Haskell Chapter 4

Recursion Like other languages Base case Recursive call Author programs a number of built-in functions as examples Quick: maximum [3,5,4] => 5 What’s the base case? replicate 4 3 => [3,3,3,3] What’s the base case? take 3 [1..10] => [1,2,3] What’s the base case? take 5 [1..3] => [1,2,3] What’s the base case?

Maximum – Recursion w/ pattern matching 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 – Recursion w/ guards 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 – Recursion w/ guard + patterns 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 – (turn in) take' 20 [1..4] take' 3 [1..4]

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] Quick trace (turn in) reverse' [1, 2, 3, 4]

Repeat Haskell supports infinite lists repeat' :: a -> [a] repeat' x = x:repeat' x repeat 3 3: repeat 3 etc, Don’t bother tracing, we don’t have all day. 

Think it through – don’t look ahead! Prelude> zip [1,2,3] [4,5,6] [(1,4),(2,5),(3,6)] How would you write zip? How many base cases? What’s the recursive call? Patterns or guards? give patterns a try! Add to your class participation

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 <- xs, a > x] in quicksort smallerOrEqual ++ [x] ++ quicksort larger Add to your class participation

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