Ocaml: Equivalence of two lists - functional-programming
New to Ocaml and have been working on a problem I haven't seen answered yet.
I'm working on a function where there is a tuple of 2 lists that are checked for equivalence.
Example:
# equivalence ([1;2],[1;2]);;
- : bool = true
# equivalence ([1;2],[2;1]);;
- : bool = true
# equivalence ([1;2],[1]);;
- : bool = false
Code I have:
let rec equivalent(a,b) = match a, b with
| [], [] -> true
| [], _
| _, [] -> false
| c::cc, d::dd -> if c = d then equivalent(cc,dd) else false;;
I know the problem lies with the last line. I can get a result of true if all elements are in the same order, but out of order it is a result of false. I'm having trouble going through one list to see if the other list has the element. I've tried to use List.nth, .hd, and .tl (not allowed to use .map or .itr) and have also tried to avoid the imperative features of Ocaml. Any suggestions or somewhere I should look? Thanks.
I assume this is some homework where you have to treat lists as poor mans set.
As Bergi mentioned performance wise the best would be to first sort both list and then compare them with your existing code. But that is probably not how you are supposed to solve this.
Assuming the lists are mend to be sets with not duplicate entries then equivalence of a and b means that a contains every element of b and b contains every element of a.
Start by writing a function contains: 'a -> 'a list -> bool that checks if an item is in a list. Next a function contains_all: 'a list -> 'a list -> bool that uses contains to check if every item in the first list is in the second. That then gives you
let equivalence a b = (contains_all a b) && (contains_all b a)
If your lists can contains duplicates then go with sorting the lists first. Anything else is just too impractical.
Note: contains exists as List.mem if you are allowed to use that.
Related
F# define search function
I am new to F# and am having trouble with my code. Its a simple problem to define a function, search, with that take a boolean function and a list and return an index. So for example: > search (fun x -> x > 10) [ 2; 12; 3; 23; 62; 8; 2 ];; val it : int = 1 > search (fun s -> s < "horse") [ "pig"; "lion"; "horse"; "cow"; "turkey" ];; val it : int = 3 What I have as of right now finds the right match but what I cant figure out is how to return a number instead of the rest of the list. I know I'm getting the list instead of a value back because I wrote "if f head then list". What I don't know is what I should put there instead or if what I have is not going to get the result I want. Below is the code I have written. let rec search f list = match list with | [] -> [-1] | head::tail -> if f head then list else search f tail
Returning a number is easy, you just... return it. Your problem is that you don't have a number to return, because you can't derive it directly from the current state. You have to keep track of the number yourself, using some internal state variable. When using recursion you change state by calling your function recursively with "modified" arguments. You're already doing that with the list here. To keep internal state in a recursive function you have to introduce another argument, but not expose it outside. You can solve that by using an internal recursive helper function. Here's one that keeps track of the previous item and returns that when it encounters a match: let search f list = let rec loop list prev = match list with | [] -> None | head::tail -> if f head then prev else loop tail (Some head) in loop list None That's a silly example, but I don't want to just solve your homework for you, because then you wouldn't learn anything. Using this you should be able to figure out how to keep a counter of which position the current item is in, and return that when it matches. Good luck!
You typically define an inner recursive function to help you carry state as you loop, and then call the inner function with an initial state. let search predicate list = let rec loop list index = match list with | [] -> -1 | head::tail -> if predicate head then index else loop tail (index + 1) loop list 0
expression has type 'a list -> 'b list but an expression was expected of type 'b list
This is my function let rec helper inputList = function | [] -> [] | a :: b :: hd -> if a = b then helper ([b::hd]) else a :: helper (b::hd) It's not complete, however I can't see why I keep getting the error in the title at helper ([b::hd]). I've tried helper (b::hd) or helper (b::hd::[]) however all come up with errors. How do I make it so that it works?
When you use function you are supplying a pattern for the parameter of the function. But you already have a parameter named inputList. So this function helper is expecting two parameters (but it ignores the first). You can fix this by removing inputList. You also have a problem in your first recursive call to helper. Your expression [b :: hd] is a list of lists. I suspect that you want something more like just b :: hd here. There is at least one other problem, but I hope this helps get you started.
There are multiple errors here. One is that the keyword function means we have an implicit parameter over which we are working. So the pattern matching happens on that "invisible" parameter. But here you defined probably the explicit one: inputList. So we can remove that one: let rec helper = function | [] -> [] | a :: b :: hd -> if a = b then helper ([b::hd]) else a :: helper (b:: hd) Next there is a problem with the types: in the recursion, you use: helper ([b::hd]); and a :: helper (b:: hd) But you put these on the same line, and that makes no sense, since the first one passes a list of lists of elements, and the second a list of elements. So the result of the first one would be a list of list of elements, and the second one a list of elements. It does not make sense to merge these. If I understood correctly that you want to ensure that no two consecutive elements should occur that are equal, then we should rewrite it to: let rec helper = function | [] -> [] | a :: b :: hd -> if a = b then helper (b::hd) else a :: helper (b:: hd) You have defined two patterns here: one for the empty list; and one for a list with at least two elements. The second one will perform recursion on the tail of the list b :: hd. So that means that eventually when we pass it a list with n elements, it will recursively work on a list with n-1 elements, n-2 elements, etc. But eventually it will have one element. And there is no case for that. So we need to add a case for the one element pattern: let rec helper = function | [] -> [] | h :: [] -> h :: [] | a :: b :: hd -> if a = b then helper (b::hd) else a :: helper (b:: hd)
how return a new type with an update value
If I want to change a value on a list, I will return a new list with the new value instead of changing the value on the old list.
Now I have four types. I need to update the value location in varEnd, instead of changing the value, I need to return a new type with the update value
type varEnd = {
v: ctype;
k: varkind;
l: location;
}
;;
type varStart = {
ct: ctype;
sy: sTable;
n: int;
stm: stmt list;
e: expr
}
and sEntry = Var of varEnd | Fun of varStart
and sTable = (string * sEntry) list
type environment = sTable list;;
(a function where environment is the only parameter i can use)
let allocateMem (env:environment) : environment =
I tried to use List.iter, but it changes the value directly, which type is also not mutable. I think List.fold will be a better option.
The biggest issue i have is there are four different types.
I think you're saying that you know how to change an element of a list by constructing a new list.
Now you want to do this to an environment, and an environment is a list of quite complicated things. But this doesn't make any difference, the way to change the list is the same. The only difference is that the replacement value will be a complicated thing.
I don't know what you mean when you say you have four types. I see a lot more than four types listed here. But on the other hand, an environment seems to contain things of basically two different types.
Maybe (but possibly not) you're saying you don't know a good way to change just one of the four fields of a record while leaving the others the same. This is something for which there's a good answer. Assume that x is something of type varEnd. Then you can say:
{ x with l = loc }
If, in fact, you don't know how to modify an element of a list by creating a new list, then that's the thing to figure out first. You can do it with a fold, but in fact you can also do it with List.map, which is a little simpler. You can't do it with List.iter.
Update
Assume we have a record type like this:
type r = { a: int; b: float; }
Here's a function that takes r list list and adds 1.0 to the b fields of those records whose a fields are 0.
let incr_ll rll =
let f r = if r.a = 0 then { r with b = r.b +. 1.0 } else r in
List.map (List.map f) rll
The type of this function is r list list -> r list list.
F# Recursive Tree Validation
This is a somewhat beginner question. I have been trying to validate the following type of FamilyTree. I can't find a simple way to do this. All help would be appreciated. type BirthYear = int;; type Tree = Person of BirthYear * Children and Children = Tree list;; I want to validate a given family tree such that every Person is older than their Children and furthermore check if the list of Children is sorted in order of their age (eldest first). Preferably done with a function that return a boolean. Something along the lines of this: let rec validate (Person(x,child)) = let vali = child |> List.forall (fun (y,_) -> y < x)
I'd do something like this: let rec checkAges minBirth = function | Person(b, _) :: t -> b >= minBirth && checkAges b t | [] -> true let rec validate (Person(b, c)) = List.forall validate c && checkAges (b + minParentAge) c where minParentAge is set to a reasonable minimum age to have children at. I'd expect checkAges to be the more difficult part here: the function checks whether the first child it sees is younger than the limit it is given, then recursively checks the next child, with the current child's age as the new limit. Note some techniques: The function that checks child ages takes the minimum birthday as input; this is used to validate that the parent is old enough for the first child to be reasonable. List.forall checks a predicate for all items in a list, and early-outs if a predicate is not fulfilled function is a shorthand to create a function that does pattern matching on its parameter. Therefore, checkAges actually has two arguments.
Here's a very simple solution using a single recursive function. It's not relying on built-in functions like List.forall but I think it's very declarative and (hopefully) easy to follow. Rule 1: Every Person is older than their Children Rule 2: List of Children is sorted in order of their age (eldest first) Code: let rec isValid = function | Person ( _ , []) -> true // Person alone without childs -> always valid | Person (minYear, Person (year, childs) :: brothers) -> year > minYear && // Validate Rules (either 1 or 2) isValid (Person (year, childs)) && // Enforce Rule 1 isValid (Person (year, brothers)) // Enforce Rule 2 I personally don't feel List.forall fits well here, it helps to solve a part of the problem but not the whole, so you need to combine it with more stuff (see the other answers) and in the end you can't avoid a recursive function. List functions are good for lists but for trees I feel recursion more natural unless your tree provides already a way to traverse it.
