I found this question from CS 217.
Divide a list into one or more sublists so that each sublist contains integers in nondecreasing (sorted) order.
[3,5,1,8,9,2,1,0] returns [[3,5],[1,8,9],[2],[1],[0]]
[1,2,3,4,5,6] returns [[1,2,3,4,5,6]]
[5,4,3,2,1] returns [[5],[4],[3],[2],[1]]
below code works:
val Q1 = [ 3, 5, 1, 8, 9, 2, 1, 0 ]
val A1 = foldl (
fn (x, a) =>
if x > hd (hd a) then (x::hd a)::tl a
else [x]::a
) [ [ hd Q1 ] ] (tl Q1)
val A1 = map rev (rev A1)
or like this: use 2 temporary list to collect.
fun split l = let
fun split' tmp subset =
fn [] => []
| [x] => (x::tmp)::subset
| (a::(c as b::_)) =>
if a < b then split' (a::tmp) subset c
else split' [] ((a::tmp)::subset) c
in (rev o map rev) (split' [] [] l) end
So many solutions for this question,
But I still want to know how to code it as a pattern match function?
maybe something like below:
(Not sure if it is possible?)
fun split [] = [[]]
| split [x] = [[x]]
| split [a, b] = if a < b then (* here *) else (* here *)
| split (a::b) = if a < hd b then (* here *) else (* here *)
This question really stuck me.
Under the assumption that this is homework, I hesitate to give a complete answer, but here are a few hints:
1) In the empty basis case I think that you want to return [[]] rather than []. Your specification doesn't address this, but since the empty list is the longest list of nondecreasing integers which can be pulled from the front of the empty list, the return value should be the list consisting of the empty list. This is somewhat similar to the fact that the powerset (set of all subsets) of the empty set is the set containing the empty set rather than the empty set itself. It shouldn't really matter how you define this particular case, since the real basis case is ...
2) In the [x] case you really need to return [[x]] rather than [x] since the type of the function that you are trying to write is int list -> int list list
3) In the remaining case you can write the pattern like
| split (x::y::zs) = (* fill this in *)
then test if x <= y to decide what to do. Since both x <= y and x > y will involve split (y::zs) you could compute this once, giving this a name in a let binding and have the if in the scope of that binding, though that is mostly a matter of taste.
Note how the pattern works in this last case. Explicit use of hd should be fairly rare in function definitions which use pattern-matching (though if you flesh out the last case without using a pattern-matching let binding you will be forced to use it in at least one of the branches of the if).
On Edit: Since this isn't homework, here is a complete implementation:
fun split [] = [[]]
| split [x] = [[x]]
| split (x::y::zs) =
let val first::rest = split (y::zs) in
if x <= y then
(x::first) :: rest
else
[x]::first::rest
end;
Related
I tried with something like this but it doesn't work how I wanted it to do. I'm new kinda new to Haskell, and I don't really know how to do it, and what's wrong.
insert a (x:xs) = insert2 a (x:xs) []
where insert2 el (x:xs) hd =
if (x:xs) == []
then []
else if ( a>=x && a < head(xs))
then hd ++ [x] ++ [a] ++ xs
else insert2 a xs hd++[x]
main = do
let list =[1 ,2 ,3 ,4 ,5 ,6]
let out = insert 2 list
print out
The output I get is [2,2,3,4,5,6,1]
First a couple of cosmetics:
Ensure indentation is right. When copy/pasting into StackOverflow, it's generally best to use ctrl+k to get it in code-block style.
There's no point matching (x:xs) only to pass the entire thing into your local function.
Omit unnecessary parentheses and use standardised spacing.
With that, your code becomes
insert a allxs = insert2 a allxs []
where insert2 el (x:xs) hd =
if x:xs == []
then []
else if a >= x && a < head xs
then hd ++ [x] ++ [a] ++ xs
else insert2 a xs hd ++ [x]
main = do
let list = [1, 2, 3, 4, 5, 6]
let out = insert 2 list
print out
Algorithmically speaking, there's no point in using an “accumulator argument” here. It's easier and actually more efficient to directly recurse on the input, and simply pass on the remaining tail after done with the insertion. Also remember to have a base case:
insert a [] = [a]
insert a (x:xs) = ...
