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
Related
I have finally found an excellent entry point into functional programming with elm, and boy, do I like it, yet I still lack some probably fundamental elegance concerning a few concepts.
I often find myself writing code similar to the one below, which seems to be doing what it should, but if someone more experienced could suggest a more compact and direct approach, I am sure that could give some valuable insights into this so(u)rcery.
What I imagine this could boil down to, is something like the following
(<-> is a vector subtraction operator):
edgeDirections : List Vector -> List Vector
edgeDirections corners = List.map2 (\p v -> p <-> v) corners (shiftr 1 corners)
but I don't really have a satisfying approach to a method that would do a shiftr.
But the rules of stackoverflow demand it, here is what I tried. I wrote an ugly example of a possible usage for shiftr (I absolutely dislike the Debug.crash and I am not happy about the Maybe):
Given a list of vectors (the corner points of a polygon), calculate the directional vectors by calculating the difference of each corner-vector to its previous one, starting with the diff between the first and the last entry in the list.
[v1,v2,v3] -> [v1-v3,v2-v1,v3-v2]
Here goes:
edgeDir : Vector -> ( Maybe Vector, List Vector ) -> ( Maybe Vector, List Vector )
edgeDir p ( v, list ) =
case v of
Nothing ->
Debug.crash ("nono")
Just vector ->
( Just p, list ++ [ p <-> vector ] )
edgeDirections : List Vector -> List Vector
edgeDirections corners =
let
last =
List.head <| List.reverse corners
in
snd <| List.foldl edgeDir ( last, [] ) corners
main =
show <| edgeDirections [ Vector -1 0, Vector 0 1, Vector 1 0 ]
I appreciate any insight into how this result could be achieved in a more direct manner, maybe using existing language constructs I am not aware of yet, or any pointers on how to lessen the pain with Maybe. The latter may Just not be possible, but I am certain that the former will a) blow me away and b) make me scratch my head a couple times :)
Thank you, and many thanks for this felicitous language!
If Elm had built-in init and last functions, this could be cleaner.
You can get away from all those Maybes by doing some pattern matching. Here's my attempt using just pattern matching and an accumulator.
import List exposing (map2, append, reverse)
shiftr list =
let shiftr' acc rest =
case rest of
[] -> []
[x] -> x :: reverse acc
(x::xs) -> shiftr' (x::acc) xs
in shiftr' [] list
edgeDirections vectors =
map2 (<->) vectors <| shiftr vectors
Notice also the shortened writing of the mapping function of (<->), which is equivalent to (\p v -> p <-> v).
Suppose Elm did have an init and last function - let's just define those quickly here:
init list =
case list of
[] -> Nothing
[_] -> Just []
(x::xs) -> Maybe.map ((::) x) <| init xs
last list =
case list of
[] -> Nothing
[x] -> Just x
(_::xs) -> last xs
Then your shiftr function could be shortened to something like:
shiftr list =
case (init list, last list) of
(Just i, Just l) -> l :: i
_ -> list
Just after I "hung up", I came up with this, but I am sure this can still be greatly improved upon, if it's even correct (and it only works for n=1)
shiftr : List a -> List a
shiftr list =
let
rev =
List.reverse list
in
case List.head rev of
Nothing ->
list
Just t ->
[ t ] ++ (List.reverse <| List.drop 1 rev)
main =
show (shiftr [ 1, 2, 3, 4 ] |> shiftr)
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.
I know how to do the equivalent of Scheme's (or Python's) map and filter functions with the list monad using only the "bind" operation.
Here's some Scala to illustrate:
scala> // map
scala> List(1,2,3,4,5,6).flatMap {x => List(x * x)}
res20: List[Int] = List(1, 4, 9, 16, 25, 36)
scala> // filter
scala> List(1,2,3,4,5,6).flatMap {x => if (x % 2 == 0) List() else List(x)}
res21: List[Int] = List(1, 3, 5)
and the same thing in Haskell:
Prelude> -- map
Prelude> [1, 2, 3, 4, 5, 6] >>= (\x -> [x * x])
[1,4,9,16,25,36]
Prelude> -- filter
Prelude> [1, 2, 3, 4, 5, 6] >>= (\x -> if (mod x 2 == 0) then [] else [x])
[1,3,5]
Scheme and Python also have a reduce function that's often grouped with map and filter. The reduce function combines the first two elements of a list using the supplied binary function, and then combines that result the the next element, and then so on. A common use to to compute the sum or product of a list of values. Here's some Python to illustrate:
>>> reduce(lambda x, y: x + y, [1,2,3,4,5,6])
21
>>> (((((1+2)+3)+4)+5)+6)
21
Is there any way to do the equivalent of this reduce using just the bind operation on a list monad? If bind can't do this on its own, what's the most "monadic" way to perform this operation?
