply asap.
We only can know the private key of the puzzles that are expected to be on low bit ranges. The method is that I already expose and you quoted if you don't know how to read is not my problem
|
|
|
|
|
brainless
Member


Activity: 491
Merit: 35
|
 |
April 22, 2024, 10:55:48 AM |
|
Anyone have seen or experience with Bitcrack / fork for load starting points by file ?
|
13sXkWqtivcMtNGQpskD78iqsgVy9hcHLF
|
|
|
WanderingPhilospher
Sr. Member
  

Activity: 1498
Merit: 286
Shooters Shoot...
|
 |
April 22, 2024, 03:57:07 PM |
|
Anyone have seen or experience with Bitcrack / fork for load starting points by file ?
Not sure what you are exactly wanting, but the easiest way is to create a Python script to feed the program starting ranges, and to keep track of said starting ranges. Not sure if starting ranges = starting points, but that is what I would do.
|
|
|
|
|
brainless
Member


Activity: 491
Merit: 35
|
 |
April 22, 2024, 05:09:55 PM Last edit: May 01, 2024, 08:27:52 PM by Mr. Big |
|
Anyone have seen or experience with Bitcrack / fork for load starting points by file ?
Not sure what you are exactly wanting, but the easiest way is to create a Python script to feed the program starting ranges, and to keep track of said starting ranges. Not sure if starting ranges = starting points, but that is what I would do. Yes starting ranges by default is sequel, One an other forked Bitcrack today I see in my system, where generated random starting points, onword it's start from each generated point to series Thinking if we load by file starting points, then maybe betterment, By python we can assign single starting point for go on, but only when we need to search in trillions keys And start/stop action waste time for small ranges for million etc
Anyone have seen or experience with Bitcrack / fork for load starting points by file ?
Yes starting ranges by default is sequel, One an other forked Bitcrack today I see in my system, where generated random starting points, onword it's start from each generated point to series Thinking if we load by file starting points, then maybe betterment, By python we can assign single starting point for go on, but only when we need to search in trillions keys And start/stop action waste time for small ranges for million etc Not sure what you are exactly wanting, but the easiest way is to create a Python script to feed the program starting ranges, and to keep track of said starting ranges. Not sure if starting ranges = starting points, but that is what I would do. just see its forked name url = https://github.com/pikachunakapika/BitCrack.gitpikachunakapika its uses random ver [Info] Initializing GeForce GTX 460 [Info] Generating 4,143,104 starting points (158.0MB) these generating point could be load by file ? ReadMe note -r, --random Each point will start in random KEYSPACE
|
13sXkWqtivcMtNGQpskD78iqsgVy9hcHLF
|
|
|
|
|
|
albert0bsd
|
 |
April 23, 2024, 01:44:14 PM |
|
I removed the space between the private keys. See how the first 4 digits of the private key are known? Now guess 66!
Bullshit
|
|
|
|
|
supika
Newbie

Activity: 43
Merit: 0
|
 |
April 23, 2024, 02:19:02 PM |
|
So the private key is starting with 37 in hex?
|
|
|
|
|
|
citb0in
|
 |
April 23, 2024, 02:55:45 PM |
|
I removed the space between the private keys. See how the first 4 digits of the private key are known? Now guess 66!  So the private key is starting with 37 in hex? Nope. It's starting with 34
|
Some signs are invisible, some paths are hidden - but those who see, know what to do. Follow the trail - Follow your intuition - [bc1qqnrjshpjpypepxvuagatsqqemnyetsmvzqnafh]
|
|
|
|
albert0bsd
|
 |
April 23, 2024, 03:04:14 PM |
|
Nope. It's starting with 34
TBH this is also BS, There is no way to know that unless you already solved it.
|
|
|
|
|
|
citb0in
|
 |
April 23, 2024, 05:10:27 PM |
|
Alberto, of course it was BS. That's the point, just emphasizing it 
|
Some signs are invisible, some paths are hidden - but those who see, know what to do. Follow the trail - Follow your intuition - [bc1qqnrjshpjpypepxvuagatsqqemnyetsmvzqnafh]
|
|
|
|
nomachine
|
 |
April 23, 2024, 05:12:40 PM |
|
Nope. It's starting with 34
And the public key starts with 03? 
|
BTC: bc1qdwnxr7s08xwelpjy3cc52rrxg63xsmagv50fa8
|
|
|
|
citb0in
|
 |
April 23, 2024, 05:19:17 PM |
|
No, 05 because of its special compression level 
|
Some signs are invisible, some paths are hidden - but those who see, know what to do. Follow the trail - Follow your intuition - [bc1qqnrjshpjpypepxvuagatsqqemnyetsmvzqnafh]
|
|
|
|
albert0bsd
|
 |
April 23, 2024, 05:19:59 PM |
|
Alberto, of course it was BS. That's the point, just emphasizing it  Nice to know, in this kind of comment I am bad getting the sarcarm of it
|
|
|
|
|
WanderingPhilospher
Sr. Member
  

