U N I V E R S I T Y O F W E L L I N G T O N EXAMINATIONS 2017 TRIMESTER 1 *** WITH SOLUTIONS *** COMP 261 ALGORITHMS and DATA STRUCTURES

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1 T E W H A R E W Ā N A N G A O T E Ū P O K O O T E I K A A M Ā U I VUW VICTORIA U N I V E R S I T Y O F W E L L I N G T O N EXAMINATIONS 2017 TRIMESTER 1 *** WITH SOLUTIONS *** COMP 261 ALGORITHMS and DATA STRUCTURES Time Allowed: TWO HOURS CLOSED BOOK Permitted materials: Only silent non-programmable calculators or silent programmable calculators with their memories cleared are permitted in this examination. Instructions: Attempt ALL Questions. Answer in the appropriate boxes if possible if you write your answer elsewhere, make it clear where your answer can be found. The exam will be marked out of 120 marks. Non-electronic foreign to English language dictionaries are permitted. Alphabetic order: A B C D E F G H I J K L M N O P Q R S T U V W X Y Z Questions Marks 1. Path Finding [12] 2. Articulation Points [20] 3. 3D Graphics [8] 4. Parsing [28] 5. String Searching [17] 6. Text Compression [15] 7. B+ Trees [20] COMP 261 Page 1 of 23

2 Question 1. Path Finding [12 marks] (a) [2 marks] State one data structure appropriate for the fringe in Dijkstra s Algorithm. Priority queue. (b) Consider the following undirected graph, where the numbers on the edges are the edge lengths, and the numbers in the brackets by the nodes are the estimated heuristic values for A* search. The start node is node S, and the goal node is node G. (i) [3 marks] Describe what a heuristic must satisfy to be admissible. An admissible heuristic never overestimates the distance to the goal node. (ii) [2 marks] State whether the heuristic in the above graph is admissible or not. The heuristic is admissible. For all the nodes, the heuristic value is no greater than the true min cost to the goal node. COMP 261 Page 2 of 23

3 (iii) [5 marks] The heuristic in the graph above is not consistent. Find one counter example in the graph. h(s) h(b) = 11 > cost(s, B) h(c) h(d) = 9 > cost(c, D) h(c) h(e) = 7 > cost(c, E) Giving one of the above three is enough. COMP 261 Page 3 of 23

4 Question 2. Articulation Points [20 marks] (a) [3 marks] Circle all the articulation points in the following graph. B, D, F, H COMP 261 Page 4 of 23

5 (b) [7 marks] Finding all the articulation points requires traversing the graph using depthfirst search and setting counter numbers for the nodes. Given the following graph, draw the depth-first search tree rooted at node A, and put the counter number of each node next to it. Draw the edges that are in the graph but not in the search tree using dashed lines. The counter number starts from 0, and the root of the search tree is given for you to start with. If a node has multiple neighbours, the search will visit the neighbours in alphabetic order. COMP 261 Page 5 of 23

6 (c) [10 marks] In the following depth-first-search tree, the count numbers (count) of the nodes are given. Calculate the reachback values of each node. Mark the articulation points in the graph. The algorithm is given on the facing page for your reference. COMP 261 Page 6 of 23

7 FindArticulationPoints (graph, start ): for each node: node.count, articulationpoints { } start.count 0, numsubtrees 0 for each neighbour of start if neighbour.count = then RecursiveArtPts(neighbour, 1, start) numsubtrees ++ if numsubtrees > 1 then add start to articulationpoints RecursiveArtPts(node, count, fromnode): node.count count, reachback count, for each neighbour of node other than fromnode if neighbour.count < then reachback min(neighbour.count, reachback) else childreach RecursiveArtPts(neighbour, count + 1, node) reachback min(childreach, reachback ) if childreach count then add node to articulationpoints return reachback COMP 261 Page 7 of 23

