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I am trying to write a program in sml that takes in the length of a list, the max number that will appear on the list and the list of course. It then calculates the length of the smallest "sub-list" that contains all numbers.
I have tried to use the sliding window approach , with two indexes , front and tail. The front scans first and when it finds a number it writes into a map how many times it has already seen this number. If the program finds all numbers then it calls the tail. The tail scans the list and if it finds that a number has been seen more times than 1 it takes it off.
The code I have tried so far is the following:
structure Key=
struct
type ord_key=int
val compare=Int.compare
end
fun min x y = if x>y then y else x;
structure mymap = BinaryMapFn ( Key );
fun smallest_sub(n,t,listall,map)=
let
val k=0
val front=0
val tail=0
val minimum= n;
val list1=listall;
val list2=listall;
fun increase(list1,front,k,ourmap)=
let
val number= hd list1
val elem=mymap.find(ourmap,number)
val per=getOpt(elem,0)+1
fun decrease(list2,tail,k,ourmap,minimum)=
let
val number=hd list2
val elem=mymap.find(ourmap,number)
val per=getOpt(elem,0)-1
val per1=getOpt(elem,0)
in
if k>t then
if (per1=1) then decrease(tl list2,tail+1,k-1,mymap.insert(ourmap,number,per),min minimum (front-tail))
else decrease(tl list2,tail+1,k,mymap.insert(ourmap,number,per),min minimum (front-tail))
else increase (list1, front,k,ourmap)
end
in
if t>k then
if (elem<>NONE) then increase (tl list1,front+1,k,mymap.insert(ourmap,number,per))
else increase(tl list1,front+1,k+1,mymap.insert(ourmap,number,per))
else (if (n>front) then decrease(list2,tail,k,ourmap,minimum) else minimum)
end
in
increase(list1,front,k,map)
end
fun solve (n,t,acc)= smallest_sub(n,t,acc,mymap.empty)
But when I call it with this smallest_sub(10,3,[1,3,1,3,1,3,3,2,2,1]); it does not work. What have I done wrong??
Example: if input is 1,3,1,3,1,3,3,2,2,1 the program should recognize that the parto of the list that contains all numbers and is the smallest is 1,3,3,2 and 3,2,2,1 so the output should be 4
This problem of "smallest sub-list that contains all values" seems to recur in
new questions without a successful answer. This is because it's not a minimal,
complete, and verifiable example.
Because you use a "sliding window" approach, indexing the front and the back
of your input, a list taking O(n) time to index elements is not ideal. You
really do want to use arrays here. If your input function must have a list, you
can convert it to an array for the purpose of the algorithm.
I'd like to perform a cleanup of the code before answering, because running
your current code by hand is a bit hard because it's so condensed. Here's an
example of how you could abstract out the book-keeping of whether a given
sub-list contains at least one copy of each element in the original list:
Edit: I changed the code below after originally posting it.
structure CountMap = struct
structure IntMap = BinaryMapFn(struct
type ord_key = int
val compare = Int.compare
end)
fun count (m, x) =
Option.getOpt (IntMap.find (m, x), 0)
fun increment (m, x) =
IntMap.insert (m, x, count (m, x) + 1)
fun decrement (m, x) =
let val c' = count (m, x)
in if c' <= 1
then NONE
else SOME (IntMap.insert (m, x, c' - 1))
end
fun flip f (x, y) = f (y, x)
val fromList = List.foldl (flip increment) IntMap.empty
end
That is, a CountMap is an int IntMap.map where the Int represents the
fixed key type of the map, being int, and the int parameter in front of it
represents the value type of the map, being a count of how many times this
value occurred.
When building the initialCountMap below, you use CountMap.increment, and
when you use the "sliding window" approach, you use CountMap.decrement to
produce a new countMap that you can test on recursively.
If you decrement the occurrence below 1, you're looking at a sub-list that
doesn't contain every element at least once; we rule out any solution by
letting CountMap.decrement return NONE.
With all of this machinery abstracted out, the algorithm itself becomes much
easier to express. First, I'd like to convert the list to an array so that
indexing becomes O(1), because we'll be doing a lot of indexing.
fun smallest_sublist_length [] = 0
| smallest_sublist_length (xs : int list) =
let val arr = Array.fromList xs
val initialCountMap = CountMap.fromList xs
fun go countMap i j =
let val xi = Array.sub (arr, i)
val xj = Array.sub (arr, j)
val decrementLeft = CountMap.decrement (countMap, xi)
val decrementRight = CountMap.decrement (countMap, xj)
in
case (decrementLeft, decrementRight) of
(SOME leftCountMap, SOME rightCountMap) =>
Int.min (
go leftCountMap (i+1) j,
go rightCountMap i (j-1)
)
| (SOME leftCountMap, NONE) => go leftCountMap (i+1) j
| (NONE, SOME rightCountMap) => go rightCountMap i (j-1)
| (NONE, NONE) => j - i + 1
end
in
go initialCountMap 0 (Array.length arr - 1)
end
This appears to work, but...
Doing Int.min (go left..., go right...) incurs a cost of O(n^2) stack
memory (in the case where you cannot rule out either being optimal). This is a
good use-case for dynamic programming because your recursive sub-problems have a
common sub-structure, i.e.
go initialCountMap 0 10
|- go leftCountMap 1 10
| |- ...
| `- go rightCountMap 1 9 <-.
`- go rightCountMap 0 9 | possibly same sub-problem!
|- go leftCountMap 1 9 <-'
`- ...
So maybe there's a way to store the recursive sub-problem in a memory array and not
perform a recursive lookup if you know the result to this sub-problem. How to
do memoization in SML is a good question in and of itself. How to do purely
functional memoization in a non-lazy language is an even better one.
Another optimization you could make is that if you ever find a sub-list the
size of the number of unique elements, you need to look no further. This number
is incidentally the number of elements in initialCountMap, and IntMap
probably has a function for finding it.
I have a problem with writing a function which result is as shown below:
split([7;1;4;3;6;8;2], 4) = ([1;3;2;4], [7;6;8])
my current code i managed to write:
let split(list, number)=
let split1(list, number, lesser, greater)=
if list = [] then lesser::greater
else if List.hd list <= element then (List.hd list)::(lesser)
else (List.hd list)::(greater)
in
(List.tl lista, element, [], []);;
Thanks in advance for your help.
For the future, it helps to be more specific when asking a question on SO. What exactly is your problem? SO users will be skeptical of someone who wants others to help them, but won't help themselves.
Your code has nearly the correct structure, but there are a few errors in there that seem to be getting in your way.
Firstly lesser::greater is wrong, since the left hand side of a cons operator must be a list itself, but what you really want is a list where both of these are elements. So instead try [lesser;greater].
Secondly, if you think through your code, you will notice that it suddenly stops. You checked the first element, but you didn't look at the rest of the list. Since you want to keep splitting the list, you need your code to keep executing till the end of the list. To achieve this, we use recursion. Recursion mean that your function split1 will call itself again. It can be very confusing the first time you see it - each time split1 runs it will take the first element off, and then split the remainder of the list.
What does (List.hd list)::(lesser) actually mean? The lesser here really means all of the lesser elements in the rest of the list. You need to keep taking an element out of the list and putting it in either lesser or greater.
Finally avoid using List.hd excessively - it is neater to find the head and tail using pattern matching.
Here's a working version of the code:
let split(list, number)=
let rec split1(list, number, lesser, greater)=
match list with
| [] -> [List.rev lesser;List.rev greater]
| head::tail ->
match (head <= number) with
true -> split1(tail,number,head::lesser,greater)
| false -> split1(tail,number,lesser,head::greater)
in split1(list, number, [], []);;
split([1;2;3;4;5],3);;
The split1 function takes the elements off one at a time, and adds them to the lists.
