QA #

Linked here.

Check-in Exercise #

Linked here.

Overview #

Summary. The sort problem is to take a sequence of objects and put them into the correct order. The search problem is to store a collection of objects such that they can be rapidly retrieved (i.e. how do we implement a Map or Set). We made the obersvation that BST maps are roughly analagous to comparison based sorting, and hash maps are roughly analagous to counting based (a.k.a. integer) sorting. We observed that we have a 3rd type of sort, which involves sorting by digit, which raised the question: What sort of data structure is analogous to LSD or MSD sort?

Terminology.

• Length of string key usually represented by L.
• Alphabet size usually represented by R.

Tries. Analogous to LSD sort. Know how to insert and search for an item in a Trie. Know that Trie nodes typically do not contain letters, and that instead letters are stored implicitly on edge links. Know that there are many ways of storing these links, and that the fastest but most memory hungry way is with an array of size R. We call such tries R-way tries.

Advantages of Tries. Tries have very fast lookup times, as we only ever look at as many characters as they are in the data we’re trying to retrieve. However, their chief advantage is the ability to efficiently support various operations not supported by other map/set implementations including:

• longestPrefixOf
• prefixMatches
• spell checking

Recommended Problems #

Throughout, we define R as the size of the alphabet, and N as the number of items in a trie.

C level #

1. Problem 5 from Princeton’s Spring 2008 final.

2. Problem 8 from Princeton’s Spring 2011 final.

3. Problem 8 from Princeton’s Spring 2012 final.

4. Draw the R-way trie that result after inserting the strings: sam, sad, sap, same, a, awls.

B level #

1. When looking for a single character string in a Trie, what is the worst case time to find that string in terms of R and N?

2. Problem 5 from Princeton’s Fall 2009 final.

3. Problem 9 from Princeton’s Fall 2012 final.

4. Problem 1 from my Fall 2013 final.

5. True or false: The number of character compares required to construct an R-way trie is always less than or equal to the number required to construct an LLRB.

6. True or false: The number of character compares required to construct an R-way trie is always less than or equal to the number of character accesses needed to construct a hash table.

A level #

1. There are many different data structures that we can use to store the children of an R-way trie node. The simplest but most memory hungry choice is to create an array of length R, with one link per character in the trie. A second choice is to have each node contain a TreeMap that maps a given character to the appropriate node. A third choice is to use HashMaps instead of TreeMaps. is to have each node contain a TreeMaps, some of you used R-way trie nodes that contained HashMaps. For the “average” node, which of these three choices is likely to use the most memory? Why does this mean that this type of node takes longer to construct?

2. Suppose we use an R-way trie to implement a set. For each of the three implementations of an R-way trie node described in problem 2, what are the best and worst case key lookup times in terms of R an N?

3. Problem 9 from my Fall 2014 midterm 2.