ch1 progress

ch01.lhs

@@ -1,6 +1,9 @@
 Chapter 1: Functional Programming
 ===
 
+> import Prelude hiding (foldl, foldr)
+> import Control.Exception
+
 Exercise 1.2
 ---
 
@@ -36,3 +39,162 @@ number of times in the rest of this book. Give appropriate definitions.
 >
 > single [x] = True
 > single _ = False
+
+Exercise 1.4
+---
+
+Write down a definition of reverse that takes linear time. One possibility is to use a foldl.
+
+> -- This takes linear time because the list basically builds up backwards
+> reverse :: [a] -> [a]
+> reverse = foldl (\a x -> x : a) []
+
+Exercise 1.5
+---
+
+Express both map and filter as an instance of foldr.
+
+> map2 :: (a -> b) -> [a] -> [b]
+> map2 f = foldr (\x a -> f x : a) []
+> 
+> filter2 :: (a -> Bool) -> [a] -> [a]
+> filter2 f = foldr (\x a -> if f x then x : a else a) []
+
+Exercise 1.6
+---
+
+Express `foldr f e . filter p` as an instance of foldr.
+
+> composed :: (a -> b -> b) -> (a -> Bool) -> b -> [a] -> b
+> composed f p e = foldr f e . filter p
+> 
+> composed2 :: (a -> b -> b) -> (a -> Bool) -> b -> [a] -> b
+> composed2 f p = foldr (\x a -> if p x then f x a else a)
+
+Exercise 1.7
+---
+
+The function takeWhile returns the longest initial segment of a list all of
+whose elements satisfy a given test. Moreover, its running time is proportional
+to the length of the result, not to the length of the input. Express takeWhile
+as an instance of foldr, thereby demonstrating once again that a foldr need not
+process the whole of its argument before terminating.
+
+> -- I think the fact that foldr doesn't consume the whole thing here is only
+> -- due to laziness, since foldr builds up the expression from the back.
+> takeWhile :: (a -> Bool) -> [a] -> [a]
+> takeWhile f = foldr (\x a -> if f x then x : a else []) []
+
+Exercise 1.8
+---
+
+The Data.List library contains a function dropWhileEnd which drops the longest
+suffix of a list all of whose elements satisfy a given boolean test. For
+example,
+
+> _ = dropWhileEnd even [1, 4, 3, 6, 2, 4] == [1, 4, 3]
+
+Define dropWhileEnd as an instance of foldr.
+
+> dropWhileEnd :: Eq a => (a -> Bool) -> [a] -> [a]
+> dropWhileEnd f = foldr (\x a -> if f x && a == [] then [] else x : a) []
+
+Exercise 1.9
+---
+
+An alternative definition of foldr is
+
+> foldr f e xs = if null xs then e else f (head xs) (foldr f e (tail xs))
+
+Dually, an alternative definition of foldl is
+
+> foldl f e xs = if null xs then e else f (foldl f e (init xs)) (last xs)
+
+where last and init are dual to head and tail. What is the problem with this
+definition of foldl?
+
+> -- init and last are both O(n) operations since they require traversing to
+> -- the end as opposed to head and tail which are O(1) which means overall this
+> -- alternate definition of foldl is really expensive.
+
+Exercise 1.10
+---
+
+Bearing the examples
+
+> -- foldr (⊕) e [x, y, z] == x ⊕ (y ⊕ (z ⊕ e))
+> -- foldl (⊕) e [x, y, z] == ((e ⊕ x) ⊕ y) ⊕ z
+
+in mind, under what simple conditions on (⊕) and e do we have
+
+> -- foldr (⊕) e xs = foldl (⊕) e xs
+
+for all finite lists xs?
+
+> -- (⊕) just needs to be associative and e the identity.
+>
+> -- This will give us x ⊕ (y ⊕ (z)) for foldr and (x ⊕ y) ⊕ z for foldl, which
+> -- still preserves the order of x, y, and z so if (⊕) is associative then
+> -- the statements are equivalent.
+
+Exercise 1.11
+---
+
+Given a list of digits representing a natural number, construct a function
+integer which converts the digits into that number. For example,
+
+> _ = integer [1, 4, 8, 4, 9, 3] == 148493
+
+Next, given a list of digits representing a real number r in the range 0 <= r <
+1, construct a function fraction which converts the digits into the
+corresponding fraction. For example,
+
+> _ = fraction [1, 4, 8, 4, 9, 3] == 0.148493
+
+> integer :: [Int] -> Int
+> integer = foldl (\a x -> 10 * a + x) 0
+> 
+> fraction :: [Int] -> Double
+> fraction = (/ 10) . foldr (\x a -> fromIntegral x + a / 10) 0
+
+Exercise 1.12
+---
+
+Complete the right hand sides of:
+
+> -- map (foldl f e) . inits = ???
+> -- map (foldr f e) . tails = ???
+
+> -- TODO
+
+Exercise 1.13
+---
+
+Define the function
+
+> apply :: Int -> (a -> a) -> a -> a
+
+that applies a function a specified number of times to a value.
+
+> apply 0 f x = x
+> apply n f x = apply (n - 1) f (f x)
+
+Exercise 1.14
+---
+
+Can the function inserts associated with the inductive definition of perms be
+expressed as an instance of foldr?
+
+> -- TODO
+
+Exercise 1.15
+---
+
+Give a definition of remove for which
+
+> -- perms3 [] = [[]]
+> -- perms3 xs = [x : ys | x <- xs, y <- perms3 (remove x xs)]
+
+computes the permutations of a list. Is the first clause necessary? What is the
+type of perms3, and can one generate the permutations of a list of functions
+with this definition?