You must use recursion to define rmax2 and you must do so from “scratch”. That is, other than the cons operator, head, tail, and comparisons, you should not use any functions from the Haskell library.
I created a function that removes all instances of the largest item, using list comprehension. How do I remove the last instance of the largest number using recursion?
ved :: Ord a => [a] -> [a]
ved [] =[]
ved as = [ a | a <- as, m /= a ]
where m= maximum as
An easy way to split the problem into two easier subproblems consists in:
get the position index of the rightmost maximum value
write a general purpose function del that eliminates the element of a list at a given position. This does not require an Ord constraint.
If we were permitted to use regular library functions, ved could be written like this:
ved0 :: Ord a => [a] -> [a]
ved0 [] = []
ved0 (x:xs) =
let
(maxVal,maxPos) = maximum (zip (x:xs) [0..])
del k ys = let (ys0,ys1) = splitAt k ys in (ys0 ++ tail ys1)
in
del maxPos (x:xs)
where the pairs produced by zip are lexicographically ordered, thus ensuring the rightmost maximum gets picked.
We need to replace the library functions by manual recursion.
Regarding step 1, that is finding the position of the rightmost maximum, as is commonly done, we can use a recursive stepping function and a wrapper above it.
The recursive step function takes as arguments the whole context of the computation, that is:
current candidate for maximum value, mxv
current rightmost position of maximum value, mxp
current depth into the original list, d
rest of original list, xs
and it returns a pair: (currentMaxValue, currentMaxPos)
-- recursive stepping function:
findMax :: Ord a => a -> Int -> Int -> [a] -> (a, Int)
findMax mxv mxp d [] = (mxv,mxp)
findMax mxv mxp d (x:xs) = if (x >= mxv) then (findMax x d (d+1) xs)
else (findMax mxv mxp (d+1) xs)
-- top wrapper:
lastMaxPos :: Ord a => [a] -> Int
lastMaxPos [] = (-1)
lastMaxPos (x:xs) = snd (findMax x 0 1 xs)
Step 2, eliminating the list element at position k, can be handled in very similar fashion:
-- recursive stepping function:
del1 :: Int -> Int -> [a] -> [a]
del1 k d [] = []
del1 k d (x:xs) = if (d==k) then xs else x : del1 k (d+1) xs
-- top wrapper:
del :: Int -> [a] -> [a]
del k xs = del1 k 0 xs
Putting it all together:
We are now able to write our final recursion-based version of ved. For simplicity, we inline the content of wrapper functions instead of calling them.
-- ensure we're only using authorized functionality:
{-# LANGUAGE NoImplicitPrelude #-}
import Prelude (Ord, Eq, (==), (>=), (+), ($), head, tail,
IO, putStrLn, show, (++)) -- for testing only
ved :: Ord a => [a] -> [a]
ved [] = []
ved (x:xs) =
let
findMax mxv mxp d [] = (mxv,mxp)
findMax mxv mxp d (y:ys) = if (y >= mxv) then (findMax y d (d+1) ys)
else (findMax mxv mxp (d+1) ys)
(maxVal,maxPos) = findMax x 0 1 xs
del1 k d (y:ys) = if (d==k) then ys else y : del1 k (d+1) ys
del1 k d [] = []
in
del1 maxPos 0 (x:xs)
main :: IO ()
main = do
let xs = [1,2,3,7,3,2,1,7,3,5,7,5,4,3]
res = ved xs
putStrLn $ "input=" ++ (show xs) ++ "\n" ++ " res=" ++ (show res)
If you are strictly required to use recursion, you can use 2 helper functions: One to reverse the list and the second to remove the first largest while reversing the reversed list.
This result in a list where the last occurrence of the largest element is removed.
We also use a boolean flag to make sure we don't remove more than one element.
This is ugly code and I really don't like it. A way to make things cleaner would be to move the reversal of the list to a helper function outside of the current function so that there is only one helper function to the main function. Another way is to use the built-in reverse function and use recursion only for the removal.
removeLastLargest :: Ord a => [a] -> [a]
removeLastLargest xs = go (maximum xs) [] xs where
go n xs [] = go' n True [] xs
go n xs (y:ys) = go n (y:xs) ys
go' n f xs [] = xs
go' n f xs (y:ys)
| f && y == n = go' n False xs ys
| otherwise = go' n f (y:xs) ys
Borrowing the implementation of dropWhileEnd from Hackage, we can implement a helper function splitWhileEnd:
splitWhileEnd :: (a -> Bool) -> [a] -> ([a], [a])
splitWhileEnd p = foldr (\x (xs, ys) -> if p x && null xs then ([], x:ys) else (x:xs, ys)) ([],[])
splitWhileEnd splits a list according to a predictor from the end. For example:
ghci> xs = [1,2,3,4,3,2,4,3,2]
ghci> splitWhileEnd (< maximum xs) xs
([1,2,3,4,3,2,4],[3,2])
With this helper function, you can write ven as:
ven :: Ord a => [a] -> [a]
ven xs =
let (x, y) = splitWhileEnd (< maximum xs) xs
in init x ++ y
ghci> ven xs
[1,2,3,4,3,2,3,2]
For your case, you can refactor splitWhileEnd as:
fun p = \x (xs, ys) -> if p x && null xs then ([], x:ys) else (x:xs, ys)
splitWhileEnd' p [] = ([], [])
splitWhileEnd' p (x : xs) = fun p x (splitWhileEnd' p xs)
ven' xs = let (x, y) = splitWhileEnd' (< maximum xs) xs in init x ++ y
If init and ++ are not allowed, you can implement them manually. It's easy!
BTW, I guess this may be your homework for Haskell course. I think it's ridiculous if your teacher gives the limitations. Who is programming from scratch nowadays?
Anyway, you can always work around this kind of limitations by reimplementing the built-in function manually. Good luck!
I have to split the given list into non-empty sub-lists each of which
is either in strictly ascending order, in strictly descending order, or contains all equal elements. For example, [5,6,7,2,1,1,1] should become [[5,6,7],[2,1],[1,1]].
Here is what I have done so far:
splitSort :: Ord a => [a] -> [[a]]
splitSort ns = foldr k [] ns
where
k a [] = [[a]]
k a ns'#(y:ys) | a <= head y = (a:y):ys
| otherwise = [a]:ns'
I think I am quite close but when I use it it outputs [[5,6,7],[2],[1,1,1]] instead of [[5,6,7],[2,1],[1,1]].
Here is a kinda ugly solution, with three reverse in one line of code :).
addElement :: Ord a => a -> [[a]] -> [[a]]
addElement a [] = [[a]]
addElement a (x:xss) = case x of
(x1:x2:xs)
| any (check a x1 x2) [(==),(<),(>)] -> (a:x1:x2:xs):xss
| otherwise -> [a]:(x:xss)
_ -> (a:x):xss
where
check x1 x2 x3 op = (x1 `op` x2) && (x2 `op` x3)
splitSort xs = reverse $ map reverse $ foldr addElement [] (reverse xs)
You can possibly get rid of all the reversing if you modify addElement a bit.
EDIT:
Here is a less reversing version (even works for infinite lists):
splitSort2 [] = []
splitSort2 [x] = [[x]]
splitSort2 (x:y:xys) = (x:y:map snd here):splitSort2 (map snd later)
where
(here,later) = span ((==c) . uncurry compare) (zip (y:xys) xys)
c = compare x y
EDIT 2:
Finally, here is a solution based on a single decorating/undecorating, that avoids comparing any two values more than once and is probably a lot more efficient.
splitSort xs = go (decorate xs) where
decorate :: Ord a => [a] -> [(Ordering,a)]
decorate xs = zipWith (\x y -> (compare x y,y)) (undefined:xs) xs
go :: [(Ordering,a)] -> [[a]]
go ((_,x):(c,y):xys) = let (here, later) = span ((==c) . fst) xys in
(x : y : map snd here) : go later
go xs = map (return . snd) xs -- Deal with both base cases
Every ordered prefix is already in some order, and you don't care in which, as long as it is the longest:
import Data.List (group, unfoldr)
foo :: Ord t => [t] -> [[t]]
foo = unfoldr f
where
f [] = Nothing
f [x] = Just ([x], [])
f xs = Just $ splitAt (length g + 1) xs
where
(g : _) = group $ zipWith compare xs (tail xs)
length can be fused in to make the splitAt count in unary essentially, and thus not be as strict (unnecessarily, as Jonas Duregård rightly commented):
....
f xs = Just $ foldr c z g xs
where
(g : _) = group $ zipWith compare xs (tail xs)
c _ r (x:xs) = let { (a,b) = r xs } in (x:a, b)
z (x:xs) = ([x], xs)
The initial try turned out to be lengthy probably inefficient but i will keep it striked for the sake of integrity with the comments. You best just skip to the end for the answer.
Nice question... but turns out to be a little hard candy. My approach is in segments, those of each i will explain;
import Data.List (groupBy)
splitSort :: Ord a => [a] -> [[a]]
splitSort (x:xs) = (:) <$> (x :) . head <*> tail $ interim
where
pattern = zipWith compare <$> init <*> tail
tuples = zipWith (,) <$> tail <*> pattern
groups = groupBy (\p c -> snd p == snd c) . tuples $ (x:xs)
interim = groups >>= return . map fst
*Main> splitSort [5,6,7,2,1,1,1]
[[5,6,7],[2,1],[1,1]]
The pattern function (zipWith compare <$> init <*> tail) is of type Ord a => [a] -> [Ordering] when fed with [5,6,7,2,1,1,1] compares the init of it by the tail of it by zipWith. So the result would be [LT,LT,GT,GT,EQ,EQ]. This is the pattern we need.
The tuples function will take the tail of our list and will tuple up it's elements with the corresponding elements from the result of pattern. So we will end up with something like [(6,LT),(7,LT),(2,GT),(1,GT),(1,EQ),(1,EQ)].
The groups function utilizes Data.List.groupBy over the second items of the tuples and generates the required sublists such as [[(6,LT),(7,LT)],[(2,GT),(1,GT)],[(1,EQ),(1,EQ)]]
Interim is where we monadically get rid of the Ordering type values and tuples. The result of interim is [[6,7],[2,1],[1,1]].
Finally at the main function body (:) <$> (x :) . head <*> tail $ interim appends the first item of our list (x) to the sublist at head (it has to be there whatever the case) and gloriously present the solution.
Edit: So investigating the [0,1,0,1] resulting [[0,1],[0],[1]] problem that #Jonas Duregård discovered, we can conclude that in the result there shall be no sub lists with a length of 1 except for the last one when singled out. I mean for an input like [0,1,0,1,0,1,0] the above code produces [[0,1],[0],[1],[0],[1],[0]] while it should [[0,1],[0,1],[0,1],[0]]. So I believe adding a squeeze function at the very last stage should correct the logic.
import Data.List (groupBy)
splitSort :: Ord a => [a] -> [[a]]
splitSort [] = []
splitSort [x] = [[x]]
splitSort (x:xs) = squeeze $ (:) <$> (x :) . head <*> tail $ interim
where
pattern = zipWith compare <$> init <*> tail
tuples = zipWith (,) <$> tail <*> pattern
groups = groupBy (\p c -> snd p == snd c) $ tuples (x:xs)
interim = groups >>= return . map fst
squeeze [] = []
squeeze [y] = [y]
squeeze ([n]:[m]:ys) = [n,m] : squeeze ys
squeeze ([n]:(m1:m2:ms):ys) | compare n m1 == compare m1 m2 = (n:m1:m2:ms) : squeeze ys
| otherwise = [n] : (m1:m2:ms) : squeeze ys
squeeze (y:ys) = y : squeeze s
*Main> splitSort [0,1, 0, 1, 0, 1, 0]
[[0,1],[0,1],[0,1],[0]]
*Main> splitSort [5,6,7,2,1,1,1]
[[5,6,7],[2,1],[1,1]]
*Main> splitSort [0,0,1,0,-1]
[[0,0],[1,0,-1]]
Yes; as you will also agree the code has turned out to be a little too lengthy and possibly not so efficient.
The Answer: I have to trust the back of my head when it keeps telling me i am not on the right track. Sometimes, like in this case, the problem reduces down to a single if then else instruction, much simpler than i had initially anticipated.
runner :: Ord a => Maybe Ordering -> [a] -> [[a]]
runner _ [] = []
runner _ [p] = [[p]]
runner mo (p:q:rs) = let mo' = Just (compare p q)
(s:ss) = runner mo' (q:rs)
in if mo == mo' || mo == Nothing then (p:s):ss
else [p] : runner Nothing (q:rs)
splitSort :: Ord a => [a] -> [[a]]
splitSort = runner Nothing
My test cases
*Main> splitSort [0,1, 0, 1, 0, 1, 0]
[[0,1],[0,1],[0,1],[0]]
*Main> splitSort [5,6,7,2,1,1,1]
[[5,6,7],[2,1],[1,1]]
*Main> splitSort [0,0,1,0,-1]
[[0,0],[1,0,-1]]
*Main> splitSort [1,2,3,5,2,0,0,0,-1,-1,0]
[[1,2,3,5],[2,0],[0,0],[-1,-1],[0]]
For this solution I am making the assumption that you want the "longest rally". By that I mean:
splitSort [0, 1, 0, 1] = [[0,1], [0,1]] -- This is OK
splitSort [0, 1, 0, 1] = [[0,1], [0], [1]] -- This is not OK despite of fitting your requirements
Essentially, There are two pieces:
Firstly, split the list in two parts: (a, b). Part a is the longest rally considering the order of the two first elements. Part b is the rest of the list.