You also don't need to use head. You've already pattern-matched the head element with the x:xs pattern. If you did need another list element, you should match that right there too, like
insert a (x:x':xs) = ...
...but you don't in fact need that, x is enough to determine what to do. Namely,
insert a (x:xs)
| a<=x = -- if the list was ordered, this implies that now _all_
-- its elements must be greater or equal a. Do you
-- need any recursion anymore?
| otherwise = -- ok, `x` was smaller, so you need to insert after it.
-- Recursion is needed here.
Here are some hints. It's a lot simpler than you're making it. You definitely don't need a helper function.
insert a [] = ??
insert a (x : xs)
| a <= x = ???
| otherwise = ???
Two things:
Prepending to a list is more efficient than appending to one.
Haskell lets you write separate definitions to avoid having to write single, nested conditional expressions.
There are two kinds of list you can insert into: empty and non-empty. Each can be handled by a separate definition, which the compiler will use to define a single function.
insert a [] = [a]
insert a (x:xs) = ...
The first case is easy: inserting into an empty list produces a singleton list. The second case is tricker: what you do depends on whether a is smaller than x or not. You can use a conditional expression
insert a (x:xs) = if a < x then a : insert x xs else x : insert a xs
thought you may see guards used instead:
insert a (x:xs) | a < x = a : insert x xs
| otherwise = x : insert a xs
In both cases, we know (because the list argument is already sorted) that insert x xs == x : xs, so we can write that directly to "short-circuit" the recursion:
insert a (x:xs) = if a < x then a : x : xs else x : insert a xs
don't complicate! , make simple ...
insertme a list = takeWhile (<a) list ++ [a] ++ dropWhile (<a) list
I'm trying to learn haskell by solving some online problems and training exercises.
Right now I'm trying to make a function that'd remove adjacent duplicates from a list.
Sample Input
"acvvca"
"1456776541"
"abbac"
"aabaabckllm"
Expected Output
""
""
"c"
"ckm"
My first though was to make a function that'd simply remove first instance of adjacent duplicates and restore the list.
module Test where
removeAdjDups :: (Eq a) => [a] -> [a]
removeAdjDups [] = []
removeAdjDups [x] = [x]
removeAdjDups (x : y : ys)
| x == y = removeAdjDups ys
| otherwise = x : removeAdjDups (y : ys)
*Test> removeAdjDups "1233213443"
"122133"
This func works for first found pairs.
So now I need to apply same function over the result of the function.
Something I think foldl can help with but I don't know how I'd go about implementing it.
Something along the line of
removeAdjDups' xs = foldl (\acc x -> removeAdjDups x acc) xs
Also is this approach the best way to implement the solution or is there a better way I should be thinking of?
Start in last-first order: first remove duplicates from the tail, then check if head of the input equals to head of the tail result (which, by this moment, won't have any duplicates, so the only possible pair is head of the input vs. head of the tail result):
main = mapM_ (print . squeeze) ["acvvca", "1456776541", "abbac", "aabaabckllm"]
squeeze :: Eq a => [a] -> [a]
squeeze (x:xs) = let ys = squeeze xs in case ys of
(y:ys') | x == y -> ys'
_ -> x:ys
squeeze _ = []
Outputs
""
""
"c"
"ckm"
I don't see how foldl could be used for this. (Generally, foldl pretty much combines the disadvantages of foldr and foldl'... those, or foldMap, are the folds you should normally be using, not foldl.)
What you seem to intend is: repeating the removeAdjDups, until no duplicates are found anymore. The repetition is a job for
iterate :: (a -> a) -> a -> [a]
like
Prelude> iterate removeAdjDups "1233213443"
["1233213443","122133","11","","","","","","","","","","","","","","","","","","","","","","","","","","",""...
This is an infinite list of ever reduced lists. Generally, it will not converge to the empty list; you'll want to add some termination condition. If you want to remove as many dups as necessary, that's the fixpoint; it can be found in a very similar way to how you implemented removeAdjDups: compare neighbor elements, just this time in the list of reductions.
bipll's suggestion to handle recursive duplicates is much better though, it avoids unnecessary comparisons and traversing the start of the list over and over.
