Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 266.

brainless

member

Activity: 348

Merit: 34

Quote from: Jolly Jocker on March 11, 2022, 12:56:16 PM

Quote from: brainless on March 11, 2022, 12:45:11 PM

Quote from: Jolly Jocker on March 11, 2022, 10:31:25 AM

start the scrypt with double click .. simply to use in Python...
You need:
faster "bit" there is a library from "ice" https://github.com/iceland2k14/secp256k1 there it is necessary to throw its libraries into the folder with the script.

donate BTC: 1DonateZNR9BUaCqJTgXCoyyCpRSosFujR

you dont have GPU ? any GPU ?

This code is only for CPU, working with two Threads...

Sure, I use GPU too, but I don't can coding for Cuda applications like, vanitygen, Kangaroo, etc...

i will sugggest same working your code at c and gpu, above link use git scripts, working 10005 better then you python code , ttoal same function,
even if your cpu have gpu inside, like i3-6100 process have gpu, its also work on this rotar, and auto use gpu at cpu based command,
https://github.com/phrutis/Rotor-Cuda, for fully random at gpu like bitcrack is BitCrack2, use that, you will forget python
dont forget merit me....

Jolly Jocker

jr. member

Activity: 49

Merit: 1

Quote from: brainless on March 11, 2022, 12:45:11 PM

Quote from: Jolly Jocker on March 11, 2022, 10:31:25 AM

start the scrypt with double click .. simply to use in Python...
You need:
faster "bit" there is a library from "ice" https://github.com/iceland2k14/secp256k1 there it is necessary to throw its libraries into the folder with the script.

donate BTC: 1DonateZNR9BUaCqJTgXCoyyCpRSosFujR

you dont have GPU ? any GPU ?

This code is only for CPU, working with two Threads...

Sure, I use GPU too, but I don't can coding for Cuda applications like, vanitygen, Kangaroo, etc...

brainless

member

Activity: 348

Merit: 34

Quote from: Jolly Jocker on March 11, 2022, 10:31:25 AM

start the scrypt with double click .. simply to use in Python...
You need:
faster "bit" there is a library from "ice" https://github.com/iceland2k14/secp256k1 there it is necessary to throw its libraries into the folder with the script.

donate BTC: 1DonateZNR9BUaCqJTgXCoyyCpRSosFujR

you dont have GPU ? any GPU ?

Jolly Jocker

jr. member

Activity: 49

Merit: 1

start the scrypt with double click .. simply to use in Python...
You need:
faster "bit" there is a library from "ice" https://github.com/iceland2k14/secp256k1 there it is necessary to throw its libraries into the folder with the script.

donate BTC: 1DonateZNR9BUaCqJTgXCoyyCpRSosFujR

willi9974

legendary

Activity: 3654

Merit: 3003

Enjoy 500% bonus + 70 FS

how can i start the script and what do i need?

Jolly Jocker

jr. member

Activity: 49

Merit: 1

Test Code for the 64 bit range... ca. 310000000 keys/hr...

Code:

import random
from threading import Thread
import secp256k1 as ice
from time import sleep
print('\n ========= ================= ==ADDRESS HUNTER== ================= ==========\n\n')
y=1000000
count=-2000000
def process1(number):
   line_count=0
   num = random.randrange(0x8000000000000000,0xffffffffffffffff)
   for i in range(number):
   num+=i
   addr = ice.privatekey_to_address(0, True, num)
   lineA = (addr, hex(num))
   line_count+=1

   if '16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN' in lineA:
   file=open(u"BTH.Target.Info.txt","a")
   file.write('\n ' + str(lineA))
   file.close()
   wait = input("Press Enter to Exit.")
   exit()

   if line_count == y:
   print(' current line: ', str(lineA))
   line_count=0

while True:
   count+=(2*y)
   number = y
   threads = []
   print('\n Total Scanned Lines: ' + str(count) + '\n')
   for n in range(0,2):
   t = Thread(target=process1, args=(number,))
   threads.append(t)
   t.start()

   for t in threads:
   t.join()

donate BTC: 1DonateZNR9BUaCqJTgXCoyyCpRSosFujR

Newbienewnewcoin

newbie

Activity: 9

Merit: 5

I’ve tried to use bitcrack and kangaroo and a few other windows based apps. I never thought about parallels but thanks so! You’re talking about M1 Mac right? I obviously should utilize the time to learn lol

