(**************************************************************************) (* *) (* Copyright (C) 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 2.1, with the special exception on linking *) (* described in file LICENSE. *) (* *) (* 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. *) (* *) (**************************************************************************) (*i $Id$ i*) open Hashcons (*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 'a elt = 'a hash_consed (*i*) type 'a t = Empty Leaf of 'a hash_consed Branch of int * int * 'a t * 'a 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.tag == j.tag Branch (_, m, l, r) -> mem k (if zero_bit k.tag 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.tag == k.tag then t else join (k.tag, Leaf k, j.tag, t) Branch (p,m,t0,t1) as t -> if match_prefix k.tag p m then if zero_bit k.tag m then Branch (p, m, ins t0, t1) else Branch (p, m, t0, ins t1) else join (k.tag, 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.tag == j.tag then Empty else t Branch (p,m,t0,t1) as t -> if match_prefix k.tag p m then if zero_bit k.tag 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. When comparing branching bits, one has to be careful with the leftmost bit (which is negative), so we introduce function [unsigned_lt] below. *) let unsigned_lt n m = n >= 0 && (m < 0 || n < m) 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 unsigned_lt 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 unsigned_lt n m && 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 unsigned_lt m2 m1 && 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 unsigned_lt m1 m2 && match_prefix p2 p1 m1 then inter (if zero_bit p2 m1 then l1 else r1) s2 else if unsigned_lt m2 m1 && 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 unsigned_lt 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 unsigned_lt m2 m1 && 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 rec filter pr = function Empty -> Empty Leaf k as t -> if pr k then t else Empty Branch (p,m,t0,t1) -> branch (p, m, filter pr t0, filter pr t1) 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 unsigned_lt m1 m2 && match_prefix p2 p1 m1 then intersect (if zero_bit p2 m1 then l1 else r1) s2 else if unsigned_lt m2 m1 && match_prefix p1 p2 m2 then intersect s1 (if zero_bit p1 m2 then l2 else r2) else false

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