In today’s Programming Praxis exercise we have to write an algorithm to find the integer logarithm of a number, i.e. the largest power the base can be raised to that does not exceed the number. Let’s get started.

First, the O(n) solution, which works the same as the Scheme version.

ilog :: Integral a => a -> a -> Integer ilog b n = if n == 0 then -1 else ilog b (div n b) + 1

For the O(log n) version, we use the until function to determine the bounds rather than using explicit recursion. Other than that, there’s not much to be had in the way of improvements.

ilognew :: (Ord a, Num a) => a -> a -> Integer ilognew b n = f (div ubound 2) ubound where ubound = until (\e -> b ^ e >= n) (* 2) 1 f lo hi | mid == lo = if b ^ hi == n then hi else lo | b ^ mid < n = f mid hi | b ^ mid > n = f lo mid | otherwise = mid where mid = div (lo + hi) 2

Like the Scheme solution, we check the equivalence of the two methods by testing a few different bases and the numbers 1 to a million.

main :: IO () main = print $ and [ilognew b n == ilog b n | b <- [2,3,5,7], n <- [1..1000000]]

Only a 30% reduction in lines this time compared to the Scheme solution, since most of the code is checking conditions. Oh well, it’ll do.