Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 250.

liupengpeng

newbie

Activity: 2

Merit: 0

Quote from: Bulista on December 28, 2015, 11:12:47 AM

Hi guys,

In continuation to this thread: https://bitcointalk.org/index

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU77MfhviY5
1EhqbyUMvvs7BfL8goY6qcPbD6YKfPqb7e
Biginteger PVK
Address 5:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU7Dq8Au4Pv
1E6NuFjCi27W5zoXg8TRdcSRq84zJeBW3k
Biginteger 1PitScNLyp2HCygzadCh7FveTnfmpPbfp8
Big
Hex PVK value:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU7hDgvu64y
1McVt1vMtCC7yn5b9wgX1833yCcLXzu1L
Address 11:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUGxXgtm63M
1PgQVLmst3Z314JrQn5TNiys8Hc38TcXJu
Biginteger PVK value: 1155
Hex PVK value: 483

Address 12:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUW5RtS2JN1
1DBaumZxUkM4qMQRt2LVWyFJq5kDtSZQot
Biginteger PVK value: 2683
Hex PVK value: a7b

Address 13:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUspniiQZds
1Pie8JkxBT6MGPz9Nvi3fsPkr2D8q3GBc1
Biginteger PVK value: 5216
Hex PVK value: 1460

Address 14:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFVfZyiN5iEG
1ErZWg5cFCe4Vw5BzgfzB74VNLaXEiEkhk
Biginteger PVK value: 10544
Hex PVK value: 2930

and so on...

until the addresses 50 (1MEzite4ReNuWaL5Ds17ePKt2dCxWEofwk) it was already cracked by someone.

Any ideas what's the formula behind the generation of these addresses?

Address 2, pvk decimal value: 3
Address 3, pvk decimal value: 7
Address 4, pvk decimal value: 8
Address 5, pvk decimal value: 21
Address 6, pvk decimal value: 49
Address 7, pvk decimal value: 76
Address 8, pvk decimal value: 224
Address 9, pvk decimal value: 467
Address 10, pvk decimal value: 514
Address 11, pvk decimal value: 1155
Address 12, pvk decimal value: 2683
Address 13, pvk decimal value: 5216
Address 14, pvk decimal value: 10544
Address 15 and after, pvk decimal value: ?

The prize would be ~32 BTC

EDIT: If you find the solution feel free to leave a tip

1DPUhjHvd2K4ZkycVHEJiN6wba79j5V1u3

Desyationer

newbie

Activity: 34

Merit: 0

I am convinced that 64 is a hoax, otherwise the key would have been found. How many pools with capacities were involved!.... And how many losers like me! I spent a month with 1400mk for 64 in random and sequential mode. I gave up this idea with 64 and then tried to make money on the stock exchange, but in the end, out of my $3000, now there are less than $300 left!!!! These are all my savings that I saved up for about a year!!! Bmtcoin and manipulators I hate you!!! Bitcoin was created to make people poor!

Homeless_PhD

newbie

Activity: 9

Merit: 1

Hi!

I've discovered this "32 BTC" Puzzle nearly a year ago when i'd fallen into "desperacy trap" - that was really bad times for myself. I have tried a few simple things with this puzzle just trying to manage my own "pain" through working on some mind-taking task, and that worked - the hope to get a prize and the task taking my mind away from my real life problems softened my feelings. Later i'd backed to work on my PhD thesis and have not returned to this Puzzle since that. The last 4 months was even worse for myself: i've made a huge mistake on investing (had invested my savings into scam i am not even want to talked about it, very stupid - now i see) and the full war against my country started (live on the south of Ukraine). Now i am at nearly zero wallet and under occupation: have no internet for the last few weeks (they have returned us mobile network today - but who knows for how long), most of the time we have no GSM,radio,TV, lots of Russian propaganda calling us a nacists - we just peaceful Ukrainian people that just want to live in democratic country (so ridiculous, so stupid and so humiliating).

What i think of the puzzle - sure the 64\120 are next, sure the scenario where creator still has the private keys is very probable (he has made Kangaroo search path for 5th keys by those small transactions). Do i want, same as you, to get those funds (at least from 1 remained wallet ~ 1 BTC)? SURE - in normal life where no war goes on my countries lands those ~40k$ could had dramatically (TREMENDOUSLY) change my life. I could actually bought a nice comfy flat for that BTC.

