This is the code I have to make a palindrome function. I already created the listReverse and explode function before that I use to make the palindrome. Can someone help me finnish the palindrome function?
let rec listReverse l = match l with
|[] -> []
|head :: tail -> (listReverse tail) # [head]
(* explode : string -> char list
* (explode s) is the list of characters in the string s in the order in
* which they appear
* e.g. (explode "Hello") is ['H';'e';'l';'l';'o']
*)
let explode s =
let rec _exp i =
if i >= String.length s then [] else (s.[i])::(_exp (i+1)) in
_exp 0
let rec palindrome w =
let a = explode w in
let b = listReverse a in
if c :: d
else false
You should use the List.rev standard function to reverse lists. Ocaml being a free software, you should look at its implementation (file stdlib/list.ml)
Try to explain in plain English (not code) what you are trying to achieve when you write
if c :: d
else false
Also, note that
if foo = bar then true else false
should be simplified to
foo = bar
You can replace your if statement with this:
(* tells wheter its a palindrome or not; most is (List.length a)/2*)
let rec same l1 l2 cur most =
match l1, l2 with
| h1::t1, h2::t2 when h1 = h2 ->
if cur < most then same t1 t2 (cur+1) most
else true
| _ -> false in
same a b 0 ((List.length a)/2)
Related
I'm a beginner in OCaml and algorithms.
I'm trying to get the number of 5 digits numbers with no repeating digits bigger than 12345.
Here is what I did in OCaml, I tried to make as tail recursive as possible, and I also used streams. But still, due to size, it stack overflowed:
type 'a stream = Eos | StrCons of 'a * (unit -> 'a stream)
let rec numberfrom n= StrCons (n, fun ()-> numberfrom (n+1))
let nats = numberfrom 1
let rec listify st n f=
match st with
|Eos ->f []
|StrCons (m, a) ->if n=1 then f [m] else listify (a ()) (n-1) (fun y -> f (m::y))
let rec filter (test: 'a-> bool) (s: 'a stream) : 'a stream=
match s with
|Eos -> Eos
|StrCons(q,w) -> if test q then StrCons(q, fun ()->filter test (w ()))
else filter test (w ())
let rec check_dup l=
match l with
| [] -> false
| h::t->
let x = (List.filter (fun x -> x = h) t) in
if (x == []) then
check_dup t
else
true;;
let digits2 d =
let rec dig acc d =
if d < 10 then d::acc
else dig ((d mod 10)::acc) (d/10) in
dig [] d
let size a=
let rec helper n aa=
match aa with
|Eos-> n
|StrCons (q,w) -> helper (n+1) (w())
in helper 0 a
let result1 = filter (fun x -> x<99999 && x>=12345 && (not (check_dup (digits2 x)))) nats
(* unterminating : size result1 *)
(*StackOverflow: listify result1 10000 (fun x->x) *)
I can't reproduce your reported problem. When I load up your code I see this:
# List.length (listify result1 10000 (fun x -> x));;
- : int = 10000
# List.length (listify result1 26831 (fun x -> x));;
- : int = 26831
It's possible your system is more resource constrained than mine.
Let me just say that the usual way to code a tail recursive function is to build the list up in reverse, then reverse it at the end. That might look something like this:
let listify2 st n =
let rec ilist accum st k =
match st with
| Eos -> List.rev accum
| StrCons (m, a) ->
if k = 1 then List.rev (m :: accum)
else ilist (m :: accum) (a ()) (k - 1)
in
if n = 0 then []
else ilist [] st n
You still have the problem that listify doesn't terminate if you ask for more elements than there are in the stream. It might be better to introduce a method to detect the end of the stream and return Eos at that point. For example, the filter function might accept a function that returns three possible values (the element should be filtered out, the element should not be filtered out, the stream should end).
The problem is that the size of your stream result1 is undefined.
Indeed, nats is an never-ending stream: it never returns Eos.
