I'm trying to make sense of this Caml function to calculate the n-th power of a number.
let rec iterate n f d =
if n = 0 then d
else iterate (n-1) f (f d)
let power i n =
let i_times a = a * i in
iterate n i_times 1
I understand what it does conceptually, but I am having trouble understanding the i_times bit.
Per my understanding, i_times takes a value a and returns a*i, whre i is passed onto power upon calling it. In the expression where i_times is defined, it's followed by 1, so wouldn't this mean that i_times 1, all together, is evaluated to 1*i?
In this case, I fail to see how 3 parameters are being passed to iterate. Why doesn't it just end up being 2, that is n and i_times 1?
I know this is a pretty basic function, but I'm just getting started with functional programming and I want to make sense of it.
Basically you're asking how OCaml parses a series of juxtaposed values.
This expression:
f a b c
is iterpreted as a call to a function f that passes 3 separate parameters. It is not parsed like this:
f a (b c)
which would be passing 2 parameters to f.
So indeed, three parameters are being passed to iterate. There are no subexpressions in the list of parameters, just three separate parameters.
Why didn't you figure that the subexpression n i_times would be evaluated before calling iterate? The reason (I suspect) is that you know that n isn't a function. But the parsing of the language doesn't depend on the types of things. If you wrote (n i_times) (with parentheses) it would be parsed as a call to n as a function. (This would, of course, be an error.)
I would like to make this functions recursive but I don't know where to start.
let rec rlist r n =
if n < 1 then []
else Random.int r :: rlist r (n-1);;
let rec divide = function
h1::h2::t -> let t1,t2 = divide t in
h1::t1, h2::t2
| l -> l,[];;
let rec merge ord (l1,l2) = match l1,l2 with
[],l | l,[] -> l
| h1::t1,h2::t2 -> if ord h1 h2
then h1::merge ord (t1,l2)
else h2::merge ord (l1,t2);;
Is there any way to test if a function is recursive or not?
If you give a man a fish, you feed him for a day. But if you give him a fishing rod, you feed him for a lifetime.
Thus, instead of giving you the solution, I would better teach you how to solve it yourself.
A tail-recursive function is a recursive function, where all recursive calls are in a tail position. A call position is called a tail position if it is the last call in a function, i.e., if the result of a called function will become a result of a caller.
Let's take the following simple function as our working example:
let rec sum n = if n = 0 then 0 else n + sum (n-1)
It is not a tail-recursive function as the call sum (n-1) is not in a tail position because its result is then incremented by one. It is not always easy to translate a general recursive function into a tail-recursive form. Sometimes, there is a tradeoff between efficiency, readability, and tail-recursion.
The general techniques are:
use accumulator
use continuation-passing style
Using accumulator
Sometimes a function really needs to store the intermediate results, because the result of recursion must be combined in a non-trivial way. A recursive function gives us a free container to store arbitrary data - the call stack. A place, where the language runtime, stores parameters for the currently called functions. Unfortunately, the stack container is bounded, and its size is unpredictable. So, sometimes, it is better to switch from the stack to the heap. The latter is slightly slower (because it introduces more work to the garbage collector), but is bigger and more controllable. In our case, we need only one word to store the running sum, so we have a clear win. We are using less space, and we're not introducing any memory garbage:
let sum n =
let rec loop n acc = if n = 0 then acc else loop (n-1) (acc+n) in
loop n 0
However, as you may see, this came with a tradeoff - the implementation became slightly bigger and less understandable.
We used here a general pattern. Since we need to introduce an accumulator, we need an extra parameter. Since we don't want or can't change the interface of our function, we introduce a new helper function, that is recursive and will carry the extra parameter. The trick here is that we apply the summation before we do the recursive call, not after.
Using continuation-passing style
It is not always the case when you can rewrite your recursive algorithm using an accumulator. In this case, a more general technique can be used - the continuation-passing style. Basically, it is close to the previous technique, but we will use a continuation in the place of an accumulator. A continuation is a function, that will actually postpone the work, that is needed to be done after the recursion, to a later time. Conventionally, we call this function return or simply k (for the continuation). Mentally, the continuation is a way of throwing the result of computation back into the future. "Back" is because you returning the result back to the caller, in the future, because, the result will be used not now, but once everything is ready. But let's look at the implementation:
let sum n =
let rec loop n k = if n = 0 then k 0 else loop (n-1) (fun x -> k (x+n)) in
loop n (fun x -> x)
You may see, that we employed the same strategy, except that instead of int accumulator we used a function k as a second parameter. If the base case, if n is zero, we will return 0, (you can read k 0 as return 0). In the general case, we recurse in a tail position, with a regular decrement of the inductive variable n, however, we pack the work, that should be done with the result of the recursive function into a function: fun x -> k (x+n). Basically, this function says, once x - the result of recursion call is ready, add it to the number n and return. (Again, if we will use name return instead of k it could be more readable: fun x -> return (x+n)).
