I'm new to ocaml and tryin to write a continuation passing style function but quite confused what value i need to pass into additional argument on k
for example, I can write a recursive function that returns true if all elements of the list is even, otherwise false.
so its like
let rec even list = ....
on CPS, i know i need to add one argument to pass function
so like
let rec evenk list k = ....
but I have no clue how to deal with this k and how does this exactly work
for example for this even function, environment looks like
val evenk : int list -> (bool -> ’a) -> ’a = <fun>
evenk [4; 2; 12; 5; 6] (fun x -> x) (* output should give false *)
The continuation k is a function that takes the result from evenk and performs "the rest of the computation" and produces the "answer". What type the answer has and what you mean by "the rest of the computation" depends on what you are using CPS for. CPS is generally not an end in itself but is done with some purpose in mind. For example, in CPS form it is very easy to implement control operators or to optimize tail calls. Without knowing what you are trying to accomplish, it's hard to answer your question.
For what it is worth, if you are simply trying to convert from direct style to continuation-passing style, and all you care about is the value of the answer, passing the identity function as the continuation is about right.
A good next step would be to implement evenk using CPS. I'll do a simpler example.
If I have the direct-style function
let muladd x i n = x + i * n
and if I assume CPS primitives mulk and addk, I can write
let muladdk x i n k =
let k' product = addk x product k in
mulk i n k'
And you'll see that the mulptiplication is done first, then it "continues" with k', which does the add, and finally that continues with k, which returns to the caller. The key idea is that within the body of muladdk I allocated a fresh continuation k' which stands for an intermediate point in the multiply-add function. To make your evenk work you will have to allocate at least one such continuation.
I hope this helps.
Whenever I've played with CPS, the thing passed to the continuation is just the thing you would normally return to the caller. In this simple case, a nice "intuition lubricant" is to name the continuation "return".
let rec even list return =
if List.length list = 0
then return true
else if List.hd list mod 2 = 1
then return false
else even (List.tl list) return;;
let id = fun x -> x;;
Example usage: "even [2; 4; 6; 8] id;;".
Since you have the invocation of evenk correct (with the identity function - effectively converting the continuation-passing-style back to normal style), I assume that the difficulty is in defining evenk.
k is the continuation function representing the rest of the computation and producing a final value, as Norman said. So, what you need to do is compute the result of v of even and pass that result to k, returning k v rather than just v.
You want to give as input the result of your function as if it were not written with continuation passing style.
Here is your function which tests whether a list has only even integers:
(* val even_list : int list -> bool *)
let even_list input = List.for_all (fun x -> x mod 2=0) input
Now let's write it with a continuation cont:
(* val evenk : int list -> (bool -> 'a) -> 'a *)
let evenk input cont =
let result = even_list input in
(cont result)
You compute the result your function, and pass resultto the continuation ...
Related
I'm a beginner in functional programming but I'm famaliar with imperative programming. I'm having trouble translating a piece of cpp code involving updatating two objects at the same time (context is n-body simulation).
It's roughly like this in c++:
for (Particle &i: particles) {
for (Particle &j: particles) {
collide(i, j) // function that mutates particles i and j
}
}
I'm translating this to Ocaml, with immutable objects and immutable Lists. The difficult part is that I need to replace two objects at the same time. So far I have this:
List.map (fun i ->
List.map (fun j ->
let (new_i, new_j) = collide(i, j) in // function that returns new particles i, j
// how do i update particles with new i, j?
) particles
) particles
How do I replace both objects in the List at the same time?
The functional equivalent of the imperative code is just as simple as,
let nbody f xs =
List.map (fun x -> List.fold_left f x xs) xs
It is a bit more generic, as a I abstracted the collide function and made it a parameter. The function f takes two bodies and returns the state of the first body as affected by the second body. For example, we can implement the following symbolic collide function,
let symbolic x y = "f(" ^ x ^ "," ^ y ^ ")"
so that we can see the result and associativity of the the collide function application,
# nbody symbolic [
"x"; "y"; "z"
];;
- : string list =
["f(f(f(x,x),y),z)"; "f(f(f(y,x),y),z)"; "f(f(f(z,x),y),z)"]
So, the first element of the output list is the result of collision of x with x itself, then with y, then with z. The second element is the result of collision of y with x, and y, and z. And so on.
Obviously the body shall not collide with itself, but this could be easily fixed by either modifying the collide function or by filtering the input list to List.fold and removing the currently being computed element. This is left as an exercise.
List.map returns a new list. The function you supply to List.map may transform the elements from one type to another or just apply some operation on the same type.
For example, let's assume you start with a list of integer tuples
let int_tuples = [(1, 3); (4, 3); (8, 2)];;
and let's assume that your update function takes an integer tuple and doubles the integers:
let update (i, j) = (i * 2, j * 2) (* update maybe your collide function *)
If you now do:
let new_int_tuples = List.map update int_tuples
You'll get
(* [(2, 6); (8, 6); (16, 4)] *)
Hope this helps
How does one get the first key,value pair from F# Map without knowing the key?
I know that the Map type is used to get a corresponding value given a key, e.g. find.
I also know that one can convert the map to a list and use List.Head, e.g.
List.head (Map.toList map)
I would like to do this
1. without a key
2. without knowing the types of the key and value
3. without using a mutable
4. without iterating through the entire map
5. without doing a conversion that iterates through the entire map behind the seen, e.g. Map.toList, etc.
I am also aware that if one gets the first key,value pair it might not be of use because the map documentation does not note if using map in two different calls guarantees the same order.
If the code can not be written then an existing reference from a site such as MSDN explaining and showing why not would be accepted.
TLDR;
How I arrived at this problem was converting this function:
let findmin l =
List.foldBack
(fun (_,pr1 as p1) (_,pr2 as p2) -> if pr1 <= pr2 then p1 else p2)
(List.tail l) (List.head l)
which is based on list and is used to find the minimum value in the associative list of string * int.
An example list:
["+",10; "-",10; "*",20; "/",20]
The list is used for parsing binary operator expressions that have precedence where the string is the binary operator and the int is the precedence. Other functions are preformed on the data such that using F# map might be an advantage over list. I have not decided on a final solution but wanted to explore this problem with map while it was still in the forefront.
Currently I am using:
let findmin m =
if Map.isEmpty m then
None
else
let result =
Map.foldBack
(fun key value (k,v) ->
if value <= v then (key,value)
else (k,v))
m ("",1000)
Some(result)
but here I had to hard code in the initial state ("",1000) when what would be better is just using the first value in the map as the initial state and then passing the remainder of the map as the starting map as was done with the list:
(List.tail l) (List.head l)
Yes this is partitioning the map but that did not work e.g.,
let infixes = ["+",10; "-",10; "*",20; "/",20]
let infixMap = infixes |> Map.ofList
let mutable test = true
let fx k v : bool =
if test then
printfn "first"
test <- false
true
else
printfn "rest"
false
let (first,rest) = Map.partition fx infixMap
which results in
val rest : Map<string,int> = map [("*", 20); ("+", 10); ("-", 10)]
val first : Map<string,int> = map [("/", 20)]
which are two maps and not a key,value pair for first
("/",20)
Notes about answers
For practical purposes with regards to the precedence parsing seeing the + operations before - in the final transformation is preferable so returning + before - is desirable. Thus this variation of the answer by marklam
let findmin (map : Map<_,_>) = map |> Seq.minBy (fun kvp -> kvp.Value)
achieves this and does this variation by Tomas
let findmin m =
Map.foldBack (fun k2 v2 st ->
match st with
| Some(k1, v1) when v1 < v2 -> st
| _ -> Some(k2, v2)) m None
The use of Seq.head does return the first item in the map but one must be aware that the map is constructed with the keys sorted so while for my practical example I would like to start with the lowest value being 10 and since the items are sorted by key the first one returned is ("*",20) with * being the first key because the keys are strings and sorted by such.
For me to practically use the answer by marklam I had to check for an empty list before calling and massage the output from a KeyValuePair into a tuple using let (a,b) = kvp.Key,kvp.Value
I don't think there is an answer that fully satisfies all your requirements, but:
You can just access the first key-value pair using m |> Seq.head. This is lazy unlike converting the map to list. This does not guarantee that you always get the same first element, but realistically, the implementation will guarantee that (it might change in the next version though).
For finding the minimum, you do not actually need the guarantee that Seq.head returns the same element always. It just needs to give you some element.
You can use other Seq-based functons as #marklam mentioned in his answer.
You can also use fold with state of type option<'K * 'V>, which you can initialize with None and then you do not have to worry about finding the first element:
m |> Map.fold (fun st k2 v2 ->
match st with
| Some(k1, v1) when v1 < v2 -> st
| _ -> Some(k2, v2)) None
Map implements IEnumerable<KeyValuePair<_,_>> so you can treat it as a Seq, like:
let findmin (map : Map<_,_>) = map |> Seq.minBy (fun kvp -> kvp.Key)
It's even simpler than the other answers. Map internally uses an AVL balanced tree so the entries are already ordered by key. As mentioned by #marklam Map implements IEnumerable<KeyValuePair<_,_>> so:
let m = Map.empty.Add("Y", 2).Add("X", 1)
let (key, value) = m |> Seq.head
// will return ("X", 1)
It doesn't matter what order the elements were added to the map, Seq.head can operate on the map directly and return the key/value mapping for the min key.
Sometimes it's required to explicitly convert Map to Seq:
let m = Map.empty.Add("Y", 2).Add("X", 1)
let (key, value) = m |> Map.toSeq |> Seq.head
The error message I've seen for this case says "the type 'a * 'b does not match the type Collections.Generic.KeyValuePair<string, int>". It may also be possible add type annotations rather than Map.toSeq.
I have the following excercise to do:
Code a function that will be a summation of a list of functions.
So I think that means that if a function get list of functions [f(x);g(x);h(x);...] it must return a function that is f(x)+g(x)+h(x)+...
I'm trying to do code that up for the general case and here's something I came up with:
let f_sum (h::t) = fold_left (fun a h -> (fun x -> (h x) + (a x))) h t;;
The problem is I'm using "+" operator and that means it works only when in list we have functions of type
'a -> int
So, can it be done more "generally", I mean can we write a function, that is a sum of ('a -> 'b) functions, given in a list?
yes, you can make plus function to be a parameter of your function, like
let f_sum plus fs =
let (+) = plus in
match fs with
| [] -> invalid_arg "f_sum: empty list"
| f :: fs -> fold_left ...
You can generalize even more, and ask a user to provide a zero value, so that you can return a function, returning zero if the list is empty. Also you can use records to group functions, or even first class modules (cf., Commutative_group.S in Core library).
Long story short, I came up with this funny function set, that takes a function, f : 'k -> 'v, a chosen value, k : 'k, a chosen result, v : 'v, uses f as the basis for a new function g : 'k -> 'v that is the exact same as f, except for that it now holds that, g k = v.
Here is the (pretty simple) F# code I wrote in order to make it:
let set : ('k -> 'v) -> 'k -> 'v -> 'k -> 'v =
fun f k v x ->
if x = k then v else f x
My questions are:
Does this function pose any problems?
I could imagine repeat use of the function, like this
let kvs : (int * int) List = ... // A very long list of random int pairs.
List.fold (fun f (k,v) -> set f k v) id kvs
would start building up a long list of functions on the heap. Is this something to be concerned about?
Is there a better way to do this, while still keeping the type?
I mean, I could do stuff like construct a type for holding the original function, f, a Map, setting key-value pairs to the map, and checking the map first, the function second, when using keys to get values, but that's not what interests me here - what interest me is having a function for "modifying" a single result for a given value, for a given function.
Potential problems:
The set-modified function leaks space if you override the same value twice:
let huge_object = ...
let small_object = ...
let f0 = set f 0 huge_object
let f1 = set f0 0 small_object
Even though it can never be the output of f1, huge_object cannot be garbage-collected until f1 can: huge_object is referenced by f0, which is in turn referenced by the f1.
The set-modified function has overhead linear in the number of set operations applied to it.
I don't know if these are actual problems for your intended application.
If you wish set to have exactly the type ('k -> 'v) -> 'k -> 'v -> 'k -> 'v then I don't see a better way(*). The obvious idea would be to have a "modification table" of functions you've already modified, then let set look up a given f in this table. But function types do not admit equality checking, so you cannot compare f to the set of functions known to your modification table.
(*) Reflection not withstanding.
Is it possible to write recursive anonymous functions in SML? I know I could just use the fun syntax, but I'm curious.
I have written, as an example of what I want:
val fact =
fn n => case n of
0 => 1
| x => x * fact (n - 1)
The anonymous function aren't really anonymous anymore when you bind it to a
variable. And since val rec is just the derived form of fun with no
difference other than appearance, you could just as well have written it using
the fun syntax. Also you can do pattern matching in fn expressions as well
as in case, as cases are derived from fn.