However, filtering a never-ending stream results in another never-ending stream
since a filtered stream only returns Eos after the underlying stream does so:
let rec filter (test: 'a-> bool) (s: 'a stream) : 'a stream=
match s with
| Eos -> Eos
| StrCons(q,w) -> if test q then StrCons(q, fun ()->filter test (w ()))
else filter test (w ())
Consequently, size result1 is stuck trying to reach the end of integers.
Note also that, in recent version of the standard library, your type stream is called Seq.node.
let rec isolate (l:'a list) =
match l with
| [] -> []
| x::xs ->
if memberof(x,xs)
then remove (x,l)
else isolate xs
I've already created functions memberof and remove, the only problem is that when line 6 remove(x,l) executes it doesn't continue with isolate(xs) for continued search through the list.
Is there a way to say,
if x then f(x) and f(y)
?
As you are using F# immutable lists, the result of remove needs to be stored somewhere:
let rec isolate (l:'a list) =
match l with
| [] -> []
| x::xs ->
if memberof(x,xs)
then
let xs = remove (x,l)
isolate xs
else isolate xs
To answer your more general question:
let f _ = ()
let f' z = z
let x = true
let y = 42
let z = 3.141
if x then
f y
f' z |> ignore
The ignore is needed here because in F# there are no statements, just expressions, so you can think of if x then f' z as
if x then
f' z
else
()
and thus the first branch needs to return () as well.
In addition to CaringDev's answer.
You may look at this simple solution.
It is worth note, that it's not a fastest way to do this.
let rec isolate (acc : 'a list) (l : 'a list) =
match l with
| [] -> acc
| head :: tail ->
if memberof (head, tail)
then remove (head, tail) |> isolate (acc # [head])
else isolate (acc # [head]) tail
let recursiveDistinct = isolate []
let uniqValues = recursiveDistinct [ 1; 1; 2; 3] //returns [1;2;3]
let isolate list =
let rec isolateInner searchList commonlist =
match searchList with
| x::xs ->
if (memberof commonlist x) then
isolateInner xs commonlist
else
let commonlist = (x :: commonlist)
isolateInner xs commonlist
| [] -> reverse commonlist
isolateInner list []
This is part of an answer to your larger problem.
Notice that this does not use remove. Since you have to pass over each item in the original list and list are immutable, it is better to create a new list and only add the unique items to the new list, then return the new list.
the code :
open Hashtbl;;
type 'a option = None | Some of 'a;;
let ht = create 0;;
let rec charCount fd =
let x =
try Some (input_char fd)
with End_of_file -> None
in
match x with
| Some c ->
let v =
try find ht c
with Not_found -> 0
in
replace ht c (v+1);
charCount fd
| None -> ();;
let loadHisto fn =
let fd = open_in fn in
charCount fd;;
let rec printList l = match l with
| [] -> print_newline ()
| h::t -> print_char h; print_string " "; printList t;;
let hashtbl_keys h = Hashtbl.fold (fun key _ l -> key :: l) h [];;
let compare_function a b = compare (find ht b) (find ht a);;
let akeys = List.sort compare_function (hashtbl_keys ht);;
printList (List.sort compare_function (hashtbl_keys ht));;
printList akeys;;
and the result :
ocaml histo.ml
e t s u a i n p j (*here is the first printList*)
(*and here should be the second one, but there is only a blank*)
Here is the problem :
I sorted a list, and tried to display it's content but as you can see, it doesn't seem like I can give a name to my resulting List
Edit :
I don't think it's a buffer problem because even if I only print akeys, there is nothing
Edit : I added the code asked for below
And ht is a hashtbl, and it contains what it should contain (I checked this)
What would be the syntax (if possible at all) for returning the list of lists ([[a]]) but without the use of empty list ([]:[a])?
