I want to find the max element in a int list.
The idea is to call find_max only once and let support do the dirty job.
The function support returns an int, the first value is 0 then when a new max is found it's value is saved, added to the result ,the previous max is changed and it's value deducted from the result.
This because :
old_max = x
result = 0 + old_max
A new max is found :
new_max= y
result = result - oldmax + new_max
So I'll save the value of new_max :
0 + old_max - old_max + new_max = new_max`.
Obviously the code above is explicative, this is my real code :
let find_max random =
let rec support rlist max =
if rlist==[] then 0
else
if (List.hd rlist) > max
then -max + (List.hd rlist) + support (List.tl rlist) (List.hd rlist)
else support (List.tl rlist) max ;;
let return = support random 0 0 ;
!return;;
let a = [1;2;3];
print_string "max element in list is : "
print_int (find_max a);
print_string "\n"
The error is on line 9 !return;;, syntax error (obviously :/ ) on ;;
There is no construct let ... = ...; in OCaml, local definition use let .. = ... in ... . You probably want to avoid using ;; altogether as a beginner too.
Also, structural equality is = and not ==. Similarly, you should never useList.hd and List.tl in your code as a beginner. Pattern matching is always the superior option.
Typically, all uses of those functions can be replaced by a simple:
match rlist with
| [] -> 0
| hd :: tl -> ...
which is shorter, clearer, and eliminate any possibility to mishandle the empty list.
Your logic is also unnecessarily complex rather than computing max - initial_value with
-max + hd + support tl hd
you can compute the maximum directly with
hd
You are then calling support support with too many argument.
We may want to use let () = ... when computing effect, rather than using ;;.
Related
In SML is it possible to find the number of occurrences of the min number in a list?
I have code to find the number of occurrences of a number but i am stumped on how to find the min and use it to find how many of the minimum num there is.
fun occurrences(nil, n)=0
| occurrences(ls, n) =
if hd(ls)=n then occurrences(tl(ls),n) + 1
else occurrences(tl(ls),n) + 0;
Thank you!
You can write a function that keeps track of the min value and its count as you iterate through the list.
We can do this by implementing a tail-recursive function which helper, which maintains the value of the current minimum and a count of the number of times that item has appeared.
We can then wrap this in another function min_count via a let-in-end block.
For example:
fun min_count [] = 0 (* the empty list has zero items in it *)
| min_count (x :: xs) =
let
(* when we reach the end of the list, return the accumulated count *)
fun helper (_, n) [] = n
| helper (m, n) (y :: ys) =
(* if we find a new minimum, reset the count *)
if y < m then helper (y, 1) ys
(* if the current list item is larger than the min, ignore it *)
else if y > m then helper (m, n) ys
(* if we've found another instance of the min, add one to the count *)
else helper (m, n + 1) ys
in
(* first item appears once *)
helper (x, 1) xs (* first item appears once *)
end;
This problem is a good test for using folds on a list.
Finding the mininum
If we want to find the minimum in a list we need to iterate over the list checking each element against a predetermined starting minimum value. If that element is less than that known minimum, we continue to iterate using that value instead. When we're done, we have the minimum value.
If the list is empty, there is no minimum value. If only one value is in the list, the minimum is obviously that. If there are more values, the starting minimum value is the first element.
We can use foldl to handle the iteration in this last case.
fun min([]) = NONE
| min([x]) = SOME x
| min(first::rest) =
SOME (foldl (fn (x, min) => if x < min then x else min)
first rest)
Finding occurrences
You've already done this, but this can be done in terms of a fold as well.
fun occurrences(lst, v) =
foldl (fn (x, count) => if x = v then count + 1 else count)
0 lst
Putting this together
We could use these two functions to find the number of times the minimum occurs in a list.
let
val numbers = [1, 4, 7, 2, 9, 0, 1, 6, 0]
val min = min(numbers)
val occ = case min of
NONE => NONE
| SOME x => SOME (occurrences(numbers, x))
in
case (min, occ) of
(NONE, NONE) => print("No minimum found.")
| (SOME m, SOME t) => print("Min: " ^ Int.toString(m) ^ "; times: " ^ Int.toString(t))
end
Can we do it in a single pass?
