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<p>Programming in Haskell Solutions to ExercisesGraham Hutton University of Nottingham</p> <p>ContentsChapter 1 - Introduction Chapter 2 - First steps Chapter 3 - Types and classes Chapter 4 - Dening functions Chapter 5 - List comprehensions Chapter 6 - Recursive functions Chapter 7 - Higher-order functions Chapter 8 - Functional parsers Chapter 9 - Interactive programs Chapter 10 - Declaring types and classes Chapter 11 - The countdown problem Chapter 12 - Lazy evaluation Chapter 13 - Reasoning about programs 1 3 4 5 7 9 13 15 19 21 23 24 26</p> <p>Chapter 1 - IntroductionExercise 1= = = = = or = = = = = = double (double 2) { applying the outer double } (double 2) + (double 2) { applying the second double } (double 2) + (2 + 2) { applying the second + } (double 2) + 4 { applying double } (2 + 2) + 4 { applying the rst + } 4+4 { applying + } 8 double (double 2) { applying the inner double } double (2 + 2) { applying double } (2 + 2) + (2 + 2) { applying the rst + } 4 + (2 + 2) { applying the second + } 4+4 { applying + } 8</p> <p>There are a number of other answers too.</p> <p>Exercise 2sum [x ] { applying sum } x + sum [ ] = { applying sum } x +0 = { applying + } x =</p> <p>Exercise 3(1) product [ ] = product (x : xs) = 1 x product xs</p> <p>1</p> <p>(2) product [2, 3, 4] { applying product } 2 (product [3, 4]) { applying product } 2 (3 product [4]) { applying product } 2 (3 (4 product [ ])) { applying product } 2 (3 (4 1)) { applying } 24</p> <p>= = = = =</p> <p>Exercise 4Replace the second equation by qsort (x : xs) = qsort larger + [x ] + qsort smaller + + That is, just swap the occurrences of smaller and larger .</p> <p>Exercise 5Duplicate elements are removed from the sorted list. For example: = = = = = qsort [2, 2, 3, 1, 1] { applying qsort } qsort [1, 1] + [2] + qsort [3] + + { applying qsort } (qsort [ ] + [1] + qsort [ ]) + [2] + (qsort [ ] + [3] + qsort [ ]) + + + + + + { applying qsort } ([ ] + [1] + [ ]) + [2] + ([ ] + [3] + [ ]) + + + + + + { applying + } + [1] + [2] + [3] + + { applying + } + [1, 2, 3]</p> <p>2</p> <p>Chapter 2 - First stepsExercise 1(2 3) 4 (2 3) + (4 5) 2 + (3 (4 5))</p> <p>Exercise 2No solution required.</p> <p>Exercise 3n = a div length xs where a = 10 xs = [1, 2, 3, 4, 5]</p> <p>Exercise 4last xs or last xs = xs !! (length xs 1) = head (reverse xs)</p> <p>Exercise 5init xs or init xs = reverse (tail (reverse xs)) = take (length xs 1) xs</p> <p>3</p> <p>Chapter 3 - Types and classesExercise 1[Char ] (Char , Char , Char ) [(Bool , Char )] ([Bool ], [Char ]) [[a ] [a ]]</p> <p>Exercise 2[a ] a (a, b) (b, a) a b (a, b) Num a a a Eq a [a ] Bool (a a) a a</p> <p>Exercise 3No solution required.</p> <p>Exercise 4In general, checking if two functions are equal requires enumerating all possible argument values, and checking if the functions give the same result for each of these values. For functions with a very large (or innite) number of argument values, such as values of type Int or Integer, this is not feasible. However, for small numbers of argument values, such as values of type of type Bool , it is feasible.</p> <p>4</p> <p>Chapter 4 - Dening functionsExercise 1halve xs or halve xs = (take n xs, drop n xs) where n = length xs div 2 = splitAt (length xs div 2) xs</p> <p>Exercise 2(a) safetail xs (b) safetail xs | null xs | otherwise (c) safetail [ ] safetail xs or safetail [ ] = safetail ( : xs) = [] xs = = [] tail xs = = [] tail xs = if null xs then [ ] else tail xs</p> <p>Exercise 3(1) False False False True True False True True (2) False False (3) False b True (4) b c | b == c | otherwise = = b True = = b True = False = True = = = = False True True True</p> <p>5</p> <p>Exercise 4ab = if a then if b then True else False else False</p> <p>Exercise 5ab = if a then b else False</p> <p>Exercise 6mult = x (y (z x y z ))</p> <p>6</p> <p>Chapter 5 - List comprehensionsExercise 1sum [x 2 | x [1 . . 