Source file setPt.ml

(*
 * Ptset: Sets of integers implemented as Patricia trees.
 * Copyright (C) 2000 Jean-Christophe FILLIATRE
 * 
 * This software is free software; you can redistribute it and/or
 * modify it under the terms of the GNU Library General Public
 * License version 2, as published by the Free Software Foundation.
 * 
 * This software is distributed in the hope that it will be useful,
 * but WITHOUT ANY WARRANTY; without even the implied warranty of
 * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
 * 
 * See the GNU Library General Public License version 2 for more details
 * (enclosed in the file LGPL).
 *)

(*i ptset.ml 1.8 2001/06/28 07:05:55 filliatr Exp  i*)

(*s Sets of integers implemented as Patricia trees, following Chris
    Okasaki and Andrew Gill's paper {\em Fast Mergeable Integer Maps}
    ({\tt\small http://www.cs.columbia.edu/\~{}cdo/papers.html\#ml98maps}).
    Patricia trees provide faster operations than standard library's
    module [Set], and especially very fast [union], [subset], [inter]
    and [diff] operations. *)

(*s The idea behind Patricia trees is to build a {\em trie} on the
    binary digits of the elements, and to compact the representation
    by branching only one the relevant bits (i.e. the ones for which
    there is at least on element in each subtree). We implement here
    {\em little-endian} Patricia trees: bits are processed from
    least-significant to most-significant. The trie is implemented by
    the following type [t]. [Empty] stands for the empty trie, and
    [Leaf k] for the singleton [k]. (Note that [k] is the actual
    element.) [Branch (m,p,l,r)] represents a branching, where [p] is
    the prefix (from the root of the trie) and [m] is the branching
    bit (a power of 2). [l] and [r] contain the subsets for which the
    branching bit is respectively 0 and 1. Invariant: the trees [l]
    and [r] are not empty. *)

(*i*)
type elt = int
(*i*)

type t =
  | Empty
  | Leaf of int
  | Branch of int * int * t * t

(*s Example: the representation of the set $\{1,4,5\}$ is
    $$\mathtt{Branch~(0,~1,~Leaf~4,~Branch~(1,~4,~Leaf~1,~Leaf~5))}$$
    The first branching bit is the bit 0 (and the corresponding prefix
    is [0b0], not of use here), with $\{4\}$ on the left and $\{1,5\}$ on the
    right. Then the right subtree branches on bit 2 (and so has a branching 
    value of $2^2 = 4$), with prefix [0b01 = 1]. *)

(*s Empty set and singletons. *)

let empty = Empty

let is_empty = function Empty -> true | _ -> false

let singleton k = Leaf k

(*s Testing the occurrence of a value is similar to the search in a
    binary search tree, where the branching bit is used to select the
    appropriate subtree. *)

let zero_bit k m = (k land m) == 0

let rec mem k = function
  | Empty -> false
  | Leaf j -> k == j
  | Branch (_, m, l, r) -> mem k (if zero_bit k m then l else r)

(*s The following operation [join] will be used in both insertion and
    union. Given two non-empty trees [t0] and [t1] with longest common
    prefixes [p0] and [p1] respectively, which are supposed to
    disagree, it creates the union of [t0] and [t1]. For this, it
    computes the first bit [m] where [p0] and [p1] disagree and create
    a branching node on that bit. Depending on the value of that bit
    in [p0], [t0] will be the left subtree and [t1] the right one, or
    the converse. Computing the first branching bit of [p0] and [p1]
    uses a nice property of twos-complement representation of integers. *)

let lowest_bit x = x land (-x)

let branching_bit p0 p1 = lowest_bit (p0 lxor p1)

let mask p m = p land (m-1)

let join (p0,t0,p1,t1) =
  let m = branching_bit p0 p1 in
  if zero_bit p0 m then 
    Branch (mask p0 m, m, t0, t1)
  else 
    Branch (mask p0 m, m, t1, t0)

(*s Then the insertion of value [k] in set [t] is easily implemented
    using [join].  Insertion in a singleton is just the identity or a
    call to [join], depending on the value of [k].  When inserting in
    a branching tree, we first check if the value to insert [k]
    matches the prefix [p]: if not, [join] will take care of creating
    the above branching; if so, we just insert [k] in the appropriate
    subtree, depending of the branching bit. *)

let match_prefix k p m = (mask k m) == p

let add k t =
  let rec ins = function
    | Empty -> Leaf k
    | Leaf j as t -> 
        if j == k then t else join (k, Leaf k, j, t)
    | Branch (p,m,t0,t1) as t ->
        if match_prefix k p m then
          if zero_bit k m then 
            Branch (p, m, ins t0, t1)
          else
            Branch (p, m, t0, ins t1)
        else
          join (k, Leaf k, p, t)
  in
  ins t

