Programming Praxis – A Programming Puzzle

In today’s Programming Praxis exercise, our goal is to write a function such that f (f n) = -n. Let’s get started, shall we?

My first instinct was to define f as a 90 degree rotation using the complex plane, but decided that that was against the spirit of the exercise. The input is plain integers, so the output shouldn’t consist of complex numbers. After that I mucked around for a good while with doubling and halving numbers, which worked on everything except multiples of four. The basic idea (reversing parity and/or signs) was correct, but halving numbers always produced cases where the same condition holds for n and f(n). About an hour and a half after starting on the exercise the solution finally hit me: simply swap each pair of adacent numbers and reverse the sign for the even ones. I replaced the first draft of the correct solution, which had the four conditions written out, with the following more elegant formula.

f :: Integer -> Integer
f n = n * (2 * mod n 2 - 1) + signum n

A test to see if everything is working properly:

main :: IO ()
main = print $ all (\n -> f (f n) == -n) [-1000..1000]

Tags: , , , , , , ,

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: