I'm doing an independent study on Elm, and I feel like I'm learning to program all over again! As a learn-the-language project, I'm trying to get an easy blackjack up and running, but once I started I realized how much I still don't grasp. I have as far as drawing cards from a deck and adding them to a list:
import Random
import Mouse
import Array
--Build the deck
faces = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
suits = ['H', 'D', 'C', 'S']
allCards faces suits =
case suits of
x :: xs -> family faces x ++ allCards faces xs
_ -> []
family faces suit =
case faces of
x :: xs -> (,) x suit :: family xs suit
_ -> []
deck = allCards faces suits
rand : Signal Int
rand = Random.range 0 (length deck-1) Mouse.clicks
pickCard n = head <| drop n deck
nextCard = lift pickCard rand
yourHand = foldp (::) [] nextCard
main = lift asText yourHand
My questions are mostly on how to continue. Looking at completed Elm projects helps a little, but many of them are hard for me to parse as a beginner. Any kind of direction helps!
One of the first problems I had was trying to figure out how to remove cards from the deck once they're drawn, using something like dropCard deck card = filter (\card /= nextCard) deck to filter out the drawn card from the list. But my understanding of Elm is that every time a signal changes, the program re-evaluates, meaning that the deck is recreated in full every time a card is drawn. Would I need to foldp the original deck as well?
What's the proper way to remove an element from one list and add it to another in functional programming? Function composition, like toHand . dropCard card?
For adding card faces to determine winning/losing, I'm not sure how to get the integer value out of the list. I tried doing fst (head deck), but I got type errors, probably because deck is itself a signal of some kind. Is there something I'm not seeing?
That said, I've really enjoyed Elm thus far!
Questions
For simple programs, the easiest way to think about signals is by thinking about what kinds of things in your program can change. In your case that would be the deck and the hand. Then you take these things and make a datastructure in which to store them. Then you do a foldp over that whole datastructure, so instead of just keeping track of the hand you also keep track of the deck.
You can write a function that takes a list and an index, and returns the item at that index in the list and the list with that item removed. Then you add the item to the other list and you're done.
fst (head deck) should work. Perhaps you forgot to remove lift in the main definition when you tried it?
Example code
-- This first part is your code:
import Random
import Mouse
import Array
--Build the deck
faces = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
suits = ['H', 'D', 'C', 'S']
allCards faces suits =
case suits of
x :: xs -> family faces x ++ allCards faces xs
_ -> []
family faces suit =
case faces of
x :: xs -> (,) x suit :: family xs suit
_ -> []
-- Here come my additions/changes:
-- Naming some types for clarity
type Card = (Int,Char)
type ProgramState = { deck : [Card], hand : [Card] }
getFromList : Int -> [a] -> (a,[a])
getFromList index list =
let prefix = take index list
(item :: postfix) = drop index list
in (item, prefix ++ postfix)
startState : ProgramState
startState = { deck = allCards faces suits, hand = [] }
rand : Signal Float
rand = Random.float Mouse.clicks
rFloatToInt : Float -> Int -> Int -> Int
rFloatToInt rnd lo hi = round ((rnd + toFloat lo) * toFloat hi)
pickCard : Float -> ProgramState -> ProgramState
pickCard rnd {deck,hand} =
let index = rFloatToInt rnd 0 (length deck - 1)
(item, newDeck) = getFromList index deck
in { deck = newDeck, hand = item :: hand }
programState : Signal ProgramState
programState = foldp pickCard startState rand
main : Signal Element
main = lift asText programState
If anything is unclear, let me know.
