How to implement the head function using fst function - haskell

I admit this is my homework. But I really couldn't find a good solution after working hard on it.
There might be some stupid ways to accomplish this, like:
myHead (x:[]) = x
myHead (x:y:xs) = fst (x, y)
But I don't think that's what the teacher wants.
BTW, error-handling is not required.
Thanks in advance!

There's a very natural function that's not in the prelude called "uncons" which is the inverse of uncurried cons.
cons :: a -> [a] -> [a]
uncurry cons :: (a, [a]) -> [a]
uncons :: [a] -> (a, [a])
uncons (x:xs) = (x, xs)
You can use it to implement head as
head = fst . uncons
Why is uncons natural?
You can think of a list as the datatype that's defined through the use of two constructor functions
nil :: [a]
nil = []
cons :: (a, [a]) -> [a]
cons (a,as) = a:as
You can also think of it as the data type which is deconstructed by a function
destruct :: [a] -> Maybe (a, [a])
destruct [] = Nothing
destruct (a:as) = Just (a, as)
It's well beyond this answer to explain why those are so definitively tied to the list type, but one way to look at it is to try to define
nil :: f a
cons :: (a, f a) -> f a
or
destruct :: f a -> Maybe (a, f a)
for any other container type f. You'll find that they all have very close relationships with lists.
You can almost already see uncons in the second case of the definition of destruct, but there's a Just in the way. This is uncons is better paired with head and tail which are not defined on empty lists
head [] = error "Prelude.head"
so we can adjust the previous answer to work for infinite streams. Here we can think of infinite streams as being constructed by one function
data Stream a = Next a (Stream a)
cons :: (a, Stream a) -> Stream a
cons (a, as) = Next a as
and destructed by one function
uncons :: Stream a -> (a, Stream a)
uncons (Next a as) = (a, as)
-- a. k. a.
uncons stream = (head stream, tail stream)
the two being inverses of one another.
Now we can get head for Streams by getting the first element of the return tuple from uncons
head = fst . uncons
And that's what head models in the Prelude, so we can pretend like lists are infinite streams and define head in that way
uncons :: [a] -> (a, [a])
uncons (a:as) = (a, as)
-- a. k. a.
uncons list = (head list, tail list)
head = fst . uncons

Perhaps you're expected write to your own cons List type, then it might make more sense. Although type synonyms can't be recursive, so you end up using a non-tuple data constructor, making the tuple superfluous.. it would look like:
data List a = Nil | List (a, List a)
deriving( Show )
head :: List a -> a
head (List c) = fst c

Like already said in the comments, this is just a silly task and you won't get something you could call a good implementation of head.
Your solution, for those requirements, is just fine – as the only change I would replace (x:y:xs) with (x:y:_) since xs isn't used at all (which would actually cause a compiler warning in some settings). In fact, you could do that with y as well:
myHead (x:_:_) = fst (x, undefined)
There would be alternatives that look perhaps not quite so useless use of fst, i.e. don't just build a tuple by hand and immediately deconstruct it again:
myHead' [x] = x
myHead' xs = myHead' . fst $ splitAt 1 xs
myHead'' = foldr1 $ curry fst
myHead''' = fromJust . find ((==0) . fst) . zip [0..]
but you could rightfully say that these are just ridiculous.

