Question

I’m trying to learn Python by working through the problems on the Project Euler website. I know exactly what I want my code to do, and my method works on paper, but I can’t make the code work.

GitHub link: https://github.com/albyr/euler-python/blob/master/euler3.py

I have created two functions, one which works out the factors of the target number, and one which checks whether a given number is prime.

# Function that finds all the factors of a given number
def findfactors(n):
    # for i in range(1,int(sqrt(n)+1)):
    for i in range(1,n+1):
        if n/i == int(n/i):
            factors.append(i)

# Function that checks if a number is prime
def checkprime(n):
    # Trial division
    for i in range(2,int(sqrt(n)+1)):
        if n/i == int(n/i):
            # Number gives a remainder upon division and therefore is not prime
            isprime = False
            break
        else:
            isprime = True
    if isprime == True:
        return True
    elif isprime == False:
        return False

I’m sure that to experts that code looks horrible. But it works if I use the Python shell:

>>> checkprime(9)
False
>>> checkprime(79)
True
>>> checkprime(factors[3])
True

But when I run the program with F5 I get:

Traceback (most recent call last):
  File "/home/alby/euler-python/euler3.py", line 45, in <module>
    checkprime(factors[i])
  File "/home/alby/euler-python/euler3.py", line 32, in checkprime
    if isprime == True:
UnboundLocalError: local variable 'isprime' referenced before assignment

If I call the checkprime function from within the program with a hardcoded number (e.g. checkprime(77)) I get no output at all. I’m certain that this is something basic about the way that Python works that I don’t understand, but I cannot for the life of me work out what.

Any suggestions?

Answer 1

In your Github code, we can see that you’re trying to call checkprime(1) (on the first iteration through your last loop).

# Check each factor to see if it is prime or compound
for i in range(0,len(factors)):
    print (factors[i])
    # Why can't I call checkprime here, like this? It works in the console.
    checkprime(factors[i])

But look at your code:

def checkprime(n):
    # Trial division
    for i in range(2,int(sqrt(n)+1)):
        if n/i == int(n/i):
            # Number gives a remainder upon division and therefore is not prime
            isprime = False
            break
        else:
            isprime = True

If n = 1, then range(2, int(sqrt(1)+1)) is range(2,2) which is empty… so isprime never gets set, because the loop body never gets run.

Keep in mind that the arguments to range() are a half-open interval – range(x,y) is “integers starting at x and ending before y”. Thus range(2,3) = [2] and range(2,2) = [].

Another issue here is that findfactors() is returning 1 as the first factor – this is probably not what you want:

def findfactors(n):
    # for i in range(1,int(sqrt(n)+1)):
    for i in range(1,n+1):

For prime factorization checking, you probably want to start at 2, not 1 (since everything is divisible by 1).

---

Also, this code is redundant:

if isprime == True:
        return True
    elif isprime == False:
        return False

You can really just write this as…

return isprime

Or you can go one step better and never use isprime in the first place – just replace isprime = True with return True and isprime = False with return False.

Finally, a shorthand for int(n/i) is n // i – Python’s // operator does integer division.