Related
This is what i have at the moment
(string -> int list)
let read filename = ....
this is working as intended, returning a list of integers from a textfile looking like this:
530070000
600195000
098000060
800600003
400803001
700020006
060000280
000419005
000080079
Yes you are correct, it is a sudoku board. This is what i have to work with:
type vertex = int * int (*Cells in the sudoku board*)
type gamma = int (*representing colors 1-9*)
(* [Vertex = Map.Make(Vertex)] *)
module Vertex = Map.Make(struct
type t = vertex
let compare = Stdlib.compare
end)
(* [Gamma = Set.Make(Gamma)] *)
module Gamma = Set.Make(struct
type t = gamma
let compare = Stdlib.compare
end)
The gamma set is for solving the sudoku board using graph coloring. I need help understanding how i can convert the list of integers to a suitable map for this kind of task. According to the structure i provided, so i can access each element in the map using it coordinates (x, y). Hope you understand, otherwise i will try to provide more info. I'm reaaally bad at OCaml but trying to learn. I'm sorry for body errors etc, first time posting here.
As far as I can understand your task, the text file contains a grid of digits with the initial disposition for sudoku. So you shouldn't interpret a line in the file as a single integer but rather as a list of integers. You can either change your read function so that it returns int list list instead of int list and then use List.fold_left over the list that will also count the position of an element in the list, but it is tedious. It is much easier to read the grid directly from the file, e.g.,
let read_matrix chan =
let rec loop i j grid =
match input_char chan with
| exception End_of_file -> grid
| '\n' -> loop (i+1) 0 grid
| '0'..'9' as c ->
loop i (j+1) ##
Vertex.add (i,j) (ascii_digit c) grid
| _ -> invalid_arg "invalid input" in
loop 0 0 Vertex.empty
where ascii_digit is defined as,
let ascii_digit c = Char.code c - Char.code '0'
The read_matrix function takes the channel as input so to read the grid from a file you can define,
let matrix_from_file file =
let chan = open_in file in
let r = read_matrix chan in
close_in chan;
r
Hint: you probably also don't want to include positions with 0 in your grid. It is easy to achieve, just add another case to the pattern in the loop function that will skip it, e.g.,
...
| '0' -> loop i (j+1) grid
...
I have a tree type:
type tree = Vertex of int * tree list;;
My recursive equality definition is that two trees are equal if their ints are equal and all of their children are equal.
How do I build the function
topo: tree -> tree list
that creates a list of all of the trees in depth first search order with each tree appearing once and only once (according to the equality definition)? I want to do this in a computationally efficient way. Maybe use lazy or a hashmap?
Here is my attempt, the code blows up when the length is too large:
type tree = Vertex of int * (tree list)
let rec base = function
| 0 -> Vertex (0, [])
| i -> Vertex (i, [base (i - 1)])
let rec range = function
| 0 -> [0]
| i -> i :: range (i - 1)
let agg i = Vertex (-1, List.map base (range i))
let rec equals (a: tree) (b: tree) : bool =
let rec deep_match a_dep b_dep = match a_dep, b_dep with
| [], [] -> true
| [], _
| _, [] -> false
| x::xs, y::ys -> equals x y && deep_match xs ys
in
let Vertex (ai, al) = a in
let Vertex (bi, bl) = b in
ai = bi && deep_match al bl
let rec in_list (a: tree) (l: tree list) : bool = match l with
| [] -> false
| hd::tl -> equals a hd || in_list a tl
let rec topological (pool: tree list) (t: tree) : tree list =
if in_list t pool then pool else
t::match t with
| Vertex(_, []) -> pool
| Vertex(_, deps) -> List.fold_left topological pool deps
let big_agg = agg 100_000
let topo_ordered = topological [] big_agg;;
Printf.printf "len %i\n" (List.length topo_ordered)
To make it efficient you need to implement ordering and hash-consing. With total ordering, you can store your trees in a balanced tree or even a hashtable, thus turning your in_list into O(logN) or even O(1). Adding hash-consing will enable O(1) comparison of your trees (at the cost of less efficient tree construction).
Instead of having both, depending on your design constraints, you can have only one. For the didactic purposes, let's implement hash-consing for your particular representation
To implement hash-consing you need to make your constructor private and hide data constructors behind an abstraction wall (to prevent users from breaking you hash-consing properties):
module Tree : sig
type t = private Vertex of int * t list
val create : int -> t list -> t
val equal : t -> t -> bool
end = struct
type t = Vertex of int * t list
let repository = Hashtbl.create 64
let create n children =
let node = Vertex (n,children) in
try Hashtbl.find repository node
with Not_found -> Hashtbl.add repository node node; node
let equal x y = x == y
end
Since we guaranteed that structurally equal trees are physically equal during the tree creation (i.e., if there exists an equal tree in our repository then we return it), we are now able to substitute structural equality with physical equality, i.e., with pointer comparison.