Related

Changing recursive guards into higher order functions

I'm trying to convert basic functions into higher order functions (specifically map, filter, or foldr). I was wondering if there are any simple concepts to apply where I could see old functions I've written using guards and turn them into higher order.
I'm working on changing a function called filterFirst that removes the first element from the list (second argument) that does not satisfy a given predicate function (first argument).
filterFirst :: (a -> Bool) -> [a] -> [a]
filterFirst _ [] = []
filterFirst x (y:ys)
| x y = y : filterFirst x ys
| otherwise = ys
For an example:
greaterOne :: Num a=>Ord a=>a->Bool
greaterOne x = x > 1
filterFirst greaterOne [5,-6,-7,9,10]
[5,-7,9,10]
Based on the basic recursion, I was wondering if there might be a way to translate this (and similar functions) to higher order map, filter, or foldr. I'm not very advanced and these functions are new to me.
There is a higher-order function that's appropriate here, but it's not in the base library. What's the trouble with foldr? If you just fold over the list, you'll end up rebuilding the whole thing, including the part after the deletion.
A more appropriate function for the job is para from the recursion-schemes package (I've renamed one of the type variables):
para :: Recursive t => (Base t (t, r) -> r) -> t -> r
In the case of lists, this specializes to
para :: (ListF a ([a], r) -> r) -> [a] -> r
where
data ListF a b = Nil | Cons a b
deriving (Functor, ....)
This is pretty similar to foldr. The recursion-schemes equivalent of foldr is
cata :: Recursive t => (Base t r -> r) -> t -> r
Which specializes to
cata :: (ListF a r -> r) -> [a] -> r
Take a break here and figure out why the type of cata is basically equivalent to that of foldr.
The difference between cata and para is that para passes the folding function not only the result of folding over the tail of the list, but also the tail of the list itself. That gives us an easy and efficient way to produce the rest of the list after we've found the first non-matching element:
filterFirst :: (a -> Bool) -> [a] -> [a]
filterFirst f = para go
where
--go :: ListF a ([a], [a]) -> [a]
go (Cons a (tl, r))
| f a = a : r
| otherwise = tl
go Nil = []
para is a bit awkward for lists, since it's designed to fit into a more general context. But just as cata and foldr are basically equivalent, we could write a slightly less awkward function specifically for lists.
foldrWithTails
:: (a -> [a] -> b -> b)
-> b -> [a] -> b
foldrWithTails f n = go
where
go (a : as) = f a as (go as)
go [] = n
Then
filterFirst :: (a -> Bool) -> [a] -> [a]
filterFirst f = foldrWithTails go []
where
go a tl r
| f a = a : r
| otherwise = tl
First, let's flip the argument order of your function. This will make a few steps easier, and we can flip it back when we're done. (I'll call the flipped version filterFirst'.)
filterFirst' :: [a] -> (a -> Bool) -> [a]
filterFirst' [] _ = []
filterFirst' (y:ys) x
| x y = y : filterFirst' ys x
| otherwise = ys
Note that filterFirst' ys (const True) = ys for all ys. Let's substitute that in place:
filterFirst' :: [a] -> (a -> Bool) -> [a]
filterFirst' [] _ = []
filterFirst' (y:ys) x
| x y = y : filterFirst' ys x
| otherwise = filterFirst' ys (const True)
Use if-else instead of a guard:
filterFirst' :: [a] -> (a -> Bool) -> [a]
filterFirst' [] _ = []
filterFirst' (y:ys) x = if x y then y : filterFirst' ys x else filterFirst' ys (const True)
Move the second argument to a lambda:
filterFirst' :: [a] -> (a -> Bool) -> [a]
filterFirst' [] = \_ -> []
filterFirst' (y:ys) = \x -> if x y then y : filterFirst' ys x else filterFirst' ys (const True)
And now this is something we can turn into a foldr. The pattern we were going for is that filterFirst' (y:ys) can be expressed in terms of filterFirst' ys, without using ys otherwise, and we're now there.
filterFirst' :: Foldable t => t a -> (a -> Bool) -> [a]
filterFirst' = foldr (\y f -> \x -> if x y then y : f x else f (const True)) (\_ -> [])
Now we just need to neaten it up a bit:
filterFirst' :: Foldable t => t a -> (a -> Bool) -> [a]
filterFirst' = foldr go (const [])
where go y f x
| x y = y : f x
| otherwise = f (const True)
And flip the arguments back:
filterFirst :: Foldable t => (a -> Bool) -> t a -> [a]
filterFirst = flip $ foldr go (const [])
where go y f x
| x y = y : f x
| otherwise = f (const True)
And we're done. filterFirst implemented in terms of foldr.
Addendum: Although filter isn't strong enough to build this, filterM is when used with the State monad:
{-# LANGUAGE FlexibleContexts #-}
import Control.Monad.State
filterFirst :: (a -> Bool) -> [a] -> [a]
filterFirst x ys = evalState (filterM go ys) False
where go y = do
alreadyDropped <- get
if alreadyDropped || x y then
return True
else do
put True
return False
If we really want, we can write filterFirst using foldr, since foldr is kind of "universal" -- it allows any list transformation we can perform using recursion. The main downside is that the resulting code is rather counter-intuitive. In my opinion, explicit recursion is far better in this case.
Anyway here's how it is done. This relies on what I consider to be an antipattern, namely "passing four arguments to foldr". I call this an antipattern since foldr is usually called with three arguments only, and the result is not a function taking a fourth argument.
