Resolução do teste de Programação Funcioal de 31/3/2014 sg ntCrislânio Macêdo, 2014--____________________________________________________________________________Um pouco mais sobre dobras(folds 'r''l''r1''l1')Primeiro vamos dar uma olhada na função foldl, também chamada de "left fold" (dobra esquerda). Isto dobra a lista a partir do lado esquerdo

A função foldr dobra a lista a partir do lado direito.

As funções foldl1 e foldr1 funcionam da mesma forma que foldl e foldr, só que você não precisa informar explicitamente qual o valor inicial. Ele assume que o primeiro (ou último) elemento da lista é o valor inicial da dobra e então inicia a dobra com ele.

foldr (+) 0 [1..10] valor inicial 0, 0+10+...1foldl (+) 0 [1..10] valor inicial 0, 0+0+...10foldr1(+) [1..10] valor inical 10,10+9...1foldl1(+) [1..10] valor inical 1,1+2...10

Prelude> putStrLn $ foldr (\x y -> concat ["(",x,"+",y,")"]) "0" (map show [1..13])(1+(2+(3+(4+(5+(6+(7+(8+(9+(10+(11+(12+(13+0))))))))))))) Prelude> putStrLn $ foldl (\x y -> concat ["(",x,"+",y,")"]) "0" (map show [1..13])(((((((((((((0+1)+2)+3)+4)+5)+6)+7)+8)+9)+10)+11)+12)+13)

Prelude> putStrLn $ foldr1 (\x y -> concat ["(",x,"+",y,")"]) (map show [1..13])(1+(2+(3+(4+(5+(6+(7+(8+(9+(10+(11+(12+13))))))))))))

Prelude> putStrLn $ foldl1 (\x y -> concat ["(",x,"+",y,")"]) (map show [1..13])((((((((((((1+2)+3)+4)+5)+6)+7)+8)+9)+10)+11)+12)+13)

Prelude> foldr (\x y->x:'.':y) "" []:"ABC":[] ["","ABC"]Prelude> foldr (\x y->x:'.':y) "" "ABC":[] ["A.B.C."]Prelude> foldr (\x y->x:'.':y) "" "ABC" "A.B.C."Prelude> foldr (\x y->x:'.':y) "" ['a']:"ABC":[] ["a.","ABC"]

--____________________________________________________________________________--1° NIVEL 3

Prelude> length (filter (/='a') "banana") 3Prelude> length (filter (/='n') "banana")4Prelude> length ([1,2]:[3,4]:[])2

Prelude> length [1,3..7]4Prelude> length (drop 6 "abcdefg123")4Prelude> length ([1,2]:[3]:[])2Prelude> length [(x,y)| x<-"abc", y<-[0,1,2]]9Prelude> head [n | n<-[1..], n*n>15]44Prelude> head [n | n<-[1..], n*n>20]5Prelude> head([]++[1,2]++[3])1Prelude> [x | x<- "ab", n<-[1,2,3]]"aaabbb"Prelude> map (`div` 2) [1,2,3][0,1,1]Prelude> (reverse [3,2])++ [4,5][2,3,4,5]Prelude> reverse ([4,5]++[1,2,3])[3,2,1,5,4]--Prelude> :t ([1,2,3], "banana")([1,2,3], "banana") :: Num t => ([t], [Char])Prelude> :t [(1,"banana"), (2,"maçã")][(1,"banana"), (2,"maçã")] :: Num t => [(t, [Char])]

--2° NIVEL 5Prelude> foldr (-) (-2) [1,2,3,4]-4Prelude> foldr (-) (2) [1,2,3,4]0Prelude> foldl (-) (2) [1,2,3,4]-8Prelude> foldl (-) (-2) [1,2,3,4]-12--3° NIVEL 6Prelude> foldr (+) 2 (map (+1) [1,2,3]) 11Prelude> foldl (+) 2 (map (+1) [1,2,3]) 11Prelude> foldr (-) 2 (map ((-)1) [1,2,2]) -2Prelude> foldl (-) 2 (map ((-)1) [1,2,2]) 4Prelude> foldr(\x y->x:'.':y) "" "PAULO""P.A.U.L.O."Prelude> foldl(\x y->x:'.':y) "" "ITALOS"

<interactive>:92:25:

Couldn't match expected type `[Char]' with actual type `Char' Expected type: [[Char]] Actual type: [Char] In the third argument of `foldl', namely `"ITALOS"' In the expression: foldl (\ x y -> x : '.' : y) "" "ITALOS"

--4° NIVEL 7Prelude> foldl (+) 0 (filter (>1) [sum(map ((-)1) [1,2,2])])0Prelude> sum(map (-1) [1,2,2])

<interactive>:96:10: No instance for (Num (a1 -> a0)) arising from a use of syntactic negation Possible fix: add an instance declaration for (Num (a1 -> a0)) In the first argument of `map', namely `(- 1)' In the first argument of `sum', namely `(map (- 1) [1, 2, 2])' In the expression: sum (map (- 1) [1, 2, 2])

Prelude> [map (>2) [1,2,3]] ++ ([]) ++ [[True]][[False,False,True],[True]]Prelude> [map (>2) [1,6,6]] ++ ([]) ++ [[]][[False,True,True],[]]Prelude> [map (>2) [-1, 2,4]] ++([])[[False,False,True]]--5° NIVEL MASTERPrelude> foldr1 (-) [4,3,2]3Prelude> foldl1 (-) [-2,3,4]-9