We got a fast comparison with the price - we now leaking memory, since we need to store all ever created trees and the create function is now O(N). We can alleviate the first problem by using ephemerons, but the latter problem will persist, of course.
Another issue, is that we're not able to put our trees into ordered structure, like a map or a set. We can of course use regular polymorphic compare, but since it will be O(N), inserting to such structure will become quadratic. Not an option for us. Therefore we need to add total ordering on our trees. We can theoretically do this without changing the representation (using ephemerons), but it is easier just to add an order parameter to our tree representation, e.g.,
module Tree : sig
type order (* = private int *) (* add this for debuggin *)
type t = private Vertex of order * int * t list
val create : int -> t list -> t
val equal : t -> t -> bool
val compare : t -> t -> int
end = struct
type order = int
type t = Vertex of order * int * t list
type tree = t
module Repository = Hashtbl.Make(struct
type t = tree
let max_hash = 16
let rec equal (Vertex (_,p1,x)) (Vertex (_,p2,y)) =
match compare p1 p2 with
| 0 -> equal_trees x y
| n -> false
and equal_trees xs ys = match xs, ys with
| [],[] -> true
| [],_ | _,[] -> false
| x :: xs, y::ys -> equal x y && equal_trees xs ys
let rec hash (Vertex (_,p,xs)) =
hash_trees (Hashtbl.hash p) max_hash xs
and hash_trees hash depth = function
| x :: xs when depth > 0 ->
hash_trees (Hashtbl.hash x) (depth-1) xs
| _ -> hash
end)
let repository = Repository.create 64
let create n children =
try Repository.find repository (Vertex (0,n,children))
with Not_found ->
let order = Repository.length repository + 1 in
let node = Vertex (order,n,children) in
Repository.add repository node node; node
let equal x y = x == y
let order (Vertex (order,_,_)) = order
let compare x y = compare (order x) (order y)
end
We had to manually implement the structural variants of equal and hash for our trees because we need to ignore the order in comparison, when we store a new tree in the repository. It looks like a bit of work, but in the real-life you can do this using derivers.
Anyways, now we got a comparable version of a tree with a comparison function which is O(1), so we can put our trees in sets and maps, and implement your topo efficiently.
A nice feature of both implementations is a tight representation of a tree, since sharing is guaranteed by the create function. E.g.,
# let t1 = Tree.create 42 [];;
val t1 : Tree.t = Tree.Vertex (1, 42, [])
# let t3 = Tree.create 42 [t1; t1];;
val t3 : Tree.t =
Tree.Vertex (2, 42, [Tree.Vertex (1, 42, []); Tree.Vertex (1, 42, [])])
# let t5 = Tree.create 42 [t1; t3; t1];;
val t5 : Tree.t =
Tree.Vertex (3, 42,
[Tree.Vertex (1, 42, []);
Tree.Vertex (2, 42, [Tree.Vertex (1, 42, []); Tree.Vertex (1, 42, [])]);
Tree.Vertex (1, 42, [])])
#
In this example, t1 in t5 and t3 will be the same pointer.
For optimal performance, one possibility would be to use hashconsing. However, in your current example, both the generation and the unicity test are quadratic in n. Fixing both points seems to already improve performance a lot.
First, we can avoid the quadratic tree generation by adding a lot of sharing:
let range max =
let rec range elt l n =
if n > max then elt::l
else
let next = Vertex(n,[elt]) in
range next (elt::l) (n+1) in
range (Vertex(0,[])) [] 1
let agg i = Vertex (-1, range i)
With this change, it is become reasonable to generate a tree with 1010 elements (but only 105 unique elements).
