UPDATE May 13, 2015: Migrated the code syntax to match Jekyll’s syntax.
Project Euler is an interesting website that offers about 400 different mathematics and computer programming questions. They range from easy (finding the sum of some set of big numbers) to impossible (navigating through Rudin-Shapiro sequences). Just recently, I solved the 179th question with the help of some Java code. While my program gave me the correct answer, the execution time on my Macbook Pro laptop was 80 seconds — and there is an informal “60 seconds” rule that implies that code should be able to solve a problem in fewer than 60 seconds. So I wanted to determine in what ways I could optimize my code.
Here was the question: Find the number of integers 1 < n < 10^7, for which n and n + 1 have the same number of positive divisors. For example, 14 has the positive divisors 1, 2, 7, 14 while 15 has 1, 3, 5, 15.
This wasn’t too bad for me. I already had a method that could compute the sum of the divisors of a number based on problem 23, so I revised it to add up the number of divisors, rather than the sum. Then I just iterated through each number from 1 to 10 million. Here was the first version of my code:
I’m not going to say what the answer was, but as mentioned before, the execution time (endTime – startTime) was about 80 seconds. Looking at the code, the limiting factor is the 10 million calls I make to the method numOfDivisors(). So how can I improve this? In other words, how can I avoid making all those calls to my static method here?
To start, I initialized an array of 10,000,001 elements, called divs, where divs[x] refers to the number of divisors of x. Then, I used two nested for loops to make sure that each entry of divs[x] did hold the number of divisors of x. The outer for loop went from int i = 1 to 10,000,000, and the inner for loop went as far from int j = 1 to as large a number such that i*j <= x. This implies that all divisors for a number are counted! For instance, if we had the number 2, which has the divisors of 1 and 2, then the entry divs should be incremented twice — which it is, because of i=1 and j=2 first, then i=2, j=1 second.
Here, I avoid all the testing of “is a number is a factor of another number?”, as I do in my old code, because if I consider i*j = n, then I know that n has at least those two factors!
The updated code is as follows:
It gave me the right answer. And the runtime was an amazingly quick 1.9 seconds — much, much better! I don’t claim full credit for this second code, as I read the discussion forum for that problem after I solved it the first time, but it’s still nice to know how to optimize a program.