List comprehensions are often overlooked. They are, of course syntactic sugar but some, like me are addicted. First off, strings are lists as they are. This functions could handle any list, too as well as singletons and empty lists. You can us map to process many lists in a list.
(\l -> [ x | (x,y) <- zip l $ (tail l) ++ " ", x /= y]) "abcddeeffa"
"abcdefa"
I don't see either how to use foldl. It's maybe because, if you want to fold something here, you have to use foldr.
main = mapM_ (print . squeeze) ["acvvca", "1456776541", "abbac", "aabaabckllm"]
-- I like the name in #bipll answer
squeeze = foldr (\ x xs -> if xs /= "" && x == head(xs) then tail(xs) else x:xs) ""
Let's analyze this. The idea is taken from #bipll answer: go from right to left. If f is the lambda function, then by definition of foldr:
squeeze "abbac" = f('a' f('b' f('b' f('a' f('c' "")))
By definition of f, f('c' "") = 'c':"" = "c" since xs == "". Next char from the right: f('a' "c") = 'a':"c" = "ac" since 'a' != head("c") = 'c'. f('b' "ac") = "bac" for the same reason. But f('b' "bac") = tail("bac") = "ac" because 'b' == head("bac"). And so forth...
Bonus: by replacing foldr with scanr, you can see the whole process:
Prelude> squeeze' = scanr (\ x xs -> if xs /= "" && x == head(xs) then tail(xs) else x:xs) ""
Prelude> zip "abbac" (squeeze' "abbac")
[('a',"c"),('b',"ac"),('b',"bac"),('a',"ac"),('c',"c")]
I was trying to implement k-out-of-N at SML so "pick(3,[1,2,3,4])" will return [[1,2,3],[1,3,4]...] (all the K-size picks out of N elements)
I used List.map which I figured it calls the function and apply it on each element.
Really can't figure out why when typing the input "pick(3,[1,2,3,4,5])" ,for example, it return an empty list.
My first thought was that it's because of the initial terms (choose (_,[]) = [])
But changing it didn't work as well.
The signature is ok (val pick = fn : int * 'a list -> 'a list list).
fun pick (_,[]) = []
| pick (0,_) = []
| pick (n,hd::tl) =
let
val with_hd = List.map (fn x => hd::x) (pick(n-1,tl))
val without_hd = pick(n,tl)
in
with_hd#without_hd
end;
The problem is related to your suspicion – the base cases are incorrect in that they always produce the empty list, and mapping fn x => hd::x onto the empty list produces the empty list.
Picking zero elements from anything should succeed, and produce the empty list.
That is, pick (0, _) = [[]] — a list with one element, which is the empty list.
You also need to rearrange the cases since pick(n, []) succeeds for n = 0 but not for any other n.
In summary,
fun pick (0, _) = [[]]
| pick (_, []) = []
with the rest of the function exactly as before.
I have seen some similar questions, but nothing that really helped me. Basically the title says it all. Using SML I want to take a string that I have, and make a list containing each letter found in the string. Any help would be greatly appreciated.
One possibility is to use the basic logic of quicksort to sort the letters while removing duplicates at the same time. Something like:
fun distinctChars []:char list = []
| distinctChars (c::cs) =
let val smaller = List.filter (fn x => x < c) cs
val bigger = List.filter (fn x => x > c) cs
in distinctChars smaller # [c] # distinctChars bigger
end
If the < and > in the definitions of smaller and bigger were to be replaced by <= and >= then it would simply be an implementation of quicksort (although not the most efficient one since it makes two passes over cs when a suitably defined auxiliary function could split into smaller and bigger in just one pass). The strict inequalities have the effect of throwing away duplicates.
To get what you want from here, do something like explode the string into a list of chars, remove non-alphabetical characters from the resulting list, while simultaneously converting to lower case, then invoke the above function -- ideally first refined so that it uses a custom split function rather than List.filter twice.