Then, the unicity test can be done with a set (or a hashtable):
module S = Set.Make(struct type t = tree let compare = compare end)
let rec topological (set, pool) t =
if S.mem t set then (set, pool) else
let set = S.add t set in
let set, pool =
match t with
| Vertex(_, []) -> set, pool
| Vertex(_, deps) -> List.fold_left topological (set,pool) deps in
set, t::pool
I'm trying to write some code in a functional paradigm for practice. There is one case I'm having some problems wrapping my head around. I am trying to create an array of 5 unique integers from 1, 100. I have been able to solve this without using functional programming:
let uniqueArray = [];
while (uniqueArray.length< 5) {
const newNumber = getRandom1to100();
if (uniqueArray.indexOf(newNumber) < 0) {
uniqueArray.push(newNumber)
}
}
I have access to lodash so I can use that. I was thinking along the lines of:
const uniqueArray = [
getRandom1to100(),
getRandom1to100(),
getRandom1to100(),
getRandom1to100(),
getRandom1to100()
].map((currentVal, index, array) => {
return array.indexOf(currentVal) > -1 ? getRandom1to100 : currentVal;
});
But this obviously wouldn't work because it will always return true because the index is going to be in the array (with more work I could remove that defect) but more importantly it doesn't check for a second time that all values are unique. However, I'm not quite sure how to functionaly mimic a while loop.
Here's an example in OCaml, the key point is that you use accumulators and recursion.
let make () =
Random.self_init ();
let rec make_list prev current max accum =
let number = Random.int 100 in
if current = max then accum
else begin
if number <> prev
then (number + prev) :: make_list number (current + 1) max accum
else accum
end
in
make_list 0 0 5 [] |> Array.of_list
This won't guarantee that the array will be unique, since its only checking by the previous. You could fix that by hiding a hashtable in the closure between make and make_list and doing a constant time lookup.
Here is a stream-based Python approach.
Python's version of a lazy stream is a generator. They can be produced in various ways, including by something which looks like a function definition but uses the key word yield rather than return. For example:
import random
def randNums(a,b):
while True:
yield random.randint(a,b)
Normally generators are used in for-loops but this last generator has an infinite loop hence would hang if you try to iterate over it. Instead, you can use the built-in function next() to get the next item in the string. It is convenient to write a function which works something like Haskell's take:
def take(n,stream):
items = []
for i in range(n):
try:
items.append(next(stream))
except StopIteration:
return items
return items
In Python StopIteration is raised when a generator is exhausted. If this happens before n items, this code just returns however much has been generated, so perhaps I should call it takeAtMost. If you ditch the error-handling then it will crash if there are not enough items -- which maybe you want. In any event, this is used like:
>>> s = randNums(1,10)
>>> take(5,s)
[6, 6, 8, 7, 2]
of course, this allows for repeats.
To make things unique (and to do so in a functional way) we can write a function which takes a stream as input and returns a stream consisting of unique items as output:
def unique(stream):
def f(s):
items = set()
while True:
try:
x = next(s)
if not x in items:
items.add(x)
yield x
except StopIteration:
raise StopIteration
return f(stream)
this creates an stream in a closure that contains a set which can keep track of items that have been seen, only yielding items which are unique. Here I am passing on any StopIteration exception. If the underlying generator has no more elements then there are no more unique elements. I am not 100% sure if I need to explicitly pass on the exception -- (it might happen automatically) but it seems clean to do so.
Used like this:
>>> take(5,unique(randNums(1,10)))
[7, 2, 5, 1, 6]
take(10,unique(randNums(1,10))) will yield a random permutation of 1-10. take(11,unique(randNums(1,10))) will never terminate.
This is a very good question. It's actually quite common. It's even sometimes asked as an interview question.