filterFirst :: (a->Bool)->[a]->[a]
filterFirst p xs = foldr go (\_ -> []) xs True
where
go y ys True
| p y = y : ys True
| otherwise = ys False
go y ys False = y : ys False
Clear? Not very much. The trick here is to exploit foldr to build a function Bool -> [a] which returns the original list if called with False, and the filtered-first list if called with True. If we craft that function using
foldr go baseCase xs
the result is then obviously
foldr go baseCase xs True
Now, the base case must handle the empty list, and in such case we must return a function returning the empty list, whatever the boolean argument is. Hence, we arrive at
foldr go (\_ -> []) xs True
Now, we need to define go. This takes as arguments:
a list element y
the result of the "recursion" ys (a function Bool->[a] for the rest of the list)
and must return a function Bool->[a] for the larger list. So let's also consider
a boolean argument
and finally make go return a list. Well, if the boolean is False we must return the list unchanged, so
go y ys False = y : ys False
Note that ys False means "the tail unchanged", so we are really rebuilding the whole list unchanged.
If instead the boolean is true, we query the predicate as in p y. If that is false, we discard y, and return the list tail unchanged
go y ys True
| p y = -- TODO
| otherwise = ys False
If p y is true, we keep y and we return the list tail filtered.
go y ys True
| p y = y : ys True
| otherwise = ys False
As a final note, we cold have used a pair ([a], [a]) instead of a function Bool -> [a], but that approach does not generalize as well to more complex cases.
So, that's all. This technique is something nice to know, but I do not recommend it in real code which is meant to be understood by others.
Joseph and chi's answers already show how to derive a foldr implementation, so I'll try to aid intuition.
map is length-preserving, filterFirst is not, so trivially map must be unsuited for implementing filterFirst.
filter (and indeed map) are memoryless - the same predicate/function is applied to each element of the list, regardless of the result on other elements. In filterFirst, behaviour changes once we see the first non-satisfactory element and remove it, so filter (and map) are unsuited.
foldr is used to reduce a structure to a summary value. It's very general, and it might not be immediately obvious without experience what sorts of things this may cover. filterFirst is in fact such an operation, though. The intuition is something like, "can we build it in a single pass through the structure, building it up as we go(, with additional state stored as required)?". I fear Joseph's answer obfuscates a little, as foldr with 4 parameters, it may not be immediately obvious what's going on, so let's try it a little differently.
filterFirst p xs = snd $ foldr (\a (deleted,acc) -> if not deleted && not (p a) then (True,acc) else (deleted,a:acc) ) (False,[]) xs
Here's a first attempt. The "extra state" here is obviously the bool indicating whether or not we've deleted an element yet, and the list accumulates in the second element of the tuple. At the end we call snd to obtain just the list. This implementation has the problem, however, that we delete the rightmost element not satisfying the predicate, because foldr first combines the rightmost element with the neutral element, then the second-rightmost, and so on.
filterFirst p xs = snd $ foldl (\(deleted,acc) a -> if not deleted && not (p a) then (True,acc) else (deleted,a:acc) ) (False,[]) xs
Here, we try using foldl. This does delete the leftmost non-satisfactory element, but has the side-effect of reversing the list. We can stick a reverse at the front, and this would solve the problem, but is somewhat unsatisfactory due to the double-traversal.
Then, if you go back to foldr, having realized that (basically) if you want transform a list whilst preserving order that foldr is the correct variant, you play with it for a while and end up writing what Joseph suggested. I do however agree with chi that straightforward recursion is the best solution here.
Your function can also be expressed as an unfold, or, more specifically, as an apomorphism. Allow me to begin with a brief explanatory note, before the solution itself.
The apomorphism is the recursion scheme dual to the paramorphism (see dfeuer's answer for more about the latter). Apomorphisms are examples of unfolds, which generate a structure from a seed. For instance, Data.List offers unfoldr, a list unfold.
unfoldr :: (b -> Maybe (a, b)) -> b -> [a]
The function given to unfoldr takes a seed and either produces a list element and a new seed (if the maybe-value is a Just) or terminates the list generation (if it is Nothing). Unfolds are more generally expressed by the ana function from recursion-schemes ("ana" is short for "anamorphism").
ana :: Corecursive t => (a -> Base t a) -> a -> t
Specialised to lists, this becomes...
ana #[_] :: (b -> ListF a b) -> b -> [a]
... which is unfoldr in different clothing.
An apomorphism is an unfold in which the generation of the structure can be short-circuited at any point of the process, by producing, instead of a new seed, the rest of the structure in a fell swoop. In the case of lists, we have:
apo #[_] :: (b -> ListF a (Either [a] b)) -> b -> [a]
Either is used to trigger the short-circuit: with a Left result, the unfold short-circuits, while with a Right it proceeds normally.
The solution in terms of apo is fairly direct:
{-# LANGUAGE LambdaCase #-}
import Data.Functor.Foldable
filterFirst :: (a -> Bool) -> [a] -> [a]
filterFirst p = apo go
where
go = \case
[] -> Nil
a : as
| p a -> Cons a (Right as)
| otherwise -> case as of
[] -> Nil
b : bs -> Cons b (Left bs)
It is somewhat more awkward than dfeuer's para-based solution, because if we want to short-circuit without an empty list for a tail we are compelled to emit one extra element (the b in the short-circuiting case), and so we have to look one position ahead. This awkwardness would grow by orders of magnitude if, rather than filterFirst, we were to impĺement plain old filter with an unfold, as beautifully explained in List filter using an anamorphism.
This answer is inspired by a comment from luqui on a now-deleted question.
filterFirst can be implemented in a fairly direct way in terms of span:
filterFirst :: (a -> Bool) -> [a] -> [a]
filterFirst p = (\(yeas, rest) -> yeas ++ drop 1 rest) . span p
span :: (a -> Bool) -> [a] -> ([a], [a]) splits the list in two at the first element for which the condition doesn't hold. After span, we drop the first element of the second part of the list (with drop 1 rather than tail so that we don't have to add a special case for []), and reassemble the list with (++).
As an aside, there is a near-pointfree spelling of this implementation which I find too pretty not to mention:
filterFirst :: (a -> Bool) -> [a] -> [a]
filterFirst p = uncurry (++) . second (drop 1) . span p
While span is a higher order function, it would be perfectly understandable if you found this implementation disappointing in the context of your question. After all, span is not much more fundamental than filterFirst itself. Shouldn't we try going a little deeper, to see if we can capture the spirit of this solution while expressing it as a fold, or as some other recursion scheme?
I believe functions like filterFirst can be fine demonstrations of hylomorphisms. A hylomorphism is an unfold (see my other answer for more on that) that generates an intermediate data structure followed by a fold which turns this data structure into something else. Though it might look like that would require two passes to get a result (one through the input structure, and another through the intermediate one), if the hylomorphism implemented properly (as done in the hylo function of recursion-schemes) it can be done in a single pass, with the fold consuming pieces of the intermediate structure as they are generated by the unfold (so that we don't have to actually build it all only to tear it down).
Before we start, here is the boilerplate needed to run what follows:
{-# LANGUAGE LambdaCase #-}
{-# LANGUAGE DeriveFunctor #-}
{-# LANGUAGE DeriveFoldable #-}
{-# LANGUAGE DeriveTraversable #-}
{-# LANGUAGE TypeFamilies #-}
{-# LANGUAGE TemplateHaskell #-}
import Data.Functor.Foldable
import Data.Functor.Foldable.TH
The strategy here is picking an intermediate data structure for the hylomorphism that expresses the essence of what we want to achieve. In this case, we will use this cute thing:
data BrokenList a = Broken [a] | Unbroken a (BrokenList a)
-- I won't actually use those instances here,
-- but they are nice to have if you want to play with the type.
deriving (Eq, Show, Functor, Foldable, Traversable)
makeBaseFunctor ''BrokenList
BrokenList is very much like a list (Broken and Unbroken mirror [] and (:), while the makeBaseFunctor incantation generates a BrokenListF base functor analogous to ListF, with BrokenF and UnbrokenF constructors), except that it has another list attached at its end (the Broken constructor). It expresses, in a quite literal way, the idea of a list being divided in two parts.
With BrokenList at hand, we can write the hylomorphism. coalgSpan is the operation used for the unfold, and algWeld, the one used for the fold.
filterFirst p = hylo algWeld coalgSpan
where
coalgSpan = \case
[] -> BrokenF []
x : xs
| p x -> UnbrokenF x xs
| otherwise -> BrokenF xs
algWeld = \case
UnbrokenF x yeas -> x : yeas
BrokenF rest -> rest
coalgSpan breaks the list upon hitting a x element such that p x doesn't hold. Not adding that element to the second part of the list (BrokenF xs rather than BrokenF (x : xs)) takes care of the filtering. As for algWeld, it is used to concatenate the two parts (it is very much like what we would use to implement (++) using cata).
(For a similar example of BrokenList in action, see the breakOn implementation in Note 5 of this older answer of mine. It suggests what it would take to implement span using this strategy.)
There are at least two good things about this hylo-based implementation. Firstly, it has good performance (casual testing suggests that, if compiled with optimisations, it is at least as good as, and possibly slightly faster than, the most efficient implementations in other answers here). Secondly, it reflects very closely your original, explicitly recursive implementation of filterFirst (or, at any rate, more closely than the fold-only and unfold-only implementations).

