python 代碼中用於查找素因子的語法錯誤。 如果有人可以幫助我，我將不勝感激

Question

我一直在使用進行素數分解的代碼時遇到語法錯誤

這是這個代碼

from sys import argv
from os import system, get_terminal_size
from math import sqrt

number = int(argv[1])
width = get_terminal_size().columns
prime_numbers = []
prime_factors = []
_ = system('clear')
print() 

def is_prime(n):
    for i in range(2, n):
        if n % i == 0:
            return False

    return True

if is_prime(number):
    print(f"It is a prime number \nIts only factors are 1 and itself \n1, {number}")
    exit()

x = len(str(number))
for i in range(2, int(sqrt(number))):
    if is_prime(i):
            prime_numbers.append(i)

            #print(f"found ")
#print(prime_numbers)

i = 0
while True:
    if (number % prime_numbers[i] != 0):
        i += 1
        continue
    
    prime_factors.append(prime_numbers[i])
    print("%2d  | %3d".center(width) % (prime_numbers[i], number))
    print("_________".center(width))                                
    number /= prime_numbers[i]
    if number == 1:
        break
print("1".center(width))

print("Answer ")

i = len(prime_factors)
j = 1

for k in prime_factors:
    if j == i:
        print(k)
        break

    print(f"{k}", end=" X ")
    j += 1

這適用於小於 4 或 5 位的小數字，但對於較大的數字會產生索引錯誤。 如果我在第 24 行刪除 sqrt function ，它開始花費太長時間。

錯誤看起來像這樣

Traceback (most recent call last):
  File "prime-factor.py", line 33, in <module>
    if (number % prime_numbers[i] != 0):
IndexError: list index out of range

real    0m0.049s
user    0m0.030s
sys 0m0.014s
(base) Souravs-MacBook-Pro-5:Fun-Math-Algorithms aahaans$ time python3 prime-factor.py 145647

我無法解決此問題，如果您能幫助我，我將不勝感激。

Answer 1

無需重建已經可用的primePy

from primePy import primes
primes.factors(101463649)

output

[7, 23, 73, 89, 97]

Answer 2

代碼有兩個基本問題。 對於素數的 for 循環之一，您必須檢查直到 int(sqrt(number))+1。 並且，在那之后的while循環中，當數字低於原始數字的sqrt時，您必須中斷，應該使用另一個變量。 更正后的代碼是：

from sys import argv
from os import system, get_terminal_size
from math import sqrt

number = int(argv[1])
width = get_terminal_size().columns
prime_numbers = []
prime_factors = []
_ = system('clear')
print() 

def is_prime(n):
  for i in range(2, n):
    if n % i == 0:
      return False

  return True

if is_prime(number):
  print(f"It is a prime number \nIts only factors are 1 and itself \n1, {number}")
  exit()

x = len(str(number))
limit = int(sqrt(number))
for i in range(2, limit+1):
  if is_prime(i):
    prime_numbers.append(i)

i = 0
while True:
  if i == len(prime_numbers)-1:
    # prime_factors.append(int(number))
    break
  if (number % prime_numbers[i] != 0):
    i += 1
    continue
  prime_factors.append(prime_numbers[i])
  print("%2d  | %3d".center(width) % (prime_numbers[i], number))
  print("_________".center(width))                                
  number /= prime_numbers[i]
prime_factors.append(int(number))
print("%2d  | %3d".center(width) % (number, number))
print("_________".center(width))
print("1".center(width))

print("Answer ")
i = len(prime_factors)
j = 1
for k in prime_factors:
  if j == i:
    print(k)
    break
  print(f"{k}", end=" X ")
  j += 1

如果我的解釋不清楚，請查看代碼中的更改。

Answer 3

我寫了一個可以分解數字的小數分解引擎。

import math

def LLL(N):
   p = 1<<N.bit_length()-1
   if N == 2:
     return 2
   if N == 3:
     return 3
   s = 4
   M = pow(p, 2) - 1
   for x in range (1, 100000):
     s = (((s * N ) - 2 )) % M
     xx = [math.gcd(s, N)] + [math.gcd(s*p+x,N) for x in range(7)] + [math.gcd(s*p-x,N) for x in range(1,7)] 
     try:
        prime = min(list(filter(lambda x: x not in set([1]),xx)))
     except:
        prime = 1
     if prime == 1:
        continue
     else:
        break
   #print (s, x, prime, xx)
   return prime

因素：

In [219]: LLL(10142789312725007)                                                                                                                                                       
Out[219]: 100711423

from https://stackoverflow.com/questions/4078902/cracking-short-rsa-keys

如果您想要在 python 中進行分解（超過 60 位級別），我還讓 Alpertons ECM SIQs 引擎工作： https://github.com/oppressionslayer/primalitytest