SML with lazy list function - sml

I'm trying to make a function which can return the specific nth element of lazylist.
Here is what I made:
datatype 'a lazyList = nullList
| cons of 'a * (unit -> 'a lazyList)
fun Nth(lazyListVal, n) = (* lazyList * int -> 'a option *)
let fun iterator (laztListVal, cur, target) =
case lazyListVal of
nullList => NONE
| cons(value, tail) => if cur = target
then SOME value
else iterator (tail(), cur+1, target)
in
iterator(lazyListVal,1,n)
end
I expected the result that as recusing proceeds, eventually the variable cur gets same as the variable target, and then the function iterator returns SOME value so it will return the final nth element.
But when I compile it and run, it only returns the very first element however I test with the lazylist objects.
Please figure what is the problem. I have no idea...
cf) I made another function which is relevant to this problem, the function that transforms lazylist into SML original list containing the first N values. Codes above:
fun firstN (lazyListVal, n) = (* lazyList * int -> 'a list *)
let fun iterator (lazyListVal, cur, last) =
case lazyListVal of
nullList => []
| cons(value, tail) => if cur = last
then []
else value::iterator(tail(),cur+1,last)
in
iterator(lazyListVal,0,n)
end
The strange thing is the function firstN is properly working.

The problem is that your iterator function does case lazyListVal of ..., but the recursive tail is called laztListVal, so for every iteration, it keeps looking at the first list. Use better variable names to avoid this kind of "invisible" bug.
For a simpler definition of nth:
datatype 'a lazyList = NullList | Cons of 'a * (unit -> 'a lazyList)
fun nth (NullList, _) = NONE
| nth (Cons (x, xs), 0) = SOME x
| nth (Cons (_, xs), n) = nth (xs (), n-1)
val nats = let fun nat n = Cons (n, fn () => nat (n+1)) in nat 0 end
val ten = nth (nats, 10)
Edit: While function pattern matching is ideal here, you could also have used a case ... of ... here. A helper function seems unnecessary, though, since you can simply use the input argument n as the iterator:
fun nth (L, n) =
case (L, n) of
(NullList, _) => NONE
| (Cons (x, xs), 0) => SOME x
| (Cons (_, xs), n) => nth (xs (), n-1)
You may however want to make the function more robust:
fun nth (L, n) =
let fun nth' (NullList, _) = NONE
| nth' (Cons (x, xs), 0) = SOME x
| nth' (Cons (_, xs), n) = nth' (xs (), n-1)
in if n < 0 then NONE else nth' (L, n) end
Here having a helper function ensures that n < 0 is only checked once.
(You could also raise Domain, since negative indices are not well-defined.)

Related

How to write a function in SML/NJ that in a given list, counts successive equal elements and returns a list of pairs (value, count)?

I have to write a function in SML/NJ that in a given list, counts successive equal elements and returns a list of pairs (value, count). Function should look like this:
fun group (xs: ''a list): (''a * int) list
I can only use anonymous functions and the structures List, ListPair and Math.
I have no idea how to do it. Can somebody please help me?
A simple but inefficient way is to write this version of group:
val group : ''a list -> ''a list list
and transform the output further into:
val group_count : ''a list -> (''a * int) list
in the following way:
fun group_count xss = map (fn xs => (hd xs, length xs)) xss
But a more efficient way would be to write a function span_count:
fun span_count p x [] count = (count, [])
| span_count p x (y::xs) count =
if p (x, y)
then span_count p x xs (count+1)
else (count, y::xs)
and use it recursively:
fun group_count [] = []
| group_count (x::xs) =
case span_count op= x xs 1 of
(count, ys) => (x, count) :: group_count ys

How to add elements to a list every nth round?

I need to
Write a function separate of type int * 'a * 'a list -> 'a lst such that
separate (k, x, l) returns the list that inserts element x after each k elements of list l (counting from
the end of the list). For example, separate (1, 0, [1,2,3,4]) should return [1,0,2,0,3,0,4] and
separate (3, 0, [1,2,3,4]) should return [1,0,2,3,4].
So far, this is what I have, but it is causing an error. Can anyone help me?
fun separate (k: int, x: 'a, l: 'a list) : 'a list =
let val count:int = k
in foldr(
(fn (h, t) =>
if count = 0
then count := 1 in
x::h::t
else count = count + 1 : int
h::t
)
Actually the logic is quite right, but it should be implemented by passing changed state into another iteration of foldr due to immutability:
fun separate (k: int, x: 'a, l: 'a list) : 'a list =
#2 (foldr (fn (h, (count, t)) =>
if count = 0
then (k - 1, h::x::t)
else (count - 1, h::t)
) (k, []) l);
Thus, instead of initiating count as a variable, we initiate foldr with tuple (k, []) (where k is the initial value of count and [] is the resulting list) and then decrease the count every step of the iteration.

