21 --Primitive recursion constructor
22 pr :: ([Int] -> Int) -> ([Int] -> Int) -> ([Int] -> Int)
23 pr f g = \xs 0 -> f xs
24 pr f g = \xs (y+1) -> g xs y ((pr f g) xs y)
I want the function this function creates to act differently on different inputs, so that it can create a recursive function. As expected, the above code doesn't work. How do I do something like pattern matching, but for the function it creates?
pr f g = \xs y' -> case y' of 0 -> f xs
(y+1) -> g xs y ((pr f g) xs y)
or simply
pr f g xs 0 = f xs
pr f g xs (y+1) = g xs y ((pr f g) xs y)
(Remember that f a b = ... is basically a shortcut for f a = \b -> ... which is a shortcut for f = \a -> \b -> ....)
Note that n+1 patterns are deprecated and that the type you specified for pr does not match your (and mine) definition.
Specifically according to your type the function takes an [Int] -> Int (f), then a function that takes another [Int] -> Int (g), then a function that takes an [Int] (xs) and then returning an Int. However you call g with three arguments and the last function that you return takes two arguments, so presumably you want something like ([Int] -> Int) -> ([Int] -> Int -> Int -> Int) -> [Int] -> Int -> Int.
Another solution is to use the lenguaje extension LambdaCase:
{-# LANGUAGE LambdaCase #-}
pr f g = \xs -> \case
0 -> f xs
(y+1) -> g xs y ((pr f g) xs y)
In these case you have two use two \ because in a \case you can have only one argument.
Related
I want a higher-order function, g, that will apply another function, f, to a list of integers such that
g = [f x1, f(f x2), f(f(f x3)), … , f^n(xn)]
I know I can map a function like
g :: (Int -> Int) -> [Int] -> [Int]
g f xs = map f xs
and I could also apply a function n-times like
g f xs = [iterate f x !! n | x <- xs]
where n the number of times to apply the function. I know I need to use recursion, so I don't think either of these options will be useful.
Expected output:
g (+1) [1,2,3,4,5] = [2,4,6,8,10]
You can work with explicit recursion where you pass each time the function to apply and the tail of the list, so:
g :: (Int -> Int) -> [Int] -> [Int]
g f = go f
where go _ [] = []
go fi (x:xs) = … : go (f . fi) xs
I here leave implementing the … part as an exercise.
Another option is to work with two lists, a list of functions and a list of values. In that case the list of functions is iterate (f .) f: an infinite list of functions that can be applied. Then we can implement g as:
g :: (Int -> Int) -> [Int] -> [Int]
g f = zipWith ($) (iterate (f .) f)
Sounds like another use for foldr:
applyAsDeep :: (a -> a) -> [a] -> [a]
applyAsDeep f = foldr (\x xs -> f x : map f xs) []
λ> applyAsDeep (+10) [1,2,3,4,5]
[11,22,33,44,55]
If you want to go a bit overkill ...
import GHC.Exts (build)
g :: (a -> a) -> [a] -> [a]
g f xs0 =
build $ \c n ->
let go x r fi = fi x `c` r (f . fi)
in foldr go (const n) xs0 f
So I'm learning about Haskell at the moment, and I came across this question:
The type of a function f is supposed to be a->[a]->a. The
following definitions of f are incorrect because their types are all
different from a->[a]->a:
i. f x xs = xs
ii. f x xs = x+1
iii. f x xs = x ++ xs
iv. f x (y:ys) = y
My answers as I see it are:
i) f :: a -> a -> a
This is because x or xs can be of any value and is not a list as it does not contain the ':' operator.
ii) f :: Int -> a -> Int
This is because the + operator is used on x, meaning x is of type Int.
iii) f :: Eq a => a -> a -> a
The ++ operators are used, therefore in order to concatenate they must be of the same type..?
iv) f :: a -> [a] -> a
f returns an element from the list.
The last one is definitely wrong, because it can't be of type a -> [a] -> a. Are there any others I did wrong, and why? I'm hoping I can fully understand types and how to find out the types of functions.
i) f :: a -> a -> a
f x xs = xs
This is because x or xs can be of any value and is not a list as it does not contain the ':' operator.
True, but it also does not have to be of the same type!
So, it's actually f :: a -> b -> b.
ii) f :: Int -> a -> Int
f x xs = x+1
This is because the + operator is used on x, meaning x is of type Int.
Correct. (Actually, in Haskell we get Num b => b -> a -> b which generalized the Int to any numeric type, but it's not that important.)
iii) f :: Eq a => a -> a -> a
f x xs = x ++ xs
The ++ operators are used, therefore in order to concatenate they must be of the same type..?
True, but they must be lists. Also, Eq is only needed if you use == or something which compares values.
Here, f :: [a] -> [a] -> [a].
iv) f :: a -> [a] -> a
f x (y:ys) = y
f returns an element from the list.