But anyway - taking out some BTC from "32 BTC puzzle" by author will ruin all the mood of this contest, so i hope he\she is not planning anything like that.

Time to time i will try to brute the keys with hope to change my life in easy way, will try to guess the hidden pattern on the private keys but not spending much time to not stimulate the schizophrenia like the "Mind Games" hero (yeah, i know that based on a real mathematician) - that what we could in our situation.

P.S. I am physicist not a mathematician, and so i really believe the "straight cryptographic problem" has its "inverse solution" - the inverse problem i assume has much easier solution than simple (or not so simple) brute.

(https://github.com/HomelessPhD/BTC32)

Jolly Jocker

jr. member

Activity: 49

Merit: 1

Quote

Hi,
I have been using a couple of your codes that you have posted here and your ideas are very fascinating, I have become a big fan of yours. i found a code on github and i am very excited to show you, my hope on this share is that you might come up with your own way to improve the code effency in a way that it can be used for puzzle 66 and 69. i hope that you would share the improved version if this code i would greatly appreciate it if you would. so here is the link:-
https://github.com/Isaacdelly/Plutus
i hope to hear from you, will be waiting in anticipation to a positive response

hey hey shelby0930!!

A big thanks for your nice words, this gives me impetus for new projects in the future! Cool

A small tool for you personally.!!
Happy hunting!! XD

For more infos and link, send me PM..

greetings Jolly! Wink

shelby0930

newbie

Activity: 27

Merit: 4

Quote from: Jolly Jocker on March 26, 2022, 01:13:19 PM

little experiment...

this code creates a sha256 key from str(random.random()), which is fed to the next "ice.get_sha256().hex()".
This is repeated, in this case + 2 times.
The created keys are concatenated ( keys = key1+key2+key3 ).
In the next step, the entire key is divided into 16-digit (64-bit) pieces and read out.
a filter keeps the keys in the desired range...

Happy hunting! XD

Code:

import time
import random
import secp256k1 as ice
from ctypes import c_int
from multiprocessing import Value, Lock, Process

y=1000000
cores=4
counter = Value(c_int)
counter_lock = Lock()
def process1(number,counter,):
while True:
t = str(random.random())
key1 = ice.get_sha256(t).hex()
key2 = ice.get_sha256(key1).hex()
key3 = ice.get_sha256(key2).hex()
keys = key1+key2+key3
for i in range(0, len(keys), 16):
line = keys[i:i+16]
if line.startswith("8") or line.startswith("9") or line.startswith("a") or line.startswith("b") or line.startswith("c") or line.startswith("d") or line.startswith("e") or line.startswith("f"):
with counter_lock:
counter.value += cores
cv = str(counter.value)
addr = ice.privatekey_to_address(0, True, int(line, 16))

if addr.startswith("16jY7q"):
print('\n\n Pattern:',('0x'+line),addr,'\n')

if addr == "16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN":
print ('\n\n TARGET FOUND!!:',addr,'\n\n')
file=open(u"16jY.Info.txt","a")
file.write('\n ' + ('0x'+line) + ' | ' + addr)
file.close()
wait = input("Press Enter to Exit.")
exit()

if (counter.value) % y == 0:
print(' CNT:',cv.zfill(10),' Recent:',('0x'+line),addr,end='\r')

if __name__ == '__main__':
t = time.ctime()
print('',t)
number = y
workers = []
print('\n K E Y - C H O P P E R - 64 \n')
print('\n ===============| |===== KEYS =======|============ ADDR ================| \n\n')
for r in range(0,cores):
p = Process(target=process1, args=(number,counter,))
workers.append(p)
p.start()

for worker in workers:
worker.join()

donate BTC: 1DonateZNR9BUaCqJTgXCoyyCpRSosFujR

the generation of the keys on a time basis (ms) is too slow. the result was double, triple and multiple repetition of the keys in a row.
For this reason I switched to random.random()...
the better way ^^

Sat Mar 26 20:40:24 2022

K E Y - C H O P P E R - 64

=============| |===== KEYS =======|============ ADDR ================|

CNT: 0032000000 Recent: 0xaaf4c88e4a40a299 175NYisjDfbxUyoonY8rEbghWvBbyMGPRj

Pattern: 0x9a5da0765a81d90f 16jY7qNwsgykvkAzm8oiwVSnC71KsZsXqM

CNT: 0304000000 Recent: 0xb338bf39e2ba9922 1EfbHH9BPXSDuibzn3gCao9aQB6Chr7C3G

[moderator's note: consecutive posts merged]

