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solutions - Programming in Haskell Solutions to Exercises...

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Programming in Haskell Solutions to Exercises Graham Hutton University of Nottingham Contents Chapter 1 - Introduction 1 Chapter 2 - First steps 3 Chapter 3 - Types and classes 4 Chapter 4 - Defining functions 5 Chapter 5 - List comprehensions 7 Chapter 6 - Recursive functions 9 Chapter 7 - Higher-order functions 13 Chapter 8 - Functional parsers 15 Chapter 9 - Interactive programs 19 Chapter 10 - Declaring types and classes 21 Chapter 11 - The countdown problem 23 Chapter 12 - Lazy evaluation 24 Chapter 13 - Reasoning about programs 26
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Chapter 1 - Introduction Exercise 1 double ( double 2) = { applying the inner double } double (2 + 2) = { applying double } (2 + 2) + (2 + 2) = { applying the first + } 4 + (2 + 2) = { applying the second + } 4 + 4 = { applying + } 8 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 first + } 4 + 4 = { applying + } 8 There are a number of other answers too. Exercise 2 sum [ x ] = { applying sum } x + sum [ ] = { applying sum } x + 0 = { applying + } x Exercise 3 (1) product [ ] = 1 product ( x : xs ) = x product xs 1
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(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 Exercise 4 Replace the second equation by qsort ( x : xs ) = qsort larger ++ [ x ] ++ qsort smaller That is, just swap the occurrences of smaller and larger . Exercise 5 Duplicate 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] 2
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Chapter 2 - First steps Exercise 1 (2 3) 4 (2 3) + (4 5) 2 + (3 (4 5)) Exercise 2 No solution required. Exercise 3 n = a div length xs where a = 10 xs = [1 , 2 , 3 , 4 , 5] Exercise 4 last xs = head ( reverse xs ) or last xs = xs !! ( length xs 1) Exercise 5 init xs = take ( length xs 1) xs or init xs = reverse ( tail ( reverse xs )) 3
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Chapter 3 - Types and classes Exercise 1 [ Char ] ( Char , Char , Char ) [( Bool , Char )] ([ Bool ] , [ Char ]) [[ a ] [ a ]] Exercise 2 [ a ] a ( a , b ) ( b , a ) a b ( a , b ) Num a a a Eq a [ a ] Bool ( a a ) a a Exercise 3 No solution required. Exercise 4 In general, checking if two functions are equal requires enumerating all possible ar- gument values, and checking if the functions give the same result for each of these values. For functions with a very large (or infinite) 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. 4
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Chapter 4 - Defining functions Exercise 1 halve xs = splitAt ( length xs div ‘ 2) xs or halve xs = ( take n xs , drop n xs ) where n = length xs div ‘ 2 Exercise 2 (a) safetail xs = if null xs then [ ] else tail xs (b) safetail xs | null xs = [ ] | otherwise = tail xs (c) safetail [ ] = [ ] safetail xs = tail xs or safetail [ ] = [ ] safetail ( : xs ) = xs Exercise 3 (1) False False = False False True = True True False = True True True = True (2) False False = False = True (3) False b = b True = True (4) b c | b == c = b |
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