How to concatenate a value to a list when both are inside Maybes?

I'm reading realworldhaskell and I've come to chapter 4 where the book talks about "Safely and sanely working with crashy functions". I'm trying to write a safe version of init :: [a] -> [a]. This is as far as I got:
mySafeInit :: [a] -> Maybe [a]
mySafeInit [] = Nothing
mySafeInit (_:[]) = Just []
mySafeInit (x:xs) = ???
I get that I can't do something like x : mySafeInit xs, because that would be like doing a : Just [a] (right?). I've read about fmap though, so I tried this:
mySafeInit :: [a] -> Maybe [a]
mySafeInit [] = Nothing
mySafeInit (_:[]) = Just []
mySafeInit (x:xs) = fmap (:) (Just x) (mySafeInit xs)
Well that's guaranteed to work, right? Well no, and I'm struggling to understand why. Here I have a Just a and a Just [a], right? Shouldn't I be able to use fmap, give it the function (:) and my two maybes, and get a Just [a] back?
After some googling I did figure out that this works:
mySafeInit :: [a] -> Maybe [a]
mySafeInit [] = Nothing
mySafeInit (_:[]) = Just []
mySafeInit (x:xs) = fmap (:) (Just x) <*> (mySafeInit xs)
but I don't understand why. Could someone walk me through this? :)
Let's look at
fmap (:) (Just x) (mySafeInit xs)
For Maybe, fmap is defined as
fmap _ Nothing = Nothing
fmap f (Just x) = Just (f x)
Therefore
fmap (:) (Just x)
is
Just ((:) x)
Note that this is not a function; it's a Maybe value of some type (in fact, it's Maybe ([a] -> [a])). That's why
(Just ((:) x)) (mySafeInit xs)
is a type error (you can only apply functions).
In the second version of your code, you did
fmap (:) (Just x) <*> (mySafeInit xs)
, which is
(Just ((:) x)) <*> (mySafeInit xs)
Looking at the type of <*>, we get:
(<*>) :: (Applicative f) => f (a -> b) -> f a -> f b
Here f is Maybe, and a and b are both [a]:
(<*>) :: Maybe ([a] -> [a]) -> Maybe [a] -> Maybe [a]
That's why this version checks out.
Shouldn't I be able to use fmap, give it the function (:) and my two maybes, and get a Just [a] back?
No, because fmap only works with functions of one argument. There is another function for two arguments, though:
liftA2 :: (Applicative f) => (a -> b -> c) -> f a -> f b -> f c
Contrast with
fmap :: (Functor f) => (a -> b) -> f a -> f b
liftA2 (:) (Just x) (mySafeInit xs) should indeed work.
But it's needlessly complicated because one of your maybes is artificial: You explicitly wrapped your x in Just. You can simply do this instead:
fmap ((:) x) (mySafeInit xs)
By the definition of fmap, this is either Nothing (if mySafeInit xs is Nothing) or Just ((:) x y) (if mySafeInit xs is Just y).
Simple solution
I think you make things too hard/complex. You can simply consider two cases:
the empty list, in which case you return a Nothing; and
the non-empty list, in which case you can use the old init:
Or putting it into code:
mySafeInit :: [a] -> Maybe [a]
mySafeInit [] = Nothing
mySafeInit l = Just (init l)
Now of course we have to trust that there are only two constructors for a list; and that init will only error when we provide it an empty list. Both assumptions hold, but can (although very unlikely) change in the future.
We can however use our own init function:
mySafeInit :: [a] -> Maybe [a]
mySafeInit [] = Nothing
mySafeInit (l:ls) = Just (init l ls)
where init _ [] = []
init x (x2:xs) = x : init x2 xs
Now the Haskell compiler can verify that both functions are syntactically total. Note that functions can be total, but not syntactically: for instance because we can make assumptions regarding the structure of the data that the compiler can not check.
Working with fmap
You provide a solution like:
-- ...
mySafeInit (x:xs) = fmap (:) (Just x) <*> (mySafeInit xs)
A first aspect we have to understand is that Maybe is a Functor. A Fuctor supports a function fmap where we map the item(s) inside the functor so to speak. For instance a list is a functor as well, and we can thus map all elements in the list with fmap f somelist. You can see Maybe as some sort of list with zero elements (the Nothing) or one element (the Just x). So in case we perform fmap on a Nothing, we get a Nothing back, in case we perform this on a Just x, we get Just (f x).
Here we see fmap (:) (Just x). Since we thus apply it on a Just x, we have actually constructed:
fmap (:) Just x
-> Just ((x :))
This is thus a Just, but it carries a function (x : ), not a value.
Next we see a <*> operator. (<*>) :: Applicative f => f (a -> b) -> f a -> f b is some sort of "multiplication" over an Applicative (a Maybe is an applicative as well). On the left side we have an applicative of functions a -> b, on the right side an applicative of as. The result is an applicative of bs: for every tuple of functions and values.
In case we perform this on a list, we thus obtain a list that starts with the values all applied to the first function, then all values applied to the second function, etc.
For a Maybe we thus have a Maybe (a -> b) left and a Maybe a on the right. In case these are of the form Just f and Just x, we obtain Just (f x), in all other cases we obtain Nothing.
Since we perform a recursive call on mySafeInit xs, we thus obtain the init of the remaining list wrapped in a Just. In that case, we thus prepend that result with xs.
An equivalent code definition is thus:
mySafeInit :: [a] -> Maybe [a]
mySafeInit [] = Nothing
mySafeInit (_:[]) = Just []
mySafeInit (x:xs) | Just ys <- mySafeInit xs = Just (x:ys)
| otherwise = Nothing