Secondly, apply splitSort on b and put all list into one list of list
Taking the longest rally is surprisingly messy but straight. Given the list x:y:xs: by construction x and y will belong to the rally. The elements in xs belonging to the rally depends on whether or not they follow the Ordering of x and y. To check this point, you zip every element with the Ordering is has compared against its previous element and split the list when the Ordering changes. (edge cases are pattern matched) In code:
import Data.List
import Data.Function
-- This function split the list in two (Longest Rally, Rest of the list)
splitSort' :: Ord a => [a] -> ([a], [a])
splitSort' [] = ([], [])
splitSort' (x:[]) = ([x],[])
splitSort' l#(x:y:xs) = case span ( (o ==) . snd) $ zip (y:xs) relativeOrder of
(f, s) -> (x:map fst f, map fst s)
where relativeOrder = zipWith compare (y:xs) l
o = compare y x
-- This applies the previous recursively
splitSort :: Ord a => [a] -> [[a]]
splitSort [] = []
splitSort (x:[]) = [[x]]
splitSort (x:y:[]) = [[x,y]]
splitSort l#(x:y:xs) = fst sl:splitSort (snd sl)
where sl = splitSort' l
I wonder whether this question can be solve using foldr if splits and groups a list from
[5,6,7,2,1,1,1]
to
[[5,6,7],[2,1],[1,1]]
instead of
[[5,6,7],[2],[1,1,1]]
The problem is in each step of foldr, we only know the sorted sub-list on right-hand side and a number to be processed. e.g. after read [1,1] of [5,6,7,2,1,1,1] and next step, we have
1, [[1, 1]]
There are no enough information to determine whether make a new group of 1 or group 1 to [[1,1]]
And therefore, we may construct required sorted sub-lists by reading elements of list from left to right, and why foldl to be used. Here is a solution without optimization of speed.
EDIT:
As the problems that #Jonas Duregård pointed out on comment, some redundant code has been removed, and beware that it is not a efficient solution.
splitSort::Ord a=>[a]->[[a]]
splitSort numList = foldl step [] numList
where step [] n = [[n]]
step sublists n = groupSublist (init sublists) (last sublists) n
groupSublist sublists [n1] n2 = sublists ++ [[n1, n2]]
groupSublist sublists sortedList#(n1:n2:ns) n3
| isEqual n1 n2 = groupIf (isEqual n2 n3) sortedList n3
| isAscen n1 n2 = groupIfNull isAscen sortedList n3
| isDesce n1 n2 = groupIfNull isDesce sortedList n3
| otherwise = mkNewGroup sortedList n3
where groupIfNull check sublist#(n1:n2:ns) n3
| null ns = groupIf (check n2 n3) [n1, n2] n3
| otherwise = groupIf (check (last ns) n3) sublist n3
groupIf isGroup | isGroup = addToGroup
| otherwise = mkNewGroup
addToGroup gp n = sublists ++ [(gp ++ [n])]
mkNewGroup gp n = sublists ++ [gp] ++ [[n]]
isEqual x y = x == y
isAscen x y = x < y
isDesce x y = x > y
My initial thought looks like:
ordruns :: Ord a => [a] -> [[a]]
ordruns = foldr extend []
where
extend a [ ] = [ [a] ]
extend a ( [b] : runs) = [a,b] : runs
extend a (run#(b:c:etc) : runs)
| compare a b == compare b c = (a:run) : runs
| otherwise = [a] : run : runs
This eagerly fills from the right, while maintaining the Ordering in all neighbouring pairs for each sublist. Thus only the first result can end up with a single item in it.
The thought process is this: an Ordering describes the three types of subsequence we're looking for: ascending LT, equal EQ or descending GT. Keeping it the same every time we add on another item means it will match throughout the subsequence. So we know we need to start a new run whenever the Ordering does not match. Furthermore, it's impossible to compare 0 or 1 items, so every run we create contains at least 1 and if there's only 1 we do add the new item.
We could add more rules, such as a preference for filling left or right. A reasonable optimization is to store the ordering for a sequence instead of comparing the leading two items twice per item. And we could also use more expressive types. I also think this version is inefficient (and inapplicable to infinite lists) due to the way it collects from the right; that was mostly so I could use cons (:) to build the lists.
