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 am trying to write a function in SML which when given a list of general elements, reorders its elements into equivalent classes and returns a list of these classes (type "a list list).
Leave the elements in the classes in the same order as in the original list.
A given function defines the equivalence of the elements and it returns true if the elements are equivalent or false otherwise.
I cannot seem to get a grip on the solution.
fun sample x y = x = y
Required type: fn : (''a -> ''a -> bool) -> ''a list -> ''a list list
Thank you very much for the help.
The helper function does not work correctly, all I want to do with it is see if a given element belongs to any of the classes and put it accordingly inside or create a new sublist which contains it.
fun srt listoflists func new =
case listoflists of [] => [[]]
| a::b => if func (new, hd a) = true then (new::a)::b
else if func (new, hd a) = false then a::(srt b func new) else [new]::a::b
The sample functions checks equivalence of two elements when divided by 11.
Tests are not all working, it is not adding 17 into a new class.
srt [[7,7,7],[5,5,5],[11,11,11],[13,13,13]] eq 7;
val it = [[7,7,7,7],[5,5,5],[11,11,11],[13,13,13]] : int list list
- srt [[7,7,7],[5,5,5],[11,11,11],[13,13,13]] eq 5;
val it = [[7,7,7],[5,5,5,5],[11,11,11],[13,13,13]] : int list list
- srt [[7,7,7],[5,5,5],[11,11,11],[13,13,13]] eq 11;
val it = [[7,7,7],[5,5,5],[11,11,11,11],[13,13,13]] : int list list
- srt [[7,7,7],[5,5,5],[11,11,11],[13,13,13]] eq 13;
val it = [[7,7,7],[5,5,5],[11,11,11],[13,13,13,13]] : int list list
- srt [[7,7,7],[5,5,5],[11,11,11],[13,13,13]] eq 17;
val it = [[7,7,7],[5,5,5],[11,11,11],[13,13,13],[]] : int list list
- srt [[7,7,7],[5,5,5],[11,11,11],[13,13,13],[111,111,111]] eq 111;
val it = [[7,7,7],[5,5,5],[11,11,11],[13,13,13],[111,111,111,111]]
How to correct this and also once this helper function works, how to encorporate it exactly into the main function that is required.
Thank you very much.
Your example code seems like you are getting close, but has several issues
1) The basis cases is where new should be added, so in that case you should return the value [[new]] rather than [[]]
2) Your problem description suggests that func be of type ''a -> ''a -> bool but your code for srt seems to be assuming it is of type (''a * ''a) -> bool. Rather than subexpressions like func (new, hd a) you need func new (hd a) (note the parentheses location).
3) if func returns a bool then comparing the output to true is needlessly verbose, instead of if func new (hd a) = true then ... simply have if func new (hd a) then ...
4) Since you are adding [new] in the basis cases, your second clause is needlessly verbose. I see no reason to have any nested if expressions.
Since this seems to be homework, I don't want to say much more. Once you get the helper working correctly it should be fairly straightforward to use it (in the recursive case) of the overall function. Note that you could use (a # [new])::b rather than (new::a)::b if you want to avoid the need for a final mapping of rev across the final return value. # is more expensive than :: (it is O(n) rather than O(1)), but for small examples it really doesn't matter and could even be slightly better since it would avoid the final step of reversing the lists.
Following my previous post here , I tried to do what was suggested and convert the code
into a Tail-recursion method with let .