The type of x does not have to be the same. We get f :: b -> [a] -> a.
i. f x xs = xs
(...)
i) f :: a -> a -> a
Although this can be a type signature, you make it too restrictive. The function takes two parameters x and xs. Initially we can reason that x and xs can have different types, so we say that x :: a, and xs :: b. Since the function returns xs, the return type is b as well, so the type is:
f :: a -> b -> b
f x xs = xs
ii. f x xs = x+1
(...)
ii) f :: Int -> a -> Int
Again you make the function too restrictive. Let us again assume that x :: a and xs :: b have different types. We see that we return x + 1 (or in more canonical form (+) x 1. Since (+) has signature (+) :: Num c => c -> c -> c (we here use c since a is already used), and 1 has signature 1 :: Num d => d, we thus see that we call (+) with x and 1, as a result we know that a ~ c (a and c are the same type), and c ~ d, so as a result we obtain the signature:
f :: Num c => c -> b -> c
f x xs = x+1
iii. f x xs = x ++ xs
(...)
iii) f :: Eq a => a -> a -> a
This is wrong: we here see that f has two parameters, x :: a and xs :: b. We see that we return (++) x xs. Since (++) has signature (++) :: [c] -> [c] -> [c], we thus know that a ~ [c] and b ~ [c], so the type is:
f :: [c] -> [c] -> [c]
f x xs = x ++ xs
iv. f x (y:ys) = y
(...)
iv) f :: a -> [a] -> a
This is again too restrictive. Here we see again two parameters: x and (y:ys). We first generate a type a for x :: a, and (y:ys) :: b, since the pattern of the second parameter is (y:ys), this is a list constructor with as parameters (:) :: c -> [c] -> [c]. As a result we can derive that y :: c, and ys :: [c], furthermore the pattern (y:ys) has type [c]. Since the function returns y, we know that the return type is c, so:
f :: a -> [c] -> c
f x (y:ys) = y
Note: you can let Haskell derive the type of the function itself. In GHCi you can use the :t command to query the type of an expression. For example:
Prelude> f x (y:ys) = y
Prelude> :t f
f :: t1 -> [t] -> t
I am writing a function my_map which takes a unary function and a list and returns the list resulting from mapping the function over all elements of the input list.
Main> my_map (^3) [1..5]
[1,8,27,64,125]
I tried it like this:
my_map :: (a -> b) -> [a] -> [b]
my_map f [] = []
my_map f (x:xs) = foldr (\x xs -> (f x):xs) [] xs
But after running above, I get only [8,27,64,125]. the first number 1 is not displaying in output.
Can anybody help me?
You are using the (x:xs) pattern in your arguments, but when you apply the fold, you only apply it to the xs part, which means your first element i.e. the one that x represents never gets processed. You need to change it to this:
my_map :: (a -> b) -> [a] -> [b]
my_map f xs = foldr (\y ys -> (f y):ys) [] xs
Since you are using foldr, you do not need to explicitly handle the empty list case. Moreoever, you do not need to specify the list in (x:xs) format.
Finally, my own preference is to avoid using the same name for function inputs and any helper functions or expressions in the function definition.That is why, I have used xs for the input list and y and ys for the parameters passed to the lambda.
"shree.pat18" is perfectly right, and also the comments are valuable. I learned a lot from that. Just make it better visible, and to explain the alternatives...
Answer
-- The problem is here ....................... vv
my_map f (x:xs) = foldr (\x xs -> (f x):xs) [] xs
-- --
The remaining part xs is aplied to foldr.
To fix just this, apply the whole list. This can be done by placing xx# before (x:xs). By that, the whole list is bound to xx.
-- vvv ........... see here ............... vv
my_map f xx#(x:xs) = foldr (\x xs -> (f x):xs) [] xx
-- --- --
Recommended impovement
Note: foldr can already deal with [] as input. Hence, my_map f [] = [] is not needed. But foldr would not be called when you apply [] to my_map. To get rid of my_map f [] = [], you need to remove the pattern matching, because (x:xs) matches only to lists with at least one element.
main :: IO ()
main = print $ my_map (^(3 :: Int)) ([1..5] :: [Integer])
my_map :: (a -> b) -> [a] -> [b]
my_map f xx = foldr (\x xs -> (f x):xs) [] xx
The answer is complete here. The rest below is for pleasure.
Further reductions
Simple expression instead of lambda expression
If you want to reduce the lambda expression (\x xs -> (f x):xs), as suggested by "Aadit M Shah"...
(:) is equal to (\x xs -> x:xs), because : is an operator and its function is (:)
. can be used to combine the function f with (:), hence (\x xs -> (f x):xs) is equal to ((:) . f)
main :: IO ()
main = print $ my_map (^(3 :: Int)) ([] :: [Integer])
my_map :: (a -> b) -> [a] -> [b]
my_map f xx = foldr ((:) . f) [] xx
Currying
A function of the form
-- v v
f a b c = .... c
can be reduced to
-- v v
f a b = ....
and a function of the form
-- v v v v
f a b c = .... b c
can be reduced to
-- v v v v
f a = ....
and so on, by currying.
