Haskell Chapter 4

Recursion  Like other languages  Base case  Recursive call  Author programs a number of built-in functions as examples

Maximum maximum' :: (Ord a) => [a] -> a maximum' [] = error "can't take maximum of empty list“ -- base case, if only one item, return it maximum' [x] = x -- else return max of first element or recursive call maximum' (x:xs) = max x (maximum' xs)

Replicate replicate' :: Int -> a -> [a] replicate' n x -- base case returns empty list | n <= 0 = [] -- else cons element to result of recursive call | otherwise = x : replicate' (n-1) x  Used guards because boolean condition, not a pattern  replicate' 3 5  5 : replicate’ 2 5 => [5,5,5]  5 : replicate’ 1 5 => [5, 5]  5: replicate’ 0 5 => [5]  n == 0, => []

Take  Two base cases (n=0, or empty list)  Guard without otherwise will fall through to patterns take' :: (Num i, Ord i) => i -> [a] -> [a] take' n _ | n <= 0 = [] take' _ [] = [] take' n (x:xs) = x : take' (n-1) xs Quick trace: take' 20 [1..10] take' 3 [1..10]

Reverse  Note ++ rather than : because both are lists  : works at front of list, not back reverse' :: [a] -> [a] reverse' [] = [] reverse' (x:xs) = reverse' xs ++ [x]

Repeat  Haskell supports infinite lists repeat' :: a -> [a] repeat' x = x:repeat' x  repeat 3  3: repeat 3  etc,

Think it through  Prelude> zip [1,2,3] [4,5,6]  [(1,4),(2,5),(3,6)]  How would you write zip?  How many base cases?  Patterns or guards?

Zip zip' :: [a] -> [b] -> [(a,b)] zip' _ [] = [] zip' [] _ = [] zip' (x:xs) (y:ys) = (x,y):zip' xs ys

Quick Exercise  With a partner, trace this code with this list of numbers:  [5,2,6,7,1,3,9,4] quicksort :: (Ord a) => [a] -> [a] quicksort [] = [] quicksort (x:xs) = let smallerOrEqual = [a | a <- xs, a <= x] larger = [a | a x] in quicksort smallerOrEqual ++ [x] ++ quicksort larger Turn in for class participation credit

Play and Share?  No, let’s start on the homework.