Quote from: PrivatePerson on March 10, 2022, 01:25:07 AM

Quote from: Newbienewnewcoin on March 09, 2022, 09:52:07 PM

I have nothing but free time on my hands lately and I've been like obsessed with it, it's fascinating. I was trying to take a jab at it with some computer experience and I find that all (well most) of the tools out there for cracking aren't compatible with my macbook air (lastest m1 model). I was wondering if anyone had any recommendations fo software or anything considering my circumstance and using a macbook air, I know it's nothing compared to what's out there but it's all I have. Any recommendations because I'm all ears!

I've used MacOS before, I've never had any compatibility issues with these programs:
Parallels Desktop - https://www.parallels.com/
Homebrew - https://brew.sh/

PrivatePerson

member

Activity: 174

Merit: 12

Quote from: Newbienewnewcoin on March 09, 2022, 09:52:07 PM

I have nothing but free time on my hands lately and I've been like obsessed with it, it's fascinating. I was trying to take a jab at it with some computer experience and I find that all (well most) of the tools out there for cracking aren't compatible with my macbook air (lastest m1 model). I was wondering if anyone had any recommendations fo software or anything considering my circumstance and using a macbook air, I know it's nothing compared to what's out there but it's all I have. Any recommendations because I'm all ears!

I've used MacOS before, I've never had any compatibility issues with these programs:
Parallels Desktop - https://www.parallels.com/
Homebrew - https://brew.sh/

Newbienewnewcoin

newbie

Activity: 9

Merit: 5

I have nothing but free time on my hands lately and I've been like obsessed with it, it's fascinating. I was trying to take a jab at it with some computer experience and I find that all (well most) of the tools out there for cracking aren't compatible with my macbook air (lastest m1 model). I was wondering if anyone had any recommendations fo software or anything considering my circumstance and using a macbook air, I know it's nothing compared to what's out there but it's all I have. Any recommendations because I'm all ears!

Jolly Jocker

jr. member

Activity: 49

Merit: 1

The problem with Python is the "print" output, which costs time/processing power, without this, 12000 to 13500 keys/s are generated.. (12000/s = 1036800000 keys/24 hr)
With the print output only 1455.26it/s are generated, which is much too slow.. even 50000 or 100000 keys/s are too slow with a range of 9223372036854775807 possible keys..
Even GPU scanners with more than 150000000 keys/s need more than 1000 years ( 1949.80 ) for this range... (4730400000000000 keys/ year)

cbdcbf

newbie

Activity: 1

Merit: 0

кинь кoд пocлeдoвaтeльнo

wipall

newbie

Activity: 10

Merit: 0

Quote from: Jolly Jocker on February 25, 2022, 04:26:14 PM

Quote from: wipall on February 25, 2022, 03:35:08 PM

Quote from: Jolly Jocker on February 25, 2022, 01:05:36 PM

I have revised this code for 64 bit, this code changes the starting point every 1000000 counters..
the problem in this code, is the truncation of the "zeros" what causes the skipping...it has pros and cons...
the advantage is that you can use it as a filter, the average of the zeros for the address you are looking for is between 22 and 43...

!! zeros = bina.count("100") .. this function only counts the zeros in binary code!!! also zeros = bina.count("0") or you count the ones ... ones = bina.count("1") !!

It's important to understand the functions so that you know what the code does and you can make conscious changes that work...

