Bitcoin puzzle transaction ~32 BTC prize to who solves it

[wipall]

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 my friend, i want to ask you a private question but i can't dm you because i am a new member. i need a code to search in the range i want. The codes here are only for puzzle 64 and I can't edit spacing.
I want to search by typing the range I want between these two ranges, with the range 0 to 115792089237316195423570985008687907853269984665640564039457584007913129639935. I'll be glad, if you help me.

kötü ingilizcem için özür dilerim.

[1715430749]

1715430749

[1715430749]

1715430749

[1715430749]

1715430749

[1715430749]

1715430749

[1715430749]

1715430749

[1715430749]

1715430749

[MrFreeDragon]

wipall, Bitcrack would help you:

https://github.com/brichard19/BitCrack
https://bitcointalk.org/index.php?topic=4453897.0

As for the specific range you can use the following command:

Code:

xxBitCrack.exe --keyspace 80000000:ffffffff 1FshYsUh3mqgsG29XpZ23eLjWV8Ur3VwH

[GR Sasa]

Hello there! I was just born, and will try my best to crack at least puzzle 64! Any Tips, on how to start?

Much regards!

[zahid888]

[PrivatePerson]

wtf?

[ligor]

Tetris? 

[zahid888]

MrFreedragon i want to block those red marked binary numbers too.. is there any way to do this ??

[Jolly Jocker]

User 'Jolly Jocker' has not chosen to allow messages from newbies. You should post in their relevant thread to remind them to enable this setting.

been trying to pm you. please do the needful so i can pm you

Sorry, I overlooked that in the settings..

I have changed it, so it's now possible.. sorry again!! 

[Jolly Jocker]

hi my friend, i want to ask you a private question but i can't dm you because i am a new member. i need a code to search in the range i want. The codes here are only for puzzle 64 and I can't edit spacing.
I want to search by typing the range I want between these two ranges, with the range 0 to 115792089237316195423570985008687907853269984665640564039457584007913129639935. I'll be glad, if you help me.

kötü ingilizcem için özür dilerim.

The range for 64 bit is 9223372036854775808 to 18446744073709551615


this is easy to show.. an example for it:

use this web-calculator
https://web2.0rechner.de/

and just enter 2^64 for 64bit
2^66 for 66bit.. etc..

greetings!

[wipall]

hi my friend, i want to ask you a private question but i can't dm you because i am a new member. i need a code to search in the range i want. The codes here are only for puzzle 64 and I can't edit spacing.
I want to search by typing the range I want between these two ranges, with the range 0 to 115792089237316195423570985008687907853269984665640564039457584007913129639935. I'll be glad, if you help me.

kötü ingilizcem için özür dilerim.

The range for 64 bit is 9223372036854775808 to 18446744073709551615


this is easy to show.. an example for it:

use this web-calculator
https://web2.0rechner.de/

and just enter 2^64 for 64bit
2^66 for 66bit.. etc..

greetings!

Thank u brother.

Have you checked which address type the codes you have written give you? compressed? not compressed?

I'm new to this puzzle. but
For 4 years I started to think that's why puzzle 64 was not solved.

[Jolly Jocker]

Thank u brother.

Have you checked which address type the codes you have written give you? compressed? not compressed?

I'm new to this puzzle. but
For 4 years I started to think that's why puzzle 64 was not solved.

You're welcome!

all addresses we are looking for are compressed...

[bikkesbakker]

https://i.imgur.com/VGuN8R1.png

I was checking exactly the same the other day, hi everyone im new into this puzzle, and im obsessed with it. Spend hours and hours looking for a pattern, and there is no pattern as the "owner" mentioned it but i was looking for the already found keys and i think i found something.

If someone is brute forcing this, you need to start from eb851eb851eb8000 to ffffffffffffffff, thats why this is not still be found, the hex is almost at the end of the range. Im bruteforcing this from feb851eb851eb800 to ffffffffffffffff at 25 MKey/s (yes is a little slow but is honest work ) and i will be checking from the back after finish a range.

I hope this helps anyone, and if i do please share something as i will share to 2 users of this forum (if i find the key) who gives me the idea


Regards from AR and sorry about my english.

[BitDominator00]

https://i.imgur.com/VGuN8R1.png

I was checking exatly the same the other day, hi everyone im new into this puzzle, and im obsessed with it. Spend hours and hours looking for a pattern, and there is no pattern as the "owner" mentioned it but i was looking for the already found keys and i think i found something.

If someone is brute forcing this, you need to start from eb851eb851eb8000 to ffffffffffffffff, thats why this is not still be found, the hex is almost at the end of the range. Im bruteforcing this from feb851eb851eb800 to ffffffffffffffff at 25 MKey/s (yes is a little slow but is honest work ) and i will be checking from the back after finish a range.

I hope this helps anyone, and if i do please share something as i will share to 2 users of this forum (if i find the key) who gives me the idea


Regards from AR and sorry about my english.

based on what you say to start from that point of the range ?

[bikkesbakker]

https://i.imgur.com/VGuN8R1.png

I was checking exatly the same the other day, hi everyone im new into this puzzle, and im obsessed with it. Spend hours and hours looking for a pattern, and there is no pattern as the "owner" mentioned it but i was looking for the already found keys and i think i found something.

If someone is brute forcing this, you need to start from eb851eb851eb8000 to ffffffffffffffff, thats why this is not still be found, the hex is almost at the end of the range. Im bruteforcing this from feb851eb851eb800 to ffffffffffffffff at 25 MKey/s (yes is a little slow but is honest work ) and i will be checking from the back after finish a range.

