Haskell create an n-ary tuple from given input - list

To put it straigth, I'm fairly new to Haskell and trying to solve a problem (programming exercise) I came over. Where it says I should create a function
com :: Int -> [t] -> [[t]]
that returns all possible choices of n elements, where n and list are the first and second arguments, respectively. Elements can be picked over again and in a different order. A result would be like:
com 2 [1,2,3] = [[1,1], [1,2]..[3,3]]
For the cases n = 1 and n = 2, I manage to solve the cases. The case n = 1 is quite simple, and, for the case n = 2, I would use concatenation and build it up. However, I don't understand how it can be made n-ary and work for all n. Like if suddenly a function call would be like com 10 ...

Is this what you want?
> sequence (replicate 3 "abc")
["aaa","aab","aac","aba","abb","abc","aca","acb","acc"
,"baa","bab","bac","bba","bbb","bbc","bca","bcb","bcc"
,"caa","cab","cac","cba","cbb","cbc","cca","ccb","ccc"]
The above exploits the fact that sequence, in the list monad, builds the cartesian product of a list of lists. So, we can simply replicate our list n times, and then take the product.
(Note that "abc" above is a shorthand for the list of charatcters ['a','b','c'])
So, a solution could be
com n xs = sequence (replicate n xs)
or equivalently, as Daniel Wagner points out below,
com = replicateM
A final note: I do realize that this is probably not very helpful for actually learning how to program. Indeed, I pulled two "magic" functions from the library which solved the task. Still, it shows how the problem can be reduced to two subproblems: 1) replicating a value n times and 2) building a cartesian product. The second task is a nice exercise on its own, if you don't want to use the library. You may wish to solve that starting from:
sequence :: [[a]] -> [[a]]
sequence [] = [[]]
sequence (x:xs) = ...
where ys = sequence xs

First: [] is a list constructor, not a tuple. I don't know any general way to build n-ary tuple.
However, sticking to lists, if you have n = 1 case solved and n = 2 case solved try to express the latter in term of the former. Then generalize to any n in terms of n-1:
com n xs = concat [map (x:) (com (n-1) xs) | x <- xs ]

A more verbose way to write it, but potentially more helpful when trying to understand List non-determinism and trying to understand exactly what the Haskell comprehension syntactic sugar really means, is to write with do notation:
com :: Int -> [a] -> [[a]]
com 0 _ = []
com 1 xs = [[x] | x <- xs]
com n xs = do
x <- xs
let ys = com (n - 1) xs
map (x:) ys

Related

I have difficulties with two problems with the langage OCaml

Important: I am only allowed to use List.head, List.tail and List.length
No List.map List.rev ...........etc
Only List.hd, List.tl and List.length
How to duplicate the elements of a list in a list of lists only if the length of the list is odd
Here is the code I tried:
let rec listes_paires x =
if x=[] then []
else [List.hd (List.hd x)]
# (List.tl (List.hd x))
# listes_paires (List.tl x);;
(* editor's note: I don't know where this line is supposed to go*)
if List.length mod 2 = 1 then []
For exemple:
lists_odd [[]; [1];[1;2];[1;2;3];[];[5;4;3;2;1]];;
returns
[[]; [1; 1]; [1; 2]; [1; 2; 3; 1; 2; 3]; []; [5; 4; 3; 2; 1; 5; 4; 3; 2; 1]]
Any help would be very appreciated
thank you all
It looks like that your exercise is about writing recursive functions on lists so that you can learn how to write functions like List.length, List.filter, and so on.
Start with the most simple recursive function, the one that computes the length to the list. Recall, that you can pattern match on the input list structure and make decisions on it, e.g.,
let rec length xs = match xs with
| [] -> 0 (* the empty list has size zero *)
| hd :: tl ->
(* here you can call `length` and it will return you
the length of the list hing how you can use it to
compute the length of the list that is made of `tl`
prepended with `hd` *)
???
The trick is to first write the simple cases and then write the complex cases assuming that your recursive function already works. Don't overthink it and don't try to compute how recursion will work in your head. It will make it hurt :) Just write correctly the base cases (the simple cases) and make sure that you call your function recursively and correctly combine the results while assuming that it works correctly. It is called the induction principle and it works, believe me :)
The above length function was easy as it was producing an integer as output and it was very easy to build it, e.g., you can use + to build a new integer from other integers, something that we have learned very early in our lives so it doesn't surprise us. But what if we want to build something more complex (in fact it is not more complex but just less common to us), e.g., a list data structure? Well, it is the same, we can just use :: instead of + to add things to our result.
So, lets try writing the filter function that will recurse over the input list and build a new list from the elements that satisfy the given predicate,
let rec filter xs keep = match xs with
| [] -> (* the simple case - no elements nothing to filter *)
[]
| x :: xs ->
(* we call filter and it returns the correctly filtered list *)
let filtered = filter xs keep in
(* now we need to decide what to do with `x` *)
if keep x then (* how to build a list from `x` and `filtered`?*)
else filtered (* keep filtering *)
The next trick to learn with recursive functions is how to employ helper functions that add an extra state (also called an accumulator). For example, the rev function, which reverses a list, is much better to define with an extra accumulator. Yes, we can easily define it without it,
let rec rev xs = match xs with
| [] -> []
| x :: xs -> rev xs # [x]
But this is an extremely bad idea as # operator will have to go to the end of the first list and build a completely new list on the road to add only one element. That is our rev implementation will have quadratic performance, i.e., for a list of n elements it will build n list each having n elements in it, only to drop most of them. So a more efficient implementation will employ a helper function that will have an extra parameter, an accumulator,
let rev xs =
(* we will pump elements from xs to ys *)
let rec loop xs ys = match xs with
| [] -> ys (* nothing more to pump *)
| x :: xs ->
let ys = (* push y to ys *) in
(* continue pumping *) in
loop xs []
This trick will also help you in implementing your tasks, as you need to filter by the position of the element. That means that your recursive function needs an extra state that counts the position (increments by one on each recursive step through the list elements). So you will need a helper function with an extra parameter for that counter.