Using the above approach, we have to iterate over the list twice. This is a more general, but less efficient way of getting both pieces of information the minimum and the number of occurrences of it. We can use foldl to get both pieces of information, and it's going to look at lot like the definition of min.
We just need to pass a function to foldl that keeps a running tally of the number of times it has found the minimum value, and we need to pass it a tuple with both an initial minimum value and an initial count of 1.
fun minCount([]) = NONE
| minCount([x]) = SOME (x, 1)
| minCount(first::rest) =
SOME (foldl (fn (x, init as (min, count)) =>
case Int.compare(x, min) of
EQUAL => (min, count + 1)
| LESS => (x, 1)
| _ => init)
(first, 1)
rest)
With this function defined, our previous code can be rewritten as:
let
val numbers = [1, 4, 7, 2, 9, 0, 1, 6, 0]
val mc = minCount(numbers)
in
case mc of
NONE => print("No minimum found.")
| SOME (m, t) => print("Min: " ^ Int.toString(m) ^ "; times: " ^ Int.toString(t))
end
Assuming that you are supposed to use your occurrences function in the solution, write a function that finds the minimum,
fun minimum [x] = x
| minimum (x::xs) = let
val min = minimum xs
in
if x < min then x else min
end
Note that this does not handle the empty list.
You need to decide whether to leave the missing pattern as a runtime error, or add it and handle the error, for instance by raising an exception or by changing the return type to int option.
If you're taking a course, use one of the methods you've learned so far.
Then you can use that function,
occurrences(the_list, minimum the_list)
I am trying to write a simple function in OCaml
let rec pell (i: int) =
(if i <= 2 then i (*if given n is less tahn 2 then return 2, else return previous n-1 th term and n-2 nd term recursively*)
else if i>2 then
2 * pell i - 1 + pell i - 2
else failwith "unimplemented" (*else fail with unimplemented message*)
);;
Write an infinite precision version of the pell function from before
pell2 0 = []
pell2 1 = [1]
pell2 7 = [9; 6; 1]
pell2 50 = [2; 2; 5; 3; 5; 1; 4; 2; 9; 2; 4; 6; 2; 5; 7; 6; 6; 8; 4]
I have written below code for this:
let rec pell2 i =
(if i <= 2 then
[] -> i;
else if i=0 then [];
else if i>2 then (*finding pell number and using sum function to
output list with infinite precision...*)
[] -> pell2 i-1 + pell2 i-2;
else failwith "unimplemented"
);;
but still has some syntax errors. Can someone help me with this please.
if i <= 2 then
[] -> i
In snippets like this, the -> is invalid. It looks like you might be mixing pattern matching with match ... with ... and if/else up.
Also, you're first checking if i is less than or equal to 2, but then you have an else to test for i being equal to zero. The first check means the second is never going to happen.
First, let's look at the examples for the output of pell2. We see that pell2 has a single integer parameter, and returns a list of integers. So, we know that the function we want to create has the following type signature:
pell2: int -> int list
Fixing (some but not all of) the syntax errors and trying to maintain your logic,
let rec pell2 i =
if i=0 then []
else if i <= 2 then i
else if i>2 then pell2 i-1 + pell2 i-2
Note that I removed the semicolons at the end of each expression since OCaml's use of a semicolon in its syntax is specifically for dealing with expressions that evaluate to unit (). See ivg's excellent explanation on this. The major flaw with this code is that it does not type check. We see that we conditionally return a list, and otherwise return an int. Notice how above we defined that pell2 should return an int list. So, we can begin fixing this by wrapping our int results in a list:
let rec pell2 n =
if n = 0 then []
else if n <= 2 then [n]
else ... something that will return the Pell number as a list ...
As you have already written, the else branch can be written using recursive calls to the pell2 function. However, we can't write it as you did previously, because pell2 evaluates to a list, and the binary operator + only works on two integers. So, we will have to define our own way of summing lists. Calling this sum_lists, we are left with the following code:
We can now fully define our function pell2:
let rec pell2 n =
if n = 0 then []
else if n <= 2 then [n]
else (* Pell(n) = (2 * Pell(n-1)) + Pell(n-2) *)
let half_of_first_term = pell2 n-1 in
let first_term = sum_lists half_of_first_term half_of_first_term in
let second_term = pell2 n-2 in
sum_lists first_term second_term
So, all that is left is to define sum_lists, so that we are properly summing together two lists of the same format as the return type of pell2. The signature for sum_lists would be
sum_lists: int list -> int list -> int list
I'll give a basic outline of the implementation, but will leave the rest for you to figure out, as this is the main crux of the assignment problem.
let sum_lists lst1 lst2 =
let rec sum_lists_helper lst1 lst2 carry =
match lst1, lst2 with
| [], [] -> if carry = 1 then [1] else []
| h::t, []
| [], h::t -> ...
| h1::t1, h2::t2 -> ...
in
sum_lists_helper lst1 lst2 0
I have the following problem: given a max(max) apacity, and given a list of values(listOfValues) i need to return a list with values from the listOfValues. The sum of the elements must be <= max and i need to prioritize the higher values.
Example: typing solvingProblem 103 [15, 20, 5, 45, 34] i must get: [45, 45, 5, 5]
To solve the problem i create the following code:
solvingProblem max [] = 0
solvingProblem max listOfValues | max == 0 = 0
| otherwise = createList max listOfValues []
createList max [] result = -1
createList max listOfValues result | smaller listOfValues > max = -1
| higher listOfValues > max = createList max (remove (higher listOfValues) listOfValues) result
| otherwise = createList (max - higher listOfValues) listOfValues (insert (higher listOfValues) result)
higher [a] = a
higher (a:b:x) | a > b = higher (a:x)
| otherwise = higher (b:x)
smaller [a] = a
smaller (a:b:x) | a < b = smaller (a:x)
| otherwise = smaller (b:x)
remove x [] = []
remove x (h:t) | x == h = remove x t
| otherwise = h : remove x t
insert x (h:t) = x : h : t
In the two lines where i'll returning "-1" should be the parameter "result", but if i change "-1" to "result" the code don't load on ghci.
Can someone help me?
Thank you and sorry for my bad english.
If I may begin with a bit of a side note, some of your functions already exist in Haskell (now that I come to think of it you might have written them for an exercise, but just in case it wouldn't be the case, let's discuss that): your higher is maximum, your smaller is minimum and your insert is just (:), beacause like you write it yourself insert x list = x:list. Note that your version will fail if you give it the empty list because the pattern matching is non-exhaustive. Also you could write remove in terms of filter: remove x list = filter (== x) list.
Now why doesn't your code load properly? ghci tells you:
• Non type-variable argument in the constraint: Num [a]
(Use FlexibleContexts to permit this)
• When checking the inferred type
solvingProblem :: forall a.
(Ord a, Num [a], Num a) =>
a -> [a] -> [a]
Which I agree is pretty cryptic, but what it's saying is that the return type of solvingProblem is a list of a and for some reason it is also an instance of the Num type class. The reason why it says it's an instance of Num is because one of the return value of solvingProblem is 0 which is a number, which is a bit odd because it is also a list. Changing the 0 with [] makes the code compile and work (if you change insert with (:) otherwise you get the non-exhaustive pattern matching I was talking about earlier).
λ> solvingProblem 103 [15,20, 5, 45, 34]
[5,5,45,45]
it :: (Ord t, Num t) => [t]
The problem is with the last guard clause in createList.
The type you intended for createList seems to be:
createList :: Int -> [Int] -> Int -> Int
but if you look at the last guard clause you have:
| otherwise = createList (max - ...) listOfValues (insert ...)
^^^^^^^^^^^ ^^^^^^^^^^^^ ^^^^^^^^^^^^
Int [Int] [Int]
Even though GHC is very good at inferring types, always adding
type signatures to your code is a good way of catching these kinds
of errors early.
I have trouble with these proofs that seem almost trivially obvious.
For instance, in the inductive case if I assume the property in the title and I want to show:
length (h'::h::l) = 1 + length (h::l)
Where do I go from here? It is so obviously true but I don't know what steps I can take without proving some kind of lemma. For instance I could say
length ([h']#(h::l)) = 1 + length (h::l)
But now I have to prove something along the lines of
length (l1#l2) = length l1 + length l2
I'm having trouble understanding when I need to prove lemmas, especially in the proofs that seem almost trivial.
