python prime number bug

Question

The question asks for the prime number from 1-25 and this is my code, it came out to be wrong. Can anyone help me plz???Thank you so much!

n=1
f=0
while n<=25:
    for i in range(2,n):
        if n%i==0:
            f+=1
        else:
            f
    if f==0:
        print str(n) + "is a prime number"
    else:
        print str(n) + "is not a prime number"
    n+=1

You can get rid of that `else: f`, it does nothing useful. BTW, this is a _very_ inefficient way to test for prime numbers. And you should start at `n = 2`, since 1 is neither prime nor composite: it's a unit. — PM 2Ring, Feb 18 '16 at 17:34
@PM2Ring So what kind of way is an efficient way of testing for prime numbers??Can you plz show an example??? Thanks! — robynscheetos, Feb 18 '16 at 17:45
There are a variety of fast ways here: [Fastest way to list all primes below N](http://stackoverflow.com/q/2068372/4014959), but they might be a bit too advanced for you at this stage. The standard way to make a list of primes starting from 2 is called the [Sieve of Eratosthenes](https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes). See if you can figure out a way to do the Sieve in Python. — PM 2Ring, Feb 18 '16 at 17:55

score 2 · Answer 1 · edited Feb 18 '16 at 17:58

You are not re-setting the value of f, so once you hit a non-prime number all the numbers following that will be reported as not prime.

n=1
f=0
while n<=25:
    for i in range(2,n):
        if n%i==0:
            f+=1
        else:
            f
    if f==0:
        print str(n) + "is a prime number"
    else:
        print str(n) + "is not a prime number"
    f = 0
    n+=1

Also your way is inefficient, and you should start from 2, not 1. Also, all the prime numbers after 3 are of the form of 6n-1 or 6n+1 (Where n is a integer). So you do not need to test all the numbers. Just test the numbers 6n-1 and 6n+1 if they are prime or not.

score 0 · Answer 2 · answered Feb 18 '16 at 18:28

I've re-arranged your code to make it a little more efficient. It first tests if n is even, and then it tests for odd factors up to the square root of n, since if n is not prime it must have a factor <= its square root. (Can you see why?)

To simplify the logic, I've moved the testing into a function.

When a factor of n is found we immediately return from the function since there's no point looking for more factors.

def prime_test(n):
    if n % 2 == 0:
        return False
    for i in range(3, int(n**0.5 + 1), 2):
        if n % i == 0:
            return False
    return True

print 2, "is a prime number"

for n in range(3, 26):
    if prime_test(n):
        print n, "is a prime number"
    else:
        print n, "is not a prime number"

output

2 is a prime number
3 is a prime number
4 is not a prime number
5 is a prime number
6 is not a prime number
7 is a prime number
8 is not a prime number
9 is not a prime number
10 is not a prime number
11 is a prime number
12 is not a prime number
13 is a prime number
14 is not a prime number
15 is not a prime number
16 is not a prime number
17 is a prime number
18 is not a prime number
19 is a prime number
20 is not a prime number
21 is not a prime number
22 is not a prime number
23 is a prime number
24 is not a prime number
25 is not a prime number

python prime number bug

2 Answers2