Taking a list of lists and generating all variants with one element replaced [closed]

Closed. This question needs to be more focused. It is not currently accepting answers.
Want to improve this question? Update the question so it focuses on one problem only by editing this post.
Closed 5 years ago.
Improve this question
I have types for a two-dimensional map of characters:
type Row = [Char]
type Mappy = [Row]
I'd like to write a function that takes a Mappy like:
[['o','o'],['o','o']]
and generates a list of all Mappys with a single 'o' element replaced with 'i':
[ [['i','o'],['o','o']]
, [['o','i'],['o','o']]
, [['o','o'],['i','o']]
, [['o','o'],['o','i']]
]
Here's what I've tried: I think I need to use the map function, because I need to go over each element, but I don't know how, because a map function doesn't keep track of the position it is working on.
type Row = [Char]
type Mappy = [Row]
func :: Mappy -> [Mappy]
func a = map (map someFunc a) a
someFunc :: Mappy -> Char -> Mappy
someFunc a b = if b == "o"
then undefined
else a
Obviously, I should change the undefined, but i have no idea how. Thanks in advance.
Zippers are great, and there's an interesting blog post about
implementing Conway's Game of Life using zippers and comonads in Haskell. On the other
hand, if this is still your first week learning Haskell, you might
want to save Comonads for Thursday, right?
Here's another approach that uses simple recursion and list
comprehensions and no complex Haskell features.
First, imagine we have an awesome function:
varyOne :: (a -> [a]) -> [a] -> [[a]]
varyOne = undefined
that works as follows. Given a function f that produces zero or
more variants of an element a, the function call varyOne f xs
generates all variants of the list xs that result from taking
exactly one element of xs, say x in the middle of the list, and replacing it with all the
variants given by f x.
This function is surprisingly flexible. It can generate the list of all variants resulting from forcibly replacing an element by a constant:
> varyOne (\x -> [3]) [1,2,3,4]
[[3,2,3,4],[1,3,3,4],[1,2,3,4],[1,2,3,3]]
By returning a singleton variant for a specific value and an empty list of variants for other values, it can generate all variants that replace an 'o' with an 'i' while suppressing the "variants" where no replacement is possible:
> let varyRow = varyOne (\c -> if c == 'o' then ['i'] else [])
> varyRow "ooxo"
["ioxo","oixo","ooxi"]
and, because varyRow itself generates variants of a row, it can be used with varyOne to generate variants of tables where a particular row has been replaced by its possible variants:
> varyOne varyRow ["ooo","oox","ooo"]
[["ioo","oox","ooo"],["oio","oox","ooo"],["ooi","oox","ooo"],
["ooo","iox","ooo"],["ooo","oix","ooo"],
["ooo","oox","ioo"],["ooo","oox","oio"],["ooo","oox","ooi"]]
It turns out that this awesome function is surprisingly easy to write:
varyOne :: (a -> [a]) -> [a] -> [[a]]
varyOne f (x:xs)
= [y:xs | y <- f x] ++ [x:ys | ys <- varyOne f xs]
varyOne _ [] = []
The first list comprehension generates all the variants for the current element. The second list comprehension generates variants that involve changes to the right of the current element using a recursive varyOne call.
Given varyOne, we can define:
replaceOne :: Char -> Char -> Mappy -> [Mappy]
replaceOne old new = varyOne (varyOne rep1)
where rep1 x = if x == old then [new] else []
and:
> replaceOne 'o' 'i' ["ooo","oox","ooo"]
[["ioo","oox","ooo"],["oio","oox","ooo"],["ooi","oox","ooo"]
,["ooo","iox","ooo"],["ooo","oix","ooo"]
,["ooo","oox","ioo"],["ooo","oox","oio"],["ooo","oox","ooi"]]
is probably the function you're looking for.
If you prefer to unconditionally replace a single element with i, no matter what the old element was, then this will work:
> varyOne (varyOne (const ['i'])) ["ooo","oox","ooo"]
[["ioo","oox","ooo"],["oio","oox","ooo"],["ooi","oox","ooo"]
,["ooo","iox","ooo"],["ooo","oix","ooo"],["ooo","ooi","ooo"]
,["ooo","oox","ioo"],["ooo","oox","oio"],["ooo","oox","ooi"]]
What you want, young BaasBartMans, is a Zipper.
data Zipper a = Zipper [a] a [a]
ofList :: [a] -> Maybe (Zipper a)
ofList [] = Nothing
ofList (a:as) = Just (Zipper [] a as)
A zipper gives you the context for a position in a list, so you
can easily modify them one at a time, step forward and backward and such.
We can recover a list from a zipper:
instance Foldable Zipper where
foldr f c (Zipper ls a rs) = foldl' (flip f) (f a (foldr f c rs)) ls
We can modify every position in a Zipper simultaneously:
instance Functor Zipper where
fmap f (Zipper ls a rs) = Zipper (fmap f ls) (f a) (fmap f rs)
Or just the focused element:
here :: Functor f => (a -> f a) -> Zipper a -> f (Zipper a)
here f (Zipper ls a rs) = fmap (\a' -> Zipper ls a' rs) (f a)
And as a Zipper is a Comonad, we can modify each element in context:
instance Comonad Zipper where
extract (Zipper _ a _) = a
extend f z#(Zipper ls a rs) = Zipper ls' a' rs' where
a' = f z
ls' = unfoldr (fmap (\z' -> (f z', z')) . goLeft) z
rs' = unfoldr (fmap (\z' -> (f z', z')) . goRight) z
Using that, we can build a function that modifies each element of a list in context:
everywhere :: Alternative f => (a -> f a) -> [a] -> f [a]
everywhere f as = case ofList as of
Nothing -> pure []
Just z -> asum $ extend (fmap toList . here f) z
Which works for simple lists:
λ everywhere (\a -> [a+1]) [10,20,30]
[[11,20,30]
,[10,21,30]
,[10,20,31]]
And nested lists:
λ everywhere (everywhere (\a -> [a+1])) [[10], [20,20], [30,30,30]]
[[[11],[20,20],[30,30,30]]
,[[10],[21,20],[30,30,30]]
,[[10],[20,21],[30,30,30]]
,[[10],[20,20],[31,30,30]]
,[[10],[20,20],[30,31,30]]
,[[10],[20,20],[30,30,31]]]