So in all its simpleness you could have written your function as
val rec fact = fn 0 => 1
| x => x * fact (x - 1)
but this is the exact same as the below more readable (in my oppinion)
fun fact 0 = 1
| fact x = x * fact (x - 1)
As far as I think, there is only one reason to use write your code using the
long val rec, and that is because you can easier annotate your code with
comments and forced types. For examples if you have seen Haskell code before and
like the way they type annotate their functions, you could write it something
like this
val rec fact : int -> int =
fn 0 => 1
| x => x * fact (x - 1)
As templatetypedef mentioned, it is possible to do it using a fixed-point
combinator. Such a combinator might look like
fun Y f =
let
exception BlackHole
val r = ref (fn _ => raise BlackHole)
fun a x = !r x
fun ta f = (r := f ; f)
in
ta (f a)
end
And you could then calculate fact 5 with the below code, which uses anonymous
functions to express the faculty function and then binds the result of the
computation to res.
val res =
Y (fn fact =>
fn 0 => 1
| n => n * fact (n - 1)
)
5
The fixed-point code and example computation are courtesy of Morten Brøns-Pedersen.
Updated response to George Kangas' answer:
In languages I know, a recursive function will always get bound to a
name. The convenient and conventional way is provided by keywords like
"define", or "let", or "letrec",...
Trivially true by definition. If the function (recursive or not) wasn't bound to a name it would be anonymous.
The unconventional, more anonymous looking, way is by lambda binding.
I don't see what unconventional there is about anonymous functions, they are used all the time in SML, infact in any functional language. Its even starting to show up in more and more imperative languages as well.
Jesper Reenberg's answer shows lambda binding; the "anonymous"
function gets bound to the names "f" and "fact" by lambdas (called
"fn" in SML).
The anonymous function is in fact anonymous (not "anonymous" -- no quotes), and yes of course it will get bound in the scope of what ever function it is passed onto as an argument. In any other cases the language would be totally useless. The exact same thing happens when calling map (fn x => x) [.....], in this case the anonymous identity function, is still in fact anonymous.
The "normal" definition of an anonymous function (at least according to wikipedia), saying that it must not be bound to an identifier, is a bit weak and ought to include the implicit statement "in the current environment".
This is in fact true for my example, as seen by running it in mlton with the -show-basis argument on an file containing only fun Y ... and the val res ..
val Y: (('a -> 'b) -> 'a -> 'b) -> 'a -> 'b
val res: int32
From this it is seen that none of the anonymous functions are bound in the environment.
A shorter "lambdanonymous" alternative, which requires OCaml launched
by "ocaml -rectypes":
(fun f n -> f f n)
(fun f n -> if n = 0 then 1 else n * (f f (n - 1))
7;; Which produces 7! = 5040.
It seems that you have completely misunderstood the idea of the original question:
Is it possible to write recursive anonymous functions in SML?
And the simple answer is yes. The complex answer is (among others?) an example of this done using a fix point combinator, not a "lambdanonymous" (what ever that is supposed to mean) example done in another language using features not even remotely possible in SML.
All you have to do is put rec after val, as in
val rec fact =
fn n => case n of
0 => 1
| x => x * fact (n - 1)
Wikipedia describes this near the top of the first section.
let fun fact 0 = 1
| fact x = x * fact (x - 1)
in
fact
end
This is a recursive anonymous function. The name 'fact' is only used internally.
Some languages (such as Coq) use 'fix' as the primitive for recursive functions, while some languages (such as SML) use recursive-let as the primitive. These two primitives can encode each other:
fix f => e
:= let rec f = e in f end
let rec f = e ... in ... end
:= let f = fix f => e ... in ... end
In languages I know, a recursive function will always get bound to a name. The convenient and conventional way is provided by keywords like "define", or "let", or "letrec",...
The unconventional, more anonymous looking, way is by lambda binding. Jesper Reenberg's answer shows lambda binding; the "anonymous" function gets bound to the names "f" and "fact" by lambdas (called "fn" in SML).
A shorter "lambdanonymous" alternative, which requires OCaml launched by "ocaml -rectypes":
(fun f n -> f f n)
(fun f n -> if n = 0 then 1 else n * (f f (n - 1))
7;;
Which produces 7! = 5040.