{- This program is an exercise in imperative sequential programming which uses pointers and mutable storage. It nicely separates the aspects of programming that are imperative -- updating values in place in the storage -- from aspects that are functional -- algorithms on trees described recursively. The program implements a set as a binary search tree which stores a set of values. See Wikipedia entry http://en.wikipedia.org/wiki/Binary_search_tree for a desription of binary search tree. The following interface is supported: insert(key): inserts key into the set. Returns true if key was not in the set prior to this operation, and false otherwise. delete(key): deletes key from the set. Returns true if key was in the set prior to this operation, and false otherwise. search(key): Returns true if key is in the set, and false otherwise. sort(): Returns a list of the set items in increasing order. Each node of a binary search tree (BST) includes value, a pointer to its left child (if it exists), and a pointer to its right child (if it exists). We simplify the algorithm by adding sentinels as follows. Below, tsent (for "top sentinel") is a node whose left pointer points to the root of the BST; its remaining fields are irrelevant. And, bsent stands for "bottom sentinel". Every node in the BST that should point to a null child points to bsent instead (thus, our structure is not really a tree). Initially, the tree is empty and the left child of tsent is bsent. Search for key starts by storing key in the value part of bsent, and applying the traditional algorithm starting at the root. There is no need to test for null pointers because the search is guaranteed to succeed, either by encountering a genuine value, or the stored value in bsent. The search result is then computed by testing if the final node in the search is bsent. A similar algorithm is used to insert a value into the set. In the following definitions, we often use a triple (p,d,q) to transmit a pair of nodes; here, q is p's child, it is the left child if d is false and right child if d is true. Since the tree has at least two nodes at all times (bsent and tsent) such a scheme is always feasible. Further, this interface allows us to insert and delete nodes more easily. -} val bsent = Ref() val tsent = Ref() >r> r.write((0,bsent,0)) >> r -- direction of pointer: false for left and true for right. def update (p,d,q) = -- redirect d pointer of p to q. val (v,l,r) = p.read() if d then p.write((v,l,q)) else p.write((v,q,r)) def searchstart(key) = -- return (p,d,q) where p.d = q and q.val = key def searchloop(p,d,q,key) = {- given q is p's d-child. Start search from q. Return (s,d,t) where s.d = t and t.val = key -} val (v,l,r) = q.read() if(key < v) >> searchloop(q,false,l,key) | if(key = v) >> (p,d,q) | if(key > v) >> searchloop(q,true,r,key) {- Goal for searchstart -} val (_,root,_) = tsent.read() bsent.write((key,0,0)) >> searchloop(tsent,false,root,key) def search(key) = -- return true or false searchstart(key) >(_,_,q)> (q /= bsent) def insert(key) = -- return true if value was inserted, false if it was there searchstart(key) >(p,d,q)> if q = bsent then Ref() >r> r.write((key,bsent,bsent)) >> update(p,d,r) >> true else false def delete(key) = def isucc(p) = {- in-order successor of p. p has genuine left and right sons. Returns (s,d,t) where t is the d-child of s. t is the in-order sucessor of s, t.left = bsent t is the leftmost genuine (non-sentinel) node in the right subtree of p. -} def leftmost(p,d,r) = -- given r is the d-child of p and r /= bsent. -- Return (p',d,r') where r' is the d-child of p' and r'.left = bsent -- Either (p,r) = (p',r') or (p',r') is in the leftmost path in -- the subtree rooted at r. val (_,l,_) = r.read() if l = bsent then (p,d,r) else leftmost(r,false,l) {- Goal of isucc: -} val (_,_,r) = p.read() leftmost(p,true,r) {- Goal of delete: -} val (p,d,q) = searchstart(key) val (_,l,r) = q.read() -- Below, nc is the number of children of q. val nc = (if l = bsent then 0 else 1) + (if r = bsent then 0 else 1) if(q = bsent) then {- key is not in -} false else ( if (nc /= 2) then -- q has zero or one genuine child (if l /= bsent then l else r) >t> update(p,d,t) >> true else -- q has two genuine children isucc(q) >(s,d,t)> t.read() >(v,_,r')> q.write((v,l,r)) >> update(s,d,r') >> true ) def sort() = -- do an in order traversal of the BST {- An explicit append operation on lists -} def append([],ys) = ys def append(xs,[]) = xs def append(x:xs, ys) = x:append(xs,ys) def traverse(p) = val (v,l,r) = p.read() if(p = bsent) then [] else append(traverse(l),v:traverse(r)) val (_,root,_) = tsent.read() traverse(root) -- Test insert(30) >> insert(20) >> insert(24) >> insert(35) >> insert(33) >> insert(38) >> delete(35)>> sort() {- insert(30) >> insert(20) >> insert(34) >> delete(30)>> sort() -}

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