1. Functional Programming Lecture 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 method to define a huge variety of lists many problems can be solved just 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 the generator variable takes on each value in the list [x*10 | x [10,20,30,40,50]

5. Comprehension Filters Used to omit some of the values produced by a generator A filter is an expression of type Bool using a generator variable If filter is True, the corresponding generator value is kept If filter is False, the value is thrown away E.g. [x | x [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 [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 =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>