100]]</p> <p>Exercise 2replicate n x = [x | [1 . . n ]]</p> <p>Exercise 3pyths n = [(x , y, z ) | x y z x [1 . . n ], [1 . . n ], [1 . . n ], 2 + y 2 == z 2]</p> <p>Exercise 4perfects n = [x | x [1 . . n ], sum (init (factors x )) == x ]</p> <p>Exercise 5concat [[(x , y) | y [4, 5, 6]] | x [1, 2, 3]]</p> <p>Exercise 6positions x xs = nd x (zip xs [0 . . n ]) where n = length xs 1</p> <p>Exercise 7scalarproduct xs ys = sum [x y | (x , y) zip xs ys ]</p> <p>Exercise 8shift shift n c | isLower c | isUpper c | otherwise freqs freqs xs :: = = = :: = Int Char Char int2low ((low2int c + n) mod 26) int2upp ((upp2int c + n) mod 26) c String [Float ] [percent (count x xs ) n | x [a . . z]] where xs = map toLower xs n = letters xs Char Int ord c ord a 7</p> <p>low2int low2int c</p> <p>:: =</p> <p>int2low int2low n upp2int upp2int c int2upp int2upp n letters letters xs</p> <p>:: = :: = :: = :: =</p> <p>Int Char chr (ord a + n) Char Int ord c ord A Int Char chr (ord A + n) String Int length [x | x xs, isAlpha x ]</p> <p>8</p> <p>Chapter 6 - Recursive functionsExercise 1(1) m 0 = 1 m (n + 1) = m m n (2) = = = = = 23 { applying } 2 (2 2) { applying } 2 (2 (2 1)) { applying } 2 (2 (2 (2 0))) { applying } 2 (2 (2 1)) { applying } 8</p> <p>Exercise 2(1) = = = = = length [1, 2, 3] { applying length } 1 + length [2, 3] { applying length } 1 + (1 + length [3]) { applying length } 1 + (1 + (1 + length [ ])) { applying length } 1 + (1 + (1 + 0)) { applying + } 3</p> <p>(2) = drop 3 [1, 2, 3, 4, 5] { applying drop drop 2 [2, 3, 4, 5] = { applying drop drop 1 [3, 4, 5] = { applying drop drop 0 [4, 5] = { applying drop [4, 5] } } } }</p> <p>9</p> <p>(3) init [1, 2, 3] { applying init } 1 : init [2, 3] = { applying init } 1 : 2 : init [3] = { applying init } 1 :2 : [] = { list notation } [1, 2]</p> <p>=</p> <p>Exercise 3and [ ] and (b : bs) concat [ ] concat (xs : xss) replicate 0 replicate (n + 1) x (x : ) !! 0 ( : xs) !! (n + 1) elem x [ ] elem x (y : ys) | x == y | otherwise = = = = = = = = = = = True b and bs [] xs + concat xss + [] x : replicate n x x xs !! n False True elem x ys</p> <p>Exercise 4merge [ ] ys merge xs [ ] merge (x : xs) (y : ys) = ys = xs = if x y then x : merge xs (y : ys) else y : merge (x : xs) ys</p> <p>Exercise 5halve xs = splitAt (length xs div 2) xs</p> <p>msort [ ] = [] msort [x ] = [x ] msort xs = merge (msort ys) (msort zs) where (ys, zs) = halve xs</p> <p>10</p> <p>Exercise 6.1Step 1: dene the type sum :: [Int ] Int = = Step 2: enumerate the cases sum [ ] sum (x : xs)</p> <p>Step 3: dene the simple cases sum [ ] sum (x : xs) = 0 =</p> <p>Step 4: dene the other cases sum [ ] sum (x : xs) = 0 = x + sum xs</p> <p>Step 5: generalise and simplify sum sum :: = Num a [a ] a foldr (+) 0</p> <p>Exercise 6.2Step 1: dene the type take :: Int [a ] [a ] 0 [] 0 (x : xs) (n + 1) [ ] (n + 1) (x : xs) = = = = Step 2: enumerate the cases take take take take</p> <p>Step 3: dene the simple cases take take take take 0 [] 0 (x : xs) (n + 1) [ ] (n + 1) (x : xs) = = = = [] [] []</p> <p>Step 4: dene the other cases take take take take 0 [] 0 (x : xs) (n + 1) [ ] (n + 1) (x : xs) = = = = [] [] [] x : take n xs Int [a ] [a ] [] [] x : take n xs</p> <p>Step 5: generalise and simplify take :: = take 0 take (n + 1) [ ] = take (n + 1) (x : xs) =</p> <p>11</p> <p>Exercise 6.3Step 1: dene the type last :: [a ] [a ] Step 2: enumerate the cases last (x : xs) = Step 3: dene the simple cases last (x : xs) | null xs | otherwise Step 4: dene the other cases last (x : xs) | null xs | otherwise Step 5: generalise and simplify last :: last [x ] = last ( : xs) = [a ] [a ] x last xs = x = last xs = x =</p> <p>12</p> <p>Chapter 7 - Higher-order functionsExercise 1map f (lter p xs)</p> <p>Exercise 