# Project Euler: Problem 7

Here is the seventh installment of this series. And just to keep things interesting I'm going to do something different this time around. I still performed the solutions for Haskell, Java, and Perl (if you want to view them check out my github repo), but instead of posting the code for all four languages, I'm going to be talking about the evolution of my Python solution, showing you all three revisions, and talking briefly about how or why things changed.

Python code:

Solution 1:

 ``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25``` ```#!/usr/bin/python3 import math def is_prime(divided): divisor = 3 sqrt_divided = math.sqrt(divided) while divisor <= sqrt_divided: if divided % divisor == 0: return False divisor += 2 return True if __name__ == "__main__": prime_list =  count = 3 while len(prime_list) < 10001: if is_prime(count) == True: prime_list.append(count) count += 2 print(prime_list[-1]) ```

Looking at this first solution, we see that it follows the Haskell solution quite closely. This is of course not an accident; I'm lazy. ;) And since I'm still learning Haskell, I find it easier to learn the basics of a language through translation. So this Python solution was my first solution, which I then translated to Haskell.

Solution 2:

 ``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28``` ```#!/usr/bin/python3 import math def is_prime(divided): divisor = 3 sqrt_divided = math.sqrt(divided) while divisor <= sqrt_divided: if divided % divisor == 0: return False divisor += 2 return True if __name__ == "__main__": count = 1 #to substitute the prime number 2 in the example last = 0 iterator = 3 while count < 10001: if is_prime(iterator): count += 1 last = iterator iterator += 2 print(last) ```

So in coming up with my Java solution, I didn't want to deal with trying to come up with a linked list-based solution for number storage. At which point I had the great thought: why not just hold the last number to be prime, and not all primes. Spending all this time with lists and arrays instead of just straight Ints has really played with my thought patterns. So after I wrote up the Java solution, I translated it to Python, to see if there would be a reduction in the execution time based on only using Ints instead of a list of Ints. Paint me a little surprised when I saw that the difference was minimal if non-existent most of the time.

Solution 3:

 ``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27``` ```#!/usr/bin/python3 import math def is_prime(divided): divisor = 3 sqrt_divided = int(math.sqrt(divided)) while divisor <= sqrt_divided: if divided % divisor == 0: return False divisor += 2 return True if __name__ == "__main__": # using a list to store each int value data = [1,0,3] while data < 10001: if is_prime(data): data += 1 data = data data += 2 print(data) ```

After digging into how Python does its memory management, like Lists being mutable while Ints are not, I modified the second solution so that the Ints are now in a list. This way the Python memory manager does not have to create a new variable to store each value. After these changes I got the speed difference I was expecting to see between the first and second revision.

Also, you might have noticed that in the is_prime function, I cast math.sqrt to an Int. As math.sqrt returns a float, every time that comparison happens for the while loop there has to be a conversion of divisor from its current state of int to a float. Math.sqrt returns a float, thus sqrt_divided is a float, and divisor is an int. So every time that while loop cycles again, there has to be a conversion of an int to a float. By casting that float to an int during the variable declaration, I save myself a few thousand cast operations and shave off a couple hundredths of a second from the total run time.

Language Run Times:

Python:

solution 1: .452s

solution 2: .433s

solution 3: .376