When you prove program correctness, you're usually working with some implementation. If you will take a trivial implementation, then the proof would be also trivial. Suppose we have the following implementation:
let rec length = function
| [] -> 0
| x::xs -> 1 + length xs
We have a proof obligation:
length (x::xs) = 1 + length xs
We proof this using structural induction. I'm assuming, that list is defined as
type 'a list =
| Nil
| Cons ('a,'a list)
and [] is a syntactic sugar for Nil, while x::xs is a syntactic sugar for Cons (x,xs)
So we perform case by case analysis. We have only one applicable case, so we
take case
| x::xs -> 1 + length xs
rewrite length (x::xs) with the right hand side, and we get:
1 + legnth xs = 1 + length xs
This can be proven by reflexivity of = operator. (If it is reflexive in your logic).
Note: the above implementation is trivial. In OCaml standard library List.length is implemented as follows:
let rec length_aux len = function
[] -> len
| a::l -> length_aux (len + 1) l
let length l = length_aux 0 l
Here proof obligation length (x::xs) = 1 + length xs produces an obligation to proof that length_aux 0 (x::xs) = 1 + length_aux 0 xs. This is less trivial.
I would say first, that length is defined by induction, length of void is 0, and length(h::l) = 1 + length(l).
Then, concatenation is defined also by induction, []#l=l, and [h]#l = h::l.
length is a function that maps # to + : the proof is an induction proof using above properties.
You proceed by induction on l1 : the property length(l1#l2) = length(l1)+length(l2) when l1 is empty (induction axiom).
Then assuming the property is right with for l1 of length n, you want to prove it is correct for n+1. length(h::l1#l2) = 1 + length(l1#l2) (thx to length definition). Then by induction hypothesis, you length(l1#l2) = length(l1)+length(l2), you conclude.
I am trying to achieve the following: Finding the element at a specific index.
So if I had a list of [5; 2; 3; 6] and ask for the element at index 2, it would return 3.
let counter = 0;;
let increase_counter c = c + 1;;
let rec get_val x n = match x with
[] -> -1
| (h::t) ->
if (counter = n) then
h
else
increase_counter counter ; get_val t n
;;
But this code is giving me a bug saying that -1 is not of type 'unit'?
As Jeffrey Scofield said, you should write let counter = ref 0 to make counter mutable. Now, you can use the built in incr function to increment it (equivalent to counter := !counter + 1), and you'll get its value with !counter.
There is also a problem in your algorithm : if the counter is equal to n, you return the head of the list... you mean : if the head of the list is equal to n, you return the counter.
Your program is then :
let counter = ref 0;;
let rec get_val x n = match x with
[] -> -1
| (h::t) ->
if (h = n) then
!counter
else
begin incr counter ; get_val t n end
;;
Note that I've added begin and end around the else block so it can be interpreted as a sequence of instructions.
Your program now works, but it is not the best way to solve this problem with ocaml.
You should write something like
let get_val x n =
let rec get_val_aux x n counter = match x with
| [] -> -1
| h :: _ when h = n -> counter
| _ :: t -> get_val_aux t n (succ counter)
in
get_val_aux x n 0
;;
Here, we add a parameter to the get_val_aux function which we increment at each call. This function is nested within the get_val function to hide this additional parameter which is initialized with 0 on the first call.
Instead of using an if statement, we use the when condition to know when the element has been found, and add a new case to match the last case (not found). Note the use of the _ wildcard to avoid an unused variable.
The succ function (for successor) only adds 1 to its parameter. It is equivalent to counter + 1.
There are many problems with this code. If you ignore your immediate problem for a moment, you are treating OCaml variables like the variables of an imperative language. However, OCaml variables are immutable. This function
let increase_counter c = c + 1
Doesn't change the value of any variable. It just returns a number 1 bigger than what you give it.
The only error I get from the toplevel when I enter your code is for this expression:
increase_counter counter ; get_val t n
The compiler is warning you that the expression before ; is supposed to be executed for its side effects. I.e., it should almost always have type unit. Since (as I say) your function increase_counter returns an int, the compiler is warning you about this.