Converting a foldl into fold1

I am using the following fold to get the final monotonically decreasing sequence of a list.
foldl (\acc x -> if x<=(last acc) then acc ++ [x] else [x]) [(-1)] a
So [9,5,3,6,2,1] would return [6,2,1]
However, with foldl I needed to supply a start for the fold namely [(-1)]. I was trying to turn into to a foldl1 to be able to handle any range of integers as well as any Ord a like so:
foldl1 (\acc x -> if x<=(last acc) then acc ++ [x] else [x]) a
But I get there error:
cannot construct infinite type: a ~ [a]
in the second argument of (<=) namely last acc
I was under the impression that foldl1 was basically :
foldl (function) [head a] a
But I guess this isn't so? How would you go about making this fold generic for any Ord type?
I was under the impression that foldl1 was basically :
foldl (function) [head a] a
No, foldl1 is basically:
foldl function (head a) (tail a)
So the initial element is not a list of head a, but head a.
How would you go about making this fold generic for any Ord type?
Well a quick fix is:
foldl (\acc x -> if x<=(last acc) then acc ++ [x] else [x]) [head a] (tail a)
But there are still two problems:
in case a is an empty list, this function will error (while you probably want to return the empty list); and
the code is not terribly efficient since both last and (++) run in O(n).
The first problem can easily be addressed by using pattern matching to prevent that scenario. But for the latter you better would for instance use a reverse approach. Like for instance:
f :: Ord t => [t] -> [t]
f [] = [] -- case when the empty list is given
f a = reverse $ foldl (\acc#(ac:_) x -> if x <= ac then (x:acc) else [x]) [head a] (tail a)
Furthermore personally I am not a huge fan of if-then-else in functional programming, you can for instance define a helper function like:
f :: Ord t => [t] -> [t]
f [] = [] -- case when the empty list is given
f a = reverse $ foldl g [head a] (tail a)
where g acc#(ac:_) x | x <= ac = (x:acc)
| otherwise = [x]
Now reverse runs in O(n) but this is done only once. Furthermore the (:) construction runs in O(1) so all the actions in g run in O(1) (well given the comparison of course works efficient, etc.) making the algorithm itself O(n).
For your sample input it gives:
*Main> f [9,5,3,6,2,1]
[6,2,1]
The type of foldl1 is:
Foldable t => (a -> a -> a) -> t a -> a
Your function argument,
\acc x -> if x<=(last acc) then acc ++ [x] else [x]
has type:
(Ord a) => [a] -> a -> [a]
When Haskell's typechecker tries typechecking your function, it'll try unifying the type a -> a -> a (the type of the first argument of foldl1) with the type [a] -> a -> [a] (the type of your function).
To unify these types would require unifying a with [a], which would lead to the infinite type a ~ [a] ~ [[a]] ~ [[[a]]]... and so on.
The reason this works while using foldl is that the type of foldl is:
Foldable t => (b -> a -> b) -> b -> t a -> b
So [a] gets unified with b and a gets unified with the other a, leading to no problem at all.
foldl1 is limited in that it can only take functions which deal with only one type, or, in other terms, the accumulator needs to be the same type as the input list (for instance, when folding a list of Ints, foldl1 can only return an Int, while foldl can use arbitrary accumulators. So you can't do this using foldl1).
With regards to making this generic for all Ord values, one possible solution is to make a new typeclass for values which state their own "least-bound" value, which would then be used by your function. You can't make this function as it is generic on all Ord values because not all Ord values have sequence least bounds you can use.
class LowerBounded a where
lowerBound :: a
instance LowerBounded Int where
lowerBound = -1
finalDecreasingSequence :: (Ord a, LowerBounded a) => [a] -> [a]
finalDecreasingSequence = foldl buildSequence lowerBound
where buildSequence acc x
| x <= (last acc) = acc ++ [x]
| otherwise = [x]
You might also want to read a bit about how Haskell does its type inference, as it helps a lot in figuring out errors like the one you got.

