CS 1501 Summer 2009: Quiz 1 Solution

1) Fill in the Blanks and True/False (22 points -- 2 points each).
Complete the statements below with the MOST APPROPRIATE words/phrases.

a) (and b) Two algorithms, A and B, have the following run-time information:

i) Algorithm A has a run-time of Theta(2ⁿ) and, for n = 4 runs in 4 seconds on a computer.

ii) Algorithm B has a run-time of Theta(n²) and, for n = 4, also runs in 4 seconds on the same computer.

How long will each algorithm take to run (in seconds) a problem of size n = 8 using the same computer? Algorithm A: _______64_______________ Algorithm B:_________16_____________

c) Order the following growth rates from smallest (best) to greatest (worst): n² nlgn lgn n! 2ⁿ n³ _____lgn < nlgn < n² < n³ < 2ⁿ < n! ______________

d) Given a multiway radix search trie in which 64-bit keys are compared 8 bits at a time, the maximum height of the tree is _____8_____ and interior nodes will each have up to __2⁸=256____ children.

e) Consider an empty separate chaining hash table of size 100. If we hash 200 keys into this table, the average chain length will be _____2____ and the worst case chain length will be ___200_____.

f) An example text string and pattern string that will produce the worst case for the brute force string matching algorithm are:

A (text) = ___AAAAAAAAAAAAB__________________
P (pattern) = ___AAAB__________

Indicate whether each of the following is TRUE or FALSE, explaining why in an informative way for false answers.

g) Pruning is a technique that improves the asymptotic run-times of exhaustive search algorithms.
False. Pruning can improve the exhaustive search in practice, but can’t change the asymptotic runtime.

h) We used DLBs in Assignment 1 rather than regular multiway tries because DLBs allow for faster searches.
False. DLBs are slower than multiway tries, but they can save memory consumption.

i) The Java String operator "+" appends one String to another in constant time.
False. It requires linear time (in the length of the strings) since a new object must be created

j) Due to the Pigeonhole Principle, I cannot avoid collisions in hashing if the size of my key space is larger than the size of my hash table.
True

k) Hashing a string by adding the ASCII values of its characters results in a good hash function, since it utilizes the entire key.
False. It does not take position of the characters into account (among other things).

2) (16 points – 8 + 8) Consider an emtpy de la Briandais Tree, which uses the lower case letters (plus a string termination character) as its alphabet, using the implementation that we discussed in lecture. Also consider the following strings: run sunday runny sun sunnny

a) Draw the de la Briandais tree that results after inserting the strings shown in the order shown above.

a) In Assignment 1 the searchPrefix method for your DLB did two things:

i) Test to see if a string is contained in the DLB

ii) Test to see if a string is a prefix of some string contained in the DLB

Explain the DLB search algorithm in detail and clearly show how each of the two above determinations can be made.

Start from the first character of the given string. Sequentially search the linked list of sibling nodes pointed by root. If the character is found, follow the child link and repeat the same procedure with the next character in the string and the child’s linked list. If the search fails before reaching the end of the string, the string is neither a word nor a prefix. If the last character in the string is reached, continue to search its child’s linked list. If a string terminator is found, the search string is a word in the DLB. If some other characters are also found, it is contained in the DLB and also it is a prefix of some strings. If only other characters are found and the string terminator is not found, it is a prefix of some strings contained in the DLB but not a word.

1) (12 points – 6 + 6) Consider the two open addressing hash tables, with h(x) = x mod 13, shown below. Also consider the following keys (in order): 23, 17, 36, 24, 26, 37, 49

h2(36) = 4 h2(37) = 5 h2(49) = 6 Linear Probing		Double Hashing
Index	key	Index	key
0	26	0	26
1	37	1	36
2	49	2
3		3	37
4	17	4	17
5		5
6		6
7		7
8		8
9		9	49
10	23	10	23
11	36	11	24
12	24	12

a) Assume that linear probing is being used for collision resolution. Show the table after the keys shown above are inserted in the order shown above.

b) Assume now that double hashing is being used for collision resolution, with h₂(x) = (x mod 11) + 1. Show the table after the keys shown above are inserted in the order shown above.