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[Caml-list] how to calculate a "xor"
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Date: -- (:)
From: Damien <Damien.Pous@e...>
Subject: [Caml-list] how to calculate a "xor"
Hi algorithmers,

Given two sets A and B, I want to calculate A\B _and_ B\A.
The sets are represented by lists.

without using an order to sort the lists, 
is there something better than the following (O(n^2)) ?

<<
let omem a = 
  let rec aux acc = function
    | [] -> None
    | a'::q when a=a' -> Some (List.rev_append acc q)
    | a'::q -> aux (a'::acc) q
  in aux []

let xor x = 
  let rec aux (a',b') = function
    | la, [] -> List.rev_append a' la, b'
    | [], lb -> a', List.rev_append b' lb
    | a::qa, b::qb when a=b -> aux (a',b') (qa,qb)
    | a::qa, b::qb -> match omem a qb, omem b qa with
	| None,     None     -> aux (a::a', b::b') (qa, qb )
	| Some qb', None     -> aux (   a', b::b') (qa, qb')
	| None,     Some qa' -> aux (a::a',    b') (qa',qb )
	| Some qb', Some qa' -> aux (   a',    b') (qa',qb')
  in aux ([],[]) x
>>

# xor ([4;1;6;2;8],[3;9;5;2;8;1;7]);;
- : int list * int list = ([6; 4], [3; 9; 5; 7])



with an order, is there something better than (O(n*ln n)) :
 * sort the two lists,
 * "merge" them to extract the result
?


what's the complexity of <Set.Make(M).diff>, from the standard library ?


thanks,
damien

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