2all p any p = = and map p or map p</p> <p>takeWhile [ ] = [ ] takeWhile p (x : xs) |px = x : takeWhile p xs | otherwise = [ ] dropWhile [ ] dropWhile p (x |px | otherwise = [] : xs) = dropWhile p xs = x : xs</p> <p>Exercise 3map f lter p = = foldr (x xs f x : xs) [ ] foldr (x xs if p x then x : xs else xs) [ ]</p> <p>Exercise 4dec2nat = foldl (x y 10 x + y) 0</p> <p>Exercise 5The functions being composed do not all have the same types. For example: sum map (2) lter even :: [Int ] Int :: [Int ] [Int ] :: [Int ] [Int ]</p> <p>Exercise 6curry curry f uncurry uncurry f :: = :: = ((a, b) c) (a b c) x y f (x , y) (a b c) ((a, b) c) (x , y) f x y</p> <p>13</p> <p>Exercise 7chop8 map f iterate f = = = unfold null (take 8) (drop 8) unfold null (f head ) tail unfold (const False) id f</p> <p>Exercise 8encode encode decode decode addparity addparity bs :: = :: = :: = String [Bit ] concat map (addparity make8 int2bin ord ) [Bit ] String map (chr bin2int checkparity ) chop9 [Bit ] [Bit ] (parity bs) : bs [Bit ] Bit 1 0 [Bit ] [[Bit ]] [] take 9 bits : chop9 (drop 9 bits) [Bit ] [Bit ] bs error "parity mismatch"</p> <p>parity :: parity bs | odd (sum bs) = | otherwise = chop9 chop9 [ ] chop9 bits checkparity checkparity (b : bs) | b == parity bs | otherwise :: = = :: = =</p> <p>Exercise 9No solution required.</p> <p>14</p> <p>Chapter 8 - Functional parsersExercise 1int = do char - n nat return (n) + +nat +</p> <p>Exercise 2comment = do string "--" many (sat (= \n)) return ()</p> <p>Exercise 3(1) expr KK KK uuu zu % expr exprI + II u I$ uu zu expr expr + term term term factor factor nat 2 (2) expr I ss III yss $ exprI expr + uu III zuu $ expr expr + term factor nat 2 term factor nat 3 term factor nat 4 factor nat 3 nat 4</p> <p>15</p> <p>Exercise 4(1) expr I III uuu $ zu expr + term factor term factor nat nat 2 3 (2) expr termI t II I$ yttt factor termG GGG ww G# {ww factor nat term factor nat 2 nat 3 4 (3) expr K ss KKK yss % expr + term factorI term III uuu II $ zuuu expr ) factor ( JJJ tt JJ zttt $ expr + term nat factor nat 2 term factor nat 3 4</p> <p>16</p> <p>Exercise 5Without left-factorising the grammar, the resulting parser would backtrack excessively and have exponential time complexity in the size of the expression. For example, a number would be parsed four times before being recognised as an expression.</p> <p>Exercise 6expr = do t term do symbol "+" e expr return (t + e) + + do symbol "-" + e expr return (t e) + + return t + = do f factor do symbol "*" t term return (f t ) + + do symbol "/" + t term return (f div t ) + + return f +</p> <p>term</p> <p>Exercise 7(1) factor atom (2) factor factor :: Parser Int = do a atom do symbol "^" f factor return (a f ) + + return a + :: = Parser Int do symbol "(" e expr symbol ")" return e + + natural + ::= ::= atom ( factor | epsilon) (expr ) | nat</p> <p>atom atom</p> <p>17</p> <p>Exercise 8(a) expr nat (b) expr = do e expr symbol "-" n natural return (e n) + + natural + ::= ::= expr nat | nat 0 | 1 | 2 | </p> <p>(c) The parser loops forever without producing a result, because the rst operation it performs is to call itself recursively. (d) expr = do n natural ns many (do symbol "-" natural) return (foldl () n ns)</p> <p>18</p> <p>Chapter 9 - Interactive programsExercise 1readLine get xs = get "" = do x getChar case x of \n return xs \DEL if null xs then get xs else do putStr "\ESC[1D \ESC[1D" get (init xs) get (xs + [x ]) +</p> <p>Exercise 2No solution available.</p> <p>Exercise 3No solution available.</p> <p>Exercise 4No solution available.</p> <p>Exercise 5No solution available.</p> <p>Exercise 6type Board initial initial nished nished b valid valid b row num move move b row num newline newline = :: = :: = :: = :: = :: = [Int ] Board [5, 4, 3, 2, 1] Board Bool all (== 0) b Board Int Int Bool b !! (row 1) num Board Int Int Board [if r == row then n num else n | (r , n) zip [1 . . 