Code:

from bit import *
import random
from time import sleep

count=0

while True:
   a = random.randint(2**63, 2**64)
# while a <= 2**64:
   for x in range(1000000):
   x = x
   bina = bin(a)[2:]
   zeros = bina.count("0")
   # if x in range(1):
   if x == 0:
   print(" !!!NEW LOOP!!!")
   sleep(2)
   if zeros >= 1: # "00000000000000000000000000000011111111111111111111111111"
   if zeros <= 63: # "00000000000000000000000000000011111111111111111111111111"
   key = Key.from_int(a)
   addr = key.address
   count+=1
   print (hex(a), bina, zeros, str(count))

   if addr.startswith('16jY7qLJnxb'):
   print(' '+hex(a)+'|'+ addr)
   file=open(u"16jY.Info.txt","a")
   file.write('\n '+hex(a)+' | '+ addr)
   file.close()
   sleep(3)
   wait = input("Press Enter to continue.")
   print(" continue...")
   sleep(1)
   # else:
   # print (' Scan Nr.: ', str(count), end='\r') #pass
   a = a +1
   pass

Thank you very much for your attention, my friend.

always a pleasure dude!!

i managed to run the code sequentially.
I scan an average of 500,000,000 addresses a day.
and although the computer is very old

do you think it's too slow?

How many addresses do you scan on average per day?

heribertoX

newbie

Activity: 2

Merit: 0

Quote from: Alpaste on February 28, 2022, 03:41:33 PM

Puzzle 64 is insane. like i've been running my GPU 2080ti 24/7 and still not getting luck. even though it's only 16 HEX characters.
i'm suspecting, that the address of the puzzle 64 is not in the range that is suppose to be.
What you think guys?

quizas los programas que usas saltaron la direccion n64 hay programas con errores

Alpaste

jr. member

Activity: 37

Merit: 1

Puzzle 64 is insane. like i've been running my GPU 2080ti 24/7 and still not getting luck. even though it's only 16 HEX characters.
i'm suspecting, that the address of the puzzle 64 is not in the range that is suppose to be.
What you think guys?

heribertoX

newbie

Activity: 2

Merit: 0

saludos alguien me podria facilitar el instalador de BitCrack para windows 7 64 bites por favor.

Jolly Jocker

jr. member

Activity: 49

Merit: 1

Quote from: wipall on February 25, 2022, 03:35:08 PM

Quote from: Jolly Jocker on February 25, 2022, 01:05:36 PM

I have revised this code for 64 bit, this code changes the starting point every 1000000 counters..
the problem in this code, is the truncation of the "zeros" what causes the skipping...it has pros and cons...
the advantage is that you can use it as a filter, the average of the zeros for the address you are looking for is between 22 and 43...

!! zeros = bina.count("100") .. this function only counts the zeros in binary code!!! also zeros = bina.count("0") or you count the ones ... ones = bina.count("1") !!

It's important to understand the functions so that you know what the code does and you can make conscious changes that work...

Code:

from bit import *
import random
from time import sleep

count=0

while True:
    a = random.randint(2**63, 2**64)
 #  while a <= 2**64:
    for x in range(1000000):
        x = x
        bina = bin(a)[2:]
        zeros = bina.count("0")
     #  if x in range(1):
        if x == 0:
            print(" !!!NEW LOOP!!!")
            sleep(2)
        if zeros >= 1:     # "00000000000000000000000000000011111111111111111111111111"
            if zeros <= 63: # "00000000000000000000000000000011111111111111111111111111"
                key = Key.from_int(a)
                addr = key.address
                count+=1
                print (hex(a), bina, zeros, str(count))
                
                if addr.startswith('16jY7qLJnxb'):
                    print(' '+hex(a)+'|'+ addr)
                    file=open(u"16jY.Info.txt","a")
                    file.write('\n '+hex(a)+' | '+ addr)
                    file.close()
                    sleep(3)
                    wait = input("Press Enter to continue.")
                    print(" continue...")
                    sleep(1)
             #  else:
             #      print (' Scan Nr.: ', str(count), end='\r') #pass
        a = a +1
    pass

Thank you very much for your attention, my friend.

always a pleasure dude!!

wipall

newbie

Activity: 10

Merit: 0

Quote from: Jolly Jocker on February 25, 2022, 01:05:36 PM

I have revised this code for 64 bit, this code changes the starting point every 1000000 counters..
the problem in this code, is the truncation of the "zeros" what causes the skipping...it has pros and cons...
the advantage is that you can use it as a filter, the average of the zeros for the address you are looking for is between 22 and 43...