A faster way of generating combinations with a given length, preserving the order

TL;DR: I want the exact behavior as filter ((== 4) . length) . subsequences. Just using subsequences also creates variable length of lists, which takes a lot of time to process. Since in the end only lists of length 4 are needed, I was thinking there must be a faster way.
I have a list of functions. The list has the type [Wor -> Wor]
The list looks something like this
[f1, f2, f3 .. fn]
What I want is a list of lists of n functions while preserving order like this
input : [f1, f2, f3 .. fn]
argument : 4 functions
output : A list of lists of 4 functions.
Expected output would be where if there's an f1 in the sublist, it'll always be at the head of the list.
If there's a f2 in the sublist and if the sublist doens't have f1, f2 would be at head. If fn is in the sublist, it'll be at last.
In general if there's a fx in the list, it never will be infront of f(x - 1) .
Basically preserving the main list's order when generating sublists.
It can be assumed that length of list will always be greater then given argument.
I'm just starting to learn Haskell so I haven't tried all that much but so far this is what I have tried is this:
Generation permutations with subsequences function and applying (filter (== 4) . length) on it seems to generate correct permutations -but it doesn't preserve order- (It preserves order, I was confusing it with my own function).
So what should I do?
Also if possible, is there a function or a combination of functions present in Hackage or Stackage which can do this? Because I would like to understand the source.
You describe a nondeterministic take:
ndtake :: Int -> [a] -> [[a]]
ndtake 0 _ = [[]]
ndtake n [] = []
ndtake n (x:xs) = map (x:) (ndtake (n-1) xs) ++ ndtake n xs
Either we take an x, and have n-1 more to take from xs; or we don't take the x and have n more elements to take from xs.
Running:
> ndtake 3 [1..4]
[[1,2,3],[1,2,4],[1,3,4],[2,3,4]]
Update: you wanted efficiency. If we're sure the input list is finite, we can aim at stopping as soon as possible:
ndetake n xs = go (length xs) n xs
where
go spare n _ | n > spare = []
go spare n xs | n == spare = [xs]
go spare 0 _ = [[]]
go spare n [] = []
go spare n (x:xs) = map (x:) (go (spare-1) (n-1) xs)
++ go (spare-1) n xs
Trying it:
> length $ ndetake 443 [1..444]
444
The former version seems to be stuck on this input, but the latter one returns immediately.
But, it measures the length of the whole list, and needlessly so, as pointed out by #dfeuer in the comments. We can achieve the same improvement in efficiency while retaining a bit more laziness:
ndzetake :: Int -> [a] -> [[a]]
ndzetake n xs | n > 0 =
go n (length (take n xs) == n) (drop n xs) xs
where
go n b p ~(x:xs)
| n == 0 = [[]]
| not b = []
| null p = [(x:xs)]
| otherwise = map (x:) (go (n-1) b p xs)
++ go n b (tail p) xs
Now the last test also works instantly with this code as well.
There's still room for improvement here. Just as with the library function subsequences, the search space could be explored even more lazily. Right now we have
> take 9 $ ndzetake 3 [1..]
[[1,2,3],[1,2,4],[1,2,5],[1,2,6],[1,2,7],[1,2,8],[1,2,9],[1,2,10],[1,2,11]]
but it could be finding [2,3,4] before forcing the 5 out of the input list. Shall we leave it as an exercise?
Here's the best I've been able to come up with. It answers the challenge Will Ness laid down to be as lazy as possible in the input. In particular, ndtake m ([1..n]++undefined) will produce as many entries as possible before throwing an exception. Furthermore, it strives to maximize sharing among the result lists (note the treatment of end in ndtakeEnding'). It avoids problems with badly balanced list appends using a difference list. This sequence-based version is considerably faster than any pure-list version I've come up with, but I haven't teased apart just why that is. I have the feeling it may be possible to do even better with a better understanding of just what's going on, but this seems to work pretty well.
