I am trying to implement a Binary Search Tree in SML. I have an insert function and I am trying to implement another function that takes a list and calls the insert function on each element in the list. This is what I have so far,
fun insertB (l) = insert (hd(l), Node(insertB(tl (l)), Nil, Nil))
but i don't have a base case, so thats one problem. My input function takes an int and a Node as parameters. The error I am currently getting is error right-hand-side of clause doesn't agree with function result type [tycon mismatch]
The base case for your insertB function is the empty list []. What binary search tree would correspond to that? An empty one, of course, which in your case seems to be named Nil:
fun insertB [] = Nil
That was the base case of the recursion. You now need the recursive case, which is pretty similar to what you did, except you don't try to build the BST, but rather have the recursive call build it:
| insertB (head :: tail) = insert (head, insertB tail)
The whole function is thus:
fun insertB [] = Nil
| insertB (head :: tail) = insert (head, insertB tail)
Alternatively, you can use foldl:
fun insertB xs = foldl insert Nil xs
Related
I'm writing a quicksort function for an exercise. I already know of the 5-line functional quicksort; but I wanted to improve the partition by having it scan through the list once and return a pair of lists splitting the original list in half. So I wrote:
fun partition nil = (nil, nil)
| partition (pivot :: rest) =
let
fun part (lst, pivot, (lesseq, greater)) =
case lst of
[] => (lesseq, greater)
| (h::t) =>
if h <= pivot then part (t, pivot, (h :: lesseq, greater))
else part (t, pivot, (lesseq, h :: greater))
in
part (rest, pivot, ([pivot], []))
end;
This partitions well enough. It gives me a signature val partition = fn : int list -> int list * int list. It runs as expected.
It's when I use the quicksort below that things start to break.
fun quicksort_2 nil = nil
| quicksort_2 lst =
let
val (lesseq, greater) = partition lst
in
quicksort_2 lesseq # quicksort_2 greater
end;
I can run the above function if I eliminate the recursive calls to quicksort_2; but if I put them back in (to actually go and sort the thing), it will cease to run. The signature will be incorrect as well, giving me val quicksort_2 = fn : int list -> 'a list. The warning I receive when I call the function on a list is:
Warning: type vars not generalized because of value restriction are instantiated to dummy types (X1,X2,...)
What is the problem here? I'm not using any ref variables; the type annotation I've tried doesn't seem to help...
The main issue is that you're lacking the singleton list base case for your quicksort function. It ought to be
fun quicksort [ ] = [ ]
| quicksort [x] = [x]
| quicksort xs =
let
val (l, r) = partition xs
in
quicksort l # quicksort r
end
which should then have type int list -> int list given the type of your partition. We have to add this case as otherwise you'll never hit a base case and instead recurse indefinitely.
For some more detail on why you saw the issues you were having though:
The signature will be incorrect as well, giving me val quicksort_2 = fn : int list -> 'a list
This is because the codomain of your function was never restricted to be less general than 'a list. Taking a look at the possible branches in your original implementation we can see that in the nil branch you return nil (of most general type 'a list) and in the recursive case you get two 'a lists (per our assumptions thus far) and append them, resulting in an 'a list---this is fine so your type is not further restricted.
[Value Restriction Warning]
What is the problem here? I'm not using any ref variables
The value restriction isn't really related to refs (though can often arise when using them). Instead it is the prohibition that anything polymorphic at the top level must be a value by its syntax (and thus precludes the possibility that a computation is behind a type abstractor at the top level). Here it is because given xs : int list we (ignoring the value restriction) have quicksort_2 xs : 'a list---which would otherwise be polymorphic, but is not a syntactic value. Correspondingly it is value restricted.
I am actually sitting over a hour on a problem and don´t find a solution for it.
I have this data type:
type 'a tree = Empty | Node of 'a * 'a tree * 'a tree
And i have to find a function which converts a given tree in a ordered list. There is also no invariant like that the left child has to be less then the right. I already found a "normal" recursion solution but not a tail recursive solution. I already thought about to build a unordered list and sort it with List.sort, but this uses a merge sort which is not tail recursive. Maybe someone has a good advice.
Thank you!
If you want to traverse the tree in order and return a list, that means our function inorder must have the type 'a tree -> 'a list.
let rec inorder t =
match t with
| Empty -> []
| Node (v, l, r) -> List.append (inorder l) (v :: (inorder r)) (* ! *)
However List.append is in tail position, not inorder. Another problem is we have two calls to inorder. If we put inorder l in tail position, inorder r could not possibly be in tail position - and vice versa.
A neat way to work around this problem is continuation passing style. We take our function above and convert it into a helper function with an extra parameter for our continuation, return
(* convert to helper function, add an extra parameter *)
let rec loop t return =
match t with
| Empty -> ...
| Node (v, l, r) -> ...
The continuation represents "what to do next", so instead of sending values directly out of our function, we must hand them to the continuation instead. That means for the Empty case, we'll return [] - instead of simply []
let rec loop t return =
match t with
| Empty -> return []
| Node (v, l, r) -> ...
For the Node (v, l, r) case, now that we have an extra parameter we can write our own continuation that informs loop what to do next. So to construct our sorted list, we will need to loop l, then loop r (or vice versa), then we can append them. We'll write our program just like this.
let rec loop t return =
match t with
| Empty -> return []
| Node (v, l, r) ->
loop l ... (* build the left_result *)
loop r ... (* build the right_result *)
return (List.append left_result (v :: right_result))
In this next step, we'll fill in the actual lambda syntax for the continuations.
let rec loop t return =
match t with
| Empty -> return []
| Node (v, l, r) ->
loop l (fun left ->
loop r (fun right ->
return (List.append left (v :: right))))
Last, we define inorder which is a call to loop with the default continuation, identity.
let identity x =
x
let inorder t =
let rec loop t return =
match t with
| Empty -> return []
| Node (v, l, r) ->
loop r (fun right ->
loop l (fun left ->
return (List.append left (v :: right))))
in
loop t identity
As you can see loop r (fun right -> ...) is in tail position for the Node branch. loop l (fun left -> ...) is in tail position of the first continuation. And List.append ... is in tail position of the second continuation. Provided List.append is a tail-recursive procedure, inorder will not grow the stack.
Note using List.append could be a costly choice for big trees. Our function calls it once per Node. Can you think of a way to avoid it? This exercise is left for the reader.
I am just reading about syntax of Erlang, and read this implementation of while loop:
-module(helloworld).
-export([while/1,while/2, start/0]).
while(L) -> while(L,0).
while([], Acc) -> Acc;
while([_|T], Acc) ->
io:fwrite("~w~n",[Acc]),
while(T,Acc+1).
start() ->
X = [1,2,3,4],
while(X).
Is the semicolon a mistake? (4th line: while([], Acc) -> Acc;)
I would write the two functions like this:
while(L) -> while(L,0).
while([], Acc) -> Acc;
while([_|T], Acc) ->
io:fwrite("~w~n",[Acc]),
while(T,Acc+1).
start() ->
X = [1,2,3,4],
while(X).
Using whitespace to separate the function definitions makes it clear that two different functions are being defined: while/1 and while/2.
I had no idea about [_|T]. It just comes after this part in the
tutorial, so that's very confusing.
That's nearly equivalent to [H|T], which deconstructs a list into the Head and the Tail, where the Head is the first element of a list and the Tail is the rest of the list. The variable name _ means that you don't care about the variable, so you will not use it in the function body. In this case, it means that you don't care about the Head of the list, all you want is the Tail of the list. If a named variable is used in the head of a function clause, and you don't use the variable in the body of the function, then the compiler will give you a warning.
Here's an example of how deconstructing a list with pattern matching works:
-module(my).
-compile(export_all).
f([Head|Tail]) ->
io:format("The head of the list is: ~w~n", [Head]),
io:format("The tail of the list is: ~w~n", [Tail]).
