Prime number finder

ccc

I never used a for / else loop before but here it works well. When we find the first factor, we print a not prime message and then fast fail out of the loop with break. If the loop finishes without a break then we use the else clause to tell the use that we have a prime number.

Let's call this the 5F approach: finding first factor fast fail...

import time

print('Made by Yasas Kulatunga, Dhari Prashanth and Dhivy Prashanth')
print(' ')
number = int(input("Enter a Number: ").strip())
print(' ')
time.sleep(0.5)
print('THINKING.')
time.sleep(0.5)
print('THINKING..')
time.sleep(0.5)
print('THINKING...')
time.sleep(0.5)

print(' ')
while number:
    for i in range(2, number):
        if number % i == 0:
            print('{} is not prime.'.format(number))
            break
    else:
        print('{} is prime!'.format(number))
    number = int(input("Enter a Number: ").strip())

ramvee

More efficient way would be to check the number for divisibility, only till its' square root - 1.
Because the largest factor of any number cannot exceed it's square root, there is no need for range to go beyond that.
With minimal alteration to script given by @ccc , this would be more efficient.

# Revised as per bug reported by @cvp
# This code takes care of numbers 1 and 2 also.
import time
from math import sqrt

print('Made by Yasas Kulatunga, Dhari Prashanth and Dhivy Prashanth')
print(' ')
number = int(input("Enter a Number: ").strip())
maxfact = int(sqrt(number)+1)
print(' ')
time.sleep(0.5)
print('THINKING.')
time.sleep(0.5)
print('THINKING..')
time.sleep(0.5)
print('THINKING...')
time.sleep(0.5)

print(' ')
while number:
    for i in range(2, maxfact):
        if number % i == 0:
            print('{} is not prime.'.format(number))
            break
    else:
        print('{} is prime!'.format(number))
    number = int(input("Enter a Number: ").strip())
    maxfact = int(sqrt(number)+1)

ccc

The above implementations do not deal with the fact that all values less than 2 are not prime.

https://primes.utm.edu/notes/faq/one.html
https://www.reference.com/math/0-prime-number-1964bd6272594f3f
https://primes.utm.edu/notes/faq/negative_primes.html

cvp

Bug, try with 25

maxfact = int(sqrt(number)-1) -> maxfact = int(sqrt(number))

cvp

Bug, try with 25

for i in range(2, maxfact+1):

cvp

Check if 2 is a divisor but skip all other even divisors
(Perhaps what @ccc minded with "less than 2")

 for i in [2]+list(range(3, maxfact+1,2)):

PINEHACKS

The exact output is this, although it is from the original code I posted on top: Enter a Number: 25

THINKING.
THINKING..
THINKING...

5 times 5 is 25

25 is not a prime number, the number has factors more than 1 and the number itself.

Enter a number:

cvp

Sorry, I had tested the last code of @ramvee until √ - 1

cvp

But he was almost right, you have to check only until sqrt(number),
And, what I just added, by skipping the even divisors!

For instance, if number is 29, your code checks the division by
2 3 4 5 6 7 8 9 10 11 12 13 ...29
But "mine" (& @ramvee) only
2 3 5

ccc

@cvp, If you go the the three vertical dots at the upper right of your forum post, you can edit it to add new insights. This is probably a better approach than replying to your own posts.

cvp

@ccc, Thanks, I knew it but sometimes I forget it, and sometimes I'm not sure how other people see an update, while I'm sure they see a new response

ramvee

@cvp
Thank you for pointing the bug in my code. :)

I put in your line of code

for i in [2]+list(range(3, maxfact+1,2)):

and it works fine,
Except it shows 2 is not a prime.. If we sort this we're good.. I think..

Phuket2

I think the consecutive 3 post's from @cvp might have been better done in edit. Maybe and maybe not. If the items inside a post themselves were also threaded , then it would be clear to do like that. But they are not. Just saying it's not a clear cut choice.

cvp

@Phuket2 thanks for your support 😜 but I think @ccc is right for most cases...

Phuket2

@cvp , lol, yes he is normally right. But need to keep him on his toes. 😱😈