CSC 125 :: Final Exam May 3 & 5, 2010

Transcription

1 CSC 125 :: Final Exam May 3 & 5, 2010 Name KEY (1 5) Complete the truth tables below: p Q p q p q p q p q p q T T T T F T T T F F T T F F F T F T T T F F F F F F T T Match the following logical equivalences with the answers found in the Answer Bank. Write the correct letter just to the right of the symbol. Note: Some answers will be used more than once: 6. Identity Laws 7. Domination Laws 8. Idempotent Laws 9. Double Negation Law 10. Commutative Laws p T p p F p p T T p F F p p p p p p ( p) p p q q p p q q p 1

2 11. Associative Laws 12. Distributive Laws 13. DeMorgan's Laws 14. Absorption Laws 15. Negation Laws (p q) r p (q r) (p q) r p (q r) p (q r) (p q) (p r) p (q r) (p q) (p r) (p q) p q (p q) p q p (p q) p p (p q) p p p T p p F Fill in the missing portion of each of the rules of inference named below: 16. Modus ponens 17. Modus tollens p q p. q p q q. p 18. Hypothetical syllogism p q q r p r 19. Disjunctive syllogism p q p. q 2

3 20. Resolution p q p r q r (21 26) Let B(x,y) be the statement student x attended basketball game y, where the domain for x consists of all students in this class and the domain for y consists of all basketball games this season. Match each of the quantified expressions below with the sentences found in the Answer Bank. Write the letter of the matching sentence just to the right of the expression. Please note: Some answers may be used more than once. 21. x y B(x,y) A student from this class attended a basketball game this season. 22. x y B(x,y) x y B(x,y) x y B(x,y) Note: Equivalent to #38. No student from this class attended a basketball game this season. {For all students and all basketball games, it not true that any student attended any game} 23. x y B(x,y) A student from this class attended every basketball game this season. 24. x y B(x,y) x y B(x,y) x y B(x,y) Note: Equivalent to #37. No student from this class attended every basketball game this season. 25. x y B(x,y) Every student from this class attended a basketball game this season. 26. x y B(x,y) Every student from this class attended every basketball game this season. (27 29) Let F(x,y) be the statement x can fool y, where the domain for both x and y consists of all people in the world. Use quantifiers to express these sentences. 27. Everyone can fool Fred. x F(x,Fred) 3

4 28. Evelyn can fool everybody. y F(Evelyn,y) 29. Someone can fool everybody. x y F(x,y) (30 32) Let the domain of P(x) consist of the integers 1, 2 and 3 Write out each proposition using disjunctions, conjunctions and negations. 30. x P(x) P(1) P(2) P(3) 31. x P(x) P(1) P(2) P(3) 32. x P(x) P(1) P(2) P(3) (33 37) Determine whether the arguments below are valid or invalid, then circle either Valid or Invalid. If valid, using the list of valid argument forms found in the Answer Bank, write the letter of the rule of inference; if invalid write the letter of the logical fallacy, again using the list of invalid argument forms found in the Answer Bank. 33. If we get 10" of snow, school will be cancelled. We got 8" of snow. School will not be cancelled. Invalid denying the hypothesis 34. If we pigs can fly, Santa Claus is real. Pigs can fly. Santa Claus is real. Valid Invalid Valid Valid modus ponens {affirming the hypothesis} 35. If we pigs can fly, Santa Claus is real. Santa Claus is not real. Pigs cannot fly. Invalid 4

5 Valid Valid modus tollens {denying the conclusion} 36. If pigs can fly, elephants can sing. If elephants can sing, monkeys can compose music. If pigs can fly, monkeys can compose music. Invalid Valid hypothetical syllogism 37. Logic is either hard or it is nutty. Either logic is easy or it is impossible. Logic is either nutty or it is impossible. Valid Invalid Valid resolution (38 41) Let A = {1, 3, 5, 6} B = {1, 2, 4, 6, 7} 38. A B = {1, 2, 3, 4, 5, 6, 7} 39. A B = {1, 6} 40. A B = {3, 5} 41. A B = {2, 3, 4, 5, 7} 42. Let A = {a, b, c} Give the power set of A, P(A) = {, {a,b,c},{a}, {b}, {c}, {a,b}, {a,c}, {b,c}} (43 44) Let A = {a,b,c} B = {x,y}. Find 43. A B = {(a,x), (a,y), (b,x), (b,y), (c,x), (c,y)} Valid Invalid 5

6 44. B B = {(x,x), (x,y), (y,y), (y,x)} (45 48) Determine whether each sentence below is true or false. Circle either True or False. 45. {a, b, c} {a, b, c} True False 46. {a, b, c} {a, b, c} True False 47. {a, b} {a, b, c} True False 48. {0} True False (49 51) Determine whether each sentence below is always true or not (sometimes false). Circle either Always True or Not Always. 49. (A B) A Always True Not Always 50. (A B) A Always True Not Always 51. A B = B A Always True Not Always (52 53) Which region(s) in the Venn diagrams above represent: 52. A (B C) B C = 2,3,4,5,6,7 A (B C) = 2,4,5 6

