I am trying to write a code in python to find all prime numbers of a number. My problem is that with this line of code it does not work to return the prime numbers of 10, the list only returns 2. Now I adapted this code from this page https://www.geeksforgeeks.org/prime-factor/ as I want to make my factors come out as a list.
The code from that website works for the number 10, however I do not understand why it does not work for number 10 when I run my own slightly modified version of it.
I have tried to +10 at the end of my range function, instead of a +1 and this does solve the problem, however I am still unsure as to why I even have a problem in the first place. Secondly, will the +10 work for all numbers with no error? In theory it should as I should only have factors unto square root of n, but I am not sure again. Lastly, if if the +10 does work, won't that make the code run slower as it will iterate through unneeded loops, how can I improve the speed?
This is my code that I used.
import math
def primefact():
n = int(input('What is your number?:'))
prime_factors = []
while n % 2 == 0: # Checks if number is divisible by 2
prime_factors.append(2) #repeats it until n is no longer divisible by 2
n = n / 2
for i in range(3, int(math.sqrt(n)) + 1, 2): # Testing for odd factors
while n % i == 0:
prime_factors.append(i)
n = n / i
print(prime_factors)
return
primefact()
Here's a piece of code that I wrote:
from numpy import mod, int0, sqrt, add, multiply, subtract, greater, equal, less, not_equal, floor_divide
class findFactors:
def __init__(self, n):
self.primeFactorize(n)
def primeFactorize(self, n):
factors = self.findFactors(n)
self.factors = factors
primeFactors = []
xprimeFactors = []
for factor in factors:
if prime(factor).isPrime:
primeFactors.append(factor)
ntf = n
nprime = 0
while not_equal(ntf, 1):
while equal(mod(ntf, primeFactors[nprime]), 0):
ntf = floor_divide(ntf, primeFactors[nprime])
xprimeFactors.append(primeFactors[nprime])
nprime = add(nprime, 1)
self.primeFactors = primeFactors
self.extendedPrimeFactors = xprimeFactors
def findFactors(self, number):
if prime(number).isPrime: return [1, number]
factors = []
s = int0(sqrt(float(number)))
for v in range(1, add(s, 1)):
if equal(mod(number, v), 0):
factors.append(int(v))
factors.append(int(floor_divide(number, v)))
factors.sort()
return factors
class prime:
def __init__(self, n):
self.isPrime = self.verify(n)
def verify(self, n):
if less(n, 2):
return False
if less(n, 4):
return True
if not n & 1 or equal(mod(n, 3), 0):
return False
s = int0(sqrt(float(n)))
for k in range(1, add(s, 1)):
mul = multiply(6, k)
p = add(mul, 1)
m = subtract(mul, 1)
if greater(m, s):
return True
if equal(mod(n, p), 0) or equal(mod(n, m), 0):
return False
Imagine func = findFactors(n)
func.factors will returns a list with all the factors of the number n,
func.extendedPrimeFactors will return a list with the prime factorization of the number,
func.primeFactors will return a list with the primes appearing only once instead of x times
also, there's a really fast prime checker down there. (Prime checker usage: prime(n).isPrime )