you can select the number of cores via ' a = ' in the top of the scrypt ..
I have to deal with the upload first... I haven't done it yet..
i just see that "
a " wasn't a good choice for the cores, since a is already used in scrypt.. change from "
a = 4" to "
c =4" and in the part "
for r in range(a): " to "
for r in range(c): " important!!
Although it works with "
a" , it could lead to errors!
The corrected code !import time
import random
import secp256k1 as ice
from time import sleep
from multiprocessing import Process, Value
c = 4
y = 2097152
counter = Value('L')
def process1(number,counter,):
while True:
a = random.randint(2**63, 2**64)
for x in range(number):
x = x
bina = bin(a)[2:]
zeros = bina.count("0")
if zeros >= 16:
if zeros <= 43:
addr = ice.privatekey_to_address(0, True, a)
with counter.get_lock():
counter.value += 1
if addr.startswith('16jY7q'):
print('\n Pattern found: '+hex(a)+' | '+ addr)
sleep(1)
print("\n continue...\n")
if addr == "16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN":
file=open(u"16jY.Info.txt","a")
file.write('\n '+hex(a)+' | '+ addr)
file.close()
wait = input("Press Enter to Exit.")
sleep(1)
exit()
if x == 0:
print ('', hex(a), bina, zeros, '', str(counter.value))
a = a +1
pass
if __name__ == '__main__':
t = time.ctime()
print('',t)
number = y
workers = []
print('\n\n BINARY MULTI PY \n #64: 8000000000000000...ffffffffffffffff\n\n')
print(' zeros')
print(' == PRIVKEY HEX == |____________________________ BINARY ____________________________| | CNT == \n')
for r in range(c):
p = Process(target=process1, args=(number,counter,))
workers.append(p)
p.start()
for worker in workers:
worker.join()
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