The original code - which does not work (due to using val inside if condition) :
fun func() =
val decimal = 0 (* the final result *)
val multiple = 0 (* keeps track of multiples, eg. In XXV, X would be a multiple *)
val current = 0 (* the digit currently being processed *)
val top = 0 (* value of the last element in the list *)
val last_add = 0 (* the last digit that wasn't a multiple, or subtraction operation *)
val last_sub = 0
val problem = 0 (* if value is 1 then there is a problem with the input *)
val myList = [1,2,3,4,5] (* the list has more values *)
while (myList <> []) (* run while the list is not empty *)
val current = tl(myList) (* grab the last element from the list *)
val myList = tl(myList) (* remove the last element from the list *)
val top = tl(myList) (* grab the value at the end of the list *)
if ( myList <> []) andalso (current > top))
then
val decimal = decimal + current - top
val last_sub = top;
val myList = tl(myList)
else
if ( (myList = []) andalso (current = top))
then val decimal = decimal + current
val multiple = multiple + 1
else
if (last_sub = current)
then val problem = 1
else
val decimal = decimal + current
val multiple = 0
val last_add = current
And the code as a tail-recursion method :
fun calc [] = 0
|calc [x] = x
|calc (head::tail) =
let
val decimal = 0
val multiple = 0
val current = 0
val top = 0
val last_add = 0
val last_sub = 0
val problem = 0
val doNothing = 0
in
let
val current = hd(rev(head::tail)) (* grab the last element *)
val head::tail = rev(tl(rev(head::tail))) (* POP action - remove the last element from the list *)
val top = hd(rev(head::tail)) (* grab the new last element after removing *)
in
if (current > top) then
let
val decimal = decimal + current - top
val last_sub = top
val head::tail = rev(tl(rev(head::tail))) (* POP action - remove the last element from the list *)
in
calc(head::tail)
end
else
if ( (head::tail = []) andalso (current = top))
then let
val decimal = decimal + current
val multiple = multiple + 1
in
calc(head::tail)
end
else
if (last_sub <> current)
then let
val decimal = decimal + current
val multiple = 0
val last_add = current
in
calc(head::tail)
end
else
(* do nothing *)
val doNothing = 0
end
end;
However , when I try to enter :
calc([0,100,20,30,4,50]);
I get :
uncaught exception Bind [nonexhaustive binding failure]
raised at: stdIn:216.13-216.50
I know the code is very hard to read and pretty long , but it would be greatly appreciated
if someone could explain to me how to fix it , or help me find the reason for this output .
Thanks
You have a few issues with your code.
First of all, you can use last to grab the last element of a list. See the List documentation for more info. But unless you have a really good reason to do so, it's easier and much more efficient to simply start from the beginning of the list and pop elements off the beginning as you recurse. You already have the first element bound to head in your code using pattern matching.
Secondly, unless you use refs (which you probably don't want to do) there are no variables in Standard ML, only values. What this means is that if you want to carry state between invocations, any accumulators need to be parameters of your function. Using a helper function to initialize accumulators is a common pattern.
Third, instead of comparing a list to [] to test if it's empty, use the null function. Trust me on this. You'll get warnings using = because of subtle type inference issues. Better yet, use a pattern match on your function's parameters or use a case statement. Pattern matching allows the compiler to tell you whether you've handled all possible cases.
Fourth, SML typically uses camelCase, not snake_case, for variable names. This is more stylistic, but as you write more code and collaborate, you're going to want to fit with the conventions.
Fifth, when you do recursion on a list, don't try to look at multiple values in the list. This complicates things. Treat it as a head element and tail list, and everything will become much simpler. In my code, instead of keeping current in the list, I did this by splitting it out into a separate parameter. Have a base case where you simply return the answer from one of your accumulators, and a recursive case where you recurse with updated accumulator values and a single value popped from your list. This eliminates the problem scenario.
I'm not sure if this logic is correct since I don't know what you're trying to calculate, but check out this code which illustrates some of the things I talked about.
(* This is the helper function which takes accumulators as
parameters. You shouldn't call this directly. *)
fun calc' decimal _ _ _ _ [] =
(* We processed everything in the list. Just return the accumulator. *)
decimal
| calc' decimal multiple lastAdd lastSub current (top::tail) =
(* This case is for when there are 1 or more elements in the list. *)
if current > top then
calc' (decimal + current - top) multiple lastAdd top top tail
else if current = top then
calc' (decimal + current) (multiple + 1) lastAdd lastSub top tail
else
calc' (decimal + current) 0 current lastSub top tail
(* This is the function you should call. *)
fun calc [] = 0
| calc [_] = 0 (* Given a single-element list. *)
| calc (x::xs) =
(* Apply the helper with correct initial values. *)
calc' 0 0 0 0 x xs
In a functional language, instead of assigning to a variable when you want to change it, simply recurse and specify the new value for the correct parameter. This is how you write a "loop" in a functional language using recursion. As long as you only use tail-recursion, it will be just as efficient as a while loop in your favorite imperative language.
I need to use hashtable of mutable variable in Ocaml, but it doesn't work out.
let link = Hashtbl.create 3;;
let a = ref [1;2];;
let b = ref [3;4];;
Hashtbl.add link a b;;
# Hashtbl.mem link a;;
- : bool = true
# a := 5::!a;;
- : unit = ()
# Hashtbl.mem link a;;
- : bool = false
Is there any way to make it works?