If possible, please limit/avoid the use of syntactic sugar (ie: do notation in Haskell or sequence comprehensions in Scala) when answering.
One of the defining properties of the bind operation is that the result is still "inside" the monad¹. So when you perform bind on a list, the result will again be a list. Since the reduce operation² often results in something other than a list, it can't be expressed in terms of the bind operation.
In addition to that the bind operation on lists (i.e. concatMap/flatMap) only looks at one element at a time and offers no way of reusing the result of previous steps. So even if we're okay with getting the result wrapped in a single-element list, there's no way to do it just with monad operations.
¹ So if you have a type that allows you to perform no operations on it except the ones defined by the monad type class, you can never "break out" of the monad. That's what makes the IO monad works.
² Which is called fold in Haskell and Scala by the way.
If bind can't do this on its own, what's the most "monadic" way to perform this operation?
While the answer given by #sepp2k is correct, there is a way to do a reduce-like operation on a list monadically, but using the product or "writer" monad and an operation which corresponds to distributing the product monad over the list functor.
The definition is:
import Control.Monad.Writer.Lazy
import Data.Monoid
reduce :: Monoid a => [a] -> a
reduce xs = snd . runWriter . sequence $ map tell xs
Let me unpack:
The Writer monad has a data type Writer w a which is basically a tuple (product) of a value a and "written" value w. The type of written values w must be a monoid where the bind operation of the Writer monad is defined something like:
(w, a) >>= f = let (w', b) = f a in (mappend w w', b)
i.e. take the incoming written value, and the result written value, and combine them using the binary operation of the monoid.
The tell operation writes a value, tell :: w -> Writer w (). Thus map tell has type [a] -> [Writer a ()] i.e. a list of monadic values where each element of the original list has been "written" in the monad.
sequence :: Monad m => [m a] -> m [a] corresponds to a distributive law between lists and monads i.e. distribute the monad type over the list type; sequence can be defined in terms of bind as:
sequence [] = return []
sequnece (x:xs) = x >>= (\x' -> (sequence xs) >>= (\xs' -> return $ x':xs'))
(actually the implementation in Prelude uses foldr, a clue to the reduction-like usage)
Thus, sequence $ map tell xs has type Writer a [()]
The runWriter operation unpacks the Writer type, runWriter :: Writer w a -> (a, w),
which is composed here with snd to project out the accumulated value.
An example usage on lists of Ints would be to use the monoid instance:
instance Monoid Int where
mappend = (+)
mempty = 0
then:
> reduce ([1,2,3,4]::[Int])
10
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 ...
For an assignment, i have written the following code in recursion. It takes a list of a vector data type, and a vector and calculates to closeness of the two vectors. This method works fine, but i don't know how to do the recursive version.
let romulus_iter (x:vector list) (vec:vector) =
let vector_close_hash = Hashtbl.create 10 in
let prevkey = ref 10000.0 in (* Define previous key to be a large value since we intially want to set closefactor to prev key*)
if List.length x = 0 then
{a=0.;b=0.}
else
begin
Hashtbl.clear vector_close_hash;
for i = 0 to (List.length x)-1 do
let vecinquestion = {a=(List.nth x i).a;b=(List.nth x i).b} in
let closefactor = vec_close vecinquestion vec in
if (closefactor < !prevkey) then
begin
prevkey := closefactor;
Hashtbl.add vector_close_hash closefactor vecinquestion
end
done;
Hashtbl.find vector_close_hash !prevkey
end;;
The general recursive equivalent of
for i = 0 to (List.length x)-1 do
f (List.nth x i)
done
is this:
let rec loop = function
| x::xs -> f x; loop xs
| [] -> ()
Note that just like a for-loop, this function only returns unit, though you can define a similar recursive function that returns a meaningful value (and in fact that's what most do). You can also use List.iter, which is meant just for this situation where you're applying an impure function that doesn't return anything meaningful to each item in the list:
List.iter f x