On Edit: # is an expensive operator and probably results in the naïve SML quicksort not being all that quick. You can use the above idea of a modified sort, but one that modifies mergesort instead of quicksort:
fun split ls =
let fun split' [] (xs,ys) = (xs,ys)
| split' (a::[]) (xs, ys) = (a::xs,ys)
| split' (a::b::cs) (xs, ys) = split' cs (a::xs, b::ys)
in split' ls ([],[])
end
fun mergeDistinct ([], ys) = ys:char list
| mergeDistinct (xs, []) = xs
| mergeDistinct (x::xs, y::ys) =
if x < y then x::mergeDistinct(xs,y::ys)
else if x > y then y::mergeDistinct(x::xs,ys)
else mergeDistinct(x::xs, ys)
fun distinctChars [] = []
| distinctChars [c] = [c]
| distinctChars chars =
let val (xs,ys) = split chars
in mergeDistinct (distinctChars xs, distinctChars ys)
end
You can get a list of all the letters in a few different ways:
val letters = [#"a",#"b",#"c",#"d",#"e",#"f",#"g",#"h",#"i",#"j",#"k",#"l",#"m",#"n",#"o",#"p",#"q",#"r",#"s",#"t",#"u",#"v",#"w",#"x",#"y",#"z"]
val letters = explode "abcdefghijklmnopqrstuvwxyz"
val letters = List.tabulate (26, fn i => chr (i + ord #"a"))
Update: Looking at your question and John's answer, I might have misunderstood your intention. An efficient way to iterate over a string and gather some result (e.g. a set of characters) could be to write a "foldr for strings":
fun string_foldr f acc0 s =
let val len = size s
fun loop i acc = if i < len then loop (i+1) (f (String.sub (s, i), acc)) else acc
in loop 0 acc0 end
Given an implementation of sets with at least setEmpty and setInsert, one could then write:
val setLetters = string_foldr (fn (c, ls) => setInsert ls c) setEmpty "some sentence"
The simplest solution I can think of:
To get the distinct elements of a list:
Take the head
Remove that value from the tail and get the distinct elements of the result.
Put 1 and 2 together.
In code:
(* Return the distinct elements of a list *)
fun distinct [] = []
| distinct (x::xs) = x :: distinct (List.filter (fn c => x <> c) xs);
(* All the distinct letters, in lower case. *)
fun letters s = distinct (List.map Char.toLower (List.filter Char.isAlpha (explode s)));
(* Variation: "point-free" style *)
val letters' = distinct o (List.map Char.toLower) o (List.filter Char.isAlpha) o explode;
This is probably not the most efficient solution, but it's uncomplicated.
Consider the following code I wrote:
import Control.Monad
increasing :: Integer -> [Integer]
increasing n
| n == 1 = [1..9]
| otherwise = do let ps = increasing (n - 1)
let last = liftM2 mod ps [10]
let next = liftM2 (*) ps [10]
alternateEndings next last
where alternateEndings xs ys = concat $ zipWith alts xs ys
alts x y = liftM2 (+) [x] [y..9]
Where 'increasing n' should return a list of n-digit numbers whose numbers increase (or stay the same) from left-to-right.
Is there a way to simplify this? The use of 'let' and 'liftM2' everywhere looks ugly to me. I think I'm missing something vital about the list monad, but I can't seem to get rid of them.
Well, as far as liftM functions go, my preferred way to use those is the combinators defined in Control.Applicative. Using those, you'd be able to write last = mod <$> ps <*> [10]. The ap function from Control.Monad does the same thing, but I prefer the infix version.
What (<$>) and (<*>) goes like this: liftM2 turns a function a -> b -> c into a function m a -> m b -> m c. Plain liftM is just (a -> b) -> (m a -> m b), which is the same as fmap and also (<$>).
What happens if you do that to a multi-argument function? It turns something like a -> b -> c -> d into m a -> m (b -> c -> d). This is where ap or (<*>) come in: what they do is turn something like m (a -> b) into m a -> m b. So you can keep stringing it along that way for as many arguments as you like.
