Download COS226 Algorithms and Data Structures Fall 2015 Final Exam and more Exams Algorithms and Programming in PDF only on Docsity!
COS 226 Algorithms and Data Structures Fall 2015
Final Exam
You have 180 minutes for this exam. The exam is closed book, except that you are allowed to use one page of notes (8.5-by-11, one side, in your own handwriting). No calculators or other electronic devices are permitted. Give your answers and show your work in the space provided. You may use the back of each page for scratch space, or to continue long answers.
Name:
NetID:
Precept:
P01 9:00 Andy Guna P02 10:00 Andy Guna P02A 10:00 Elena Sizikova P03 11:00 Maia Ginsburg P03A 11:00 Nora Coler P04 12:30 Maia Ginsburg P04A 12:30 Miles Carlsten P05 1:30 Tom Wu
Write and sign: “ I pledge my honor that I have not violated the Honor Code during this examination. ”
Grading note: To ensure that guessing on true/false and multiple-choice questions does not affect your expected score, grading on these questions will be as follows:
True / False: +1 point if correct, −1 point if incorrect, 0 points if left unanswered. Multiple choice: +2 points if correct, − 0 .4 points if incorrect, 0 points if left unanswered.
Problem Score Problem Score 0 7 1 8 2 9 3 10 4 11 5 12 6 13 Sub 1 Sub 2
Total:
0. Init. (1 point)
In the space provided on the front of the exam, write your name, Princeton netID, and precept number, and write and sign the honor code.
1. Flow. (10 points)
Consider the following flow network and feasible flow f from the source vertex S to the sink vertex T.
(a) What is the value of the flow f? Circle the correct answer.
3 5 7 11 13 17
(b) Starting from the flow given above, perform one iteration of the Ford-Fulkerson algorithm. List the sequence of vertices on the augmenting path, in order from S to T.
(c) What is the value of the maximum flow? Circle the correct answer.
3 5 7 11 13 17
(d) Circle all vertices on the sink (T) side of the minimum cut.
S A B C D E F T
3. TST. (13 points)
Consider the following Ternary Search Trie (TST), where the values are shown next to the nodes of the corresponding string keys.
(a) We would like to construct the above TST by inserting six strings into an empty TST. Circle the sequences below that can produce the above TST. There may be multiple correct answers.
Sequence 1: ATT ACG T CT AGC GA
Sequence 2: ATT T CT ACG GA AGC
Sequence 3: ATT T GA ACG CT AGC
Sequence 4: ATT T AGC ACG CT GA
Sequence 5: ATT ACG AGC T CT GA
Sequence 6: ATT T AGC GA ACG CT
Sequence 7: ATT ACG T CT GA AGC
(b) Insert the three strings CA, AGA, and GAC into the TST with the associated values 0, 18, and 29, respec- tively. Update the figure above to reflect the changes.
4. KMP DFA. (13 points)
(a) Below is a partially-completed Knuth-Morris-Pratt DFA for a string s of length 6 over the alphabet { A , B }. State 6 is the accept state. Fill in all the missing spots in the table.
j 0 1 2 3 4 5
pat.charAt(j)
A 1 1
B 3 3
(b) Given the following KMP DFA:
j 0 1 2 3 4 5 6
A 1 1 3 1 5 1 5
B 0 2 0 4 0 6 7
List the string that this DFA searches for.
(c) Given each of the following strings as input, what state would the DFA in (b) end in? Circle the correct answer for each string.
BABBAA: 0 1 2 3 4 5 6 accept
ABABABA: 0 1 2 3 4 5 6 accept
BABABABA: 0 1 2 3 4 5 6 accept
BBAABBABAB: 0 1 2 3 4 5 6 accept
6. Regex. (11 points)
(a) Consider the regular expression
((A|B)DA*C)
Circle all words matched by this regular expression.
ABDAC ADAAC ABDACA BDC BDAC AACA
(b) The following NFA matches the regular expression in (a):
Which of the labeled edges correspond to ε transitions (as opposed to match transitions)? Circle the num- bers of only the ε transitions:
1 2 3 4 5 6 7 8 9 10 11 12 13 14
(c) Which of the following (if any) are true reasons why we usually prefer NFAs for matching a regular expression (RE), as opposed to DFAs? Circle the correct answer in each case.
The size of the NFA is linear in the size of the RE, while the size of the DFA might be as bad as quadratic.
True False
The size of the NFA is linear in the size of the RE, while the size of the DFA might be as bad as exponential.
True False
The running time to simulate the NFA is linear in the size of the RE, while the running time for the DFA might be as bad as quadratic.
True False
The running time to simulate the NFA is linear in the size of the RE, while the running time for the DFA might be as bad as exponential.
True False
The NFA only has two kinds of transitions (match and ε), while the DFA requires determining the correct transition for each possible input character.
True False
The DFA might require backing up in the input stream, while the NFA does not.
