I have just written a function in Haskell that will perform a binary search on a list of integers. Pass it your list, the value you’re searching for, default low and high (0 and n-1) and it will return for you the position of your value. I’m sure there are plenty of experienced Haskell coders that will tell me there’s a better way, but I haven’t heard from one yet.

binsearch :: [Int] -> Int -> Int -> Int -> Int -- list, value, low, high, return int binsearch xs value low high | high < low = -1 | xs!!mid > value = binsearch xs value low (mid-1) | xs!!mid < value = binsearch xs value (mid+1) high | otherwise = mid where mid = low + ((high - low) `div` 2)

You could make your function a tad more general:

binsearch :: (Ord a) => [a] -> a -> Int -> Int -> Maybe Int — list, value, low, high, return int

binsearch xs value low high

| high value = binsearch xs value low (mid-1)

| xs!!mid < value = binsearch xs value (mid+1) high

| otherwise = Just mid

where

mid = low + ((high – low) `div` 2)

There is no point of doing a binary search on a list, because it takes O(n*log(n)) time, whereas a linear search takes O(n) time. You should change the binary search to an array. This is simply a matter of replacing !! with !. e.g.:

import Data.Array;

binarySearch haystack needle lo hi

| hi needle = binarySearch haystack needle lo (mid-1)

| pivot < needle = binarySearch haystack needle (mid+1) hi

| otherwise = Just mid

where

mid = lo + (hi-lo) `div` 2

pivot = haystack!mid

stupid html. i meant:

import Data.Array;

binarySearch haystack needle lo hi

| hi < lo = Nothing

| pivot > needle = binarySearch haystack needle lo (mid-1)

| pivot < needle = binarySearch haystack needle (mid+1) hi

| otherwise = Just mid

where

mid = lo + (hi-lo) `div` 2

pivot = haystack!mid