(similar as the second commented guard (2) below, which is incorrect)
This is a function that works correctly:
-- Split string on every (shouldSplit == true)
splitWith :: (Char -> Bool) -> [Char] -> [[Char]]
splitWith shouldSplit list = filter (not.null) -- would like to get rid of filter
(imp' shouldSplit list)
where
imp' _ [] = [[]]
imp' shouldSplit (x:xs)
| shouldSplit x = []:imp' shouldSplit xs -- (1) this line is adding empty lists
-- | shouldSplit x = [imp' shouldSplit xs] -- (2) if this would be correct, no filter needed
| otherwise = let (z:zs) = imp' shouldSplit xs in (x:z):zs
This is the correct result
Prelude> splitWith (== 'a') "miraaaakojajeja234"
["mir","koj","jej","234"]
However, it must use "filter" to clean up its result, so I would like to get rid of function "filter".
This is the result without the use of filter:
["mir","","","","koj","jej","234"]
If "| shouldSplit x = imp' shouldSplit xs" is used instead the first guard, the result is incorrect:
["mirkojjej234"]
The first guard (1) adds empty list so (I assume) compiler can treat the result as a list of lists ([[a]]).
(I'm not interested in another/different solutions of the function, just the syntax clarification.)
.
.
.
ANSWER:
Answer from Dave4420 led me to the answer, but it was a comment, not an answer so I can't accept it as answer. The solution of the problem was that I'm asking the wrong question. It is not the problem of syntax, but of my algorithm.
There are several answers with another/different solutions that solve the empty list problem, but they are not the answer to my question. However, they expanded my view of ways on how things can be done with basic Haskell syntax, and I thank them for it.
Edit:
splitWith :: (Char -> Bool) -> String -> [String]
splitWith p = go False
where
go _ [] = [[]]
go lastEmpty (x:xs)
| p x = if lastEmpty then go True xs else []:go True xs
| otherwise = let (z:zs) = go False xs in (x:z):zs
This one utilizes pattern matching to complete the task of not producing empty interleaving lists in a single traversal:
splitWith :: Eq a => (a -> Bool) -> [a] -> [[a]]
splitWith f list = case splitWith' f list of
[]:result -> result
result -> result
where
splitWith' _ [] = []
splitWith' f (a:[]) = if f a then [] else [[a]]
splitWith' f (a:b:tail) =
let next = splitWith' f (b : tail)
in if f a
then if a == b
then next
else [] : next
else case next of
[] -> [[a]]
nextHead:nextTail -> (a : nextHead) : nextTail
Running it:
main = do
print $ splitWith (== 'a') "miraaaakojajeja234"
print $ splitWith (== 'a') "mirrraaaakkkojjjajeja234"
print $ splitWith (== 'a') "aaabbbaaa"
Produces:
["mir","koj","jej","234"]
["mirrr","kkkojjj","jej","234"]
["bbb"]
The problem is quite naturally expressed as a fold over the list you're splitting. You need to keep track of two pieces of state - the result list, and the current word that is being built up to append to the result list.
I'd probably write a naive version something like this:
splitWith p xs = word:result
where
(result, word) = foldr func ([], []) xs
func x (result, word) = if p x
then (word:result,[])
else (result, x:word)
Note that this also leaves in the empty lists, because it appends the current word to the result whenever it detects a new element that satisfies the predicate p.
To fix that, just replace the list cons operator (:) with a new operator
(~:) :: [a] -> [[a]] -> [[a]]
that only conses one list to another if the original list is non-empty. The rest of the algorithm is unchanged.
splitWith p xs = word ~: result
where
(result, word) = foldr func ([], []) xs
func x (result, word) = if p x
then (word ~: result, [])
else (result, x:word)
x ~: xs = if null x then xs else x:xs
which does what you want.