That said, Travis Brown is correct that, in this case, it seems you don't really need any of the above. In fact, you can simplify your function a great deal: For instance, both last and next can be written as single-argument functions mapped over the same list, ps, and zipWith is the same as a zip and a map. All of these maps can be combined and pushed down into the alts function. This makes alts a single-argument function, eliminating the zip as well. Finally, the concat can be combined with the map as concatMap or, if preferred, (>>=). Here's what it ends up:
increasing' :: Integer -> [Integer]
increasing' 1 = [1..9]
increasing' n = increasing' (n - 1) >>= alts
where alts x = map ((x * 10) +) [mod x 10..9]
Note that all refactoring I did to get to that version from yours was purely syntactic, only applying transformations that should have no impact on the result of the function. Equational reasoning and referential transparency are nice!
I think what you are trying to do is this:
increasing :: Integer -> [Integer]
increasing 1 = [1..9]
increasing n = do p <- increasing (n - 1)
let last = p `mod` 10
next = p * 10
alt <- [last .. 9]
return $ next + alt
Or, using a "list comprehension", which is just special monad syntax for lists:
increasing2 :: Integer -> [Integer]
increasing2 1 = [1..9]
increasing2 n = [next + alt | p <- increasing (n - 1),
let last = p `mod` 10
next = p * 10,
alt <- [last .. 9]
]
The idea in the list monad is that you use "bind" (<-) to iterate over a list of values, and let to compute a single value based on what you have so far in the current iteration. When you use bind a second time, the iterations are nested from that point on.
It looks very unusual to me to use liftM2 (or <$> and <*>) when one of the arguments is always a singleton list. Why not just use map? The following does the same thing as your code:
increasing :: Integer -> [Integer]
increasing n
| n == 1 = [1..9]
| otherwise = do let ps = increasing (n - 1)
let last = map (flip mod 10) ps
let next = map (10 *) ps
alternateEndings next last
where alternateEndings xs ys = concat $ zipWith alts xs ys
alts x y = map (x +) [y..9]
Here's how I'd write your code:
increasing :: Integer -> [Integer]
increasing 1 = [1..9]
increasing n = let allEndings x = map (10*x +) [x `mod` 10 .. 9]
in concatMap allEndings $ increasing (n - 1)
I arrived at this code as follows. The first thing I did was to use pattern matching instead of guards, since it's clearer here. The next thing I did was to eliminate the liftM2s. They're unnecessary here, because they're always called with one size-one list; in that case, it's the same as calling map. So liftM2 (*) ps [10] is just map (* 10) ps, and similarly for the other call sites. If you want a general replacement for liftM2, though, you can use Control.Applicative's <$> (which is just fmap) and <*> to replace liftMn for any n: liftMn f a b c ... z becomes f <$> a <*> b <*> c <*> ... <*> z. Whether or not it's nicer is a matter of taste; I happen to like it.1 But here, we can eliminate that entirely.
The next place I simplified the original code is the do .... You never actually take advantage of the fact that you're in a do-block, and so that code can become
let ps = increasing (n - 1)
last = map (`mod` 10) ps
next = map (* 10) ps
in alternateEndings next last
From here, arriving at my code essentially involved writing fusing all of your maps together. One of the only remaining calls that wasn't a map was zipWith. But because you effectively have zipWith alts next last, you only work with 10*p and p `mod` 10 at the same time, so we can calculate them in the same function. This leads to
let ps = increasing (n - 1)
in concat $ map alts ps
where alts p = map (10*p +) [y `mod` 10..9]
And this is basically my code: concat $ map ... should always become concatMap (which, incidentally, is =<< in the list monad), we only use ps once so we can fold it in, and I prefer let to where.
1: Technically, this only works for Applicatives, so if you happen to be using a monad which hasn't been made one, <$> is `liftM` and <*> is `ap`. All monads can be made applicative functors, though, and many of them have been.
I think it's cleaner to pass last digit in a separate parameter and use lists.
f a 0 = [[]]
f a n = do x <- [a..9]
k <- f x (n-1)
return (x:k)
num = foldl (\x y -> 10*x + y) 0
increasing = map num . f 1