Hence, my_map f xx = foldr ((:) . f) [] xx equals my_map f = foldr ((:) . f) [].
Combination and flip
flip flips the first two parameters.
Example, the following functions are equal:
f' a b c = (\c' b' a' -> ((a' - b') / c')) b a c
f'' a b c = flip (\c' b' a' -> ((a' - b') / c')) a b c
f''' = flip (\c' b' a' -> ((a' - b') / c'))
Hence, the following code works as well.
main :: IO ()
main = print $ my_map (^(3 :: Int)) ([1..5] :: [Integer])
my_map :: (a -> b) -> [a] -> [b]
my_map f = flip foldr [] ((:) . f)
But we can not get rid of f as above, because of the form in the expression flip foldr [] ((:) . f).
If we remove f ...
`((:) . f)` has type `a -> [a] -> [a]
-- v
`((:) . )` has type `(a -> a) -> a -> [a] -> [a]`
and
`flip foldr []` has type `Foldable t => (a1 -> [a2] -> [a2]) -> t a1 -> [a2]`
hence
f :: a -> a
is passed to
((:) . )
becomming
a -> [a] -> [a]
is passed to
flip foldr []
becomming
t a1 -> [a2]
Hence,
main :: IO ()
main = print $ my_map (^(3 :: Int)) ([1..5] :: [Integer])
my_map :: (a -> b) -> [a] -> [b]
my_map = flip foldr [] . ((:) . )
works nicely.
For a function that maps a function to every nth element in a list:
mapEvery :: Int -> (a -> a) -> [a] -> [a]
mapEvery n f = zipWith ($) (drop 1 . cycle . take n $ f : repeat id)
Is it possible to implement this with foldr like ordinary map?
EDIT: In the title, changed 'folder' to 'foldr'. Autocorrect...
Here's one solution
mapEvery :: Int -> (a -> a) -> [a] -> [a]
mapEvery n f as = foldr go (const []) as 1 where
go a as m
| m == n = f a : as 1
| otherwise = a : as (m+1)
This uses the "foldl as foldr" trick to pass state from the left to the right along the list as you fold. Essentially, if we read the type of foldr as (a -> r -> r) -> r -> [a] -> r then we instantiate r as Int -> [a] where the passed integer is the current number of elements we've passed without calling the function.
Yes, it can:
mapEvery :: Int -> (a -> a) -> [a] -> [a]
mapEvery n f xs
= foldr (\y ys -> g y : ys) []
$ zip [1..] xs
where
g (i, y) = if i `mod` n == 0 then f y else y
And since it's possible to implement zip in terms of foldr, you could get even more fold-y if you really wanted. This even works on infinite lists:
> take 20 $ mapEvery 5 (+1) $ repeat 1
[1,1,1,1,2,1,1,1,1,2,1,1,1,1,2,1,1,1,1,2]
This is what it looks like with even more foldr and inlining g:
mapEvery :: Int -> (a -> a) -> [a] -> [a]
mapEvery _ _ [] = []
mapEvery n f xs
= foldr (\(i, y) ys -> (if i `mod` n == 0 then f y else y) : ys) []
$ foldr step (const []) [1..] xs
where
step _ _ [] = []
step x zipsfn (y:ys) = (x, y) : zipsfn ys
Now, would I recommend writing it this way? Absolutely not. This is about as obfuscated as you can get while still writing "readable" code. But it does demonstrate that this is possible to use the very powerful foldr to implement relatively complex functions.
I'm writing a program to solve the result of primitive recursive functions:
1 --Basic functions------------------------------
2
3 --Zero function
4 z :: Int -> Int
5 z = \_ -> 0
6
7 --Successor function
8 s :: Int -> Int
9 s = \x -> (x + 1)
10
11 --Identity/Projection function generator
12 idnm :: Int -> Int -> ([Int] -> Int)
13 idnm n m = \(x:xs) -> ((x:xs) !! (m-1))
14
15 --Constructors--------------------------------
16
17 --Composition constructor
18 cn :: ([Int] -> Int) -> [([Int] -> Int)] -> ([Int] -> Int)
19 cn f [] = \(x:xs) -> f
20 cn f (g:gs) = \(x:xs) -> (cn (f (g (x:xs))) gs)
these functions and constructors are defined here: http://en.wikipedia.org/wiki/Primitive_recursive_function
The issue is with my attempt to create the compositon constructor, cn. When it gets to the base case, f is no longer a partial application, but a result of the function. Yet the function expects a function as the first argument. How can I deal with this problem?
Thanks.
Given f,
f :: [a] -> b
and g_k,
g_k :: [a] -> a
we want to produce h,
h :: [a] -> b
so the composition should be like
compo :: ([a] -> b) -> [[a] -> a] -> [a] -> b
compo f gs xs = f (map ($ xs) gs)
Example: http://codepad.org/aGIKi8dF
Edit: It can also be written in applicative style (eliminating that $) as
compo f gs xs = f (gs <*> pure xs)