Here's the general idea. Suppose we ask for ndtake 3 [1..5]. We first produce all the results ending in 3 (of which there is one). Then we produce all the results ending in 4. We do this by (essentially) calling ndtake 2 [1..3] and adding the 4 onto each result. We continue in this manner until we have no more elements.
import qualified Data.Sequence as S
import Data.Sequence (Seq, (|>))
import Data.Foldable (toList)
We will use the following simple utility function. It's almost the same as splitAtExactMay from the 'safe' package, but hopefully a bit easier to understand. For reasons I haven't investigated, letting this produce a result when its argument is negative leads to ndtake with a negative argument being equivalent to subsequences. If you want, you can easily change ndtake to do something else for negative arguments.
-- to return an empty list in the negative case.
splitAtMay :: Int -> [a] -> Maybe ([a], [a])
splitAtMay n xs
| n <= 0 = Just ([], xs)
splitAtMay _ [] = Nothing
splitAtMay n (x : xs) = flip fmap (splitAtMay (n - 1) xs) $
\(front, rear) -> (x : front, rear)
Now we really get started. ndtake is implemented using ndtakeEnding, which produces a sort of "difference list", allowing all the partial results to be concatenated cheaply.
ndtake :: Int -> [t] -> [[t]]
ndtake n xs = ndtakeEnding n xs []
ndtakeEnding :: Int -> [t] -> ([[t]] -> [[t]])
ndtakeEnding 0 _xs = ([]:)
ndtakeEnding n xs = case splitAtMay n xs of
Nothing -> id -- Not enough elements
Just (front, rear) ->
(front :) . go rear (S.fromList front)
where
-- For each element, produce a list of all combinations
-- *ending* with that element.
go [] _front = id
go (r : rs) front =
ndtakeEnding' [r] (n - 1) front
. go rs (front |> r)
ndtakeEnding doesn't call itself recursively. Rather, it calls ndtakeEnding' to calculate the combinations of the front part. ndtakeEnding' is very much like ndtakeEnding, but with a few differences:
We use a Seq rather than a list to represent the input sequence. This lets us split and snoc cheaply, but I'm not yet sure why that seems to give amortized performance that is so much better in this case.
We already know that the input sequence is long enough, so we don't need to check.
We're passed a tail (end) to add to each result. This lets us share tails when possible. There are lots of opportunities for sharing tails, so this can be expected to be a substantial optimization.
We use foldr rather than pattern matching. Doing this manually with pattern matching gives clearer code, but worse constant factors. That's because the :<|, and :|> patterns exported from Data.Sequence are non-trivial pattern synonyms that perform a bit of calculation, including amortized O(1) allocation, to build the tail or initial segment, whereas folds don't need to build those.
NB: this implementation of ndtakeEnding' works well for recent GHC and containers; it seems less efficient for earlier versions. That might be the work of Donnacha Kidney on foldr for Data.Sequence. In earlier versions, it might be more efficient to pattern match by hand, using viewl for versions that don't offer the pattern synonyms.
ndtakeEnding' :: [t] -> Int -> Seq t -> ([[t]] -> [[t]])
ndtakeEnding' end 0 _xs = (end:)
ndtakeEnding' end n xs = case S.splitAt n xs of
(front, rear) ->
((toList front ++ end) :) . go rear front
where
go = foldr go' (const id) where
go' r k !front = ndtakeEnding' (r : end) (n - 1) front . k (front |> r)
-- With patterns, a bit less efficiently:
-- go Empty _front = id
-- go (r :<| rs) !front =
-- ndtakeEnding' (r : end) (n - 1) front
-- . go rs (front :|> r)