Maybe my comments on the following code snippet would help:
let split list num =
let rec loop lesser greater list =
match list with
| [] -> (lesser, greater)
(* when your initial list is empty, you have to return the accumulators *)
| x :: xs ->
if x <= num then
(* x is lesser than num, so add x in the lesser list and
call loop on the tail of the list (xs) *)
else
(* x is greater than num, so add x in the greater list and
call loop on the tail of the list (xs) *)
in
(* here you make the first call to loop by initializing
your accumulators with empty list*)
loop [] [] list
I want to make a program insertAt where z is the place in the list, and y is the number being inserted into the list xs. Im new to haskell and this is what I have so far.
insertAt :: Int-> Int-> [Int]-> [Int]
insertAt z y xs
| z==1 = y:xs
but I'm not sure where to go from there.
I have an elementAt function, where
elementAt v xs
| v==1 = head xs
| otherwise = elementAt (v-1) (tail xs)
but I'm not sure how I can fit it in or if I even need to. If possible, I'd like to avoid append.
If this isn't homework: let (ys,zs) = splitAt n xs in ys ++ [new_element] ++ zs
For the rest of this post I'm going to assume you're doing this problem as homework or to teach yourself how to do this kind of thing.
The key to this kind of problem is to break it down into its natural cases. You're processing two pieces of data: the list you're inserting into, and the position in that list. In this case, each piece of data has two natural cases: the list you're procssing can be empty or not, and the number you're processing can be zero or not. So the first step is to write out all four cases:
insertAt 0 val [] = ...
insertAt 0 val (x:xs) = ...
insertAt n val [] = ...
insertAt n val (x:xs) = ...
Now, for each of these four cases, you need to think about what the answer should be given that you're in that case.
For the first two cases, the answer is easy: if you want to insert into the front of a list, just stick the value you're interested in at the beginning, whether the list is empty or not.
The third case demonstrates that there's actually an ambiguity in the question: what happens if you're asked to insert into, say, the third position of a list that's empty? Sounds like an error to me, but you'll have to answer what you want to do in that case for yourself.
The fourth case is most interesting: Suppose you want to insert a value into not-the-first position of a list that's not empty. In this case, remember that you can use recursion to solve smaller instances of your problem. In this case, you can use recursion to solve, for instance, insertAt (n-1) val xs -- that is, the result of inserting your same value into the tail of your input list at the n-1th position. For example, if you were trying to insert 5 into position 3 (the fourth position) of the list [100,200,300], you can use recursion to insert 5 into position 2 (the third position) of the list [200,300], which means the recursive call would produce [200,300,5].
We can just assume that the recursive call will work; our only job now is to convert the answer to that smaller problem into the answer to the original problem we were given. The answer we want in the example is [100,200,300,5] (the result of inserting 5 into position 4 of the list [100,200,300], and what we have is the list [200,300,5]. So how can we get the result we want? Just add back on the first element! (Think about why this is true.)
With that case finished, we've covered all the possible cases for combinations of lists and positions to update. Since our function will work correctly for all possibilities, and our possibilities cover all possible inputs, that means our function will always work correctly. So we're done!
I'll leave it to you to translate these ideas into Haskell since the point of the exercise is for you to learn it, but hopefully that lets you know how to solve the problem.
You could split the list at index z and then concatenate the first part of the list with the element (using ++ [y]) and then with the second part of the list. However, this would create a new list as data is immutable by default. The first element of the list by convention has the index 0 (so adjust z accordingly if you want the meaning of fist elemnt is indexed by 1).
insertAt :: Int -> Int-> [Int] -> [Int]
insertAt z y xs = as ++ (y:bs)
where (as,bs) = splitAt z xs
While above answers are correct, I think this is more concise:
insertAt :: Int -> Int-> [Int]-> [Int]
insertAt z y xs = (take z xs) ++ y:(drop z xs)
I'm new to SML and I'm attempting to get the index of an item in a list. I know that using List.nth will give me the value of an item at a index position, but I want the index value. There may even be a built in function that I'm not aware of. In my case, the list will not contain duplicates so if the item is in the list I get the index, if not it returns ~1. Here is the code I have so far. It works, but I don't think it is very clean:
val L=[1,2,3,4,5];
val m=length L-1;
fun Index(item, m, L)=if m<0 then ~1 else
if List.nth(L, m)=item then m else Index(item,m-1,L);
To elaborate on my previous comment, I suggest some changes for an implementation that fits better in the ML idiom:
fun index(item, xs) =
let
fun index'(m, nil) = NONE
| index'(m, x::xr) = if x = item then SOME m else index'(m + 1, xr)
in
index'(0, xs)
end
The individual changes are:
Have index return a value of type int option. NONE means the item is not in the list, SOME i means it is in the list, and the index of its first occurrence is i. This way, no special values (~1) need be used and the function's intended usage can be inferred from its type.
