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Functional Programming Lecture 4 - More Lists

Functional Programming Lecture 4 - More Lists. Muffy Calder. Wildcards and Pattern Matching. When we exploit the structure of an argument on the lhs of an equation, we call this pattern matching. E.g. tail x:xs = xs If we do not need the value of an argument on the

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Functional Programming Lecture 4 - More Lists

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  1. Functional ProgrammingLecture 4 - More Lists Muffy Calder

  2. Wildcards and Pattern Matching When we exploit the structure of an argument on the lhs of an equation, we call this pattern matching. E.g. tail x:xs = xs If we do not need the value of an argument on the rhs of an equation, then we can use _, the wildcard. Example: head x:_ = x tail (_:xs) = xs

  3. List Comprehensions • Based on the standard “set comprehension”notation in mathematics • Provides a simple way to define many lists • Provides an exremely powerful methodto define a huge variety of lists • many problems can be solvedjust by writing list comprehensions

  4. Comprehension Expressions and Generators Basic form of list comprehension: [ expression | generator] Form of a generator: variable <- list Result: list of expression values, where thegenerator variable takes on each value in the list [x*10 | x <- [1..5]] => [10,20,30,40,50]

  5. Comprehension Filters • Used to omit some of the values producedby a generator • A filter is an expression of type Boolusing a generator variable • If filter is True, the corresponding generatorvalue is kept • If filter is False, the value is thrown away • E.g. • [x | x <- [1..10], x `mod` 3 == 0] => [3,6,9]

  6. Multiple Generators • For the first value of the first generator,use each value of the second generator • For the second value of the first generator,use each value of the second generator, • ….. Continue until done ….. • E.g. • [x*y | x <- [1..4], y <- [1000..1002]] => [1000, 1001, 1002, 2000, 2002, 2004, 3000, 3003, 3006, 4000, 4004, 4008]

  7. More List Comprehension Examples spaces :: Int -> String spaces n = [' '|i <- [1..n]] > spaces 5 " " isPrime :: Int -> Bool isPrime (n+2) = (factors (n+2) == []) where factors m = [x|x<-[2..(m-1)], m `mod` x == 0] > isPrime 5 True > isPrime 10 False

  8. More List Comprehension Examples You can use values produced by earlier qualifiers: [n*n | n<-[1..10], even n] =>[4,16,36,64,100] [(n,m)| m<-[1..3], n <- [1..m]] • [(1,1),(1,2),(2,2),(1,3),(2,3),(3,3)] But the generated values do not have to occur in the expression: [27| n<-[1..3]] => [27,27,27] [x|x<-[1..3],y<-[1..2]] => [1,1,2,2,3,3] Generators occurring later vary more quickly, i.e. loops are nested: [(m,n)| m <-[1..3],n<-[4,5]] => [(1,4),(1,5),(2,4),(2,5),(3,4),(3,5)]

  9. Examples Define a function which returns the list of even numbers in a list: evens :: [Int] -> [Int] evens [] = [] evens x:xs | iseven x = x: evens xs | otherwise = even xs using primitive recursion or using list comprehensions evens xs = [x| x <- xs, iseven x]

  10. Examples Define a function which returns the first digit in a string: firstdig :: String -> Char isdigit :: Char -> Bool (in prelude) isdigit c= c>=‘0’ && c<=‘9’ firstdig [] = ? firstdig x:xs = if (isdigit x) then x else firstdig xs firstdig xs = hd digits xs where digits :: String -> String digits [] = [] digits x:xs = if (isdigit x) then x:(digits xs) else digits xs or using list comprehensions digits xs = [x| x <- xs, isdigit x]

  11. World’s shortest sorting algorithm? quicksort :: Ord a => [a] -> [a] quicksort [] = [] quicksort (x:xs) = quicksort[y|y<-xs,y<x] ++ [x] ++ quicksort [y|y<-xs,y>=x] quicksort [1,5,2,7] => [1,2,5,7]

  12. A puzzle A 14 digit number is to be constructed in the following way. (Guardian,7 December 2002?) • It starts with the digit 4 and ends with the digit 3. • It contains two 7’s, separated by 7 digits, two 6’s, separated by 6 digits, etc. • What is the number??

  13. -- program to solve Guardian puzzle December 2002 answer = mksolns (possibles) l :: [Int] l = [4,0,0,0,0,4,0,0,0,3,0,0,0,3] possibles :: [[(Int,Int)]] -- combinations of positions for 7,6,5,2 and 1 possibles = [[(7,x7),(6,x6),(5,x5),(2,x2),(1,x1)]| x7 <- [1..4], x6<- [1..5], x5<- [1..6], x2<- [1..9], x1<- [1..10]] mksolns :: [[(Int,Int)]] -> [Int] -- find unique solution amongst all combinations mksolns (x:xs) | p = s | otherwise = mksolns xs where (s,p) = soln x l soln :: [(Int,Int)] -> [Int] -> ([Int],Bool) -- determine whether a list of positions is a solution soln [] ys = (ys,True) soln (x:xs) ys | (noplace x ys) = (ys,False) | (toobig x) = (ys, False) | otherwise = soln xs (place x ys)

  14. place :: (Int,Int) -> [Int] -> [Int] -- place number n at position i in xs place (n,i) xs = (a ++ [n] ++ b ++ [n] ++ c) where a = take (i) xs b = drop (i+1) (take (i+n+1) xs) c = drop (i+n+2) xs noplace :: (Int,Int) -> [Int] -> Bool -- number n cannot be placed at position i in xs noplace (n,i) xs = ((xs!!i) /= 0) || ((xs!!(i+n+1)) /= 0) toobig :: (Int,Int) -> Bool toobig (n,i) = (i+n+1) >= 14 ******************************************************* Main> answer [4,6,1,7,1,4,5,2,6,3,2,7,5,3] (125 reductions, 217 cells) Main>

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