Altering A Single Value in List of Lists in Haskell [Homework]

Been having some real issues with this and haven't been able to find any guidance on doing this in any reading. Have been tasked with implementing functions to complete a Haskell version of Connect 4. The board is represented as a list of lists of pieces using Data.List.
One of the functions is to drop a piece given the piece and column number. For this I would like to just add the piece to the appropriate column and be done with it but the only way I seem to be able to do it is recursing through the list until I get to the right column and then add the piece.
Is there any way to do this better?
My horrendous code is below:
cheatPiece :: GameState -> Int -> Piece -> GameState
cheatPiece [] _ _ = []
cheatPiece (xs:xss) 0 x = (x:xs) : xss
cheatPiece (xs:xss) n x = xs : cheatPiece xss (n-1) x
I don't think your implementation is horrendous at all. That's pretty much the standard way to work with immutable, linked lists.
I think the main thing that makes it feel clumsy is that working with indices and linked lists is never going to be very natural.
So, in the context of a homework assignment, your implementation is, I think, the most correct way to implement cheatPiece. If you had control over the board presentation I might consider using, for example, a vector or an IntMap to store the columns.
There's also always lens which lets you work with nested, immutable structures using terser abstractions but if you are still new to Haskell then the lens package definitely does not have the gentlest of learning curves.
import Control.Lens
data Piece = X | O deriving Show
type GameState = [[Piece]]
cheatPiece :: GameState -> Int -> Piece -> GameState
cheatPiece st i p = st & ix i %~ (p:)
You could use the take and drop functions and the list-indexing operator !!.
cheatPiece xss n x = take n xss ++ [x : (xss !! i)] ++ drop (n + 1) xss
Or there's splitAt which combines take and drop - I'll throw in a check for when the index is too big:
cheatPiece xss n x = case splitAt n xss of
(_, []) -> error "out of range"
(yss, zs:zss) -> yss ++ [x:zs] ++ zss
But I'd be tempted to generalize that by writing a function for modifying an element at an index:
modifyAt :: Int -> (a -> a) -> [a] -> [a]
modifyAt n f xs = case splitAt n xs of
(_, []) -> error "out of range"
(ys, z:zs) -> ys ++ [f z] ++ zs
which can be used like this:
> modifyAt 3 (+1000) [0..9]
[0,1,2,1003,4,5,6,7,8,9]
Then your function would be
cheatPiece xss n x = modifyAt n (x:) xss

Need to partition a list into lists based on breaks in ascending order of elements (Haskell)

