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 the following question "Given a list of integer pairs, write a function to return a list of even numbers in that list in sml".
this is what I've achieved so far
val x = [(6, 2), (3, 4), (5, 6), (7, 8), (9, 10)];
fun isEven(num : int) =
if num mod 2 = 0 then num else 0;
fun evenNumbers(list : (int * int) list) =
if null list then [] else
if isEven(#1 (hd list)) <> 0
then if isEven(#2 (hd list)) <> 0
then #1 (hd list) :: #1 (hd list) :: evenNumbers(tl list)
else []
else if isEven(#2 (hd list)) <> 0
then #1 (hd list) :: evenNumbers(tl list)
else [];
evenNumbers(x);
the result should be like this [6,2,4,6,8,10]
any help would be appreciated.
I see two obvious problems.
If both the first and second number are even, you do
#1 (hd list) :: #1 (hd list) :: evenNumbers(tl list)
which adds the first number twice and ignores the second.
If the first number is odd and the second even, you do
#1 (hd list) :: evenNumbers(tl list)
which adds the number that you know is odd and ignores the one you know is even.
Programming with selectors and conditionals gets complicated very quickly (as you've noticed).
With pattern matching, you could write
fun evenNumbers [] = []
| evenNumber ((x,y)::xys) = ...
and reduce the risk of using the wrong selector.
However, this still makes for complicated logic, and there is a better way.
Consider the simpler problem of filtering the odd numbers out of a list of numbers, not pairs.
If you transform the input into such a list, you only need to solve that simpler problem (and there's a fair chance that you've already solved something very similar in a previous exercise).
Exercise: implement this transformation. Its type will be ('a * 'a) list -> 'a list.
Also, your isEven is more useful if it produces a truth value (if you ask someone, "is 36 even?", "36" is a very strange answer).
fun isEven x = x mod 2 = 0
Now, evenNumbers can be implemented as "just" a combination of other, more general, functions.
So running your current code,
- evenNumbers [(6, 2), (3, 4), (5, 6), (7, 8), (9, 10)];
val it = [6,6,3,5,7,9] : int list
suggests that you're not catching all even numbers, and that you're catching some odd numbers.
The function isEven sounds very much like you want to have the type int -> bool like so:
fun isEven n =
n mod 2 = 0
Instead of addressing the logic error of your current solution, I would like to propose a syntactically much simpler approach which is to use pattern matching and fewer explicit type annotations. One basis for such a solution could look like:
fun evenNumbers [] = ...
| evenNumbers ((x,y)::pairs) = ...
Using pattern matching is an alternative to if-then-else: the [] pattern is equivalent to if null list ... and the (x,y)::pairs pattern matches when the input list is non-empty (holds at least one element, being (x,y). At the same time, it deconstructs this one element into its parts, x and y. So in the second function body you can express isEven x and isEven y.
As there is a total of four combinations of whether x and y are even or not, this could easily end up with a similarly complicated nest of if-then-else's. For this I might do either one of two things:
Use case-of (and call evenNumbers recursively on pairs):
fun evenNumbers [] = ...
| evenNumbers ((x,y)::pairs) =
case (isEven x, isEven y) of
... => ...
| ... => ...
Flatten the list of pairs into a list of integers and filter it:
fun flatten [] = ...
| flatten ((x,y)::pairs) = ...
val evenNumbers pairs = ...
I'm trying to write a function that finds the longest running sub-list of increasing integers in a list.
This is what I have so far (very long variable names to help me understand it better):
let increasing list =
let rec aux inputList currentLongestList currentHighestList lastInsertedElement =
match inputList with
| [] ->
if List.length currentLongestList > List.length currentHighestList then currentLongestList
else currentLongestList
| hd :: tl ->
if hd > lastInsertedElement then aux tl (currentLongestList # hd) currentHighestList hd
else aux tl currentLongestList currentHighestList lastInsertedElement
in
aux list [] [] [min_int]
;;
My issue is that when this is entered, it says val increasing : int list list -> int list = <fun> implying 2 lists need to be passed where I only want one in the function? What am I doing wrong? Also, the function doesn't work if I take away the square brackets from min_int, even though I don't see why they're needed - what would I have to do to get rid of them?
The inferred type informs you that the argument named list is a list of lists of integers. Let us understand why, step by step.
First, the type checker has unified the type list with the type of inputList because of your call to aux at the end of the function increasing.
Second, the head of the aux argument named inputList is compared with lastInsertedElement in the second case of aux pattern-matching. Since lastInsertedElement is of type int list because of your initial call to aux, the type induces that inputList has type int list list.
For your second question, "why the brackets seem needed around min_int?". Look at the expression "... # hd". It implies that "hd" must be a list...
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.
I wrote a recursive version of index as follows
let index list value =
let rec counter num = function
| [] -> -1
| h::t ->
if h == value
then num
else (counter (num + 1)) t
in counter 0 list;;
It works, but then our professor said we should use a tail recursive version in order to not timeout on the server, so I wrote a new index function using fold, but I can't seem to figure out why if it doesn't find the element, it returns a number greater than the length of the list, even though I want it to return -1.
let index2 list value = fold (fun i v ->
if i > (length list) then -1
else if v == value then i
else i+1) 0 list;;
Here's my fold version as well:
let rec fold f a l = match l with
[] -> a
| (h::t) -> fold f (f a h) t;;
Your folded function is called once for each element of the list. So you'll never see a value of i that's greater than (length list - 1).
As a side comment, it's quite inefficient (quadratic complexity) to keep calculating the length of the list. It would be better to calculate it once at the beginning.
As another side comment, you almost never want to use the == operator. Use the = operator instead.
EDIT
Why do you redefine fold instead of using List.fold_left?
Your first version of index is already tail recursive, but you can improve its style by:
using option type instead of returning -1 if not found;
directly call index recursively instead of a count function;
use = (structural) comparator instead of == (physical);
use a guard in your pattern matching instead of an if statement.
So
let index list value =
let rec index' list value i = match list with
| [] -> None
| h :: _ when h = value -> Some i
| _ :: t -> index' t value (succ i)
in index' list value 0
And as already said, index2 does not work because you'll never reach an element whose index is greater than the length of the list, so you just have to replace i > (length list) with i = (length list) - 1 to make it work.
But index2 is less efficient than index because index stops as soon as the element is found whereas index2 always evaluate each element of the list and compare the list length to the counter each time.