To Hell in a Handbasket

After feedback from some friends of mine, and doing a little bit of background research, I am writing this update to my original post last week.  As it turns out, by checking all the numbers in the form 6k ± 1  instead of checking each number up to the input value, I have increased the speed by 7 times! determineIsPrime3 (line 48 below) is the fastest algorithm so far.  determineIsPrime2, a simple comparision against the , was twice as fast as the original algorithm.

For now, I am putting this one to bed.  It was a fun exercise, but I have got what I need from it.

'''
Checks the specified value to determine if it is a prime number.
If it is not prime the divisor will be returned instead.

@author: Eric Silva
'''

import math, time

#Change this value to whatever value you want to test for prime.
#testValue = 65027
#testValue = 155188329701
testValue = 99194853094755497
#testValue = 10888869450418352160768000001
print 'Testing %d...' % testValue

def determineIsPrime(testPrime):
    if testPrime % 2 == 0:
        return 'Divisible by 2'
    if testPrime % 3 == 0:
        return 'Divisible by 3'
    testNum = 7
    testLimit = testPrime
    while testLimit >= testNum:
        if testPrime % testNum == 0:
           return 'Divisible by %d' % testNum
        testLimit = testPrime/testNum

        testNum = testNum + 2

    return '%d is prime!' % testPrime

def determineIsPrime2(testPrime):
    if testPrime % 2 == 0:
        return 'Divisible by 2'
    if testPrime % 3 == 0:
        return 'Divisible by 3'
    testNum = 5
    sqrt = math.sqrt(testPrime)
    while testNum <= sqrt:
        if testPrime % testNum == 0:
           return 'Divisible by %d' % testNum

        testNum = testNum + 2

    return '%d is prime!' % testPrime

def determineIsPrime3(testPrime):
    if testPrime % 2 == 0:
        return 'Divisible by 2'
    if testPrime % 3 == 0:
        return 'Divisible by 3'
    testNum = 7
    sqrt = math.sqrt(testPrime)
    while ((6 * testNum) + 1 <= sqrt) or ((6 * testNum) - 1 <= sqrt):
        if testPrime % testNum == 0:
           return 'Divisible by %d' % testNum

        testNum = testNum + 2

    return '%d is prime!' % testPrime

startTime = time.time()
result = determineIsPrime(testValue)
endTime = time.time()

print result
print '1. Calculation took %f s\n' % (endTime - startTime)

startTime = time.time()
result = determineIsPrime2(testValue)
endTime = time.time()

print result
print '2. Calculation took %f s\n' % (endTime - startTime)

startTime = time.time()
result = determineIsPrime3(testValue)
endTime = time.time()

print result
print '3. Calculation took %f s\n' % (endTime - startTime)

Results:

Testing 99194853094755497...
99194853094755497 is prime!
1. Calculation took 202.609000 s

99194853094755497 is prime!
2. Calculation took 114.813000 s

99194853094755497 is prime!
3. Calculation took 28.781000 s

While working on some caching settings, I had a need to know if a number is prime. I wrote this little Python script which will tell you if the number defined in the script is indeed a prime.

'''
Checks the specified value to determine if it is a prime number.
If it is not prime the divisor will be returned instead.

@author: Eric Silva
'''

#Change this value to whatever value you want to test for prime.
testValue = 3011

def determineIsPrime(testPrime):
    if testPrime % 2 == 0:
        return 'Divisible by 2'
    testNum = 3
    testLimit = testPrime
    while testLimit >= testNum:
        if testPrime % testNum == 0:
           return 'Divisible by %d' % testNum
        testLimit = testPrime/testNum

        testNum = testNum + 2

    return '%d is prime!' % testPrime

result = determineIsPrime(testValue)

print result

I was looking for a version of PyXML for Python 2.5 and had some difficulty due to the project being unmaintained.  I was able to find someone that compiled PyXML 0.8.4 for Python 2.5.  You can find it here.