Unite multiple code lines into one function in sml/ml using "let" command

I'm having a problem grouping these code lines into one function
sumFirstEven : int * int seq -> int
such that sumFirstEven (5, s) is the sum of the first 5 even elements of the sequence s
I've been told that I need to group my lines into one function that includes "let-in commands"
these are my lines for these func:
datatype 'a seq=null|SEQ of 'a*(unit->'a seq);
fun head(SEQ(x,_))=x
|head null=raise Empty;
fun tail(SEQ(_,xf))=xf()
|tail null=raise Empty;
fun sumFirstEven(0,_)=0
|sumFirstEven(n,null)=0
|sumFirstEven(n,xs)=if(head(xs) mod 2=0)then head(xs)+sumFirstEven(n-1,tail(xs))
else sumFirstEven(n,tail(xs));
fun seqFrom i=SEQ(i,fn()=>seqFrom(i+1));
val seqStart=seqFrom 1;
sumFirstEven(5,seqStart);
My function works fine but I dont know how to regroup all of these lines
correctly
The beauty of pattern matching is that it often obviates helper functions like head and tail.
fun sumFirstEven (0, _) = 0
| sumFirstEven (_, null) = 0
| sumFirstEven (n, SEQ (h, t)) =
case h mod 2
of 0 => h + sumFirstEven (n-1, t ())
| _ => sumFirstEven (n, t ())
Since the first two clauses didn't match, there's no need in the third clause to call a function with a failure condition like head... we already know the data is in the right form!
The pattern matching in the function clause serves the same role that a let-in-end expression (not command!) would.
Given your lazy list type,
datatype 'a seq = Null | Seq of 'a * (unit -> 'a seq)
you might want the helper function,
fun even n = n mod 2 = 0
and you might want to either recurse manually,
fun sumFirstEven (0, _) = 0
| sumFirstEven (_, Null) = 0
| sumFirstEven (n, Seq (x, seqf) =
if even x
then x + sumFirstEven (n-1, seqf ())
else sumFirstEven (n, seqf ())
or you may want to accumulate the result in a function parameter,
fun sumFirstEven (n, seq) =
let fun sfe (0, _, result) = result
| sfe (_, Null, result) = result
| sfe (n, Seq (x, seqf), result) =
if even x
then sfe (n-1, seqf (), result + x)
else sfe (n, seqf (), result)
in sfe (n, seq, 0) end
to make it tail-recursive (i.e. use a fixed amount of stack frames). More conveniently, you might want to extract the recursion part of the function into a separate higher-order fold combinator,
fun foldseqn _ e 0 _ = e
| foldseqn _ e _ Null = e
| foldseqn f e n (Seq (x, seqf)) =
case f (x, e) of
(e', true) => foldseqn f e' (n-1) (seqf ())
| (e', false) => foldseqn f e' n (seqf ())
fun sumFirstEven (n, seq) =
foldseqn (fn (x, sum) => if even x
then (sum + x, true)
else (sum, false)) 0 n seq
And a quick test,
fun nats n = Seq (n, fn () => nats (n+1))
val twelve = sumFirstEven (3, nats 1) (* 2+4+6 *)
Edit: Responded to Nick's comment, thanks.

how to stop a recursive function in this case (SML)

