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20150428.hs
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-- Exercícios slide 6
--Determine, sem usar o GHCi, os tipos das seguintes expressões:
-- foldr (:)
(:) :: a -> [a] -> [a]
foldr :: (s -> z -> z) -> z -> [s] -> z
in1: (s -> z -> z)
out1: z -> [s] -> z
in1 = (:)
(s -> z -> z) = a -> [a] -> [a]
s = a
z = [a]
foldr (:) :: [a] -> [a] -> [a]
--map.(.)
(.) :: (b -> c) -> (a -> b) -> a -> c
in1: (b -> c)
out1: (a -> b) -> a -> c
map :: (x -> y) -> [x] -> [y]
in2: (x -> y)
out2: [x] -> [y]
out1 = in2
(a -> b) -> (a -> c) = (x -> y)
x = (a -> b)
y = (a -> c)
in1 -> out2
(b -> c) -> [a - > b] -> [a -> c]
map.(.) :: (b -> c) -> [a - > b] -> [a -> c]
--foldr (+).(.).map
--igual à letra a) do trabalho
--map1.map2.foldr == map1.(map2.foldr)
* map2.foldr
foldr :: (s -> z -> z) -> z -> [s] -> z
in1: (s -> z -> z)
out1: z -> [s] -> z
map2 :: (x -> y) -> [x] -> [y]
in2: (x -> y)
out2: [x] -> [y]
out1 = in2
z -> [s] -> z = (x -> y)
x = z
y = [s] -> z
in1 -> out2
(s -> z -> z) -> [x] -> [y]
(s -> z -> z) -> [z] -> [[s] -> z]
map2.foldr :: (s -> z -> z) -> [z] -> [[s] -> z]
* map1.(map2.foldr)
map2.foldr :: (s -> z -> z) -> [z] -> [[s] -> z]
in1: (s -> z -> z)
out1: [z] -> [[s] -> z]
map1 :: (t -> u) -> [t] -> [u]
in2: (t -> u)
out2: [t] -> [u]
ou1 = in2
[z] -> [[s] -> z] = (t -> u)
t = [z]
u = [[s] -> z]
in1 -> out2
(s -> z -> z) -> [t] -> [u]
(s -> z -> z) -> [[z]] -> [[[s] -> z]]
map.(map.foldr) :: (s -> z -> z) -> [[z]] -> [[[s] -> z]]
--map.((.) (foldr (++) (foldr (++) [] [[1], [2]])))
--igual à letra c) do trabalho
--(foldr).(.)$(!!)
--igual à letra d) do trabalho
-- Trabalho 10
-- a) foldr (+).(.).map == (foldr (+)).((.).map)
* (.).map
map :: (a -> b) -> [a] -> [b]
in1: (a -> b)
out1: [a] -> [b]
(.) :: (e -> f) -> (d -> e) -> d -> f
in2: (e -> f)
out2: (d -> e) -> d -> f
out1 = in2
[a] -> [b] = (e -> f)
[a] = e
[b] = f
in1 -> out2
(a -> b) -> (d -> e) -> d -> f
(a -> b) -> (d -> [a]) -> d -> [b]
(.).map :: (a -> b) -> (d -> [a]) -> d -> [b]
* foldr (+)
foldr :: (s -> z -> z) -> z -> [s] -> z
in1: (s -> z -> z)
out1: z -> [s] -> z
(+) :: Num x => x -> x -> x
in1 = (+)
(s -> z -> z) = (x -> x -> x)
s = x
z = x
foldr (+) -> x -> [x] -> x
* (foldr (+)).((.).map)
(.).map :: (a -> b) -> (d -> [a]) -> d -> [b]
in1: (a -> b)
out1: (d -> [a]) -> d -> [b]
foldr (+) -> x -> [x] -> x
in2: x
out2: [x] -> x
out1 = in2
((d -> [a]) -> d -> [b]) = x
in1 -> out2
(a -> b) -> [x] -> x
(a -> b) -> [((d -> [a]) -> d -> [b])] -> ((d -> [a]) -> d -> [b])
-- b) (\x y z -> foldr z x y).map
map :: (a -> b) -> [a] -> [b]
in1: (a -> b)
out1: [a] -> [b]
foldr :: (m -> n -> n) -> n -> [m] -> n
(\x y z) :: n -> [m] -> (m -> n -> n) -> n
in2: n
out2: [m] -> (m -> n -> n) -> n
in2 = out1
n = ([a] -> [b])
in1 -> out2
(a -> b) -> [m] -> (m -> n -> n) -> n
(a -> b) -> [m] -> (m -> ([a] -> [b]) -> [a] -> [b]) -> [a] -> [b]
(\x y z -> foldr z x y).map :: (a -> b) -> [m] -> (m -> ([a] -> [b]) -> [a] -> [b]) -> [a] -> [b]
-- c) map.((.) (foldr (++) (foldr (++) [] [[1], [2]])))
foldr (++) [] [[1], [2]] :: Num a => [a]
foldr (++) (foldr (++) [] [[1], [2]]) :: Num a => [[a]] -> [a]
(.) :: (e -> f) -> (d -> e) -> d -> f
in: (e -> f)
out: (d -> e) -> d -> f
in = foldr (++) (foldr (++) [] [[1], [2]])
(e -> f) = [[a]] -> [a]
e = [[a]]
f = [a]
(.) (foldr (++) (foldr (++) [] [[1], [2]]))
in1: (d -> e) = (d -> [[a]])
out1: (d -> f) = (d -> [a])
map :: (x -> y) -> [x] -> [y]
in2: (x -> y)
out2: [x] -> [y]
* map.((.) (foldr (++) (foldr (++) [] [[1], [2]])))
out1 = in2
(d -> [a]) = (x -> y)
x = d
y = [a]
in1 -> out2
(d -> [[a]]) -> [d] -> [[a]]
map.((.) (foldr (++) (foldr (++) [] [[1], [2]]))) :: (Num a) => (d -> [[a]]) -> [d] -> [[a]]
-- d) (foldr).(.)$(!!)
* (foldr).(.)
(.) :: (e -> f) -> (d -> e) -> d -> f
in1: (e -> f)
out1: (d -> e) -> d -> f
foldr :: (s -> z -> z) -> z -> [s] -> z
in2: (s -> z -> z)
out2: z -> [s] -> z
out1 = in2
(d -> e) -> d -> f = (s -> z -> z)
s = (d -> e)
z = d
z = f
(d = f)
in1 -> out2
(e -> f) -> z -> [s] -> z
(e -> f) -> f -> [f -> e] -> f
(foldr).(.) :: (e -> f) -> f -> [f -> e] -> f
* (foldr).(.)$(!!)
(!!) :: [a] -> Int -> a
(foldr).(.) :: (e -> f) -> f -> [f -> e] -> f
in: (e -> f)
out: f -> [f -> e] -> f
in = (!!)
(e -> f) = [a] -> (Int -> a)
e = [a]
f = (Int -> a)
(foldr).(.)$(!!) :: (Int -> a) -> [(Int -> a) -> [a]] -> Int -> a