Here's my solution to generating 5 integers from 0 to 100.
let rec take lst n =
if n = 0 then []
else
match lst with
| [] -> []
| x :: xs -> x :: take xs (n-1)
let shuffle d =
let nd = List.map (fun c -> (Random.bits (), c)) d in
let sond = List.sort compare nd in
List.map snd sond
let rec range a b =
if a >= b then []
else a :: range (a+1) b;;
let _ =
print_endline
(String.concat "\t" ("5 random integers:" :: List.map string_of_int (take (shuffle (range 0 101)) 5)))
How's this:
const addUnique = (ar) => {
const el = getRandom1to100();
return ar.includes(el) ? ar : ar.concat([el])
}
const uniqueArray = (numberOfElements, baseArray) => {
if (numberOfElements < baseArray.length) throw 'invalid input'
return baseArray.length === numberOfElements ? baseArray : uniqueArray(numberOfElements, addUnique(baseArray))
}
const myArray = uniqueArray(5, [])
Right now I have an SML function:
method([1,1,1,1,2,2,2,3,3,3]);
returns:
val it = [[2,2,2],[3,3,3]] : int list list
but I need it to return:
val it = [[1,1,1,1],[2,2,2],[3,3,3]] : int list list
This is my current code:
- fun method2(L: int list) =
= if tl(L) = [] then [hd(L)] else
= if hd(tl(L)) = hd(L) then hd(L)::method(tl(L)) else [hd(L)];
- fun method(L: int list) =
= if tl(L) = [] then [] else
= if hd(tl(L)) = hd(L) then method(tl(L)) else
= method2(tl(L))::method(tl(L));
As you can see it misses the first method2 call. Any ideas on how I can fix this? I am completely stumped.
Your problem is here if hd(tl(L)) = hd(L) then method(tl(L)) else. This is saying if the head of the tail is equal to the head, then continue processing, but don't add it to the result list. this will skip the first contiguous chunk of equal values. I would suggest separating the duties of these functions a bit more. The way to do this is to have method2 strip off the next contiguous chunk of values, and return a pair, where the first element will have the contiguous chunk removed, and the second element will have the remaining list. For example, method2([1, 1, 1, 2, 2, 3, 3]) = ([1, 1, 1], [2, 2, 3, 3]) and method2([2, 2, 3, 3]) = ([2, 2], [3, 3]). Now, you can just keep calling method2 until the second part of the pair is nil.
I'm not quite sure what you are trying to do with your code. I would recommend creating a tail recursive helper function which is passed three arguments:
1) The list of lists you are trying to build up
2) The current list you are building up
3) The list you are processing
In your example, a typical call somewhere in the middle of the computation would look like:
helper([[1,1,1,1]], [2,2],[2,3,3,3])
The recursion would work by looking at the head of the last argument ([2,3,3,3]) as well as the head of the list which is currently being built up ([2,2]) and, since they are the same -- the 2 at the end of the last argument is shunted to the list being built up:
helper([[1,1,1,1]], [2,2,2],[3,3,3])
in the next step in the recursion the heads are then compared and found to be different (2 != 3), so the helper function will put the middle list at the front of the list of lists:
helper([[2,2,2], [1,1,1,1]], [3],[3,3])
the middle list is re-initialized to [3] so it will start growing
eventually you reach something like this:
helper([[2,2,2], [1,1,1,1]], [3,3,3],[])
the [3,3,3] is then tacked onto the list of lists and the reverse of this list is returned.
Once such a helper function is defined, the main method checks for an empty list and, if not empty, initializes the first call to the helper function. The following code fleshes out theses ideas -- using pattern-matching style rather than hd and tl (I am not a big fan of using those functions explicitly -- it makes the code too Lisp-like). If this is homework then you should probably thoroughly understand how it works and then translate it to code involving hd and tl since your professor would regard it as plagiarized if you use things you haven't yet studied and haven't made it your own work:
fun helper (xs, ys, []) = rev (ys::xs)
| helper (xs, y::ys, z::zs) =
if y = z
then helper(xs, z :: y :: ys, zs)
else helper((y::ys)::xs,[z],zs);
fun method [] = []
| method (x::xs) = helper([],[x],xs);
Given this algorithm, I would like to know if there exists an iterative version. Also, I want to know if the iterative version can be faster.