There is no magic here, we still have the same tradeoff, as with accumulator, as we create a new closure (functional object) at every recursive call. And each newly created closure contains a reference to the previous one (that was passed via the parameter). For example, fun x -> k (x+n) is a function, that captures two free variables, the value n and function k, that was the previous continuation. Basically, these continuations form a linked list, where each node bears a computation and all arguments except one. So, the computation is delayed until the last one is known.
Of course, for our simple example, there is no need to use CPS, since it will create unnecessary garbage and be much slower. This is only for demonstration. However, for more complex algorithms, in particular for those that combine results of two or more recursive calls in a non-trivial case, e.g., folding over a graph data structure.
So now, armed with the new knowledge, I hope that you will be able to solve your problems as easy as pie.
Testing for the tail recursion
The tail call is a pretty well-defined syntactic notion, so it should be pretty obvious whether the call is in a tail position or not. However, there are still few methods that allow one to check whether the call is in a tail position. In fact, there are other cases, when tail-call optimization may come into play. For example, a call that is right to the shortcircuit logical operator is also a tail call. So, it is not always obvious when a call is using the stack or it is a tail call. The new version of OCaml allows one to put an annotation at the call place, e.g.,
let rec sum n = if n = 0 then 0 else n + (sum [#tailcall]) (n-1)
If the call is not really a tail call, a warning is issued by a compiler:
Warning 51: expected tailcall
Another method is to compile with -annot option. The annotation file will contain an annotation for each call, for example, if we will put the above function into a file sum.ml and compile with ocamlc -annot sum.ml, then we can open sum.annot file and look for all calls:
"sum.ml" 1 0 41 "sum.ml" 1 0 64
call(
stack
)
If we, however, put our third implementation, then the see that all calls are tail calls, e.g. grep call -A1 sum.annot:
call(
tail
--
call(
tail
--
call(
tail
--
call(
tail
Finally, you can just test your program with some big input, and see whether your program will fail with the stack overflow. You can even reduce the size of the stack, this can be controlled with the environment variable OCAMLRUNPARAM, for example, to limit the stack to one thousand words:
export OCAMLRUNPARAM='l=1000'
ocaml sum.ml
You could do the following :
let rlist r n =
let aux acc n =
if n < 1 then acc
else aux (Random.int r :: acc) (n-1)
in aux [] n;;
let divide l =
let aux acc1 acc2 = function
| h1::h2::t ->
aux (h1::acc1) (h2::acc2) t
| [e] -> e::acc1, acc2
| [] -> acc1, acc2
in aux [] [] l;;
But for divide I prefer this solution :
let divide l =
let aux acc1 acc2 = function
| [] -> acc1, acc2
| hd::tl -> aux acc2 (hd :: acc1) tl
in aux [] [] l;;
let merge ord (l1,l2) =
let rec aux acc l1 l2 =
match l1,l2 with
| [],l | l,[] -> List.rev_append acc l
| h1::t1,h2::t2 -> if ord h1 h2
then aux (h1 :: acc) t1 l2
else aux (h2 :: acc) l1 t2
in aux [] l1 l2;;
As to your question about testing if a function is tail recursive or not, by looking out for it a bit you would have find it here.
I am learning OCaml. I know that OCaml provides us with both imperative style of programming and functional programming.
I came across this code as part of my course to compute the n'th Fibonacci number in OCaml
let memoise f =
let table = ref []
in
let rec find tab n =
match tab with
| [] ->
let v = (f n)
in
table := (n, v) :: !table;
v
| (n', v) :: t ->
if n' = n then v else (find t n)
in
fun n -> find !table n
let fibonacci2 = memoise fibonacci1
Where the function fibonacci1 is implemented in the standard way as follows:
let rec fibonacci1 n =
match n with
| 0 | 1 -> 1
| _ -> (fibonacci1 (n - 1)) + (fibonacci1 (n - 2))
Now my question is that how are we achieving memoisation in fibonacci2. table has been defined inside the function fibonacci2 and thus, my logic dictates that after the function finishes computation, the list table should get lost and after each call the table will get built again and again.
I ran some a simple test where I called the function fibonacci 35 twice in the OCaml REPL and the second function call returned the answer significantly faster than the first call to the function (contrary to my expectations).
I though that this might be possible if declaring a variable using ref gives it a global scope by default.
So I tried this
let f y = let x = ref 5 in y;;
print_int !x;;
But this gave me an error saying that the value of x is unbounded.
Why does this behave this way?
The function memoise returns a value, call it f. (f happens to be a function). Part of that value is the table. Every time you call memoise you're going to get a different value (with a different table).
In the example, the returned value f is given the name fibonacci2. So, the thing named fibonacci2 has a table inside it that can be used by the function f.
There is no global scope by default, that would be a huge mess. At any rate, this is a question of lifetime not of scope. Lifetimes in OCaml last as long as an object can be reached somehow. In the case of the table, it can be reached through the returned function, and hence it lasts as long as the function does.