!! zeros = bina.count("100") .. this function only counts the zeros in binary code!!! also zeros = bina.count("0") or you count the ones ... ones = bina.count("1") !!

It's important to understand the functions so that you know what the code does and you can make conscious changes that work...

Code:

from bit import *
import random
from time import sleep

count=0

while True:
    a = random.randint(2**63, 2**64)
 #  while a <= 2**64:
    for x in range(1000000):
        x = x
        bina = bin(a)[2:]
        zeros = bina.count("0")
     #  if x in range(1):
        if x == 0:
            print(" !!!NEW LOOP!!!")
            sleep(2)
        if zeros >= 1:     # "00000000000000000000000000000011111111111111111111111111"
            if zeros <= 63: # "00000000000000000000000000000011111111111111111111111111"
                key = Key.from_int(a)
                addr = key.address
                count+=1
                print (hex(a), bina, zeros, str(count))
                
                if addr.startswith('16jY7qLJnxb'):
                    print(' '+hex(a)+'|'+ addr)
                    file=open(u"16jY.Info.txt","a")
                    file.write('\n '+hex(a)+' | '+ addr)
                    file.close()
                    sleep(3)
                    wait = input("Press Enter to continue.")
                    print(" continue...")
                    sleep(1)
             #  else:
             #      print (' Scan Nr.: ', str(count), end='\r') #pass
        a = a +1
    pass

Thank you very much for your attention, my friend.

nilasahap

newbie

Activity: 5

Merit: 0

My remark about you taking my Bitcoins was a joke. You can't take them for the numerical reasons I gave. The best way to get the following prize is to beast force search the scope of the prize. You are right that others with a lot quicker equipment will presumably beat you to the prize.There has been a ton of exertion on this string where individuals attempt to "work out" (all in all conjecture) the normal sub-range withing the following prize reach.

You could look through their speculation at the normal sub-range however again without quicker equipment you will most likely not win the following prize.
You could attempt to look through arbitrarily in the following reach (or their speculation at the normal sub-range). You will presumably still lose yet hello, you could actually luck out.

Jolly Jocker

jr. member

Activity: 49

Merit: 1

I have revised this code for 64 bit, this code changes the starting point every 1000000 counters..
the problem in this code, is the truncation of the "zeros" what causes the skipping...it has pros and cons...
the advantage is that you can use it as a filter, the average of the zeros for the address you are looking for is between 22 and 43...

!! zeros = bina.count("100") .. this function only counts the zeros in binary code!!! also zeros = bina.count("0") or you count the ones ... ones = bina.count("1") !!

It's important to understand the functions so that you know what the code does and you can make conscious changes that work...

Code:

from bit import *
import random
from time import sleep

count=0

while True:
    a = random.randint(2**63, 2**64)
 #  while a <= 2**64:
    for x in range(1000000):
        x = x
        bina = bin(a)[2:]
        zeros = bina.count("0")
     #  if x in range(1):
        if x == 0:
            print(" !!!NEW LOOP!!!")
            sleep(2)
        if zeros >= 1:     # "00000000000000000000000000000011111111111111111111111111"
            if zeros <= 63: # "00000000000000000000000000000011111111111111111111111111"
                key = Key.from_int(a)
                addr = key.address
                count+=1
                print (hex(a), bina, zeros, str(count))
                
                if addr.startswith('16jY7qLJnxb'):
                    print(' '+hex(a)+'|'+ addr)
                    file=open(u"16jY.Info.txt","a")
                    file.write('\n '+hex(a)+' | '+ addr)
                    file.close()
                    sleep(3)
                    wait = input("Press Enter to continue.")
                    print(" continue...")
                    sleep(1)
             #  else:
             #      print (' Scan Nr.: ', str(count), end='\r') #pass
        a = a +1
    pass

Jolly Jocker

jr. member

Activity: 49

Merit: 1

Quote from: wipall on February 25, 2022, 10:55:49 AM

Quote from: Jolly Jocker on February 25, 2022, 10:06:33 AM

Quote from: ?? on ??