(*s The code to remove an element is basically similar to the code of
    insertion. But since we have to maintain the invariant that both
    subtrees of a [Branch] node are non-empty, we use here the 
    ``smart constructor'' [branch] instead of [Branch]. *)

let branch = function
  | (_,_,Empty,t) -> t
  | (_,_,t,Empty) -> t
  | (p,m,t0,t1)   -> Branch (p,m,t0,t1)

let remove k t =
  let rec rmv = function
    | Empty -> Empty
    | Leaf j as t -> if k == j then Empty else t
    | Branch (p,m,t0,t1) as t -> 
        if match_prefix k p m then
          if zero_bit k m then
            branch (p, m, rmv t0, t1)
          else
            branch (p, m, t0, rmv t1)
        else
          t
  in
  rmv t

(*s One nice property of Patricia trees is to support a fast union
    operation (and also fast subset, difference and intersection
    operations). When merging two branching trees we examine the
    following four cases: (1) the trees have exactly the same
    prefix; (2/3) one prefix contains the other one; and (4) the
    prefixes disagree. In cases (1), (2) and (3) the recursion is
    immediate; in case (4) the function [join] creates the appropriate
    branching. *)

let rec merge = function
  | Empty, t  -> t
  | t, Empty  -> t
  | Leaf k, t -> add k t
  | t, Leaf k -> add k t
  | (Branch (p,m,s0,s1) as s), (Branch (q,n,t0,t1) as t) ->
      if m == n && match_prefix q p m then
        (* The trees have the same prefix. Merge the subtrees. *)
        Branch (p, m, merge (s0,t0), merge (s1,t1))
      else if m < n && match_prefix q p m then
        (* [q] contains [p]. Merge [t] with a subtree of [s]. *)
        if zero_bit q m then 
          Branch (p, m, merge (s0,t), s1)
        else 
          Branch (p, m, s0, merge (s1,t))
      else if m > n && match_prefix p q n then
        (* [p] contains [q]. Merge [s] with a subtree of [t]. *)
        if zero_bit p n then
          Branch (q, n, merge (s,t0), t1)
        else
          Branch (q, n, t0, merge (s,t1))
      else
        (* The prefixes disagree. *)
        join (p, s, q, t)

let union s t = merge (s,t)

(*s When checking if [s1] is a subset of [s2] only two of the above
    four cases are relevant: when the prefixes are the same and when the
    prefix of [s1] contains the one of [s2], and then the recursion is
    obvious. In the other two cases, the result is [false]. *)

let rec subset s1 s2 = match (s1,s2) with
  | Empty, _ -> true
  | _, Empty -> false
  | Leaf k1, _ -> mem k1 s2
  | Branch _, Leaf _ -> false
  | Branch (p1,m1,l1,r1), Branch (p2,m2,l2,r2) ->
      if m1 == m2 && p1 == p2 then
        subset l1 l2 && subset r1 r2
      else if m1 > m2 && match_prefix p1 p2 m2 then
        if zero_bit p1 m2 then 
          subset l1 l2 && subset r1 l2
        else 
          subset l1 r2 && subset r1 r2
      else
        false

(*s To compute the intersection and the difference of two sets, we
    still examine the same four cases as in [merge]. The recursion is
    then obvious. *)

let rec inter s1 s2 = match (s1,s2) with
  | Empty, _ -> Empty
  | _, Empty -> Empty
  | Leaf k1, _ -> if mem k1 s2 then s1 else Empty
  | _, Leaf k2 -> if mem k2 s1 then s2 else Empty
  | Branch (p1,m1,l1,r1), Branch (p2,m2,l2,r2) ->
      if m1 == m2 && p1 == p2 then 
        merge (inter l1 l2, inter r1 r2)
      else if m1 < m2 && match_prefix p2 p1 m1 then
        inter (if zero_bit p2 m1 then l1 else r1) s2
      else if m1 > m2 && match_prefix p1 p2 m2 then
        inter s1 (if zero_bit p1 m2 then l2 else r2)
      else
        Empty