In your second example you are testing the scope (not the lifetime) of x, and indeed the scope of x is restricted to the subexpresssion of its let. (I.e., it is meaningful only in the expression y, where it's not used.) In the original code, all the uses of table are within its let, hence there's no problem.
Although references are a little tricky, the underlying semantics of OCaml come from lambda calculus, and are extremely clean. That's why it's such a delight to code in OCaml (IMHO).
I have a project I am working on and I need to implement a histogram function in OCaml. I have to write a function that takes a list as an argument and returns a histogram, in the form of a list of tuples. It would look something like this:
histogram [1;2;3;1;1;3];;
[(1,3);(2,1);(3,2)]
I can't however get this to work. I feel like I am pretty close but I just need some help / guidance on how to finish the actual histogram function. So far I have made it check a list (nl) for anything that may already be in there. If it doesnt contain the number already I add it to the list. My problem is that I dont know how to recall the function once I have added it to the list. See my code for where the problem lies.
let check a ls = match (a,ls) with
|a,[] -> false
|a,xs -> if fst (hd xs) != a then check a (tl xs) else true
let rec count a ls = match ls with
|[] -> 0
|x::xs -> if x = a then 1 + count a xs else 0 + count a xs
let nl = []
let rec histo l = match l with
|[]-> []
|x::xs -> if check x nl then histo xs else nl # [(x,count x l)] *******
I need to re call histo where the ****** are. Any help at all will be really appreciated.
This is pretty clearly a school assignment, so I'll just give a few hints.
You seem to expect to modify nl in your histo function. But OCaml variables are immutable. The variable named nl in your code will always be bound to an empty list. This is usually the first thing to figure out in FP: how to work with immutable values.
As a hint, the essence is to pass values like nl as parameters of your function. A function's parameters can (of course) be different in different calls to the function.
I don't follow your code for when x appears in nl (check returns true). In this case you need to create a list with an incremented count. But your code just continues without doing anything in particular.
If you redefine histo to take all the required parameters, your unfinished case just needs to call histo with the right parameters.
toastedDeli, just a little advice about your function count. You can use more pattern matching:
let rec count a ls = match ls with
|[] -> 0
|x::xs when x=a -> 1 + count a xs
|_::xs -> count a xs
let rec count a = function
|[] -> 0
|x::xs when x=a -> 1 + count a xs
|_::xs -> count a xs
I am trying to write a function that returns the index of the passed value v in a given list x; -1 if not found. My attempt at the solution:
let rec index (x, v) =
let i = 0 in
match x with
[] -> -1
| (curr::rest) -> if(curr == v) then
i
else
succ i; (* i++ *)
index(rest, v)
;;
This is obviously wrong to me (it will return -1 every time) because it redefines i at each pass. I have some obscure ways of doing it with separate functions in my head, none which I can write down at the moment. I know this is a common pattern in all programming, so my question is, what's the best way to do this in OCaml?
Mutation is not a common way to solve problems in OCaml. For this task, you should use recursion and accumulate results by changing the index i on certain conditions:
let index(x, v) =
let rec loop x i =
match x with
| [] -> -1
| h::t when h = v -> i
| _::t -> loop t (i+1)
in loop x 0
Another thing is that using -1 as an exceptional case is not a good idea. You may forget this assumption somewhere and treat it as other indices. In OCaml, it's better to treat this exception using option type so the compiler forces you to take care of None every time:
let index(x, v) =
let rec loop x i =
match x with
| [] -> None
| h::t when h = v -> Some i
| _::t -> loop t (i+1)
in loop x 0
This is pretty clearly a homework problem, so I'll just make two comments.
First, values like i are immutable in OCaml. Their values don't change. So succ i doesn't do what your comment says. It doesn't change the value of i. It just returns a value that's one bigger than i. It's equivalent to i + 1, not to i++.
Second the essence of recursion is to imagine how you would solve the problem if you already had a function that solves the problem! The only trick is that you're only allowed to pass this other function a smaller version of the problem. In your case, a smaller version of the problem is one where the list is shorter.
You can't mutate variables in OCaml (well, there is a way but you really shouldn't for simple things like this)
A basic trick you can do is create a helper function that receives extra arguments corresponding to the variables you want to "mutate". Note how I added an extra parameter for the i and also "mutate" the current list head in a similar way.
let rec index_helper (x, vs, i) =
match vs with
[] -> -1
| (curr::rest) ->
if(curr == x) then
i
else
index_helper (x, rest, i+1)
;;
let index (x, vs) = index_helper (x, vs, 0) ;;
This kind of tail-recursive transformation is a way to translate loops to functional programming but to be honest it is kind of low level (you have full power but the manual recursion looks like programming with gotos...).
For some particular patterns what you can instead try to do is take advantage of reusable higher order functions, such as map or folds.