Lecture L 16 — April 19 , 2012

1 Overview In this lecture, we consider the string matching problem-finding some or all places in a text where the query string occurs as a substring. From the perspective of a one-shot approach, we can solve string matching in O(|T |) time, where |T | is the size of our text. This purely algorithmic approach has been studied extensively in the papers by Knuth-Morris-Pratt [6], Boyer-Moore [1], and Rabin-Karp [4]. However, we entertain the possibility that multiple queries will be made to the same text. This motivates the development of data structures that preprocess the text to allow for more efficient queries. We will show how to construct, use, and analyze these string data structures. Before we get to string matching we will solve an easier problem which is necessary to solve string matching. This is the predecessor problem among strings. Assume we have k strings, T 1 , T 2 , ..., T k and the query is: given some pattern P , we would like to know where P fits among the k strings in lexicographical order. More specifically we would like to know the predecessor of P among the k strings, according to the lexicographical order of the strings. Using our already developed data structures for this problem is not efficient as strings can be quite long and comparing strings is slow. So we need to use some other data structure that takes into account this fact. To solve the predecessor problem we will use a structure called a trie. A trie is a rooted tree where each child branch is labeled with letters in the alphabet Σ. We will consider any node v to store a string which represents the concatenation of all brach labels on the path from the root r to the v. More specifically given a path of increasing depth p = r, v 1 , v 2 , ..., v from the root r to a node v, the string stored at node v i is the concatenation of the string stored in v i−1 with the letter stored on the branch v i−1 v i. We will denote the strings stored in the leaves of the trie as words, and the strings stored in all other nodes as prefixes. The root r represents the empty string. We order the child branches of every node alphabetically from left to right. Note that …