Hi,
I have been using a couple of your codes that you have posted here and your ideas are very fascinating, I have become a big fan of yours. i found a code on github and i am very excited to show you, my hope on this share is that you might come up with your own way to improve the code effency in a way that it can be used for puzzle 66 and 69. i hope that you would share the improved version if this code i would greatly appreciate it if you would. so here is the link:-
https://github.com/Isaacdelly/Plutus
i hope to hear from you, will be waiting in anticipation to a positive response

m4nd4xiality

newbie

Activity: 1

Merit: 0

For the laymen or people just getting started, I am referencing some online tools that helped me when I started, should get the engine started at least for a neophyte. This is a combination of tools I have used/am using.

*This is a great tool for converting HEX into WIF or vice versa, as well as related hashes/checksums in the event that special day comes when you crack a puzzle privkey.

https://gobittest.appspot.com/PrivateKey

*Have not read any mention of this program yet, but I found Plutus on GitHub, and yes I know what you're thinking, randomly brute forcing the entire space is a waste of time, and you very well could be right. However, as a lot of this task is based on three determining factors: 1)luck/chance, 2)access to impressive hardware and/or 3)coding/computer skills, it may be sufficient for someone just wanting to dabble in the project without much knowledge.

I will mention that Plutus uses, in my opinion, an inefficient method for checking BTC wallet balances via deprecated databases. It uses 'btcposbal2csv' to dump all known BTC wallets with positive balances into a database which it checks against. I read recently that blockchain.com limited/restricted API calls for balance checking, because i'm sure with all the utilities being built and running they are taxing the hell out of blockchain.com's resources. Other than that, I like how this tool was built. This is clearly a more rogue approach as it does not search a specific private key space, but takes a lottery approach.

https://github.com/Isaacdelly/Plutus

*Original BitCrack repo on GitHub.

https://github.com/brichard19/BitCrack

*I see this tool was recently mentioned, but I am going to list it here anyway for convenience and repetition. Bitcoin Key Compression Tool from Ian Coleman, which can convert a public key into respective compressed/uncompressed addresses.

https://iancoleman.io/bitcoin-key-compression/

*Additionally, Ian Coleman provides another tool on his site which will post Derived Addresses from randomly generated or provided mneumonics for BTC and others. It also generates a QR code for the mneumonic related address for convenience.

https://iancoleman.io/bip39/?entropy-type=hexadecimal

*In the event you capture a key that has not had its funds spent, I used this tool back in 2017 to check Bitcoin Cash and other forks post-facto. From what I can tell the site still works (supports P2PKH and P2SH addresses).

http://www.findmycoins.ninja/

*For reference and historical data on the subject, Large Bitcoin Collider.

https://lbc.cryptoguru.org/

...also from that site is private key directory.

https://lbc.cryptoguru.org/dio/315469542924288#5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nqkMQzN75Kc7TP39

*For those interested in other upcoming or currently unsolved puzzles, visit the r/bitcoinpuzzles forum on Reddit.

https://www.reddit.com/r/bitcoinpuzzles/

*Binary/Decimal/Hex converter for individual cases or ease of use; this particular tool can handle large integers.

https://www.rapidtables.com/convert/number/decimal-to-hex.html

*More to the point of private keys, here is a random privkey generator.

https://learnmeabitcoin.com/technical/private-key

*If you are just starting out, and to give you an idea of how monumentally infinitesimal the scope of this project is, you can play around with generating random hex values specified by the number of digits/bits desired. This of course is a cumbersome manual solution for newcomers to get an idea of what the process looks like inside a program like BitCrack/Plutus.

https://www.browserling.com/tools/random-hex

*Another sake of convenience, if you do scrape a key worth sweeping, use this QR code generator with WIF to sweep easily in Mycelium or otherwise.

https://www.the-qrcode-generator.com/

*This article goes into detail about 'baby step, giant step' and related math.

https://andrea.corbellini.name/2015/06/08/elliptic-curve-cryptography-breaking-security-and-a-comparison-with-rsa/

*Program which brute forces a private key using 'baby step, giant step' via public key.

