I have a few problems creating a tree size function with type 'a option tree -> int
type 'a tree = Leaf of 'a
| Fork of 'a * 'a tree * 'a tree
How would I create a t_opt_size function with type 'a option tree -> int?
I know I would have to use Some and the None operate.
I have this so far, but it's complicated to match with the option type.
let rec t_size (tr: 'a tree): int =
match tr with
| Leaf _ -> 1
| Fork (_, t1, t2) -> t_size t1 + t_size t2 + 1
I assume from your comments that you want a leaf that looks like (Leaf None) not to be counted in your tree size calculation.
Seems like the key is to split this:
| Leaf _ -> 1
Into two cases:
| Leaf None -> (* Left as exercise *)
| Leaf (Some _) -> (* Left as exercise *)
Since OCaml will take the first match, you can abbreviate this as follows if you like:
| Leaf None -> (* Left as exercise *)
| Leaf _ -> (* Left as exercise *)
You should make a similar change to the Fork case, though I have to say that Fork (None, l, r) doesn't really work for constructing a search tree.
If you want to generalize, you might need to write a generic tree walker which accepts a visitor function. I recommend you try to implement fold_tree, which accepts: (1) a fold function, taking some value, a tree and producing a new result ('a -> 'b t -> 'c), (2) an initial element of type 'a as well as (3) a tree. Then, fold_tree returns a value of type 'c.
Then, you should be able to call fold_tree with a function that skips over None leaves but otherwise increment the count like you did.
If you don't want to count all values in the tree as 1, but each depending on its contents, write a function that determines the count per value and use that:
let weight = function
| _ -> 1 (* or anything else *)
let rec t_opt_size (tr: 'a tree): int = match tr with
| Leaf v -> weight v
| Fork (v, t1, t2) -> t_size t1 + t_size t2 + weight v
You even might want to generalise and pass the weight function as a parameter to t_size instead of writing different size functions that all use their own weighting.
Related
I start in ocaml and I would like to know how in a recursive function of type
'a list -> int ,
let rec int l =
match l with
| [] -> 0
| hd::tl -> 10
the list can be flattened only if necessary
for example if [0;2;3;4] just returns the int
and if [[0];2; [3;4]], then do -> [0;2;3;4] and then return the int.
Thank you in advance.
You cannot store directly either a list or a number in a list, because lists must store values of the same type.
You can, however, declare a variant type (tagged union) for both kinds of values.
Here the type 'a lisr_or_val represents values that are either a value of type 'a, denoted for example (A 3), or lists of values of type 'a lisr_or_val, for example (L [(A 3); (A 5)]):
type 'a list_or_val =
L of 'a list_or_val list
| A of 'a
Then you access the leftmost value as follows:
let rec leftmost_value term = match term with
| L ([]) -> failwith "Unexpected"
| L (x::_) -> leftmost_value x
| A v -> v;;
For example:
# leftmost_value (L [A 5; A 3]);;
- : int = 5
I am trying to write a function that filters positive integers from a list of list of integers, returning a list of only negative integers.
For example, if I have a list of list such as [[-1; 1]; [1]; [-1;-1]] it would return [[-1]; []; [-1;-1]].
I tried to use filter and transform functions, which was in my textbook.
let rec transform (f:'a -> 'b) (l:'a list) : 'b list =
begin match l with
| [] -> []
| x::tl -> (f x)::(transform f tl)
end
and for filter, I had previously written:
let rec filter (pred: 'a -> bool) (l: 'a list) : 'a list =
begin match l with
| [] -> []
| x :: tl -> if pred x then x :: (filter pred tl) else filter pred tl
end
So, using these, I wrote
let filter_negatives (l: int list list) : int list list =
transform (fun l -> (filter(fun i -> i<0)) + l) [] l
but I'm still having trouble fully understanding anonymous functions, and I'm getting error messages which I don't know what to make of.
This function has type ('a -> 'b) -> 'a list -> 'b list
It is applied to too many arguments; maybe you forgot a `;'.
(For what it's worth this transform function is more commonly called map.)
The error message is telling you a simple, true fact. The transform function takes two arguments: a function and a list. You're giving it 3 arguments. So something must be wrong.
The transformation you want to happen to each element of the list is a filtering. So, if you remove the + (which really doesn't make any sense) from your transforming function you have something very close to what you want.
Possibly you just need to remove the [] from the arguments of transform. It's not clear (to me) why it's there.
I want to print a list with different element in it (for educational purpose)
I have read a tutorial that explain how to store different type in list.
type _ list =
[] : unit list
| ( :: ) : 'b * 'a list -> ('b ->'a) list;;
1 :: "e" :: 'r' :: [];; (* this is allowed *)
how I can do something like this pseudo-code:
match typeof(my_expr) with
int -> print_int
| string -> print_string
we will have "1,e,r" printed.
Some solutions i have searched
Change my type in text and printing it
Use a different type definition maybe ('a, 'b) list ?
I ask this because the OCaml toplevel know the type of every variable and show always the type in the right format: can I call this printer ?
Is there a solution only for toplevel that we can install with the #install_printer ?
I know that compiler discard type's info after the type checking pass.
The printer of the toplevel should work fine:
[1; "one"; 1.];;
- : (int -> string -> float -> unit) list =
(::) (1, (::) ("one", (::) (1., [])))
(The unoptimal printing is an unfortunate consequence of ensuring that values printed by the toplevel can be copy-pasted back to the top-level and yields the same value)
But this is only possible outside of the language itself: the toplevel printers can inspect the typing environment which is purposefully not possible in the language itself. Indeed functions like typeof would break parametricity. There is thus no universal printer function in OCaml (without looking at the internal memory representation) and no universal heterogeneous list printer.
If you want to print an heterogeneous list, there are three possible paths:
print a specific type of the heterogeneous list
let print_concrete ppf (x::y::z::rest) = Format.fprintf ppf "%f %f %f" x y z
(Contrary to appearance, this function is total: its type makes it impossible to use on lists with fewer than three elements)
Use heterogeneous lists that always pack a printing function along its main value
type 'a printer = Format.formatter -> 'a -> unit
type _ showable_list =
| [] : unit showable_list
| (::):
('a * 'a printer) * 'b showable_list
-> ('a -> 'b) showable_list
let rec print: type a. a showable_list printer =
fun ppf l -> match l with
| [] -> ()
| (a,printer) :: rest -> Format.fprintf ppf "%a# %a" printer a print rest
provide a matching heterogeneous list of printing functions
type 'a plist =
| []: unit plist
| (::): 'a printer * 'b plist -> ('a -> 'b) plist
let rec print: type l. l plist -> l list printer = fun printers ppf values ->
match printers, values with
| [], [] -> ()
| p :: prest, a :: rest -> Format.fprintf ppf "%a# %a" p a (print prest) rest
The fact that you often need to specialize the heterogeneous list type may make it worthwhile to introduce a functor for generating them:
module Hlist(Specialization: sig type 'a t end) = struct
open Specialization
type 'a list =
| []: unit list
| (::): 'a t * 'b list -> ('a -> 'b) list
end
then the previous specialized type can be constructed with
module Showable_list = Hlist(struct type 'a t = 'a * 'a printer end)
module Printer_list = Hlist (struct type 'a t = 'a printer end)
let sample_table4 = [
[["11"];["21"];["31"];["41"]];
[["12"];["22"];["32"]];
[["13"];["23"]];
[["14"]]];;
This is where I'm stuck with writing a function to get one of these numbers
let tgvc (pos, table) =
match pos with
|[] -> []
|i::[j] -> List.nth (List.nth table (j-1)) (i-1)
|i::_ -> []
;;
val tgvc : int list * 'a list list list -> 'a list = <fun>
I'm supposed to get this signature
tgvc ([3;2],sample_table4);;
val tgvc : int list * ’a list list -> ’a = <fun>
-: string list = ["32"]
What's missing in the function?
I'm sure it has to be recursive now.
Even though it computes the right answer, it's not the right method. The ->[ ] is what's getting me
let rec tgvc (pos, table) = function
|_,[] -> []
|[i;1], h::_ -> List.nth h (i-1)
|[i;j], _::t -> tgvc ([i;j-1], t)
|_ -> []
|[i;j], _::t -> tgvc ([i;j-1], t)
^^^^^^^^^^^^^^^^^
Error: This expression has type int list * 'a list list list -> 'a list
but an expression was expected of type 'a list
What's missing in the function?
A lot of things. Your function simply returns one of many lists of initial input. You don't even use i indice.
I suggest you to think what your function need to do for the given input:
[i; 1], h::_ - you are "in front" of the desirable list
[i; j], _::t - not desirable list yet (some recursion maybe?)