Say I have any list like this:
[4,5,6,7,1,2,3,4,5,6,1,2]
I need a Haskell function that will transform this list into a list of lists which are composed of the segments of the original list which form a series in ascending order. So the result should look like this:
[[4,5,6,7],[1,2,3,4,5,6],[1,2]]
Any suggestions?
You can do this by resorting to manual recursion, but I like to believe Haskell is a more evolved language. Let's see if we can develop a solution that uses existing recursion strategies. First some preliminaries.
{-# LANGUAGE NoMonomorphismRestriction #-}
-- because who wants to write type signatures, amirite?
import Data.List.Split -- from package split on Hackage
Step one is to observe that we want to split the list based on a criteria that looks at two elements of the list at once. So we'll need a new list with elements representing a "previous" and "next" value. There's a very standard trick for this:
previousAndNext xs = zip xs (drop 1 xs)
However, for our purposes, this won't quite work: this function always outputs a list that's shorter than the input, and we will always want a list of the same length as the input (and in particular we want some output even when the input is a list of length one). So we'll modify the standard trick just a bit with a "null terminator".
pan xs = zip xs (map Just (drop 1 xs) ++ [Nothing])
Now we're going to look through this list for places where the previous element is bigger than the next element (or the next element doesn't exist). Let's write a predicate that does that check.
bigger (x, y) = maybe False (x >) y
Now let's write the function that actually does the split. Our "delimiters" will be values that satisfy bigger; and we never want to throw them away, so let's keep them.
ascendingTuples = split . keepDelimsR $ whenElt bigger
The final step is just to throw together the bit that constructs the tuples, the bit that splits the tuples, and a last bit of munging to throw away the bits of the tuples we don't care about:
ascending = map (map fst) . ascendingTuples . pan
Let's try it out in ghci:
*Main> ascending [4,5,6,7,1,2,3,4,5,6,1,2]
[[4,5,6,7],[1,2,3,4,5,6],[1,2]]
*Main> ascending [7,6..1]
[[7],[6],[5],[4],[3],[2],[1]]
*Main> ascending []
[[]]
*Main> ascending [1]
[[1]]
P.S. In the current release of split, keepDelimsR is slightly stricter than it needs to be, and as a result ascending currently doesn't work with infinite lists. I've submitted a patch that makes it lazier, though.
ascend :: Ord a => [a] -> [[a]]
ascend xs = foldr f [] xs
where
f a [] = [[a]]
f a xs'#(y:ys) | a < head y = (a:y):ys
| otherwise = [a]:xs'
In ghci
*Main> ascend [4,5,6,7,1,2,3,4,5,6,1,2]
[[4,5,6,7],[1,2,3,4,5,6],[1,2]]
This problem is a natural fit for a paramorphism-based solution. Having (as defined in that post)
para :: (a -> [a] -> b -> b) -> b -> [a] -> b
foldr :: (a -> b -> b) -> b -> [a] -> b
para c n (x : xs) = c x xs (para c n xs)
foldr c n (x : xs) = c x (foldr c n xs)
para c n [] = n
foldr c n [] = n
we can write
partition_asc xs = para c [] xs where
c x (y:_) ~(a:b) | x<y = (x:a):b
c x _ r = [x]:r
Trivial, since the abstraction fits.
BTW they have two kinds of map in Common Lisp - mapcar
(processing elements of an input list one by one)
and maplist (processing "tails" of a list). With this idea we get
import Data.List (tails)
partition_asc2 xs = foldr c [] . init . tails $ xs where
c (x:y:_) ~(a:b) | x<y = (x:a):b
c (x:_) r = [x]:r
Lazy patterns in both versions make it work with infinite input lists
in a productive manner (as first shown in Daniel Fischer's answer).
update 2020-05-08: not so trivial after all. Both head . head . partition_asc $ [4] ++ undefined and the same for partition_asc2 fail with *** Exception: Prelude.undefined. The combining function g forces the next element y prematurely. It needs to be more carefully written to be productive right away before ever looking at the next element, as e.g. for the second version,
partition_asc2' xs = foldr c [] . init . tails $ xs where
c (x:ys) r#(~(a:b)) = (x:g):gs
where
(g,gs) | not (null ys)
&& x < head ys = (a,b)
| otherwise = ([],r)
(again, as first shown in Daniel's answer).
You can use a right fold to break up the list at down-steps:
foldr foo [] xs
where
foo x yss = (x:zs) : ws
where
(zs, ws) = case yss of
(ys#(y:_)) : rest
| x < y -> (ys,rest)
| otherwise -> ([],yss)
_ -> ([],[])
(It's a bit complicated in order to have the combining function lazy in the second argument, so that it works well for infinite lists too.)
One other way of approaching this task (which, in fact lays the fundamentals of a very efficient sorting algorithm) is using the Continuation Passing Style a.k.a CPS which, in this particular case applied to folding from right; foldr.
As is, this answer would only chunk up the ascending chunks however, it would be nice to chunk up the descending ones at the same time... preferably in reverse order all in O(n) which would leave us with only binary merging of the obtained chunks for a perfectly sorted output. Yet that's another answer for another question.
chunks :: Ord a => [a] -> [[a]]
chunks xs = foldr go return xs $ []
where
go :: Ord a => a -> ([a] -> [[a]]) -> ([a] -> [[a]])
go c f = \ps -> let (r:rs) = f [c]
in case ps of
[] -> r:rs
[p] -> if c > p then (p:r):rs else [p]:(r:rs)
*Main> chunks [4,5,6,7,1,2,3,4,5,6,1,2]
[[4,5,6,7],[1,2,3,4,5,6],[1,2]]
*Main> chunks [4,5,6,7,1,2,3,4,5,4,3,2,6,1,2]
[[4,5,6,7],[1,2,3,4,5],[4],[3],[2,6],[1,2]]
In the above code c stands for current and p is for previous and again, remember we are folding from right so previous, is actually the next item to process.