https://gist.github.com/jhoenicke/2e39b3c6c49b1d7b216b8626197e4b89

*Existing threads i'm aware of, starting with original and ending with this one.

https://bitcointalksearch.org/topic/brute-force-on-bitcoin-addresses-video-of-the-action-1305887

https://bitcointalksearch.org/topic/bitcoin-puzzle-transaction-32-btc-prize-to-who-solves-it-1306983

https://bitcointalksearch.org/topic/archive-bitcoin-challenge-discusion-5166284

https://bitcointalksearch.org/topic/bitcoin-challenge-transaction-1000-btc-total-bounty-to-solvers-updated-5218972

*In case you're just bored and want to brush up on your public key cryptography skills.

https://www.math.auckland.ac.nz/~sgal018/crypto-book/crypto-book.html

*For shit and giggles, or for those who are hellbent on cracking a particular 'puzzle' address. See you in the afterlife!

https://github.com/samr7/vanitygen OR https://github.com/JeanLucPons/VanitySearch

*For calculating or dealing with big numbers, simple interface, no thrills.

https://www.calculator.net/big-number-calculator.html?cx=3416454622906707&cy=2111485077978050&cp=20&co=plus

*This is a great tool which will help you figure out the ranges for (x) puzzle, showing 2^n.

https://www.easycalculation.com/power-of-numbers.php

----------------------------------------------------------------------------------------------------------------------------------------------------------

Since i've made a day of this, I want to give a shoutout to the mod BurtW, reading your posts was both comical and informative. Also, arulbero, zielar, Bulista, holy_ship, racminer, pikachunakapika and many others, thank you for your inputs and dedication to the project. I very much enjoyed your content.

This is essentially a comprehensive list of resources for individuals who may lack the required knowledge to understand the aforementioned posts, or who need a helping hand in discovering what this 'puzzle' is all about. If you are new, please be aware that the private keys related to this are random and are devoid of pattern. The entire purpose of this, as stated by the creator, satoshi-rising, is to represent the vast space private keys reside in and the security of 256 bit keys. Many here will find it annoying if you 1) accuse them of stealing bitcoins, which does not apply here due to the nature of why these transactions exist; 2) constant theories of pattern recognition or sequence cracking, neither of which apply in this instance. With this knowledge you should have a good foundation on which to pursue more practical applications like BitCrack and related programs.

If anything I have posted is in error or misrepresented, I do apologize, as my day has been filled with wrecking my eyes with this forum and more numbers than should be digested by a human in a given evening.

I find it fascinating that #64 has yet to be cracked, given how many resources have been thrown at it for the past few years. I have been scanning various sectors of the keyspace for the past few years to no avail. On my ASUS ROG STRIX 3090 OC 24GB I can process over 2000 MKeys per second on Bitcrack, and over 4000 MKeys in VanitySearch. I have also come up with a counting methodology which works via stratification that I am hoping to implement which will be able to scan the keyspace in a more efficient manner (i'll update when that occurs). If you have any questions or need any help feel free to respond/DM me.

Best of luck brave souls!

weazy2

newbie

Activity: 2

Merit: 0

Quote from: PrivatePerson on March 30, 2022, 07:15:26 AM

Quote

64 puzzle patterns:

What to do with it?

Just for info, may be someone have ideas how we can apply this Huh

PrivatePerson

member

Activity: 174

Merit: 12

Quote

64 puzzle patterns:

What to do with it?

weazy2

newbie

Activity: 2

Merit: 0

64 puzzle patterns:

Code:

Pattern: 0xe13721aadfd2b018 16jY7qxm4TQUEdHvMi1aRwtBpdJuFq6AMC
Pattern: 0x86fa9e3ba77db2ba 16jY7qtqXD6YwUTAE4TxLg8Jx6fvaAJkzf
Pattern: 0x9ba788b9ac24c79f 16jY7qsQFJ7w19SYArVCwL9TbiaHcyN1L9
Pattern: 0x9512ba04d754016d 16jY7qCpf1xe2TbN6c69dzyswVv9e8v1hM
Pattern: 0xc95c441e99aadc0c 16jY7qzC7EEcgd9dmn92Q8KwrRKMyEqnE1
Pattern: 0xfe12d3c983bdb605 16jY7q3gUovc27puZHcicGjd4ymUVeDqYS
Pattern: 0xcadfa0f0bb0a539b 16jY7qguGeyNVvGTDd5B4pokGNGwNWvHSG
Pattern: 0x9b10822c28030a1b 16jY7qUet7iqQmCesKAqC1k64Led51gAY5
Pattern: 0x9d3a0e6daa145bcc 16jY7qkC5ejyyFoAC49J4Uq1SujPfrewwK
Pattern: 0xe1a9e01c06971b63 16jY7qzGurLj9BxdKV445gn924XwJdfVZz
Pattern: 0xd713028dd2c7140d 16jY7q1Kc6zULFsjNKZpbg2Ak8o6scNwbg
Pattern: 0xd6236b4f95dd99d7 16jY7qnPVD5xj1mCFd51B3ea3PMjHDTfRT
Pattern: 0x8513b294d2f45c49 16jY7qqhsFL7tM99NGuBcZUgUnM4pzpunU
Pattern: 0xa90e3d5d15d6bbeb 16jY7qvTW513cJ8deUjawgujHjAoMh5UNQ
Pattern: 0x811d796def3ef064 16jY7qQsUBHmZgtVNe7qDxTSZR7SeBBCRg
Pattern: 0xf554101e5daae25e 16jY7q1JmjDEYSGY8gG7Nk4EaTFHmmftFd
Pattern: 0xc5c7dc9c38f9fa92 16jY7qKt49SZvfXj8DwarHRK1rNjTU9p3d
Pattern: 0xccd9b1c700b11eb0 16jY7qnmSCEojEzDkeEjV9JTzH61ifjfoX
Pattern: 0xd041e8ba4c0f02ae 16jY7qoxXKxQuh5snz1zfoUujEYmcPJpiR
Pattern: 0x90f7e802bd352d1b 16jY7q6KQtEB1mZaZG4oos2uxrqyGktnQt
Pattern: 0xf9d049684b9b57e5 16jY7qPZz6vMPDgjEd4h5HXT3fTJtRu4Pf
Pattern: 0xdd942412ddc0f9d9 16jY7qq5sZc96813fgYUR6oSLkYTUHFY4p
Pattern: 0xaa9ed2cfca90a4bb 16jY7q7KJrKJPjjhpT7LcaayNTExNVRSKH
Pattern: 0xd331490c13237780 16jY7qPATHpumqCHsvczKyJQAqkGe3UFjP
Pattern: 0xcd64a44d8ecd659d 16jY7quoQ5ftYDgce3Qh2Vvq4Geq12HMGE
Pattern: 0xc564c76474fb6e43 16jY7qUYgS2qgcAZ2W1UQcy3MgKB8bDE2o
Pattern: 0xf36e199fdb060f75 16jY7qRfJGNuyWqhDYPnBSpFvs8DCHUhyQ
Pattern: 0xd6f6dd9d0d2da302 16jY7qyMuC3tg3inrQ9PMgVygyypi6BsmD
Pattern: 0xff26a4f5b5bbf9dd 16jY7qJ76qFAnMUcftrkFSeedt3S3egZtt
Pattern: 0xb4238c2c227618d8 16jY7qTdKdMQjUPxTRVJoP16UB3LR6rzVq
Pattern: 0x8f39f63723a6898e 16jY7qWfh5n8tPYfJSTUmavj8n968Wv3DU
Pattern: 0xc4f6bb1db868265b 16jY7q4LGYkdoxacPpung7HMHoARPj1pVE
Pattern: 0xb0294ba38ea32e01 16jY7q4VBStiHXWrr1VTp3k4PE9SHvVYR1
Pattern: 0x822cb2cae367017f 16jY7qGd3CcRQLi3sTTwPxzwm4aZdr1nNc
Pattern: 0x875f23737f087543 16jY7qx4p1tuMfehwEu9ZCDGbDgCwVtgAm

Jolly Jocker

jr. member

Activity: 49

Merit: 1

little experiment...

this code creates a sha256 key from str(random.random()), which is fed to the next "ice.get_sha256().hex()".
This is repeated, in this case + 2 times.
The created keys are concatenated ( keys = key1+key2+key3 ).
In the next step, the entire key is divided into 16-digit (64-bit) pieces and read out.
a filter keeps the keys in the desired range...