This some kind of pseudo-python...
the algorithm returns a reference to root of the tree
make_tree(array a)
if len(a) == 0
return None;
node = pick a random point from the array
calculate distances of the point against the others
calculate median of such distances
node.left = make_tree(subset of the array, such that the distance of points is lower to the median of distances)
node.right = make_tree(subset, such the distance is greater or equal to the median)
return node
A recursive function with only one recursive call can usually be turned into a tail-recursive function without too much effort, and then it's trivial to convert it into an iterative function. The canonical example here is factorial:
# naïve recursion
def fac(n):
if n <= 1:
return 1
else:
return n * fac(n - 1)
# tail-recursive with accumulator
def fac(n):
def fac_helper(m, k):
if m <= 1:
return k
else:
return fac_helper(m - 1, m * k)
return fac_helper(n, 1)
# iterative with accumulator
def fac(n):
k = 1
while n > 1:
n, k = n - 1, n * k
return k
However, your case here involves two recursive calls, and unless you significantly rework your algorithm, you need to keep a stack. Managing your own stack may be a little faster than using Python's function call stack, but the added speed and depth will probably not be worth the complexity. The canonical example here would be the Fibonacci sequence:
# naïve recursion
def fib(n):
if n <= 1:
return 1
else:
return fib(n - 1) + fib(n - 2)
# tail-recursive with accumulator and stack
def fib(n):
def fib_helper(m, k, stack):
if m <= 1:
if stack:
m = stack.pop()
return fib_helper(m, k + 1, stack)
else:
return k + 1
else:
stack.append(m - 2)
return fib_helper(m - 1, k, stack)
return fib_helper(n, 0, [])
# iterative with accumulator and stack
def fib(n):
k, stack = 0, []
while 1:
if n <= 1:
k = k + 1
if stack:
n = stack.pop()
else:
break
else:
stack.append(n - 2)
n = n - 1
return k
Now, your case is a lot tougher than this: a simple accumulator will have difficulties expressing a partly-built tree with a pointer to where a subtree needs to be generated. You'll want a zipper -- not easy to implement in a not-really-functional language like Python.
Making an iterative version is simply a matter of using your own stack instead of the normal language call stack. I doubt the iterative version would be faster, as the normal call stack is optimized for this purpose.
The data you're getting is random so the tree can be an arbitrary binary tree. For this case, you can use a threaded binary tree, which can be traversed and built w/o recursion and no stack. The nodes have a flag that indicate if the link is a link to another node or how to get to the "next node".
From http://en.wikipedia.org/wiki/Threaded_binary_tree
Depending on how you define "iterative", there is another solution not mentioned by the previous answers. If "iterative" just means "not subject to a stack overflow exception" (but "allowed to use 'let rec'"), then in a language that supports tail calls, you can write a version using continuations (rather than an "explicit stack"). The F# code below illustrates this. It is similar to your original problem, in that it builds a BST out of an array. If the array is shuffled randomly, the tree is relatively balanced and the recursive version does not create too deep a stack. But turn off shuffling, and the tree gets unbalanced, and the recursive version stack-overflows whereas the iterative-with-continuations version continues along happily.
#light
open System
let printResults = false
let MAX = 20000
let shuffleIt = true
// handy helper function
let rng = new Random(0)
let shuffle (arr : array<'a>) = // '
let n = arr.Length
for x in 1..n do
let i = n-x
let j = rng.Next(i+1)
let tmp = arr.[i]
arr.[i] <- arr.[j]
arr.[j] <- tmp
// Same random array
let sampleArray = Array.init MAX (fun x -> x)
if shuffleIt then
shuffle sampleArray
if printResults then
printfn "Sample array is %A" sampleArray
// Tree type
type Tree =
| Node of int * Tree * Tree
| Leaf
// MakeTree1 is recursive
let rec MakeTree1 (arr : array<int>) lo hi = // [lo,hi)
if lo = hi then
Leaf
else
let pivot = arr.[lo]
// partition
let mutable storeIndex = lo + 1
for i in lo + 1 .. hi - 1 do
if arr.[i] < pivot then
let tmp = arr.[i]
arr.[i] <- arr.[storeIndex]
arr.[storeIndex] <- tmp
storeIndex <- storeIndex + 1
Node(pivot, MakeTree1 arr (lo+1) storeIndex, MakeTree1 arr storeIndex hi)
// MakeTree2 has all tail calls (uses continuations rather than a stack, see
// http://lorgonblog.spaces.live.com/blog/cns!701679AD17B6D310!171.entry
// for more explanation)
let MakeTree2 (arr : array<int>) lo hi = // [lo,hi)
let rec MakeTree2Helper (arr : array<int>) lo hi k =
if lo = hi then
k Leaf
else
let pivot = arr.[lo]
// partition
let storeIndex = ref(lo + 1)
for i in lo + 1 .. hi - 1 do
if arr.[i] < pivot then
let tmp = arr.[i]
arr.[i] <- arr.[!storeIndex]
arr.[!storeIndex] <- tmp
storeIndex := !storeIndex + 1
MakeTree2Helper arr (lo+1) !storeIndex (fun lacc ->
MakeTree2Helper arr !storeIndex hi (fun racc ->
k (Node(pivot,lacc,racc))))
MakeTree2Helper arr lo hi (fun x -> x)
// MakeTree2 never stack overflows
printfn "calling MakeTree2..."
