[python] Python Prime number checker

I have been trying to write a program that will take an inputed number, and check and see if it is a prime number. The code that I have made so far works perfectly if the number is in fact a prime number. If the number is not a prime number it acts strange. I was wondering if anyone could tell me what the issue is with the code.

a=2
num=13
while num > a :
  if num%a==0 & a!=num:
    print('not prime')
    a=a+1
  else:
    print('prime')
    a=(num)+1

the result given when 24 is inputed is: not prime not prime not prime prime

How would i fix the error with the reporting prime on every odd and not prime for every even

This question is related to python primes

The answer is


The two main problems with your code are:

  1. After designating a number not prime, you continue to check the rest of the divisors even though you already know it is not prime, which can lead to it printing "prime" after printing "not prime". Hint: use the `break' statement.
  2. You designate a number prime before you have checked all the divisors you need to check, because you are printing "prime" inside the loop. So you get "prime" multiple times, once for each divisor that doesn't go evenly into the number being tested. Hint: use an else clause with the loop to print "prime" only if the loop exits without breaking.

A couple pretty significant inefficiencies:

  1. You should keep track of the numbers you have already found that are prime and only divide by those. Why divide by 4 when you have already divided by 2? If a number is divisible by 4 it is also divisible by 2, so you would have already caught it and there is no need to divide by 4.
  2. You need only to test up to the square root of the number being tested because any factor larger than that would need to be multiplied with a number smaller than that, and that would already have been tested by the time you get to the larger one.

def isprime(n):
    '''check if integer n is a prime'''

    # make sure n is a positive integer
    n = abs(int(n))

    # 0 and 1 are not primes
    if n < 2:
        return False

    # 2 is the only even prime number
    if n == 2: 
        return True    

    # all other even numbers are not primes
    if not n & 1: 
        return False

    # range starts with 3 and only needs to go up 
    # the square root of n for all odd numbers
    for x in range(3, int(n**0.5) + 1, 2):
        if n % x == 0:
            return False

    return True

Taken from:

http://www.daniweb.com/software-development/python/code/216880/check-if-a-number-is-a-prime-number-python


Begginer here, so please let me know if I am way of, but I'd do it like this:

def prime(n):
    count = 0
    for i in range(1, (n+1)): 
         if n % i == 0: 
             count += 1
    if count > 2:
        print "Not a prime"
    else:
        print "A prime"

This is a slight variation in that it keeps track of the factors.

def prime(a):
    list=[]
    x=2
    b=True

    while x<a:
        if a%x==0:
            b=False
            list.append(x)
        x+=1

    if b==False:
        print "Not Prime"
        print list
    else:
        print "Prime"

This example is use reduce(), but slow it:

def makepnl(pnl, n):
    for p in pnl:
        if n % p == 0:
            return pnl
    pnl.append(n)
    return pnl

def isprime(n):
    return True if n == reduce(makepnl, range(3, n + 1, 2), [2])[-1] else False

for i in range(20):
    print i, isprime(i)

It use Sieve Of Atkin, faster than above:

def atkin(limit):
    if limit > 2:
        yield 2
    if limit > 3:
        yield 3

    import math
    is_prime = [False] * (limit + 1)

    for x in range(1,int(math.sqrt(limit))+1):
        for y in range(1,int(math.sqrt(limit))+1):
            n = 4*x**2 + y**2

            if n<=limit and (n%12==1 or n%12==5):
                # print "1st if"                                                                                                                    
                is_prime[n] = not is_prime[n]
            n = 3*x**2+y**2
            if n<= limit and n%12==7:
                # print "Second if"                                                                                                                 
                is_prime[n] = not is_prime[n]
            n = 3*x**2 - y**2
            if x>y and n<=limit and n%12==11:
                # print "third if"                                                                                                                  
                is_prime[n] = not is_prime[n]

    for n in range(5,int(math.sqrt(limit))):
        if is_prime[n]:
            for k in range(n**2,limit+1,n**2):
                is_prime[k] = False

    for n in range(5,limit):
        if is_prime[n]: yield n

def isprime(n):
    r = list(atkin(n+1))
    if not r: return False
    return True if n == r[-1] else False

for i in range(20):
    print i, isprime(i)

# is digit prime? we will see (Coder: Chikak)

def is_prime(x): flag = False if x < 2: return False else: for count in range(2, x): if x % count == 0: flag = True break if flag == True: return False return True


After you determine that a number is composite (not prime), your work is done. You can exit the loop with break.

while num > a :
  if num%a==0 & a!=num:
    print('not prime')
    break          # not going to update a, going to quit instead
  else:
    print('prime')
    a=(num)+1

Also, you might try and become more familiar with some constructs in Python. Your loop can be shortened to a one-liner that still reads well in my opinion.

any(num % a == 0 for a in range(2, num))

This would do the job:

number=int(raw_input("Enter a number to see if its prime:"))
if number <= 1:
    print "number is not prime"
else:
    a=2
    check = True
    while a != number:
        if number%a == 0:
            print "Number is not prime"
            check = False
            break
        a+=1
    if check == True:
        print "Number is prime" 

max=int(input("Find primes upto what numbers?"))
primeList=[]
for x in range(2,max+1):
    isPrime=True
    for y in range(2,int(x**0.5)+1) :
        if x%y==0:
            isPrime=False
            break

    if isPrime:
        primeList.append(x)
print(primeList)

Prime number check.

def is_prime(x):
    if x < 2:
        return False
    else:
        if x == 2:
            return True
        else:
            for i in range(2, x):
                if x % i == 0:
                    return False
            return True
x = int(raw_input("enter a prime number"))
print is_prime(x)

a=input("Enter number:")

def isprime(): 

    total=0
    factors=(1,a)# The only factors of a number
    pfactors=range(1,a+1) #considering all possible factors


    if a==1 or a==0:# One and Zero are not prime numbers
        print "%d is NOT prime"%a


    elif a==2: # Two is the only even prime number
        print "%d is  prime"%a


    elif a%2==0:#Any even number is not prime except two
        print "%d is NOT prime"%a  



    else:#a number is prime if its multiples are 1 and itself 
         #The sum of the number that return zero moduli should be equal to the "only" factors
        for number in pfactors: 
            if (a%number)==0: 
                total+=number
        if total!=sum(factors):
            print "%d is NOT prime"%a 
        else:
             print "%d is  prime"%a
isprime()

Your problem is that the loop continues to run even thought you've "made up your mind" already. You should add the line break after a=a+1


def is_prime(n):
    return all(n%j for j in xrange(2, int(n**0.5)+1)) and n>1