You can use the functorial interface, which lets you supply your own hash and equality functions. Then you write functions that are based only on the non-mutable parts of your values. In this example, there are no non-mutable parts. So, it's not especially clear what you're expecting to find in the table. But in a more realistic example (in my experience) there are non-mutable parts that you can use.
If there aren't any non-mutable parts, you can add them specifically for use in hashing. You could add an arbitrary unique integer to each value, for example.
It's also possible to do hashing based on physical equality (==), which has a reasonable definition for references (and other mutable values). You have to be careful with it, though, as physical equality is a little tricky. For example, you can't use the physical address of a value as your hash key--an address can change at any time due to garbage collection.
Mutable variables that may happen to have the same content can still be distinguished because they are stored at different locations in memory. They can be compared with the physical equality operator (==). However, OCaml doesn't provide anything better than equality, it doesn't provide a nontrivial hash function or order on references, so the only data structure you can build to store references is an association list of some form, with $\Theta(n)$ access time for most uses.
(You can actually get at the underlying pointer if you play dirty. But the pointer can move under your feet. There is a way to make use of it nonetheless, but if you need to ask, you shouldn't use it. And you aren't desperate enough for that anyway.)
It would be easy to compare references if two distinct references had a distinct content. So make it so! Add a unique identifier to your references. Keep a global counter, increment it by 1 each time you create a reference, and store the counter value with the data. Now your references can be indexed by their counter value.
let counter = ref 0
let new_var x = incr counter; ref (!counter, x)
let var_value v = snd !v
let update_var v x = v := (fst !v, x)
let hash_var v = Hashtbl.hash (fst !v)
For better type safety and improved efficiency, define a data structure containing a counter value and an item.
let counter = ref 0
type counter = int
type 'a variable = {
key : counter;
mutable data : 'a;
}
let new_var x = incr counter; {key = !counter; data = x}
let hash_var v = Hashtbl.hash v.key
You can put the code above in a module and make the counter type abstract. Also, you can define a hash table module using the Hashtbl functorial interface. Here's another way to define variables and a hash table structure on them with a cleaner, more modular structure.
module Counter = (struct
type t = int
let counter = ref 0
let next () = incr counter; !counter
let value c = c
end : sig
type t
val next : unit -> t
val value : t -> int
end)
module Variable = struct
type 'a variable = {
mutable data : 'a;
key : Counter.t;
}
let make x = {key = Counter.next(); data = x}
let update v x = v.data <- x
let get v = v.data
let equal v1 v2 = v1 == v2
let hash v = Counter.value v.key
let compare v1 v2 = Counter.value v2.key - Counter.value v1.key
end
module Make = functor(A : sig type t end) -> struct
module M = struct
type t = A.t Variable.variable
include Variable
end
module Hashtbl = Hashtbl.Make(M)
module Set = Set.Make(M)
module Map = Map.Make(M)
end
We need the intermediate module Variable because the type variable is parametric and the standard library data structure functors (Hashtbl.Make, Set.Make, Map.Make) are only defined for type constructors with no argument. Here's an interface that exposes both the polymorphic interface (with the associated functions, but no data structures) and a functor to build any number of monomorphic instances, with an associated hash table (and set, and map) type.
module Variable : sig
type 'a variable
val make : 'a -> 'a variable
val update : 'a variable -> 'a -> unit
val get : 'a variable -> 'a
val equal : 'a -> 'a -> bool
val hash : 'a variable -> int
val compare : 'a variable -> 'b variable -> int
end
module Make : functor(A : sig type t end) -> sig
module M : sig
type t = A.t variable.variable
val make : A.t -> t
val update : t -> A.t -> unit
val get : t -> A.t
val equal : t -> t -> bool
val hash : t -> int
val compare : t -> t -> int
end
module Hashtbl : Hashtbl.S with type key = M.t
module Set : Set.S with type key = M.t
module Map : Map.S with type key = M.t
end
Note that if you expect that your program may generate more than 2^30 variables during a run, an int won't cut it. You need two int values to make a 2^60-bit value, or an Int64.t.
Note that if your program is multithreaded, you need a lock around the counter, to make the incrementation and lookup atomic.
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;
}