5] b ] IO () putChar \n</p> <p>19</p> <p>putBoard putBoard [a, b, c, d , e ]</p> <p>:: =</p> <p>Board IO () do putRow 1 a putRow 2 b putRow 3 c putRow 4 d putRow 5 e Int Int IO () do putStr (show row ) putStr ": " putStrLn (stars num) Int String concat (replicate n "* ") String IO Int do putStr prom x getChar newline if isDigit x then return (ord x ord 0) else do putStrLn "ERROR: Invalid digit" getDigit prom IO () play initial 1 Board Int IO () do newline putBoard board if nished board then do newline putStr "Player " putStr (show (next player )) putStrLn " wins!!" else do newline putStr "Player " putStrLn (show player ) r getDigit "Enter a row number: " n getDigit "Stars to remove : " if valid board r n then play (move board r n) (next player ) else do newline putStrLn "ERROR: Invalid move" play board player Int Int 2 1</p> <p>putRow putRow row num</p> <p>:: =</p> <p>stars stars n getDigit getDigit prom</p> <p>:: = :: =</p> <p>nim nim play play board player</p> <p>:: = :: =</p> <p>next next 1 next 2</p> <p>:: = =</p> <p>20</p> <p>Chapter 10 - Declaring types and classesExercise 1mult m Zero mult m (Succ n) = Zero = add m (mult m n)</p> <p>Exercise 2occurs m (Leaf n) = occurs m (Node l n r ) = m == n case compare m n of LT occurs m l EQ True GT occurs m r</p> <p>This version is more ecient because it only requires one comparison for each node, whereas the previous version may require two comparisons.</p> <p>Exercise 3leaves (Leaf ) leaves (Node l r ) = = 1 leaves l + leaves r True abs (leaves l leaves r ) 1 balanced l balanced r</p> <p>balanced (Leaf ) = balanced (Node l r ) =</p> <p>Exercise 4halve xs = splitAt (length xs div 2) xs Leaf x Node (balance ys) (balance zs) where (ys, zs) = halve xs</p> <p>balance [x ] = balance xs =</p> <p>Exercise 5data Prop eval s (Or p q) eval s (Equiv p q) vars (Or p q) vars (Equiv p q) = | Or Prop Prop | Equiv Prop Prop = eval s p eval s q = eval s p == eval s q = vars p + vars q + = vars p + vars q +</p> <p>Exercise 6No solution available.</p> <p>21</p> <p>Exercise 7data Expr type Cont data Op eval eval (Val n) ops eval (Add x y) ops eval (Mult x y) ops exec exec exec exec exec exec [] n (EVALA y : ops) n (ADD n : ops) m (EVALM y : ops) n (MUL n : ops) m = Val Int | Add Expr Expr | Mult Expr Expr = [Op ] = EVALA Expr | ADD Int | EVALM Expr | MUL Int :: = = = :: = = = = = Expr Cont Int exec ops n eval x (EVALA y : ops) eval x (EVALM y : ops) Cont Int Int n eval y (ADD n : ops) exec ops (n + m) eval y (MUL n : ops) exec ops (n m)</p> <p>value value e</p> <p>:: Expr Int = eval e [ ]</p> <p>Exercise 8instance Monad Maybe where return :: a Maybe a return x = Just x (&gt;=) &gt; Nothing &gt;= &gt; (Just x ) &gt;= f &gt; :: Maybe a (a Maybe b) Maybe b = Nothing = f x</p> <p>instance Monad [ ] where return :: a [a ] return x = [x ] (&gt;=) &gt; xs &gt;= f &gt; :: [a ] (a [b ]) [b ] = concat (map f xs)</p> <p>22</p> <p>Chapter 11 - The countdown problemExercise 1choices xs = [zs | ys subs xs, zs perms ys ]</p> <p>Exercise 2removeone x [ ] = [] removeone x (y : ys) | x == y = ys | otherwise = y : removeone x ys isChoice [ ] isChoice (x : xs) [ ] isChoice (x : xs) ys = True = False = elem x ys isChoice xs (removeone x ys)</p> <p>Exercise 3It would lead to non-termination, because recursive calls to exprs would no longer be guaranteed to reduce the length of the list.</p> <p>Exercise 4&gt; length [e | ns choices [1, 3, 7, 10, 25, 50], e exprs ns ] 33665406 &gt; length [e | ns choices [1, 3, 7, 10, 25, 50], e exprs ns, eval e = [ ]] 4672540</p> <p>Exercise 5Modifying the denition of valid by valid Sub x y valid Div x y gives &gt; length [e | ns choices [1, 3, 7, 10, 25, 50], e exprs ns , eval e = [ ]] 10839369 = True = y = 0 x mod y == 0</p> <p>Exercise 6No solution available.</p> <p>23</p>...