8 Question 3. 3D Graphics [8 marks] (a) [8 marks] Consider the triangular polygon below. The coordinates of the three vertices are v1 = (v1.x, v1.y, v1.z), v2 = (v2.x, v2.y, v2.z), v3 = (v3.x, v3.y, v3.z). A point v = (v.x, v.y,?) inside the polygon is on a horizontal line that intersects the edges (v1, v2) and (v1, v3). The calculation of v.z requires a sequence of linear interpolation steps. Each step is a call of the method Interpolate(), which is described on the facing page. Write the steps to calculate v.z. The first two steps are already given. xleft = Interpolate(v.y, v1.x, v1.y, v2.x, v2.y); zleft = Interpolate(v.y, v1.z, v1.y, v2.z, v2.y); xright = Interpolate(v.y, v1.x, v1.y, v3.x, v3.y); zright = Interpolate(v.y, v1.z, v1.y, v3.z, v3.y); v.z = Interpolate(v.x, zleft, xleft, zright, xright); COMP 261 Page 8 of 23

9 Consider an edge in a 2D x-y plane, the two end-nodes are (x1, y1) and (x2, y2). For a point (?, y) on the edge, the method Interpolate() calculates its x coordinate based on its y coordinate. double Interpolate(double y, double x1, double y1, double x2, double y2) { double slope = (x2 x1) / (y2 y1); double x = x1 + slope (y y1); return x; } COMP 261 Page 9 of 23

10 Question 4. Parsing [28 marks] Consider the following grammar for very simple robot programs, where nonterminals are in uppercase, terminals are enclosed in quotation marks, and PROG is the start symbol. Assume that tokens will be separated by spaces. Note that nonterminal COND uses a regular expression to define the terminals it accepts. PROG ::= "{" STMT+ "}" STMT ::= "if" "(" COND ")" PROG ACT ";" ACT ::= "F" "L" "R" COND ::= ID ID ::= [a-z]+ (a) [4 marks] For each of the following strings, state whether it belongs to the language defined by this grammar. (i) if ( a ) { F ; } } yes (ii) { F ; R } no (iii) { if ( a ) { if ( b ) { F; } } } yes (iv) { if ( a ) F; if { b } R; } no COMP 261 Page 10 of 23

11 (b) [4 marks] Draw a concrete syntax tree (or parse tree) for the following string, using the above grammar: { if ( a ) { L ; } } You may abbreviate nonterminals to their first letter, writing P for PROG, S for STMT, etc.. (c) [4 marks] Show how your concrete syntax tree in part (b) can be simplified to obtain an abstract syntax tree, and explain the changes you have made. (see text in latex file) Delete constant strings in the rules. Delete unnecessary (singleton) nonterminals. COMP 261 Page 11 of 23

12 (d) [10 marks] Suppose you are writing a recursive descent parser for the above grammar. Below is a heading for one of the methods you will need. Complete this method, following the style discussed in lectures, assuming that the other methods that make up the parser have been written. Briefly explain the purpose of any other methods you call (this should take no more than four words per method). Note that the parser just returns a boolean, indicating whether the input was valid or not. You may write your methods in Java, or in a suitable pseudocode. public boolean parsestmt(scanner s) { } COMP 261 Page 12 of 23

13 (e) [6 marks] Suppose you want to extend the language to allow complex conditions, with infix operators and brackets. You start by writing the following grammar rule: COND ::= ID COND "and" COND COND "or" COND "(" COND ")" Discuss briefly why you cannot write a recursive descent parser based on this grammar rule, and how you would modify the above rule so that it can be used to construct the required parser. The grammar is both ambiguous and left-recursive. (Being ambiguous means that you can t decide which branch to take based on the next token; being left-recursive means the parser will go into an infinite loop.) Change the grammar to use repetition, which turns into a loop in the parser. COND ::= ID [BOOLOP ID]+ "(" COND ")" BOOLOP ::= "and" "or or, better still, to give correct precedence:. COND ::= COND1 ["or" COND1 ]+ COND1 ::= COND2 [ "and" COND2 ]+ COND2 ::= ID "(" COND ")" COMP 261 Page 13 of 23

14 Question 5. String Searching [17 marks] (a) [5 marks] The brute force string search algorithm can be very inefficient. What is the worst case cost? Give an example that would lead to the worse case cost. The algorithm can examine characters of the text many times. Worse case is when T is n a s, and S is m 1 as followed by a b. In this case we get m.n comparisons. (Actually m(n m + 1), which tends to 1 as m reduces to 1 or increases to n, so is max when m = n/2.) (b) [6 marks] The Knuth-Morris-Pratt (KMP) algorithm uses a table, derived from the search string, to speed up string search. Explain briefly how the table is used to speed up the search, using the following example table, where S = ABACABABC is the search string, and M is the partial match table.. S A B A C A B A B C M i COMP 261 Page 14 of 23