True False
7. Huffman. (10 points)
Consider the following Huffman tree:
(a) Decode the following 24-bit bitstring: 111000110001100001011000
(b) What is the compression ratio (compressed size / uncompressed size) for the above bitstring? Assume that characters were represented by 8 bits before compression.
(c) What is the best compression ratio achievable on any string using this Huffman tree?
(d) Suppose you added another character, H, with a count of 1. After re-creating the new Huffman code, circle all the letters that acquire a different codeword.
A B C D E F G
(e) Using the Huffman code from (d), what is the worst compression ratio achievable on any string?
9. Sort. (14 points)
The column on the left is an array of strings to be sorted. The column on the right is in sorted order. The other columns are the contents of the array at some intermediate step during one of the algorithms below. Write the number of each algorithm under the corresponding column. You may use each number more than once.
mink bear bear calf crow myna crab bear bear moth calf calf lamb lamb crab toad crow calf crow crow crow hare deer lamb swan calf crab myna crab crab wasp crab toad bear crab crow swan deer hare hawk hare mule deer deer deer wolf hare kiwi ibex bear hare ibex hare hare mule hawk deer bear kiwi sole hoki hawk hawk slug hoki hawk deer calf wolf mule hoki hoki hare ibex ibex mink hawk calf sole ibex ibex bear kiwi hoki lion ibex slug wolf kiwi kiwi kiwi lion lion kiwi hoki moth calf lion lamb calf lynx lynx slug lion kiwi lamb lynx lion hawk lamb lamb toad lynx hoki myna lamb lynx ibex mink mink hoki mink mink mink mink mink oryx moth mule sole mule hawk lynx moth moth lion myna myna wolf myna swan lion myna mule sole mule moth moth moth lion crow mule myna wasp oryx wasp crab wasp wasp hare oryx oryx lynx swan sole crow sole bear wasp swan slug hoki slug oryx oryx oryx deer moth slug sole crab sole slug mule slug crow slug sole swan deer toad wolf swan wolf ibex kiwi toad toad lamb wolf toad myna toad oryx hawk wolf wasp toad wasp swan lynx swan lynx oryx wasp wolf
0 4
(0) Original input (2) LSD radix sort (4) Sorted (1) 3-way radix quicksort (3) MSD radix sort (no shuffle)
10. G^2. (10 points)
The square of a digraph G consisting of vertices V and edges E is a digraph G^2 such that:
- the vertices in G^2 are the same as the vertices in G , and
- two vertices in G^2 are connected by an edge ( u , v ) if and only if G contains edges ( u , w ) and ( w , v ), for some vertex w.
That is, vertices u and v are connected by an edge in G^2 whenever G contains a path with exactly two edges from u to v.
Describe an algorithm for computing the square of a digraph (represented using adjacency lists). For full credit, your solution should run in O ( V E ) time. To simplify the problem, you need not remove duplicates from the adjacency lists in G^2.
12. Reduction. (10 points)
(a) The FIND-42ND^ problem is to find the 42nd^ smallest item in an (initially unsorted) array. You can imple- ment this easily by sorting the array in O ( N log N ) time and returning the item in the 42nd^ position. Given this, which of the following (if any) must be true? Circle the correct answer in each case.
FIND-42ND^ reduces to sorting.
True False
Sorting reduces to FIND-42ND^.
True False
O ( N log N ) is a lower bound on FIND-42ND^.
True False
O ( N log N ) is an upper bound on FIND-42ND^.
True False
FIND-42ND^ must be NP-complete.
True False
FIND-42ND^ cannot be NP-complete unless P = NP.
True False
(b) Of course, it is also easy to implement FIND-42ND^ in O ( N ) time, using O ( 42 ) additional space. Further- more, it is possible to show that linear time is the lower bound on FIND-42ND^ , since all elements must be examined. Given this algorithm and the reduction in (a), which of the following (if any) must be true:
O ( N ) is a lower bound on sorting.
True False
O ( N ) is an upper bound on sorting.
True False
Sorting is strictly harder than FIND-42ND^ , so can never be accomplished in O ( N ) time.
True False
New developments in sorting might result in an asymptotically faster algorithm for FIND-42ND^.
True False
13. MST. (16 points)
You are given an edge-weighted undirected graph, using the adjacency list representation, together with the list of edges in its minimum spanning tree (MST). Describe an efficient algorithm for updating the MST, when each of the following operations is performed on the graph. Assume that common graph operations (e.g., DFS, BFS, finding a cycle, etc.) are available to you, and don’t describe how to re-implement them.
(a) Update the MST when the weight of an edge that was not part of the MST is decreased. Give the order-of-growth running time of your algorithm as a function of V and/or E.
(b) Update the MST when the weight of an edge that was part of the MST is decreased. Give the order-of-growth running time of your algorithm as a function of V and/or E.
