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[英]Syntax error in python machine learning sklearn code mosh Python course. Id appreciate if someone could hep me
[英]Syntax error in python code to find Prime-factors. Id appreciate if someone could help me
我一直在使用进行素数分解的代码时遇到语法错误
这是这个代码
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
这适用于小于 4 或 5 位的小数字,但对于较大的数字会产生索引错误。 如果我在第 24 行删除 sqrt function ,它开始花费太长时间。
错误看起来像这样
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
我无法解决此问题,如果您能帮助我,我将不胜感激。
代码有两个基本问题。 对于素数的 for 循环之一,您必须检查直到 int(sqrt(number))+1。 并且,在那之后的while循环中,当数字低于原始数字的sqrt时,您必须中断,应该使用另一个变量。 更正后的代码是:
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
如果我的解释不清楚,请查看代码中的更改。
我写了一个可以分解数字的小数分解引擎。
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 中进行分解(超过 60 位级别),我还让 Alpertons ECM SIQs 引擎工作: https://github.com/oppressionslayer/primalitytest
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