Code:

import sys
import time
import random
#from tqdm import tqdm
from time import sleep
import secp256k1 as ice

print('\n      B I N A R Y   S E E D   S C A N   Q U A D R O   \n')
print(' 11_________________0____________________________________________\n')
x=16000
y=131071
list=[]
x0=0b100000000000000000000000000000000000000000000
for n in range(y):
    x1 = (x0+n)
    a1 = bin(random.randrange(131072,262143))#[2:].zfill(18)
    a1 = (a1)[2:]
    for i in range(x):
        a0 = str("1")            
        a2 = str("0")
        a3 = bin(random.randrange(x1,0b111111111111111111111111111111111111111111111)) 
        a3 = (a3)[3:]#.zfill(44)
        a4 = bin(random.randrange(x1,0b111111111111111111111111111111111111111111111)) 
        a4 = (a4)[3:]#.zfill(44)
        a5 = bin(random.randrange(x1,0b111111111111111111111111111111111111111111111)) 
        a5 = (a5)[3:]#.zfill(44)
        a6 = bin(random.randrange(x1,0b111111111111111111111111111111111111111111111)) 
        a6 = (a6)[3:]#.zfill(44)
        binb = "".join(a0+a1+a2+a3)
        binc = "".join(a0+a1+a2+a4)
        bind = "".join(a0+a1+a2+a5)
        bine = "".join(a0+a1+a2+a6)
        list.append(binb)
        list.append(binc)
        list.append(bind)
        list.append(bine)
        if len(list)==x:
            line_count=0
            for bina in (list):
                bina!='\n'
                addr = ice.privatekey_to_address(0, True, int(bina,2))
                line_count+=1
                                 
                if addr.startswith('16jY7'): # 16jY7
                    ran = int(bina,2)
                    print('',bina, '\n')
                    sys.stdout.write('\r ' + hex(ran))
                    sys.stdout.write(' ' + '| ' + addr + ' ' + str(n))
                    sleep(1)
                    sys.stdout.write('\n\n continue...\n\n')
                    sleep(1)

                if addr == "16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN":
                    ran = int(bina,2)
                    print('\n\n Target found!!' + hex(ran) + ' \n ' + addr,'\n')
                    file=open(u"16j.Target.Info.txt","a")
                    file.write('\n ' + hex(ran) + ' | ' + addr)
                    file.close()
                    sleep(2)
                    wait = input("Press Enter to Exit.")
                    sleep(1)
                    exit()
                    
                if line_count==x:
                    list.clear()
                    
        if len(list)==0:
            print(' scan... ', str(n), end='\r')

Ca 42000 keys/s from c000000000000000 to ffffffffffffffff... The sector between 11_________________0 comprise a range of 131071 and will be processed.
The rest of this line are random numbers from 4 different generators..
The counter reflects the number of loops...

B I N A R Y S E E D S C A N Q U A D R O

11_________________0____________________________________________

1110010110110011011010011110001111111110100110111011100101000111

0xe5b369e3fe9bb947 | 16jY7mM6wrpuAy53vKkh7QXm1JNdWK9Myu 19

continue...

1110010011000111001010001111000110100111001000001001110001000100

0xe4c728f1a7209c44 | 16jY7Ezk5et98CVBWYEigGRwWovzhEtaQ3 121

continue...

scan... 192

B I N A R Y S E E D S C A N Q U A D R O

11_________________0____________________________________________

Traceback (most recent call last):
File "11.py", line 41, in
addr = ice.privatekey_to_address(0, True, int(bina,2))
AttributeError: module 'secp256k1' has no attribute 'privatekey_to_address'

You must use this https://github.com/iceland2k14/secp256k1 for my code.....

Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 266. (Read 243769 times)