Filter an infinite list of monadic values

Perhaps this is obvious, but I can't seem to figure out how to best filter an infinite list of IO values. Here is a simplified example:
infinitelist :: [IO Int]
predicate :: (a -> Bool)
-- how to implement this?
mysteryFilter :: (a -> Bool) -> [IO a] -> IO [a]
-- or perhaps even this?
mysteryFilter' :: (a -> Bool) -> [IO a] -> [IO a]
Perhaps I have to use sequence in some way, but I want the evaluation to be lazy. Any suggestions? The essence is that for each IO Int in the output we might have to check several IO Int values in the input.
Thank you!
Not doable without using unsafeInterleaveIO or something like it. You can't write a filter with the second type signature, since if you could you could say
unsafePerformIOBool :: IO Bool -> Bool
unsafePerformIOBool m = case mysteryFilter' id [m] of
[] -> False
(_:_) -> True
Similarly, the first type signature isn't going to work--any recursive call will give you back something of type IO [a], but then to build a list out of this you will need to perform this action before returning a result (since : is not in IO you need to use >>=). By induction you will have to perform all the actions in the list (which takes forever when the list is infinitely long) before you can return a result.
unsafeInterleaveIO resolves this, but is unsafe.
mysteryFilter f [] = return []
mysteryFilter f (x:xs) = do ys <- unsafeInterleaveIO $ mysteryFilter f xs
y <- x
if f y then return (y:ys) else return ys
the problem is that this breaks the sequence that the monad is supposed to provide. You no longer have guarantees about when your monadic actions happen (they might never happen, they might happen multiple times, etc).
Lists just do not play nice with IO. This is why we have the plethora of streaming types (Iteratees, Conduits, Pipes, etc).
The simplest such type is probably
data MList m a = Nil | Cons a (m (MList m a))
note that we observe that
[a] == MList Id a
since
toMList :: [a] -> MList Id a
toMList [] = Nil
toMList (x:xs) = Cons x $ return $ toMList xs
fromMList :: MList Id a -> [a]
fromMList Nil = []
fromMList (Cons x xs) = x:(fromMList . runId $ xs)
also, MList is a functor
instance Functor m => Functor (MList m) where
fmap f Nil = Nil
fmap f (Cons x xs) = Cons (f x) (fmap (fmap f) xs)
and it is a functor in the category of Functor's and Natural transformations.
trans :: Functor m => (forall x. m x -> n x) -> MList m a -> MList n a
trans f Nil = Nil
trans f (Cons x xs) = Cons x (f (fmap trans f xs))
with this it is easy to write what you want
mysteryFilter :: (a -> Bool) -> MList IO (IO a) -> IO (MList IO a)
mysteryFilter f Nil = return Nil
mysteryFilter f (Cons x xs)
= do y <- x
let ys = liftM (mysteryFilter f) xs
if f y then Cons y ys else ys
or various other similar functions.

Resources