13-1 Persistent dynamic sets
During the course of an algorithm, we sometimes find that we need to maintain past versions of a dynamic set as it is updated. We call such a set persistent. One way to implement a persistent set is to copy the entire set whenever it is modified, but this approach can slow down a program and also consume much space. Sometimes, we can do much better.
Consider a persistent set with the operations , , and , which we implement using binary search trees as shown in Figure 13.8(a). We maintain a separate root for every version of the set. In order to insert the key into the set, we create a new node with key . This node becomes the left child of a new node with key , since we cannot modify the existing node with key . Similarly, the new node with key becomes the left child of a new node with key whose right child is the existing node with key . The new node with key becomes, in turn, the right child of a new root with key whose left child is the existing node with key . We thus copy only part of the tree and share some of the nodes with the original tree, as shown in Figure 13.8(b).
Assume that each tree node has the attributes , , and but no parent. (See also Exercise 13.3-6.)
a. For a general persistent binary search tree, identify the nodes that we need to change to insert a key or delete a node .
b. Write a procedure that, given a persistent tree and a key to insert, returns a new persistent tree that is the result of inserting into .
c. If the height of the persistent binary search tree is , what are the time and space requirements of your implementation of ? (The space requirement is proportional to the number of new nodes allocated.)
d. Suppose that we had included the parent attribute in each node. In this case, would need to perform additional copying. Prove that would then require time and space, where is the number of nodes in the tree.
e. Show how to use red-black trees to guarantee that the worst-case running time and space are per insertion or deletion.
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