Here's a way to do it. Perhaps spending some time analyzing how this works will be helpful to you. let rec check (Person(age, children)) = match children with | [] -> true | Person(eldest, _)::_ -> Seq.pairwise children |> Seq.forall ((<||) (>)) && age > eldest && List.forall check children
Erlang sudoku solver - How to find the empty spots and try possible values recursively
I have been busy with a sudoku solver in Erlang yesterday and today. The working functionality I have now is that I can check if a sudoku in the form of a list, e.g.,
[6,7,1,8,2,3,4,9,5,5,4,9,1,7,6,3,2,8,3,2,8,5,4,9,1,6,7,1,3,2,6,5,7,8,4,9,9,8,6,4,1,2,5,7,3,4,5,7,3,9,8,6,1,2,8,9,3,2,6,4,7,5,1,7,1,4,9,3,5,2,8,6,2,6,5,7,8,1,9,3,4].
is valid or not by looking at the constraints (no duplicates in squares, rows, and columns).
This function is called valid(S) which takes a sudoku S and returns true if it is a valid sudoku and false if it is not. The function ignores 0's, which are used to represent empty values. This is an example of the same sudoku with some random empty values:
[0,7,1,8,2,3,4,0,5,5,4,9,0,7,6,3,2,8,3,0,8,5,0,9,1,6,7,1,3,2,6,5,7,8,4,9,0,8,6,4,1,2,5,7,0,4,5,7,3,9,8,6,1,0,8,9,3,2,6,4,7,5,1,7,1,4,9,3,0,2,8,6,2,6,5,7,8,1,9,3,4].
The next step is to find the first 0 in the list, and try a value from 1 to 9 and check if it produces a valid sudoku. If it does we can continue to the next 0 and try values there and see if it is valid or not. Once we cannot go further we go back to the previous 0 and try the next values et cetera until we end up with a solved sudoku.
The code I have so far looks like this (based on someone who got it almost working):
solve(First,Nom,[_|Last]) -> try_values({First,Nom,Last},pos()).
try_values(_,[]) -> {error, "No solution found"};
try_values({First,Nom,Last},[N|Pos]) ->
case valid(First++[N]++Last) of
true ->
case solve({First++[N]},Nom,Last) of
{ok,_} -> {ok, "Result"};
{error,_} -> try_values({First,N,Last},Pos)
end;
false -> try_values({First,N,Last},Pos)
end.
pos() is a list consisting of the values from 1 to 9. The idea is that we enter an empty list for First and a Sudoku list for [_|Last] in which we look for a 0 (Nom?). Then we try a value and if the list that results is valid according to our function we continue till we fail the position or have a result. When we fail we return a new try_values with remaining (Pos) values of our possibitilies.
Naturally, this does not work and returns:
5> sudoku:solve([],0,S).
** exception error: bad argument
in operator ++/2
called as {[6]}
++
[1,1,8,2,3,4,0,5,5,4,9,0,7,6,3,2,8,3,2,8,5,4,9,1,6,7,1,3,2|...]
in call from sudoku:try_values/2 (sudoku.erl, line 140)
in call from sudoku:try_values/2 (sudoku.erl, line 142)
With my inexperience I cannot grasp what I need to do to make the code logical and working. I would really appreciate it if someone with more experience could give me some pointers.
try_values([], []) -> error("No solution found");
try_values([Solution], []) -> Solution;
try_values(_, []) -> error("Bad sudoku: multiple solutions");
try_values(Heads, [0|Tail]) ->
NewHeads = case Heads of
[] -> [[P] || P <- pos()];
_ -> [Head++[P] || P <- pos(), Head <- Heads]
end,
ValidHeads = [Head || Head <- NewHeads, valid(Head++Tail)],
try_values(ValidHeads, Tail);
try_values([], [H|Tail]) -> try_values([[H]], Tail);
try_values(Heads, [H|Tail]) -> try_values([Head++[H] || Head <- Heads], Tail).
solve(Board) ->
case valid(Board) of
true -> try_values([], Board);
false -> error("No solution found")
end.
try_values does what you described. It builds solution by going through Board, trying all possible solutions (from pos()) when it finds 0 and collecting valid solutions in ValidHeads to pass them further to continue. Thus, it goes all possible ways, if at some point there are multiple valid sudoku they all will be added to Heads and will be tested on validity on following steps. solve is just a wrapper to call try_values([], Board).
Basically, the way to iterate recursively over 0's is to skip all non-zeros (2 last try_values expression) and do the job on zeros (fourth try_values expression).
First three try_values expressions check if solution is exist and single and return it in that case.