15 (c) [6 marks] Show the table that the following KMP table building algorithm would construct given the string S = "anant": computekmptable(s) initialise M to an array of integers with same length as S M[0] 1, M[1] 0, pos 2, j 0, while pos < S.length if S[pos 1] = S[j] M[pos] j+1, pos pos+1, j j+1 else if j > 0 j table[ j ] else M[pos] 0, pos pos+1 return M S a n a n t M i COMP 261 Page 15 of 23

16 Question 6. Text Compression [15 marks] (a) [10 marks] Explain the key idea of the Lempel-Ziv 77 compression algorithm, using the following text as an example: ban bananas Eliminate repeating patterns ban bananas [0,0,b][0,0,a][0,0,n][0,0, ][0,3,a]... (b) [5 marks] Explain briefly, the key idea of Huffman coding. NO ANSWER PROVIDED COMP 261 Page 16 of 23

17 SPARE PAGE FOR EXTRA ANSWERS Cross out rough working that you do not want marked. Specify the question number for work that you do want marked. COMP 261 Page 17 of 23

18 Question 7. B and B+ Trees [20 marks] (a) [8 marks] Consider the following B-tree of order 5: H L P B D E J K M N O X Y Z Show how the B-tree is modified as each of the keys Z, M, X and L is deleted. Briefly describe what you have done for each key that is deleted. Note, the empty trees below are to save you time; you may modify their structure if you choose. The B-tree after deleting key value Z: The B-tree after deleting key values Z and M: (Question 7 continued on next page) COMP 261 Page 18 of 23

19 (Question 7 continued) The B-tree after deleting key values Z, M, and X: The B-tree after deleting key values Z, M, X, and L: (Question 7 continued on next page) COMP 261 Page 19 of 23

20 (Question 7 continued) (b) [8 marks] Consider the following B+ tree, which has internal nodes holding up to 3 keys, and leaf nodes holding up to 4 key-value pairs. The keys are letters; the values are numbers. Show how the B+ tree below would be changed if the following key-value pairs were added to it: Z-5 F-18 T-2 L-34 State any assumptions you make. root: K S B-15 E-5 G-3 H-15 K-21 M-9 P-12 S-2 V-19 W-8 Assumptions (if any): WRONG AND OLD: When the leaf is too full, it could be split 2-3 or 3-2; Here, we have split it 3-2. (Question 7 continued on next page) COMP 261 Page 20 of 23

21 (Question 7 continued) (c) [4 marks] Suppose you are implementing a B+ tree in a file, storing each node of the tree in one block. The keys of the B+ tree are share market codes (for example AAPL ), and the values are company names. Each share market code is up to 10 characters long, and the company names are limited to 100 characters. The blocks are 1024 bytes long. Assume that each character takes one byte. How many key-value pairs can be stored in a leaf node? Explain your reasoning, show your working, and state any assumptions you make about the information stored in the blocks. WRONG and OLD: 6 or 7 key-value pairs in node, depending on bytes for number of pairs. Leaf needs 1 byte to store block type, (or even just one bit!) 1, 2, or 4 bytes to store the number of pairs, (or even as few as 3 bits) 4 bytes for the index of the next block, and leaving 506, 505, or 503 bytes for the K-V pairs. 72 bytes for each pair means 506/72 = 7 K-V pairs, if 1 byte for number of pairs 505/72 = 7 K-V pairs, if 2 bytes for number of pairs 503/72 = 6 K-V pairs, if 4 bytes for number of pairs * * * * * * * * * * * * * * * COMP 261 Page 21 of 23

22 SPARE PAGE FOR EXTRA ANSWERS Cross out rough working that you do not want marked. Specify the question number for work that you do want marked. COMP 261 Page 22 of 23

23 SPARE PAGE FOR EXTRA ANSWERS Cross out rough working that you do not want marked. Specify the question number for work that you do want marked. COMP 261 Page 23 of 23