Happy hunting! XD

Code:

import time
import random
import secp256k1 as ice
from ctypes import c_int
from multiprocessing import Value, Lock, Process

y=1000000
cores=4
counter = Value(c_int)
counter_lock = Lock()
def process1(number,counter,):
while True:
t = str(random.random())
key1 = ice.get_sha256(t).hex()
key2 = ice.get_sha256(key1).hex()
key3 = ice.get_sha256(key2).hex()
keys = key1+key2+key3
for i in range(0, len(keys), 16):
line = keys[i:i+16]
if line.startswith("8") or line.startswith("9") or line.startswith("a") or line.startswith("b") or line.startswith("c") or line.startswith("d") or line.startswith("e") or line.startswith("f"):
with counter_lock:
counter.value += cores
cv = str(counter.value)
addr = ice.privatekey_to_address(0, True, int(line, 16))

if addr.startswith("16jY7q"):
print('\n\n Pattern:',('0x'+line),addr,'\n')

if addr == "16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN":
print ('\n\n TARGET FOUND!!:',addr,'\n\n')
file=open(u"16jY.Info.txt","a")
file.write('\n ' + ('0x'+line) + ' | ' + addr)
file.close()
wait = input("Press Enter to Exit.")
exit()

if (counter.value) % y == 0:
print(' CNT:',cv.zfill(10),' Recent:',('0x'+line),addr,end='\r')

if __name__ == '__main__':
t = time.ctime()
print('',t)
number = y
workers = []
print('\n K E Y - C H O P P E R - 64 \n')
print('\n ===============| |===== KEYS =======|============ ADDR ================| \n\n')
for r in range(0,cores):
p = Process(target=process1, args=(number,counter,))
workers.append(p)
p.start()

for worker in workers:
worker.join()

donate BTC: 1DonateZNR9BUaCqJTgXCoyyCpRSosFujR

the generation of the keys on a time basis (ms) is too slow. the result was double, triple and multiple repetition of the keys in a row.
For this reason I switched to random.random()...
the better way ^^

Sat Mar 26 20:40:24 2022

K E Y - C H O P P E R - 64

=============| |===== KEYS =======|============ ADDR ================|

CNT: 0032000000 Recent: 0xaaf4c88e4a40a299 175NYisjDfbxUyoonY8rEbghWvBbyMGPRj

Pattern: 0x9a5da0765a81d90f 16jY7qNwsgykvkAzm8oiwVSnC71KsZsXqM

CNT: 0304000000 Recent: 0xb338bf39e2ba9922 1EfbHH9BPXSDuibzn3gCao9aQB6Chr7C3G

[moderator's note: consecutive posts merged]

BorisTheHorist

newbie

Activity: 22

Merit: 3

as per someones suggestion earlier in the thread here is a list of the currently known private keys in relation to their accompanying range sizes in percentages.

sorted = [0.0, 0.06, 0.08, 0.09, 0.1, 0.12, 0.13, 0.17, 0.18, 0.19, 0.22, 0.23, 0.27, 0.28, 0.31, 0.32, 0.33, 0.35, 0.36, 0.38, 0.4, 0.43, 0.44, 0.45, 0.46, 0.49, 0.5, 0.51, 0.53, 0.57, 0.62, 0.63, 0.64, 0.65, 0.66, 0.67, 0.68, 0.69, 0.7, 0.72, 0.75, 0.82, 0.87, 0.91, 0.92, 0.95, 0.96, 0.97] (48 unique values)

sorted (raw) = [0.0, 0.0, 0.00390625, 0.0693586707348004, 0.08560360020207014, 0.09034376966231544, 0.10738698705333893, 0.1279296875, 0.13666732696611916, 0.17072322126477957, 0.1777045763192291, 0.1875, 0.19316889756160655, 0.2273166453323929, 0.23364647093694657, 0.23667356096793157, 0.2734375, 0.287109375, 0.2887251778456132, 0.31005859375, 0.3125, 0.3166639075566309, 0.32627265442988573, 0.33485841751098633, 0.3586072690006503, 0.3638877868652344, 0.36981157668053144, 0.3876168823447197, 0.4023493269008403, 0.43391395367944174, 0.43408918380737305, 0.44051053281873465, 0.45348233016493467, 0.4588522112899227, 0.46112229085167655, 0.4621429443359375, 0.4927569553256035, 0.5, 0.5018395352846268, 0.5149424275913058, 0.5157241821289062, 0.53125, 0.57196044921875, 0.6253847479820251, 0.63983154296875, 0.6439841804320401, 0.6453061435604468, 0.6466464996337891, 0.6571150545548307, 0.6618142630904913, 0.6678542153963616, 0.6681841164827347, 0.6797900857488876, 0.6847959722838368, 0.6949864005878701, 0.6960085034370422, 0.7005655069250736, 0.7200322151184082, 0.7278327941894531, 0.75, 0.75, 0.7513186500162874, 0.8217038442259545, 0.82421875, 0.8256312850098766, 0.8285546544961626, 0.8289294721848898, 0.8725002169265093, 0.9185453074950023, 0.9244143441319466, 0.9500958897872267, 0.9580019181594253, 0.9689828764301732, 0.9780104756355286]

