Algorithmic COMP108 Foundations Examination 2004-05
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Question 1 Solution
1A. The trace table for m = 5 and n = 3. Before while loop 1st iteration 2nd iteration 3rd iteration i x s 0 1 0 1 5 5 2 25 30 3 125 155
The output of the algorithm for m = 5 and n = 3 is 155. In general, the algorithm outputs m + m2 + m3 + . . . + mn . The time complexity of the algorithm is O(n). [Mark distribution: trace table 6 marks, general output 2 marks, time complexity 2 marks] 1B. I. Table of edges in ascending order of costs. edge (a, b) (e, f ) cost 1 1 edge (b, h) (e, h) cost 3 3 PSfrag replacements II. The MST: b a (b, f ) (c, e) 2 2 (c, h) (f, h) 4 5 c d e (a, f ) (b, c) 3 3 (c, d) (d, e) 8 9
h f
Order of edges selected: (a, b), (e, f ), (b, f ), (c, e), (b, h), (c, d). (The order of the underlined pairs can be swapped, and (b, h) can be replaced by (e, h).) III. The MST is not unique. IV. Because the costs of the edges are all distinct, the MST must be unique. [Mark distribution: (I) 3 marks, (II) 7 marks, (III) 2 marks, (IV) 3 marks.] 1
Question 2 Solution
2A. count = 0 for i = 0 to n m do begin j=0 while j < m and S[i + j] == T [j] do j =j+1 if j == m then count = count + 1 end if count == 1 then output T is a unique substring of S else output T is not a unique substring of S Time complexity: The outer for-loop iterates for O(n) times, and each inner while-loop iterates for O(m) times. Therefore, the time complexity of this algorithm is O(nm). [Mark distribution : Report substring 5 marks, Report unique substring 6 marks, Time complexity 4 marks] 2B. To show that 3n2 log n + 2n2 + 5n + 1 is O(n2 log n), we show that For any integer n > 1, 3n2 log n + 2n2 + 5n + 1 cn2 log n for some constant c. Notice that for any integer n > 1, 1 log n n. Therefore, 2n2 2n2 log n (because 1 log n), 3n2 log n 3n2 log n,
1 n2 log n (because 1 log n and n > 1).
5n 5n2 log n (because 1 log n and n > 1), and
As a result, 3n2 log n + 2n2 + 5n + 1 11n2 log n for all n > 1. Since 11 is a constant, the function 3n2 log n + 2n2 + 5n + 1 is O(n2 log n). [Mark distribution : correct approach 4 marks, state that 1 log n n 2 marks, correct constant c 2 marks, correct minimum value of n 2 marks]
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Question 3 Solution
3A. I. O(n5 ) II. O(n4 ) III. O(n3 log n) IV. O( n) or O(n1/2 ) V. O( n3 ) or O(n3/2 ) [Mark distribution: 2 marks each.] 3B. Denote the statement
n k=1 (2k
+ 1) = n(n + 2) as p(n).
Basic step / Base case : For n = 1, n (2k + 1) = 3. k=1 On the other hand, n(n + 2) = 3. Therefore, p(1) is true. Induction step : Assume that p(n) true is for some positive integer n. Consider the case for n + 1.
n+1
(2k + 1)
k=1 n
=
k=1
(2k + 1) + 2(n + 1) + 1 n(n + 2) + 2n + 3 n2 + 2n + 2n + 3 n2 + 4n + 3 (n + 1)(n + 3) (by I.H.)
= = = =
Therefore, p(n + 1) is also true. By the principle of M.I., the statement p(n) is true for all positive integers n. [Mark distribution: base case 2 marks, state the induction hypothesis 2 marks, induction step 6 marks.] 3C. I. A decision problem is a computational problem for which the output is either yes or no. In an optimisation problem we try to maximise or minimise some value (the objective function). II. (a) is an optimisation problem, (b) is a decision problem. III. Any NP-complete problem: CNF-SAT, 3-SAT, Vertex Cover, etc. [Mark distribution: (I) 2 marks (II) 0.5 mark each, (III) 2 marks.]
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Question 4 Solution
4A. T (n) = = = = = = = = = 3T (n 1) + 1 3(3T (n 2) + 1) + 1 32 T (n 2) + 3 + 1 32 (3T (n 3) + 1) + 3 + 1 33 T (n 3) + 32 + 3 + 1 . . . 3n1 T (1) + 3n2 + . . . 32 + 3 + 1 3n1 + 3n2 + . . . 32 + 3 + 1 because T (1) = 1 n (3 1)/(3 1) (3n 1)/2
[Mark distribution: the rst few iterations 4 marks, expressed in terms of T (1) and use of T (1) 7 marks, correct nal formula 4 marks.] 4B. I. The degree of the graph is 3. II. One possible DFS traversal: a, b, c, d, f , e, h, g III. One possible BFS traversal: a, b, h, c, g, d, f , e [Mark distribution: I 1 mark, II & III 2 marks each.] 4C. I. The sum of in-degree is 7 and the sum of out-degree is also 7. II. The sum of in-degree over all vertices = the sum of out-degree ...