In the shell:
8> c(my).
my.erl:2: Warning: export_all flag enabled - all functions will be exported
{ok,my}
9> my:f([1, 2, 3]).
The head of the list is: 1
The tail of the list is: [2,3]
ok
10>
No. There are two functions defined here: while/1 (one argument) and while/2 (two arguments). The second one have two function bodies; which one to use is decided through pattern matching.
I am writing a common lisp program and I have a variable that can contain either a string or a function. I want to call the function if it is one and return that as well as the string. How do I test if a variable is a function?
Code so far:
(defun string-or-function (var)
(if (typep var 'simple-array)
var
(if "Function equivalent of typep goes here."
(setf temp (fn-that-does-something))
(string-or-function temp)
Edit: Code that works:
(defun string-or-function (var)
(let ((s-or-f (type-of var)))
(if (equal s-or-f 'function)
(print "function")
(if (equal (car s-or-f) 'simple-array)
(print "string")))))
Is there a better way to do it?
Common Lisp has a predicative type system. The notion that a value has a "principal" type doesn't make as much sense in Lisp. The type-of function is actually fairly infrequently used, as it makes less sense to ask "What is the type of X" and more sense to ask "Is X of type Y". This can be done with typep, or in your case more concisely with typecase, which is just a case statement for types.
(defun string-or-function (var)
(typecase var
(string (format t "string"))
(function (format t "function"))
(t (format t "something else"))))
I want to call the function if it is one and return that as well as the string.
I think you mean something like this:
(defun evaluate (arg)
"Returns something from evaluating ARG in some manner. If ARG is a string,
return it. If ARG is a function, call it with no arguments and return its
return value(s)."
(ctypecase arg
(string arg)
(function (funcall arg))))
If you need extensibility:
(defgeneric evaluate (arg)
(:documentation "Returns something from evaluating ARG in some manner."))
(defmethod evaluate ((arg string))
arg)
(defmethod evaluate ((arg function))
(funcall arg))
Here are some other ways:
(defun string-or-function-p (x)
(typep x '(or string function)))
...but you can probably also use check-type, which is not a predicate but a check which signals a restartable condition in case the value does not satisfy the type specification:
(check-type place (or string function))
If you happen to use this type a lot, define a custom type:
(deftype string-or-fun () '(or string function))
Of course, you can also use generic functions depending on your needs (silly example):
(defgeneric execute (object)
(:method ((s string)) (eval (read-from-string s)))
(:method ((f function)) (funcall f)))
But note that generic function dispatch on classes, not types, which are different things.
(eq (type-of var) 'function)
However, remember that Common Lisp keeps variables and function names in different namespaces, so (var 1 2 3) and (cons var 1) are looking in two different places. You probably cannot call var like (var), but will rather need to use (funcall var), depending on which namespace this is in.
Basically, you probably shouldn't be stuffing a function-or-maybe-a-string into one variable.
I'm looking at the code in the F# 'Tutorial' template that is provided with Visual Studio 2015 and I see this code; I'm wondering why the first function isn't tail-recursive; I think I understand it but want to confirm:
/// Computes the sum of a list of integers using recursion.
let rec sumList xs =
match xs with
| [] -> 0
| y::ys -> y + sumList ys
/// Make the function tail recursive, using a helper function with a result accumulator
let rec private sumListTailRecHelper accumulator xs =
match xs with
| [] -> accumulator
| y::ys -> sumListTailRecHelper (accumulator+y) ys
Is the first one not tail recursive in the because '+' is a function and its' two arguments are evaluated first? Therefore the actual order of evaluation would be: y, then sumList ys, then +? Whereas in the second case, the order of evaluation is: accumulator,y,+ then sumListTailRecHelper(..)?
A call is tail-recursive if there's nothing left to do after the recursive call returns. So the last call amounts to going back to the start of the function code, with modified parameters.
In the first function you still have to add y to the result.