let tree2 = MakeTree2 sampleArray 0 MAX
if printResults then
printfn "MakeTree2 yields"
printfn "%A" tree2
// MakeTree1 might stack overflow
printfn "calling MakeTree1..."
let tree1 = MakeTree1 sampleArray 0 MAX
if printResults then
printfn "MakeTree1 yields"
printfn "%A" tree1
printfn "Trees are equal: %A" (tree1 = tree2)
Yes it is possible to make any recursive algorithm iterative. Implicitly, when you create a recursive algorithm each call places the prior call onto the stack. What you want to do is make the implicit call stack into an explicit one. The iterative version won't necessarily be faster, but you won't have to worry about a stack overflow. (do I get a badge for using the name of the site in my answer?
While it is true in the general sense that directly converting a recursive algorithm into an iterative one will require an explicit stack, there is a specific sub-set of algorithms which render directly in iterative form (without the need for a stack). These renderings may not have the same performance guarantees (iterating over a functional list vs recursive deconstruction), but they do often exist.
Here is stack based iterative solution (Java):
public static Tree builtBSTFromSortedArray(int[] inputArray){
Stack toBeDone=new Stack("sub trees to be created under these nodes");
//initialize start and end
int start=0;
int end=inputArray.length-1;
//keep memoy of the position (in the array) of the previously created node
int previous_end=end;
int previous_start=start;
//Create the result tree
Node root=new Node(inputArray[(start+end)/2]);
Tree result=new Tree(root);
while(root!=null){
System.out.println("Current root="+root.data);
//calculate last middle (last node position using the last start and last end)
int last_mid=(previous_start+previous_end)/2;
//*********** add left node to the previously created node ***********
//calculate new start and new end positions
//end is the previous index position minus 1
end=last_mid-1;
//start will not change for left nodes generation
start=previous_start;
//check if the index exists in the array and add the left node
if (end>=start){
root.left=new Node(inputArray[((start+end)/2)]);
System.out.println("\tCurrent root.left="+root.left.data);
}
else
root.left=null;
//save previous_end value (to be used in right node creation)
int previous_end_bck=previous_end;
//update previous end
previous_end=end;
//*********** add right node to the previously created node ***********
//get the initial value (inside the current iteration) of previous end
end=previous_end_bck;
//start is the previous index position plus one
start=last_mid+1;
//check if the index exists in the array and add the right node
if (start<=end){
root.right=new Node(inputArray[((start+end)/2)]);
System.out.println("\tCurrent root.right="+root.right.data);
//save the created node and its index position (start & end) in the array to toBeDone stack
toBeDone.push(root.right);
toBeDone.push(new Node(start));
toBeDone.push(new Node(end));
}
//*********** update the value of root ***********
if (root.left!=null){
root=root.left;
}
else{
if (toBeDone.top!=null) previous_end=toBeDone.pop().data;
if (toBeDone.top!=null) previous_start=toBeDone.pop().data;
root=toBeDone.pop();
}
}
return result;
}