Hide the parameter m by renaming the function to index' and wrapping it into an outer function index that calls it with the appropriate arguments. The prime character (`) often indicates auxiliary values.
Use pattern matching on the list to get to the individual elements, eliminating the need for List.nth.
Also note that most commonly, function and variable names begin with a lowercase letter (index rather than Index), while capital letters are used for constructor constants (SOME) and the like.
I would like to propose a simpler and less efficient version of this index function. I agree that it is not as desirable to use exceptions rather than int option, and that it is not tail-recursive. But it is certainly easier to read and thus may serve as learning material:
fun index (x, []) = raise Subscript
| index (x, y::ys) =
if x = y then 0 else 1 + index (x, ys)
fun index(list,n)=
= if n=0 then hd(list) else index(tl(list),n-1);
val index = fn : 'a list * int -> 'a
index([1,2,3,4,5],2);
val it = 3 : int
index([1,2,3,4,5],0);
val it = 1 : int
I'm really new to F#, and I need a bit of help with an F# problem.
I need to implement a cut function that splits a list in half so that the output would be...
cut [1;2;3;4;5;6];;
val it : int list * int list = ([1; 2; 3], [4; 5; 6])
I can assume that the length of the list is even.
I'm also expected to define an auxiliary function gencut(n, xs) that cuts xs into two pieces, where n gives the size of the first piece:
gencut(2, [1;3;4;2;7;0;9]);;
val it : int list * int list = ([1; 3], [4; 2; 7; 0; 9])
I wouldn't normally ask for exercise help here, but I'm really at a loss as to where to even start. Any help, even if it's just a nudge in the right direction, would help.
Thanks!
Since your list has an even length, and you're cutting it cleanly in half, I recommend the following (psuedocode first):
Start with two pointers: slow and fast.
slow steps through the list one element at a time, fast steps two elements at a time.
slow adds each element to an accumulator variable, while fast moves foward.
When the fast pointer reaches the end of the list, the slow pointer will have only stepped half the number of elements, so its in the middle of the array.
Return the elements slow stepped over + the elements remaining. This should be two lists cut neatly in half.
The process above requires one traversal over the list and runs in O(n) time.
Since this is homework, I won't give a complete answer, but just to get you partway started, here's what it takes to cut the list cleanly in half:
let cut l =
let rec cut = function
| xs, ([] | [_]) -> xs
| [], _ -> []
| x::xs, y::y'::ys -> cut (xs, ys)
cut (l, l)
Note x::xs steps 1 element, y::y'::ys steps two.
This function returns the second half of the list. It is very easy to modify it so it returns the first half of the list as well.
You are looking for list slicing in F#. There was a great answer by #Juliet in this SO Thread: Slice like functionality from a List in F#
Basically it comes down to - this is not built in since there is no constant time index access in F# lists, but you can work around this as detailed. Her approach applied to your problem would yield a (not so efficient but working) solution:
let gencut(n, list) =
let firstList = list |> Seq.take n |> Seq.toList
let secondList = list |> Seq.skip n |> Seq.toList
(firstList, secondList)
(I didn't like my previous answer so I deleted it)
The first place to start when attacking list problems is to look at the List module which is filled with higher order functions which generalize many common problems and can give you succinct solutions. If you can't find anything suitable there, then you can look at the Seq module for solutions like #BrokenGlass demonstrated (but you can run into performance issues there). Next you'll want to consider recursion and pattern matching. There are two kinds of recursion you'll have to consider when processing lists: tail and non-tail. There are trade-offs. Tail-recursive solutions involve using an accumulator to pass state around, allowing you to place the recursive call in the tail position and avoid stack-overflows with large lists. But then you'll typically end up with a reversed list! For example,
Tail-recursive gencut solution:
let gencutTailRecursive n input =
let rec gencut cur acc = function
| hd::tl when cur < n ->
gencut (cur+1) (hd::acc) tl
| rest -> (List.rev acc), rest //need to reverse accumulator!