Using Haskell's map function to calculate the sum of a list

Haskell
addm::[Int]->Int
addm (x:xs) = sum(x:xs)
I was able to achieve to get a sum of a list using sum function but is it possible to get the sum of a list using map function? Also what the use of map function?
You can't really use map to sum up a list, because map treats each list element independently from the others. You can use map for example to increment each value in a list like in
map (+1) [1,2,3,4] -- gives [2,3,4,5]
Another way to implement your addm would be to use foldl:
addm' = foldl (+) 0
Here it is, the supposedly impossible definition of sum in terms of map:
sum' xs = let { ys = 0 : map (\(a,b) -> a + b) (zip xs ys) } in last ys
this actually shows how scanl can be implemented in terms of map (and zip and last), the above being equivalent to foldl (+) 0 xs === last $ scanl (+) 0 xs:
scanl' f z xs = let { ys = z : map (uncurry f) (zip ys xs) } in ys
I expect one can calculate many things with map, arranging for all kinds of information flow through zip.
edit: the above is just a zipWith in disguise of course (and zipWith is kind of a map2):
sum' xs = let { ys = 0 : zipWith (+) ys xs } in last ys
This seems to suggest that scanl is more versatile than foldl.
It is not possible to use map to reduce a list to its sum. That recursive pattern is a fold.
sum :: [Int] -> Int
sum = foldr (+) 0
As an aside, note that you can define map as a fold as well:
map :: (a -> b) -> ([a] -> [b])
map f = fold (\x xs -> f x : xs) []
This is because foldr is the canonical recursive function on lists.
References: A tutorial on the universality and expressiveness of fold, Graham Hutton, J. Functional Programming 9 (4): 355–372, July 1999.
After some insights I have to add another answer: You can't get the sum of a list with map, but you can get the sum with its monadic version mapM. All you need to do is to use a Writer monad (see LYAHFGG) over the Sum monoid (see LYAHFGG).
I wrote a specialized version, which is probably easier to understand:
data Adder a = Adder a Int
instance Monad Adder where
return x = Adder x 0
(Adder x s) >>= f = let Adder x' s' = f x
in Adder x' (s + s')
toAdder x = Adder x x
sum' xs = let Adder _ s = mapM toAdder xs in s
main = print $ sum' [1..100]
--5050
Adder is just a wrapper around some type which also keeps a "running sum." We can make Adder a monad, and here it does some work: When the operation >>= (a.k.a. "bind") is executed, it returns the new result and the value of the running sum of that result plus the original running sum. The toAdder function takes an Int and creates an Adder that holds that argument both as wrapped value and as running sum (actually we're not interested in the value, but only in the sum part). Then in sum' mapM can do its magic: While it works similar to map for the values embedded in the monad, it executes "monadic" functions like toAdder, and chains these calls (it uses sequence to do this). At this point, we get through the "backdoor" of our monad the interaction between list elements that the standard map is missing.
Map "maps" each element of your list to an element in your output:
let f(x) = x*x
map f [1,2,3]
This will return a list of the squares.
To sum all elements in a list, use fold:
foldl (+) 0 [1,2,3]
+ is the function you want to apply, and 0 is the initial value (0 for sum, 1 for product etc)
As the other answers point out, the "normal" way is to use one of the fold functions. However it is possible to write something pretty similar to a while loop in imperative languages:
sum' [] = 0
sum' xs = head $ until single loop xs where
single [_] = True
single _ = False
loop (x1 : x2 : xs) = (x1 + x2) : xs
It adds the first two elements of the list together until it ends up with a one-element list, and returns that value (using head).
I realize this question has been answered, but I wanted to add this thought...
listLen2 :: [a] -> Int
listLen2 = sum . map (const 1)
I believe it returns the constant 1 for each item in the list, and returns the sum!
Might not be the best coding practice, but it was an example my professor gave to us students that seems to relate to this question well.
map can never be the primary tool for summing the elements of a container, in much the same way that a screwdriver can never be the primary tool for watching a movie. But you can use a screwdriver to fix a movie projector. If you really want, you can write
import Data.Monoid
import Data.Foldable
mySum :: (Foldable f, Functor f, Num a)
=> f a -> a
mySum = getSum . fold . fmap Sum
Of course, this is silly. You can get a more general, and possibly more efficient, version:
mySum' :: (Foldable f, Num a) => f a -> a
mySum' = getSum . foldMap Sum
Or better, just use sum, because its actually made for the job.