Second thought: I could collect the lists from the left using plain recursion.
ordruns :: Ord a => [a] -> [[a]]
ordruns [] = []
ordruns [a] = [[a]]
ordruns (a1:a2:as) = run:runs
where
runs = ordruns rest
order = compare a1 a2
run = a1:a2:runcontinuation
(runcontinuation, rest) = collectrun a2 order as
collectrun _ _ [] = ([], [])
collectrun last order (a:as)
| order == compare last a =
let (more,rest) = collectrun a order as
in (a:more, rest)
| otherwise = ([], a:as)
More exercises. What if we build the list of comparisons just once, for use in grouping?
import Data.List
ordruns3 [] = []
ordruns3 [a] = [[a]]
ordruns3 xs = unfoldr collectrun marked
where
pairOrder = zipWith compare xs (tail xs)
marked = zip (head pairOrder : pairOrder) xs
collectrun [] = Nothing
collectrun ((o,x):xs) = Just (x:map snd markedgroup, rest)
where (markedgroup, rest) = span ((o==).fst) xs
And then there's the part where there's a groupBy :: (a -> a -> Bool) -> [a] -> [[a]] but no groupOn :: Eq b => (a -> b) -> [a] -> [[a]]. We can use a wrapper type to handle that.
import Data.List
data Grouped t = Grouped Ordering t
instance Eq (Grouped t) where
(Grouped o1 _) == (Grouped o2 _) = o1 == o2
ordruns4 [] = []
ordruns4 [a] = [[a]]
ordruns4 xs = unmarked
where
pairOrder = zipWith compare xs (tail xs)
marked = group $ zipWith Grouped (head pairOrder : pairOrder) xs
unmarked = map (map (\(Grouped _ t) -> t)) marked
Of course, the wrapper type's test can be converted into a function to use groupBy instead:
import Data.List
ordruns5 [] = []
ordruns5 [a] = [[a]]
ordruns5 xs = map (map snd) marked
where
pairOrder = zipWith compare xs (tail xs)
marked = groupBy (\a b -> fst a == fst b) $
zip (head pairOrder : pairOrder) xs
These marking versions arrive at the same decoration concept Jonas Duregård applied.
How can I apply a function to only a single element of a list?
Any suggestion?
Example:
let list = [1,2,3,4,3,6]
function x = x * 2
in ...
I want to apply function only to the first occurance of 3 and stop there.
Output:
List = [1,2,6,4,3,6] -- [1, 2, function 3, 4, 3, 6]
To map or not to map, that is the question.
Better not to map.
Why? Because map id == id anyway, and you only want to map through one element, the first one found to be equal to the argument given.
Thus, split the list in two, change the found element, and glue them all back together. Simple.
See: span :: (a -> Bool) -> [a] -> ([a], [a]).
Write: revappend (xs :: [a]) (ys :: [a]) == append (reverse xs) ys, only efficient.
Or fuse all the pieces together into one function. You can code it directly with manual recursion, or using foldr. Remember,
map f xs = foldr (\x r -> f x : r) [] xs
takeWhile p xs = foldr (\x r -> if p x then x : r else []) [] xs
takeUntil p xs = foldr (\x r -> if p x then [x] else x : r) [] xs
filter p xs = foldr (\x r -> if p x then x : r else r) [] xs
duplicate xs = foldr (\x r -> x : x : r) [] xs
mapFirstThat p f xs = -- ... your function
etc. Although, foldr won't be a direct fit, as you need the combining function of the (\x xs r -> ...) variety. That is known as paramorphism, and can be faked by feeding tails xs to the foldr, instead.
you need to maintain some type of state to indicate the first instance of the value, since map will apply the function to all values.
Perhaps something like this
map (\(b,x) -> if (b) then f x else x) $ markFirst 3 [1,2,3,4,3,6]
and
markFirst :: a -> [a] -> [(Boolean,a)]
markFirst a [] = []
markFirst a (x:xs) | x==a = (True,x): zip (repeat False) xs
| otherwise = (False,x): markFirst a xs
I'm sure there is an easier way, but that's the best I came up with at this time on the day before Thanksgiving.
Here is another approach based on the comment below
> let leftap f (x,y) = f x ++ y
leftap (map (\x -> if(x==3) then f x else x)) $ splitAt 3 [1,2,3,4,3,6]
You can just create a simple function which multiples a number by two:
times_two :: (Num a) => a -> a
times_two x = x * 2
Then simply search for the specified element in the list, and apply times_two to it. Something like this could work:
map_one_element :: (Eq a, Num a) => a -> (a -> a) -> [a] -> [a]
-- base case
map_one_element _ _ [] = []
-- recursive case
map_one_element x f (y:ys)
-- ff element is found, apply f to it and add rest of the list normally
| x == y = f y : ys
-- first occurence hasnt been found, keep recursing
| otherwise = y : map_one_element x f ys
Which works as follows:
*Main> map_one_element 3 times_two [1,2,3,4,3,6]
[1,2,6,4,3,6]