the missing points would be these ranges:
['8147ae147ae14800:828f5c28f5c29000', '828f5c28f5c29000:83d70a3d70a3d800', '83d70a3d70a3d800:851eb851eb852000', '851eb851eb852000:8666666666666800', '8666666666666800:87ae147ae147b000', '88f5c28f5c28f800:8a3d70a3d70a4000', '8e147ae147ae1800:8f5c28f5c28f6000', '91eb851eb851f000:9333333333333000', '9333333333333000:947ae147ae147800', '947ae147ae147800:95c28f5c28f5c000', '9999999999999800:9ae147ae147ae000', '9ae147ae147ae000:9c28f5c28f5c2800', '9eb851eb851eb800:a000000000000000', 'a000000000000000:a147ae147ae14800', 'a147ae147ae14800:a28f5c28f5c29000', 'a51eb851eb852000:a666666666666800', 'a666666666666800:a7ae147ae147b000', 'ab851eb851eb8800:acccccccccccd000', 'af5c28f5c28f6000:b0a3d70a3d70a000', 'b1eb851eb851f000:b333333333333000', 'b47ae147ae147800:b5c28f5c28f5c000', 'b5c28f5c28f5c000:b70a3d70a3d70800', 'bc28f5c28f5c2800:bd70a3d70a3d7000', 'bd70a3d70a3d7000:beb851eb851eb800', 'c28f5c28f5c29000:c3d70a3d70a3d800', 'c51eb851eb852000:c666666666666800', 'c666666666666800:c7ae147ae147b000', 'c7ae147ae147b000:c8f5c28f5c28f800', 'ca3d70a3d70a4000:cb851eb851eb8000', 'cb851eb851eb8000:ccccccccccccd000', 'ccccccccccccd000:ce147ae147ae1000', 'ce147ae147ae1000:cf5c28f5c28f6000', 'dae147ae147ae000:dc28f5c28f5c2800', 'dd70a3d70a3d7000:deb851eb851eb800', 'deb851eb851eb800:e000000000000000', 'e147ae147ae14800:e28f5c28f5c29000', 'e28f5c28f5c29000:e3d70a3d70a3d800', 'e3d70a3d70a3d800:e51eb851eb852000', 'e51eb851eb852000:e666666666666800', 'e666666666666800:e7ae147ae147b000', 'e7ae147ae147b000:e8f5c28f5c28f800', 'ea3d70a3d70a4000:eb851eb851eb8000', 'eb851eb851eb8000:ecccccccccccd000', 'ecccccccccccd000:ee147ae147ae1000', 'ee147ae147ae1000:ef5c28f5c28f6000', 'f0a3d70a3d70a000:f1eb851eb851f000', 'f1eb851eb851f000:f333333333333000', 'f333333333333000:f47ae147ae148000', 'f70a3d70a3d71000:f851eb851eb85000', 'f851eb851eb85000:f999999999999800', 'fd70a3d70a3d7000:feb851eb851eb800', 'feb851eb851eb800:ffffffffffffffff']

though that does not help much each range in the 63bit key range would take approximately 18,000 hours on Bitcrack with 1400Mk/s

COBRAS

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Quote from: Jolly Jocker on March 24, 2022, 01:14:46 PM

