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COS 226 Algorithms and Data Structures Spring 2019
Final exam: Reference solutions _.
1. Minimal spanning tree [12 points] A set of edges is highlighted in different copies of the same graph in each of the figures below. Determine whether it could possibly represent a partial spanning tree obtained from (a prematurely stopped) Prim’s algorithm, (a prematurely stopped) Kruskal’s algorithm, both, or neither. Write “Prim’s”, “Kruskal’s”, “Both”, or “Neither” the blank below each figure. Both Kruskal’s Neither Prim’s Neither Prim’s
2. 2d tree [8 points] Figure 1 Figure 2 A 2d-tree contains ten points (denoted A, B, ... J). The tree is shown in Figure 1 and the points are shown in Figure 2. A. Points W, X, Y, and Z with the coordinates shown in Figure 2 are inserted into the tree in that order. Show where the four new points will be inserted by filling in the blanks below. (Write ‘left’ or ‘right’ in the first blank of each statement). W will be the right child of D. X will be the left child of H. Y will be the left child of G. Z will be the left child of X. B. Suppose we do a nearest neighbor search for a point with the same x- and y-coordinates as D (before inserting W, X, Y, Z). How many distance computations between the query point and a point in the tree will be performed? 3 ‾‾‾ C. Complete the following sentence with a succinct phrase. In general, if we do a nearest-neighbor search for a query point that equals a point P that’s already in a kd-tree, the number of distance computations between points that will be performed is one more than the depth of P. ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
4. NFA [10 points] Consider the following NFA: A. Write the regular expression that this NFA matches:
(AB* | BC*) ⁺
B. Select all states that the NFA can be in after processing the string ABB. If the NFA gets stuck while processing the string, select None. None 0 1 2 3 4 5 6 7 8 9 10 11 12
C. Select all states that the NFA can be in after processing the string AACBA. If the NFA gets stuck while processing the string, select None. None 0 1 2 3 4 5 6 7 8 9 10 11 12
D. Does the NFA accept the string BBB? Write Yes or No. Yes E. Does the NFA accept the string ABBBBCCCC? Write Yes or No. Yes
5. Compression [12 points] A. Decode the following LZW-encoded bitstream represented as a sequence of hex characters: 46 41 44 45 44 81 43 82 84 45 86 84 80 Decoded string: F A D E D FA C AD ED E FAC ED ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ Recall that: 1. The ASCII values of the uppercase letters in hex are A: 41, B: 42, C: 43, etc. 2. We use hex 80 as a stop symbol, and codes for new codewords start at hex 81. Show the codeword table that results from decoding by filling in the cells below. Note that there may be more cells in the table than necessary. Code 81 82 83 84 85 86 87 88 89 8a 8b Codeword FA AD DE ED DF FAC CA ADE EDE EF FACE ‾‾‾ ‾‾‾ ‾‾‾ ‾‾‾ ‾‾‾ ‾‾‾ ‾‾‾ ‾‾‾ ‾‾‾ ‾‾‾ ‾‾‾ B. Construct a Huffman code for the string “SLEEPLESSNESS_LESSENS_SENSE”. Count the frequency of each of the six characters (including the underscore) that appear in the string, and write the binary code for each. Freq Code E 8 10 L 3 1111 N 3 110 P 1 11100 S 10 0 _ 2 11101 Note: many other correct answers are possible.
7. String symbol tables [11 points] A. A Ternary Search Trie (TST) is shown above. Values are indicated with a number next to the node. List all the strings that the TST contains. Note that there might be more spaces below than necessary, and that the strings might not be English words. There is a score penalty for incorrect answers. m, a, ax, min, meo, ore, oore, ost B. For each of the following three string symbol table implementations, indicate how characters and strings are stored by picking the best match from among three choices: Linear probing Hash table R-way trie TST Neither characters nor strings are stored explicitly
Characters are stored explicitly but strings are not
Strings (and characters) are stored explicitly
C. Suppose a million random strings, each of length 10, over a 256-character alphabet are inserted into a symbol table. Among the above three implementations, which one would consume the most space?
Answer: R-way trie Which of the three would consume the least space? Answer: Hash table TSTs are space efficient for typical inputs (where most strings are prefixes of other strings) but inefficient for random inputs since most characters will require separate nodes.
9. True or False [14 points] For each statement below, write ‘T’ or ‘F’ in the corresponding box to indicate whether the statement is true or false. Suppose we explore the right/top subtree before the left/bottom subtree in 2d-tree range search. Then it may return the wrong answer. Suppose we explore the right/top subtree before the left/bottom subtree in 2d-tree range search. Then it may explore more nodes. Ignoring the encoding of the trie, the Huffman encoding of a message will never be longer than the message itself (assume that the message is represented using a fixed-length code such as ASCII). Let G be a connected edge-weighted graph and G ’ be a graph obtained by adding a single edge to G. Then G has an MST T and G ’ an MST T ’ such that there is at most one edge in T that is not in T ’. For any regular expression with m characters, there is a DFA with at most m^2 states that recognizes the same set of strings matched by the regular expression. Given a flow network with V vertices and E edges, as well as a min-cut of the network, the value of the max-flow can be computed in time proportional to E + V. MSD radix sort is applicable only when the input strings all have the same length.
F F T T F T F
Some explanations: ● We know that Huffman encoding produces an optimal prefix-free code, i.e., it requires no more bits than any other prefix-free code. A fixed-length code is a prefix-free code. ● Let T be any MST of G and e the new edge in G’. There is a cycle in T ∪ { e }. Remove the max-weight edge in this cycle. This results in a MST of G’. ● The DFA may have exponentially many states. ● The value of the max-flow equals the capacity of the min-cut, which can be computed by iterating over the edges and summing the weights of the ones that cross the cut.
10. Data structure and algorithm design [20 points] By the late 19th century, most of the world’s major cities were connected by the telegraph network. Your goal in this problem is to model this network to understand how quickly news would propagate around the world. Each telegraph link connects a pair of cities. When news originates in a city, it is sent to each of the cities to which it is linked. The first time a city hears a piece of news, it will send that news to each of the cities to which it is linked. Some links are congested, which means that messages will have to wait exactly one hour before being sent through them (in either direction). There are no other delays in the system: when a city hears a piece of news, all cities linked to it via uncongested links immediately hear it as well. The network is given to you as a sequence of pairs of cities, along with congestion status: London New_York Congested New_York Atlanta Uncongested Washington Philadelphia Congested ... There are V cities and E telegraph links in the input. Assume that the lengths of city names are at most some constant number. Your answers below will be graded for correctness, efficiency, and clarity. Answer in prose, but feel free to use some pseudocode if you think it will improve clarity. You may make any standard technical assumptions that we have seen in this course. A. Describe how you will pre-process the input so that you can quickly answer queries about propagation time for news originating in any given city (i.e. the queries in parts C-E). For full credit, this preprocessing must complete in time proportional to E log V (or better) in the worst case. Hints: 1. Read the entire question before answering part A, so that you understand what is required of the preprocessing step. 2. In our reference solution, the majority of the text is in part A, whereas the solutions to parts C-E are relatively succinct.
C. Given the names of two cities, determine whether they are connected via a path of telegraph links. For full credit, this must complete in time proportional to log V (or better) in the worst case. Check whether the two cities are connected in uf_all using find operations. D. Given the names of two cities, determine whether news originating in one of them will reach the other instantly (i.e., via uncongested links). For full credit, this must complete in time proportional to log V (or better) in the worst case. Check whether the two cities are connected in uf0 using find operations.
E. Given an originating city, sort the V cities based on the time at which they will hear news originating in the given city. If two cities hear about the news at the same time, they may be sorted relative to each other in either order. For full credit, your algorithm must complete in time proportional to E + V (or better) in the worst case. Run a modified BFS on G starting from the source uf0.find(originating_city). Modify BFS so that for each vertex in G (i.e. component of the original network) that it visits, output all the cities in that component using the lookup table T. Finally, output the cities unreachable from the originating city (i.e. cities in the components not marked as visited by BFS). Alternate approach (requires no preprocessing except creating a weighted graph of cities and links). Modify BFS to run on 0/1 weighted graphs (instead of unweighted graphs), and to use a deque instead of a queue. For edges of weight 0, add the new vertex to the front of the queue. For edges of weight 1, add it to the end. Always dequeue from the front. Note that Dijkstra’s algorithm and MST algorithms are too slow to meet the performance requirements. F. What is the order-of-growth running time of your solution to part E? Briefly justify your answer. BFS runs in time proportional to E’ + V’ where E’ ≤ E and V’ ≤ V. The other steps complete in time proportional to V.