gencut 0 [] input
Non-tail-recursive gencut solution:
let gencutNonTailRecursive n input =
let rec gencut cur = function
| hd::tl when cur < n ->
let x, y = gencut (cur+1) tl //stackoverflow with big lists!
hd::x, y
| rest -> [], rest
gencut 0 input
Once you have your gencut solution, it's really easy to define cut:
let cut input = gencut ((List.length input)/2) input
Here's yet another way to do it using inbuilt library functions, which may or may not be easier to understand than some of the other answers. This solution also only requires one traversal across the input. My first thought after I looked at your problem was that you want something along the lines of List.partition, which splits a list into two lists based on a given predicate. However, in your case this predicate would be based on the index of the current element, which partition cannot handle, short of looking up the index for each element.
We can accomplish creating our own equivalent of this behavior using a fold or foldBack. I will use foldBack here as it means you won't have to reverse the lists afterward (see Stephens excellent answer). What we are going to do here is use the fold to provide our own index, along with the two output lists, all as the accumulator. Here is the generic function that will split your list into two lists based on n index:
let gencut n input =
//calculate the length of the list first so we can work out the index
let inputLength = input |> List.length
let results =
List.foldBack( fun elem acc->
let a,b,index = acc //decompose accumulator
if (inputLength - index) <= n then (elem::a,b,index+1)
else (a,elem::b,index+1) ) input ([],[],0)
let a,b,c = results
(a,b) //dump the index, leaving the two lists as output.
So here you see we start the foldBack with an initial accumulator value of ([],[],0). However, because we are starting at the end of the list, the 0 representing the current index needs to be subtracted from the total length of the list to get the actual index of the current element.
Then we simply check if the current index falls within the range of n. If it does, we update the accumulator by adding the current element to list a, leave list b alone, and increase the index by 1 : (elem::a,b,index+1). In all other cases, we do exactly the same but add the element to list b instead: (a,elem::b,index+1).
Now you can easily create your function that splits a list in half by creating another function over this one like so:
let cut input =
let half = (input |> List.length) / 2
input |> gencut half
I hope that can help you somewhat!
> cut data;;
val it : int list * int list = ([1; 2; 3], [4; 5; 6])
> gencut 5 data;;
val it : int list * int list = ([1; 2; 3; 4; 5], [6])
EDIT: you could avoid the index negation by supplying the length as the initial accumulator value and negating it on each cycle instead of increasing it - probably simpler that way :)
let gencut n input =
let results =
List.foldBack( fun elem acc->
let a,b,index = acc //decompose accumulator
if index <= n then (elem::a,b,index-1)
else (a,elem::b,index-1) ) input ([],[],List.length input)
let a,b,c = results
(a,b) //dump the index, leaving the two lists as output.
I have the same Homework, this was my solution. I'm just a student and new in F#
let rec gencut(n, listb) =
let rec cut n (lista : int list) (listb : int list) =
match (n , listb ) with
| 0, _ -> lista, listb
| _, [] -> lista, listb
| _, b :: listb -> cut (n - 1) (List.rev (b :: lista )) listb
cut n [] listb
let cut xs = gencut((List.length xs) / 2, xs)
Probably is not the best recursive solution, but it works. I think
You can use List.nth for random access and list comprehensions to generate a helper function:
let Sublist x y data = [ for z in x..(y - 1) -> List.nth data z ]
This will return items [x..y] from data. Using this you can easily generate gencut and cut functions (remember to check bounds on x and y) :)
check this one out:
let gencut s xs =
([for i in 0 .. s - 1 -> List.nth xs i], [for i in s .. (List.length xs) - 1 -> List.nth xs i])
the you just call
let cut xs =
gencut ((List.length xs) / 2) xs
with n durationn only one iteration split in two