In the last episode of Clojure Golf we saw some interesting ways to write a function that takes a couple sequences, applies a function to the pairing elements, and then returns a list of the results based on a supplied filtering function. Thanks to all who participated in that exercise.
This episode is a little tougher.
Below I have written a function that takes two numbers: any given number, and another number that acts as a starting point. This function will then calculate the largest prime factor of the given number. The implementation of this function came from a Twitter meme where people would describe songs in code within the constraints of 140 characters. As it turns out, most of the snippets were meant to be humerous and didn’t actually run or represent legal source programs at all. I wanted to break that trend while still maintaining some sense of humor about the whole thing. In any case, before I thought of an entry I knew that my snippet would deal with the Tommy Tutone song 867-5309/Jenny. However, what I didn’t know at the time was that the number 8675309 is a prime number 1, but once I realized this I knew what my #songsincode entry would be.
[sourcecode lang=”clj” gist=”183954″] ;; largest prime factor (defn lpf “Takes a number n and a starting number d > 1 and calculates the largest prime factor of n starting at number d.
usage: (lpf 364362978 2) => 8675309″ [n d] (if (> d n) (- d 1) (recur (#(if (zero? (rem % d)) (recur (/ % d)) %) n) (inc d)))) [/sourcecode]
I think the above function is likely as small as it can get 2, but it’s not very fast (e.g.
(lpf 1234567890123456789012345678901234567890 2) 3 took about 10 minutes on my machine). Therefore, this episode of Clojure Golf is devoted to making
lpf as fast as possible, while still preserving as much compactness as possible. It would also be nice to see how to squeeze a few more characters out of this particular implementation.
As always, Clojure snippets are not required — any programming language is encouraged.