let rec diff s1 s2 = match (s1,s2) with
  | Empty, _ -> Empty
  | _, Empty -> s1
  | Leaf k1, _ -> if mem k1 s2 then Empty else s1
  | _, Leaf k2 -> remove k2 s1
  | Branch (p1,m1,l1,r1), Branch (p2,m2,l2,r2) ->
      if m1 == m2 && p1 == p2 then
        merge (diff l1 l2, diff r1 r2)
      else if m1 < m2 && match_prefix p2 p1 m1 then
        if zero_bit p2 m1 then 
          merge (diff l1 s2, r1) 
        else 
          merge (l1, diff r1 s2)
      else if m1 > m2 && match_prefix p1 p2 m2 then
        if zero_bit p1 m2 then diff s1 l2 else diff s1 r2
      else
        s1

(*s All the following operations ([cardinal], [iter], [fold], [for_all],
    [exists], [filter], [partition], [choose], [elements]) are
    implemented as for any other kind of binary trees. *)

let rec cardinal = function
  | Empty -> 0
  | Leaf _ -> 1
  | Branch (_,_,t0,t1) -> cardinal t0 + cardinal t1

let rec iter f = function
  | Empty -> ()
  | Leaf k -> f k
  | Branch (_,_,t0,t1) -> iter f t0; iter f t1
      
let rec fold f s accu = match s with
  | Empty -> accu
  | Leaf k -> f k accu
  | Branch (_,_,t0,t1) -> fold f t0 (fold f t1 accu)

let rec for_all p = function
  | Empty -> true
  | Leaf k -> p k
  | Branch (_,_,t0,t1) -> for_all p t0 && for_all p t1

let rec exists p = function
  | Empty -> false
  | Leaf k -> p k
  | Branch (_,_,t0,t1) -> exists p t0 || exists p t1

let filter p s = 
  let rec filt acc = function
    | Empty -> acc
    | Leaf k -> if p k then add k acc else acc
    | Branch (_,_,t0,t1) -> filt (filt acc t0) t1
  in
  filt Empty s

let partition p s =
  let rec part (t,f as acc) = function
    | Empty -> acc
    | Leaf k -> if p k then (add k t, f) else (t, add k f)
    | Branch (_,_,t0,t1) -> part (part acc t0) t1
  in
  part (Empty, Empty) s

let rec choose = function
  | Empty -> raise Not_found
  | Leaf k -> k
  | Branch (_, _,t0,_) -> choose t0   (* we know that [t0] is non-empty *)

let elements s =
  let rec elements_aux acc = function
    | Empty -> acc
    | Leaf k -> k :: acc
    | Branch (_,_,l,r) -> elements_aux (elements_aux acc l) r
  in
  elements_aux [] s

(*s There is no way to give an efficient implementation of [min_elt]
    and [max_elt], as with binary search trees.  The following
    implementation is a traversal of all elements, barely more
    efficient than [fold min t (choose t)] (resp. [fold max t (choose
    t)]). Note that we use the fact that there is no constructor
    [Empty] under [Branch] and therefore always a minimal
    (resp. maximal) element there. *)

let rec min_elt = function
  | Empty -> raise Not_found
  | Leaf k -> k
  | Branch (_,_,s,t) -> min (min_elt s) (min_elt t)

let rec max_elt = function
  | Empty -> raise Not_found
  | Leaf k -> k
  | Branch (_,_,s,t) -> max (max_elt s) (max_elt t)

(*s Another nice property of Patricia trees is to be independent of the
    order of insertion. As a consequence, two Patricia trees have the
    same elements if and only if they are structurally equal. *)

let equal = (=)

let compare = compare

(*i*)
let make l = List.fold_right add l empty
(*i*)

(*s Additional functions w.r.t to [Set.S]. *)

let rec intersect s1 s2 = match (s1,s2) with
  | Empty, _ -> false
  | _, Empty -> false
  | Leaf k1, _ -> mem k1 s2
  | _, Leaf k2 -> mem k2 s1
  | Branch (p1,m1,l1,r1), Branch (p2,m2,l2,r2) ->
      if m1 == m2 && p1 == p2 then
        intersect l1 l2 || intersect r1 r2
      else if m1 < m2 && match_prefix p2 p1 m1 then
        intersect (if zero_bit p2 m1 then l1 else r1) s2
      else if m1 > m2 && match_prefix p1 p2 m2 then
        intersect s1 (if zero_bit p1 m2 then l2 else r2)
      else
        false