I hope this helps anyone, and if i do please share something as i will share to 2 users of this forum (if i find the key) who gives me the idea


Regards from AR and sorry about my english.

based on what you say to start from that point of the range ?

Hard to explain, but is based on all the other resolved puzzles

[enfarktus]

Thank u brother.

Have you checked which address type the codes you have written give you? compressed? not compressed?

I'm new to this puzzle. but
For 4 years I started to think that's why puzzle 64 was not solved.

You're welcome!

all addresses we are looking for are compressed...

Greetings guys, unrelated to the topic, but I have a question.
I have 3000 publickey data. and I am looking for python code to convert them to btc address in bulk. can you help.

Try this: https://github.com/Mizogg/Mizogg-Tools

[Feron]

Thank u brother.

Have you checked which address type the codes you have written give you? compressed? not compressed?

I'm new to this puzzle. but
For 4 years I started to think that's why puzzle 64 was not solved.

You're welcome!

all addresses we are looking for are compressed...

Greetings guys, unrelated to the topic, but I have a question.
I have 3000 publickey data. and I am looking for python code to convert them to btc address in bulk. can you help.

maybe this help

Code:

import secp256k1 as ice
file = open("publicklist.txt")
data = file.read().split()
for ccc in data:
 vvv = bytes.fromhex(str(ccc))
 com = ice.pubkey_to_address(0,True,vvv)
 unc = ice.pubkey_to_address(0,False,vvv)
 print(com,unc,vvv.hex())

write for text document

Code:

import secp256k1 as ice
file = open("publicklist.txt")
data = file.read().split()
for ccc in data:
 vvv = bytes.fromhex(str(ccc))
 com = ice.pubkey_to_address(0,True,vvv)
 unc = ice.pubkey_to_address(0,False,vvv)
 print(com,unc,vvv.hex())
 f=open("von.txt","a")
 f.write(str(com)+"\n")
 f.write(str(unc)+"\n")
 f.close()

little problem compressed address is not 100% correctly generated generates any sometimes incorrectly

[brainless]

Thank u brother.

Have you checked which address type the codes you have written give you? compressed? not compressed?

I'm new to this puzzle. but
For 4 years I started to think that's why puzzle 64 was not solved.

You're welcome!

all addresses we are looking for are compressed...

Greetings guys, unrelated to the topic, but I have a question.
I have 3000 publickey data. and I am looking for python code to convert them to btc address in bulk. can you help.

https://github.com/matja/bitcoin-tool

[Feron]

loop scanner
they will print the decimal what it means it means the approximate decimal only the last 2 digits will not be correct you have to take the decimal data sequentially and you have the original private key easy
you may not understand it simply
if the private key finds a decimal printout but you are scanning loop 100 then the last 2 numbers will not be correct you have to take the printed decimal to release the sequence upwards loop key-14428642-final key-14428676

Code:

import random
from tqdm import tqdm
import secp256k1 as ice
for x in tqdm(range(1000000)):
 xx = random.randrange(1000,9999)
 for c in range(922,1845):
  #xxx = random.randrange(10000000000000000000,18446744073709551615)
  x1 = ''.join(str(c))
  x2 = ''.join(str(xx))
  cc = int(x1+x2)
  ke = ice.privatekey_loop_h160(100,0,True,cc).hex()
  #print(cc,ke)
  if "0959e80121f36aea13b3bad361c15dac26189e2f" in (ke):#24 PUZLE ADDRESS TEST
   f=open("von.txt","a")
   f.write(str(cc)+"-"+(ke)+"\n")
   f.close()

simply if it finds a 64 puzzle address then 18 decimal will be correct only the last 2 decimal will be wrong just take the whole 20 decimal and run sequencial upwards
64 PUZLE LOOP SCANNER

Code:

import random
from tqdm import tqdm
import secp256k1 as ice
for x in tqdm(range(1000000)):
 xx = random.randrange(1000000000000000,9999999999999999)
 for c in range(922,1845):
  #xxx = random.randrange(10000000000000000000,18446744073709551615)
  x1 = ''.join(str(c))
  x2 = ''.join(str(xx))
  cc = int(x1+x2)
  ke = ice.privatekey_loop_h160(100,0,True,cc).hex()
  #print(cc,ke)
  if "3ee4133d991f52fdf6a25c9834e0745ac74248a4" in (ke):
   f=open("von.txt","a")
   f.write(str(cc)+"-"+(ke)+"\n")
   f.close()

[PrivatePerson]

Very slow.

[Feron]

therefore it goes slowly because the first and last bits go sequentially
removing the first sequence right away will be faster

Code:

import random
from tqdm import tqdm
import secp256k1 as ice
for x in tqdm(range(1000000)):
 xx = random.randrange(10000000,14999999)
 #xxx = random.randrange(10000000000000000000,18446744073709551615)
 #x1 = ''.join(str(c))
 #x2 = ''.join(str(xx))
 cc = int(xx)
 ke = ice.privatekey_loop_h160(100,0,True,cc).hex()
 #print(cc,ke)
 if "0959e80121f36aea13b3bad361c15dac26189e2f" in (ke):#24 PUZZLE ADDRESS TEST
  f=open("von.txt","a")
  f.write(str(cc)+"-"+(ke)+"\n")
  f.close()

what I wanted to say with that code is that the loop or bloom filter is much more efficient in terms of the chance of a collision
to that speed my 8 year old laptop with 1 core has a speed of 5000 key in one second it means 5000X100 = 500000 keys per second theoretically what is probably the fastest what can be on 1 thread