I guess I had a similar idea to Chris, I think, even if not as elegant:
splitWith shouldSplit list = imp' list [] []
where
imp' [] accum result = result ++ if null accum then [] else [accum]
imp' (x:xs) accum result
| shouldSplit x =
imp' xs [] (result ++ if null accum
then []
else [accum])
| otherwise = imp' xs (accum ++ [x]) result
This is basically just an alternating application of dropWhile and break, isn't it:
splitWith p xs = g xs
where
g xs = let (a,b) = break p (dropWhile p xs)
in if null a then [] else a : g b
You say you aren't interested in other solutions than yours, but other readers might be. It sure is short and seems clear. As you learn, using basic Prelude functions becomes second nature. :)
As to your code, a little bit reworked in non-essential ways (using short suggestive function names, like p for "predicate" and g for a main worker function), it is
splitWith :: (Char -> Bool) -> [Char] -> [[Char]]
splitWith p list = filter (not.null) (g list)
where
g [] = [[]]
g (x:xs)
| p x = [] : g xs
| otherwise = let (z:zs) = g xs
in (x:z):zs
Also, there's no need to pass the predicate as an argument to the worker (as was also mentioned in the comments). Now it is arguably a bit more readable.
Next, with a minimal change it becomes
splitWith :: (Char -> Bool) -> [Char] -> [[Char]]
splitWith p list = case g list of ([]:r)-> r; x->x
where
g [] = [[]]
g (x:xs)
| p x = case z of []-> r; -- start a new word IF not already
_ -> []:r
| otherwise = (x:z):zs
where -- now z,zs are accessible
r#(z:zs) = g xs -- in both cases
which works as you wanted. The top-level case is removing at most one empty word here, which serves as a separator marker at some point during the inner function's work. Your filter (not.null) is essentially fused into the worker function g here, with the conditional opening1 of a new word (i.e. addition1 of an empty list).
Replacing your let with where allowed for the variables (z etc.) to became accessible in both branches of the second clause of the g definition.
In the end, your algorithm was close enough, and the code could be fixed after all.
1 when thinking "right-to-left". In reality the list is constructed left-to-right, in guarded recursion ⁄ tail recursion modulo cons fashion.
I want to write a function that taking a string and return a list of char. Here is a function, but I think it is not do what I want ( I want to take a string and return a list of characters).
let rec string_to_char_list s =
match s with
| "" -> []
| n -> string_to_char_list n
Aside, but very important:
Your code is obviously wrong because you have a recursive call for which all the parameters are the exact same one you got in. It is going to induce an infinite sequence of calls with the same values in, thus looping forever (a stack overflow won't happen in tail-rec position).
The code that does what you want would be:
let explode s =
let rec exp i l =
if i < 0 then l else exp (i - 1) (s.[i] :: l) in
exp (String.length s - 1) []
Source:
http://caml.inria.fr/pub/old_caml_site/FAQ/FAQ_EXPERT-eng.html#strings
Alternatively, you can choose to use a library: batteries String.to_list or extlib String.explode
Try this:
let explode s = List.init (String.length s) (String.get s)
Nice and simple:
let rec list_car ch =
match ch with
| "" -> []
| ch -> String.get ch 0 :: list_car (String.sub ch 1 (String.length ch - 1));;
How about something like this:
let string_to_list str =
let rec loop i limit =
if i = limit then []
else (String.get str i) :: (loop (i + 1) limit)
in
loop 0 (String.length str);;
let list_to_string s =
let rec loop s n =
match s with
[] -> String.make n '?'
| car :: cdr ->
let result = loop cdr (n + 1) in
String.set result n car;
result
in
loop s 0;;
As of OCaml 4.07 (released 2018), this can be straightforwardly accomplished with sequences.
let string_to_char_list s =
s |> String.to_seq |> List.of_seq
Here is an Iterative version to get a char list from a string:
let string_to_list s =
let l = ref [] in
for i = 0 to String.length s - 1 do
l := (!l) # [s.[i]]
done;
!l;;
My code, suitable for modern OCaml:
let charlist_of_string s =
let rec trav l i =
if i = l then [] else s.[i]::trav l (i+1)
in
trav (String.length s) 0;;
let rec string_of_charlist l =
match l with
[] -> ""
| h::t -> String.make 1 h ^ string_of_charlist t;;