AMTH140 Trimester Question 1. [8 marks] Answer

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1 Question 1 Let A B = {(1, 1), (2, 2), (3, 1), (3, 2), (1, 2), (1, 4), (2, 1), (2, 4), (3, 4)}. Find the power set of B, P(B). The set B = {1, 2, 4}. Then P(B) = {, {1}, {2}, {4}, {1, 2}, {1, 4}, {2, 4}, {1, 2, 4}} Question 2 Let f (x) = x 2 + 3x + 2. Evaluate f (2) and f ( 2) by Horner s method. f (x) = x 2 + 3x + 2 = x(x + 3) + 2. Then f (2) = 2 (2 + 3) + 2 = = 12 f ( 2) = ( 2) ( 2 + 3) + 2 = ( 2) = = 0 Question 3 Use mathematical induction to show that [10 marks] n! 2 n 1 for n = 1, 2,... Basic step. Let S n be the proposition n! 2 n 1, then we must show that S 1 is true. This is easily accomplished since 1! = 1 1 = Inductive step. We must show that if S 1, S 2,..., S n are true (i! 2 i 1 for i = 1, 2,..., n) then S n+1, that is (n + 1)! 2 n is true. Assume that S 1, S 2,..., S n are true, that is i! 2 i 1 for i = 1, 2,..., n are true. Then, in particular, for i = n we have n! 2 n 1. Now, (n + 1)! = (n + 1)n! (n + 1)2 n n 1 = 2 n since n Therefore S n+1 is true and hence P.M.I. tells us that n! 2 n 1 is true for every postive integer n. Question 4 [10 marks] If f, g : N R, f 1 (n) = O(g 1 (n)) and f 2 (n) = O(g 2 (n)) then f 1 (n)+ f 2 (n) = O(max(g 1 (n), g 2 (n))) By definition of big O, there are two integers, say M 1 and M 2 and two constants C 1 and C 2 such that f 1 (n) C 1 g 1 (n) for n M 1 and f 2 (n) C 2 g 2 (n) for n M 2 Let M 0 = max(m 1, M 2 ) and let C 0 = 2 max(c 1, C 2 ) and consider the sum f 1 (n)+ f 2 (n) for n M 0 : f 1 (n)+ f 2 (n) f 1 (n) + f 2 (n) C 1 g 1 (n) +C 2 g 2 (n) C 0 (g 1 (n)+g 2 (n))/2 C 0 max(g 1 (n), g 2 (n)) Therefore f 1 (n) + f 2 (n) = O(max(g 1 (n), g 2 (n))). Question 5 2

2 In MERGE SORT algorithm, if we are given two sorted lists, LIST1 with n elements and LIST2 with m elements, what is the maximum number of comparisons required to merge these two lists into a single sorted list? Each time we compare two elements, one is removed and put in the combined list. If we have only one element left, no further comparisons are possible. So the number of comparisons is certainly no bigger than one less than the total number of elements, that is m + n 1. Question 6 Use a truth table to prove that Modus Ponens is a valid argument. Use a truth table to prove that p q, p, q. p q p q p q T T T T T T F F T F F T T F T F F T F F Question 7 Let us suppose that G is a graph with seven vertices, such that two vertices have the degree one, three vertices have the degree two and two vertices have the degree three. Can this graph be a tree? If your answer is yes, draw a tree with these properties. If your answer is no, prove that it couldn t be a tree. For any graph G, we denote by N the sum of the degrees of all vertices of G, denote by E the number of edges in G, and denote by V the total number of vertices of G. Then obviously N = 2E (*) because each edge will contribute exactly 2 degrees (by its 2 ends) to the total degrees N. Moreover a connected graph G is a tree if and only if V = E + 1 (**) N = 14 so E = 7 (edges). From (**), a connected graph with 7 vertices is a tree iff it has 6 edges. So this graph could not be a tree. Question 8 Convert to bases four, octal and hexadecimal. Since the number is of base 2 and that 4 = 2 2, we need to convert 2-digit blocks into the corresponding base 4 representation: = 0. }{{} 11 }{{} 00 }{{} 01 = Since the number is of base 2 and that 8 = 2 3, we need to convert 3-digit blocks into the corresponding base 8 representation: = 0. }{{} }{{} = Since the number is of base 2 and that 16 = 2 4, we need to convert 4-digit blocks into the corresponding base 16 representation: 3

3 = 0. }{{} 1100 }{{} 0100 C 4 Question 9 Given the adjacency matrix = 0.C4 16 A = [4 marks] draw its corresponding graph. v1 v2 v3 Question 10 Use a binary tree to sort the following list of numbers 65, 75, 6, 57, 99, 27, 0, 96 The binary tree constructed during the course of sorting reads Figure 1: Binary tree Question 11 Let A be the set of all integers greater than or equal to 2, and relation R be defined by m, n A : (m, n) R iff m n Show that 4

4 (a) 7 is a minimal element of A. (b) 6 is not a minimal element of A. (c) A does not have any maximal elements. (d) 2 is not a least element of A. (a) Because 7 is prime, no numbers in A can divide 7, i.e., x R 7 does not hold far any x A (no x precedes 7), we conclude 7 is a minimal element of A. (b) Since 2 6, we see immediately that 6 is not a minimal element of A. (c) For any x A, since 2x A and x 2x, we have x R 2x, i.e., x precedes 2x under relation R. Hence x cannot be a maximal element. Since x is an arbitrary element in A, we conclude that A does not have any maximal elements. (d) The fact that 2, 3 A and 2 is not comparable with 3 means 2 is not a least element of A. Question 12 Given S = {0, 1}, the table below specifies a Boolean function f : S S S S x y z f (x, y, z) (a) Using a Karnaugh map, simplify the Boolean expression corresponding to this function. (b) Draw a gate implementation for the simplified Boolean expression. The Boolean expression corresponding to this function is x y z + x y z + x yz + xyz. We may simplify the expression with the Karnaugh map below 5

5 z z xy xy x y x y The minimal expression is thus xyz + x y + x z. Question 13 Given that (a 1 + a a k ) n = n 1,n 2,...,n k 0 n 1 +n 2 + +n k =n n! n 1! n 2! n k! an 1 1 an 2 2 an k k [5 marks] write down (a 1 + a 2 + a 3 ) 3 =? (a 1 + a 2 + a 3 ) 3 = a a2 1 a 2 + 3a 2 1 a 3+ a a2 2 a 1 + 3a 2 2 a 3+ a a2 3 a 1 + 3a 2 3 a 2+ 6a 1 a 2 a 3 Question 14 Let a sequence (x i ) i N be determined by the recurrence relation [5 marks] x n+2 = 1 + x n+1 x n and the initial conditions x 0 = a and x 1 = b with a, b 0x. Find x 10. x 2 = 1 + x 1 x 0 1+x 1 x 0 x 3 = 1 + x 2 = 1 + = 1 + x 0 + x 1 x 1 x 1 x 0 x 1 x 4 = 1 + x 3 x 2 =... = 1 + x 0 x 1 x 5 = 1 + x 4 x 3 =... = x 0 x 6 = 1 + x 5 = 1 + x 0 x 1+x 0 = x 1 4 x 1 The recurrence relation has period 5. Hence x 10 = x 5 = x 0. 6