_, [] - empty table
_ - everything else
Edit
You have two problems with your last implementation. First of all in your first and last branches you return [], I guess you would like to exit with an error, so you can throw an exception (via failwith for example). The second problem is actually in the first line: get_table_values_cell (pos, table) = function, it means that you define get_table_values_cell as function with two arguments, you give one explicitly ((pos, table)) and the second is introduced by function keyword. So all you need is to pick only one: get_table_values_cell = function
I am often told that using the Lazy module in OCaml, one can do everything you can do in a lazy language such as Haskell. To test this claim, I'm trying to write a function that converts a regular list into a static doubly linked list in ocaml.
type 'a dlist = Dnil | Dnode of 'a dlist * 'a * 'a dlist
Given this type I can create several static doubly linked lists by hand:
let rec l1 = Dnode (Dnil,1,l2)
and l2 = Dnode (l1,2,l3)
and l3 = Dnode (l2,3,Dnil)
but I'd like to write a function of type 'a list -> 'a dlist that given any list builds a static doubly linked list in OCaml. For example given [1;2;3] it should output something equivalent to l1 above.
The algorithm is pretty straightforward to write in Haskell:
data DList a = Dnil | Dnode (DList a) a (DList a)
toDList :: [a] -> DList a
toDList l = go Dnil l
where
go _ [] = Dnil
go h (x:xs) = let r = Dnode h x (go r xs) in r
but I haven't been able to figure out where to place calls to lazy to get this to compile in OCaml.
If you build your linked list in right-to-left order (as for normal lists), then the left element of every node will only be built after that node itself is built. You need to represent this by making the left element lazy, which means "this value will be constructed later" :
type 'a dlist =
| Dnil
| Dnode of 'a dlist Lazy.t * 'a * 'a dlist
Once you have this, construct every node as a lazy value using a recursive definition which passes the lazy (still unconstructed) node to the function call that builds the next node (so that it has access to the previous node). It's actually simpler than it looks :
let dlist_of_list list =
let rec aux prev = function
| [] -> Dnil
| h :: t -> let rec node = lazy (Dnode (prev, h, aux node t)) in
Lazy.force node
in
aux (Lazy.lazy_from_val Dnil) list
You can only build a cyclic immutable strict data structure of a shape that's determined at compile time. I'm not going to define or prove this formally, but intuitively speaking, once the data structure is created, its shape isn't going to change (because it's immutable). So you can't add to a cycle. And if you create any element of the cycle, you need to create all the other elements of the cycle at the same time, because you can't have any dangling pointer.
Ocaml can do what Haskell can do, but you do have to get the Lazy module involved! Unlike Haskell's, ML's data structures are strict unless otherwise specified. A lazy data structure has pieces of type 'a Lazy.t. (ML's typing is more precise than Haskell on that particular issue.) Lazy data structures allow cycles to be built by having provisionally-dangling pointers (whose linked values are automatically created when the pointer is first dereferenced).
type 'a lazy_dlist_value =
| Dnil
| Dnode of 'a lazy_dlist_value * 'a * 'a lazy_dlist_value
and 'a lazy_dlist = 'a lazy_dlist_value Lazy.t
Another common way to have cyclic data structures is to use mutable nodes. (In fact, die-hard proponents of strict programming might see lazy data structures as a special case of mutable data structures that doesn't break referential transparency too much.)
type 'a mutable_dlist_value =
| Dnil
| Dnode of 'a mutable_dlist_value * 'a * 'a mutable_dlist_value
and 'a mutable_dlist = 'a mutable_dlist_value ref
Cyclic data structures are mostly useful when they involve at least one mutable component, one function (closure), or sometimes modules. But there'd be no reason for the compiler to enforce that — cyclic strict immutable first-order data structures are just a special case which can occasionally be useful.
type 'a dlist = Dnil | Dnode of 'a dlist Lazy.t * 'a * 'a dlist Lazy.t
let rec of_list list = match list with
[] -> Dnil
| x :: [] ->
let rec single () = Dnode (lazy (single ()), x, lazy (single ()))
in single ()
| x :: y -> Dnode (
lazy (
of_list (match List.rev list with
[] | _ :: [] -> assert false
| x :: y -> x :: List.rev y
)
),
x,
lazy (
of_list (match list with
[] | _ :: [] -> assert false
| x :: y -> y # x :: []
)
)
)
let middle dlist = match dlist with
Dnil -> raise (Failure "middle")
| Dnode (_, x, _) -> x
let left dlist = match dlist with
Dnil -> raise (Failure "left")
| Dnode (x, _, _) -> Lazy.force x
let right dlist = match dlist with
Dnil -> raise (Failure "right")
| Dnode (_, _, x) -> Lazy.force x