fun p(L) =
[L] # p( tl(L) # [hd(L)] );
If L is [1,2,3] then I want to have a [ [1,2,3], [2,3,1], [3,1,2] ].
Since every time I append the first num to the end, then if L = [] then [] doesn't work here.
How to stop the function once it has the three lists?
You can have a parameter x in the function to keep track of how many levels deep in the recursion you are.
fun p(L, x) =
if x < length(L) then [L] # p(tl(L) # [hd(L)], x+1)
else [];
Then call the function with x=0.
p([1, 2, 3], 0)
And if you don't like the extra parameter, then as you probably know you can define another function and make it equal to the p function with the parameter forced to 0.
fun p0(L) = p(L, 0);
p0([1, 2, 3]); (* same result as p([1, 2, 3], 0); *)
Let me show some more implementation variants.
First of all, let's define an auxiliary function, which rotates a list 1 position to the left:
(* operates on non-empty lists only *)
fun rot1_left (h :: tl) = tl # [h]
Then the p function could be defined as follows:
fun p xs =
let
(* returns reversed result *)
fun loop [] _ _ = []
| loop xs n res =
if n = 0
then res
else loop (rot1_left xs) (n-1) (xs :: res)
in
List.rev (loop xs (length xs) [])
end
It's usually better (performance-wise) to add new elements at the beginning of the list and then reverse the resulting list once, than to append to the end many times. Note: this version does one spurious rotate at the end and I could have optimized it out, but didn't, to make code more clear.
We have calculated the length of the given list to make its rotated "copies", but we don't have to traverse xs beforehand, we can do it as we rotate it. So, we can use xs as a kind of counter, recursively calling the loop helper function on the tail of the xs list.
fun p xs =
let
(* returns reversed result *)
fun loop [] _ _ = []
| loop xs [] res = res
| loop xs (_::tl) res =
loop (rot1_left xs) tl (xs :: res)
in
List.rev (loop xs xs [])
end
Having done that, we are now closer to implementing p as a foldl function:
fun p xs =
(List.rev o #1)
(List.foldl
(fn (_, (res, rot)) => (rot::res, rot1_left rot))
([], xs)
xs)
The second argument to the List.foldl function is our "accumulator", which is represented here as a pair of the current (partial) result as in the previous implementations and the current rotated list. That explains (List.rev o #1) part: we need to take the first component of the accumulator and reverse it. And as for the ([], xs) part -- the current result is empty at the beginning (hence []) and we start rotating the initial xs list. Also, the _ in (_, (res, rot)) means the current element of the given xs, which we don't care about, since it just serves as a counter (see the prev. variant).
Note: o stands for function composition in Standard ML.

SML list summing

I'm very new to SML and I am trying a list exercise. The goal is sum up the previous numbers of a list and create a new list. For example, an input list [1, 4, 6, 9] would return [1, 5, 11, 20].
This is my solution so far, but I think the issue is with how I'm defining the function.
fun rec sum:int list -> int list =
if tl(list) = nil then
hd(list)
else
hd :: sum((hd(tail) + hd(tl(list)))::tl(tl(list)));
Besides that you are using rec as a function name, then you have some minor issues to work on.
The explicit type annotation you have made is treated as an annotation of the function result.
Thus, according to what you have written, then it should return a function and not the expected
list. This is clearly seen from the below example:
- fun rec_ sum : int list -> int list = raise Domain;
val rec_ = fn : 'a -> int list -> int list
Your should be careful of using the head and tail functions, when you don't do any checks on the
number of elements in the list. This could be done with either the length function, or (even
easier and often better) by pattern matching the number of elements.
Your code contains sum as a function call and tail as an variable. The variable tail has never
been defined, and using sum as a function call, makes me believe that you are actually using rec
as a keyword, but don't know what it means.
The keyword rec is used, when defining functions using the val keyword. In this case, rec is
needed to be able to define recursive functions (not a big surprise). In reality, the keyword fun
is syntactic sugar (a derived form) of val rec.
The following 3 are examples of how it could have been made:
The first is a simple, straight forward solution.
fun sumList1 (x::y::xs) = x :: sumList1 (x+y::xs)
| sumList1 xs = xs
This second example, uses a helper function, with an added argument (an accumulator). The list is constructed in the reverse order, to avoid using the slow append (#) operator. Thus we reverse the list before returning it:
fun sumList2 xs =
let
fun sumList' [] acc = rev acc
| sumList' [x] acc = rev (x::acc)
| sumList' (x :: y :: xs) acc = sumList' (y+x :: xs) (x :: acc)
in
sumList' xs []
end
The last example, show how small and easy it can be, if you use the standard list functions. Here the fold left is used, to go through all elements. Again note that the list is constructed in the reverse order, thus it is reversed as the last step:
fun sumList3 [] = []
| sumList3 (x::xs) = rev (foldl (fn (a, b) => hd b + a :: b) [x] xs)
try this -
fun recList ([], index, sum) = []
| recList (li, index, sum) =
if index=0 then
hd li :: recList (tl li, index+1, hd li)
else
sum + hd li :: recList (tl li, index+1, sum + hd li)
fun recSum li = recList (li, 0, 0)
In your case -
recSum([1,4,6,9]) ;
will give
val it = [1,5,11,20] : int list
also don't use rec as fun name -it keyword .