Quote from: BorisTheHorist on March 23, 2022, 10:13:34 PM

Jolly Jocker,

From your info I have this:
[16jY7qLJnxtt8FSYMze4DRVMWQF47auU8M,8ec5edb376aeb55a]
[16jY7qLJnxht5HL2NBywuoC8jK34rz8NJv ,9357254130f9270a]
[16jY7qLJnx4enVJC2Ao3wRv3tRusseu5Yb ,af6f7a0015695aa4]
[16jY7qLJnxcZQ1ff7Y6ascX2FwkVfMEqcn ,b0d6bbb249f4d96f]
[16jY7qLJnxoBx9Zhsa1tN62rfZXX1bgKNp ,c4230300dfe089af]
[16jY7qLJnxPwmtRgVS3HBg2RtAoHe1koZF ,d461baace35db135]
[16jY7qLJnx4f4yqNvmp7LtnaatiUuwgUwW ,ebe2eb35503b43e4]
[16jY7qLJnx2ixrxCnTLSraerkgyB3YYAiT ,edf9cd60356dcc19]
[16jY7qLJnxHBp3dqwV2kzYq1LucfZzgxsH ,eeb9ab334df2770d]

my bet based on your data is that #64 is between edf9cd60356dcc19:eeb9ab334df2770d
but that is probably not true.

Edit: b is after H in the Base58 Alphabet do with that what you will

This database is sorted by Public key!!
from: 0200a89f5508695dd9c0900621681ff9b961d67772c323642a5e1f042d4cff3529
to : 03ff94c3c4030faf55a10bda211e1d4d0891e3c68089e1ace2e62ce2f2f2eeb405

it's all in the text!

Bro, check pm please

dextronomous

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Activity: 431

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what complaints? why here? who you? Roll Eyes

MrFreeDragon

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Activity: 443

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Quote from: zahid888 on March 24, 2022, 02:16:43 PM

Suppose I crack puzzle 64.. please suggest to me how can I avoid double-spending Until I take out all the coins.. like BTC, BTH, BTG ??

You should broadcast all the transactions at the same time. You also should turn off RBF options (replace-by-fee)

zahid888

member

Activity: 275

Merit: 20

the right steps towerds the goal

Suppose I crack puzzle 64.. please suggest to me how can I avoid double-spending Until I take out all the coins.. like BTC, BTH, BTG ??

Jolly Jocker

jr. member

Activity: 49

Merit: 1

Quote from: BorisTheHorist on March 23, 2022, 10:13:34 PM

Jolly Jocker,

From your info I have this:
[16jY7qLJnxtt8FSYMze4DRVMWQF47auU8M,8ec5edb376aeb55a]
[16jY7qLJnxht5HL2NBywuoC8jK34rz8NJv ,9357254130f9270a]
[16jY7qLJnx4enVJC2Ao3wRv3tRusseu5Yb ,af6f7a0015695aa4]
[16jY7qLJnxcZQ1ff7Y6ascX2FwkVfMEqcn ,b0d6bbb249f4d96f]
[16jY7qLJnxoBx9Zhsa1tN62rfZXX1bgKNp ,c4230300dfe089af]
[16jY7qLJnxPwmtRgVS3HBg2RtAoHe1koZF ,d461baace35db135]
[16jY7qLJnx4f4yqNvmp7LtnaatiUuwgUwW ,ebe2eb35503b43e4]
[16jY7qLJnx2ixrxCnTLSraerkgyB3YYAiT ,edf9cd60356dcc19]
[16jY7qLJnxHBp3dqwV2kzYq1LucfZzgxsH ,eeb9ab334df2770d]

my bet based on your data is that #64 is between edf9cd60356dcc19:eeb9ab334df2770d
but that is probably not true.

Edit: b is after H in the Base58 Alphabet do with that what you will

This database is sorted by Public key!!
from: 0200a89f5508695dd9c0900621681ff9b961d67772c323642a5e1f042d4cff3529
to : 03ff94c3c4030faf55a10bda211e1d4d0891e3c68089e1ace2e62ce2f2f2eeb405

it's all in the text!

BorisTheHorist

newbie

Activity: 22

Merit: 3

Quote from: PrivatePerson on March 23, 2022, 11:03:01 PM

Which b? Between which 2 and H?

well I have the alphabet backwards so 2 H b would be the correct order
123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz

PrivatePerson

member

Activity: 174

Merit: 12

Which b? Between which 2 and H?

BorisTheHorist

newbie

Activity: 22

Merit: 3

Quote from: PrivatePerson on March 23, 2022, 10:38:57 PM

How did you come to this?

it's probably not true but b is between 2 and H.

PrivatePerson

member

Activity: 174

Merit: 12

How did you come to this?

Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 250. (Read 230559 times)