Activity: 1498
Merit: 286
Shooters Shoot...
|
 |
April 24, 2024, 03:29:44 AM |
|
I removed the space between the private keys. See how the first 4 digits of the private key are known? Now guess 66!  What is this sorcery?!
|
|
|
|
|
mohammedlaoui
Newbie

Activity: 14
Merit: 0
|
 |
April 24, 2024, 09:26:01 AM |
|
guys any result puzzle solved 
|
|
|
|
|
BlackHatCoiner
Legendary

Activity: 2044
Merit: 9825
Avatar for rent
|
 |
April 24, 2024, 09:27:51 AM |
|
Is this puzzle really worth trying to solve? It's been close to a decade. Can someone give me a TL;DR?
|
|
|
|
mohammedlaoui
Newbie

Activity: 14
Merit: 0
|
 |
April 24, 2024, 09:45:36 AM |
|
guys i found a method to get this number -673909/1307674368000*x^15 + 5004253/87178291200*x^14 - 151337/52254720*x^13 + 9320029/106444800*x^12 - 25409989753/14370048000*x^11 + 2192506957/87091200*x^10 - 19011117413/73156608*x^9 + 1200887962891/609638400*x^8 - 3585932821063/326592000*x^7 + 647416874047/14515200*x^6 - 18586394742863/143700480*x^5 + 30899291755337/119750400*x^4 - 274137631043849/825552000*x^3 + 36933161067083/151351200*x^2 - anyone give me how i can use my method  i m realy didnt know how thst method help us to solve the puzzle ??
|
|
|
|
|
brainless
Member


Activity: 491
Merit: 35
|
 |
April 24, 2024, 10:45:18 AM |
|
guys i found a method to get this number -673909/1307674368000*x^15 + 5004253/87178291200*x^14 - 151337/52254720*x^13 + 9320029/106444800*x^12 - 25409989753/14370048000*x^11 + 2192506957/87091200*x^10 - 19011117413/73156608*x^9 + 1200887962891/609638400*x^8 - 3585932821063/326592000*x^7 + 647416874047/14515200*x^6 - 18586394742863/143700480*x^5 + 30899291755337/119750400*x^4 - 274137631043849/825552000*x^3 + 36933161067083/151351200*x^2 - anyone give me how i can use my method  i m realy didnt know how thst method help us to solve the puzzle ?? What is x value ? Is it public key x value ?
|
13sXkWqtivcMtNGQpskD78iqsgVy9hcHLF
|
|
|
mohammedlaoui
Newbie

Activity: 14
Merit: 0
|
 |
April 24, 2024, 11:06:59 AM |
|
In the context of the polynomial generated using the Lagrange interpolation method in Sage, the x value represents the variable of the polynomial. It's not directly related to a public key's x value in cryptography. Here, x is the independent variable in the polynomial. In Sage, when you construct a polynomial, x represents the variable in that polynomial. In the given code: R = PolynomialRing(QQ,'x') f = R.lagrange_polynomial([(0,1),(1,3),(2,7),(3,8),(4,21),(5,49),(6,76),(7,224),(8,467),(9,514),(10,1155),(11,2683),(12,5216),(13,10544),(14,26867),(15,51510)]) for i in range(16): print(f(i)) Here, x is the variable in the polynomial f. So, x in the polynomial formula is just a placeholder for the variable in the polynomial equation, and it doesn't have any direct relationship to a public key's x value in cryptography.
|
|
|
|
|
AlanJohnson
Member


Activity: 185
Merit: 11
|
Is this puzzle really worth trying to solve? It's been close to a decade. Can someone give me a TL;DR?
Of course not ... Think for a moment: why someone keeps so much money locked in btc puzzle instead of cashing it out ? Because he has a much more and this puzzle is used for monitoring current cracking possibilities. It's not some kind of competition someone made for fun. It's made on purpose. Nobody will crack 66 or more bit puzzle using python scripts on home computer. It's a total waste of time. The puzzle is there to meter if there is anybody who can use enough processing power to crack it. And you can only crack it randomly guessing private keys if you don't have public key available. Those with public key available can be solved using kangaroo or bsgs but still it needs too much processing power for an average person for a random guy (that's why the higher that 66 bit ranges are already solved but still nobody can't beat 130bit - FYI Every fifth puzzle here have intentionally released it's public key by the creator) Everything here can be calculated - i mean you can calculate how much processing power you need to crack it, how many and what kind of hardware you need to achieve that. You must invest hundreds of thousands $$$ to even try to crack it.
|
|
|
|
|
|