7 53. cmp(a) cmp(b) cmp(c) Note: cmp(a) cmp(b) cmp(c) = cmp(a B C) A B C = 1,2,3,4,5,6,7 cmp(a B C) = 8 = cmp(a) cmp(b) cmp(c) (54 55) Give the following combinations in simplified form, if: f(x) = (x 2 5) g(x) = (3x + 1) 54. (f + g) (x) = (x 2 5) + (3x + 1) = x 2 + 3x (f o g) (x) = f(g(x)) = ((3x+1) 2 5) = 9x 2 + 6x = 9x 2 + 6x 6 (56 57) For the questions below, the notation n j=0 x represents the summation of x as j goes from 0 to n. (56 57) What are the values of these sums? k=0 (k+1) (0+1) + (1+1) + (2+1) + (3+1) + (4+1) + (5+1) = = j= = 5*3 = 15 (58 59) What is the value of each of these geometric progressions? j=0 2 j a = 1; r = 2; n = 5 a(r n+1 1)/(r 1) = 1(2 6 1)/(2 1) = 64 1 = j=0 ( 3) j a = 1; r = 3; n = 4 a(r n+1 1)/(r 1) = 1(( 3) 5 1)/( 3 1) = ( 243 1)/( 4) = 244/ 4 = 61 (60 61) Compute each of these double sums i=1 3 j=1 (i+j) 3 i=1 (i+1) + (i+2) + (i+3) = 3 i=1 3i+6 = (3*1 + 6) + (3*2 + 6) + (3*3 + 6) = = 36 7

8 61. 3 i=0 2 j=0 (3i+2j) 3 i=0 (3i + 2*0) + (3i + 2*1) +(3i + 2*2) = 3 i=0 (3i i i + 4) = 3 i=0 (9i + 6) = (9*0 + 6) + (9*1 + 6) + (9*2 + 6) + (9*3 + 6) = 9*6 + 4*6 = = Arrange these complexity classes in ascending order of complexity: O(n n ), O(n b ), O(n), O(1), O(n!), O(n log n), O(log n), O(b n ) Answer:: O(1), O(log n), O(n), O(n log n), O(n b ), O(b n ), O(n!), O(n n ) 63. P(9,3) = 9!/6! = /720 = C(10,4) = 10!/4!6! = /24*720 = The 8 th line of Pascal s triangle is given below. Give the next line Answer:: How many different bit strings of length 8 are there? mod 17 = mod 13 = Give the prime factorization of *7*11* Give the prime factorization of *5* Give the prime factorization of 7! 2 4 *3 2 *5*7 72. List 3 consecutive odd integers that are all prime. 8

9 3, 5, How many bit strings of length 8 contain exactly 3 1's? 8*7*6/3*2*1 = What is the minimum number of students, each of whom comes from one of the 50 states, that must be enrolled in a university to guarantee that there are at least 100 that come from the same state? What are the two basic components of a recursive function? base case recursive call 76. What are the two basic components of mathematical induction? base case inductive step 77. What are the steps in proof by mathematical induction? 1. prove the base case 2. assume true for k 3. prove that it follows for k+1 OR: 1. prove base case, P(0) 2. prove P(k) P(k+1) 78. Give the recursive definition of factorial: fact(n) = 1, if n = 0 n * fact(n), otherwise 79. Why is this not a circular definition? Because: 1. n-1 < n, and therefore: 2. We are guaranteed to eventually reach the base case in the definition Prove: The sum of two odd integers is even. Proof: Let: a = 2k + 1 b = 2l + 1 Then: a + b = 2k l + 1 = 2(k + l +1) EVEN 9

10 Prove: The product of two odd integers is odd. Proof: Let: a = 2k + 1 b = 2l + 1 Then: ab = (2k + 1)*(2 l + 1) = 4kl + 2k + 2l +1 = 2(2kl + k + l) + 1 ODD Prove: For every positive integer n, there are n consecutive composite integers. [Hint: Consider the n consecutive integers starting with (n+1)! + 2.] Proof: 2 (n+1)! (n+1)! (n+1)! (n+1)! n (n+1)! + n n+1 (n+1)! + n+1 Thus, there are n consecutive composites. Extra Credit EC-1. Give the values of C and k which can be used to show that f(x) = 3x + 7 is O(x). C = 4; k = 7 EC-2. Give the values of C and k which can be used to show that f(x) = x 4 + 9x 3 + 5x 2 + 4x + 7 is O(x 4 ). C = 5; k = 9 EC-3. How many bit strings of length 10 contain at least 4 1's? 2 10 at most 3 1's = 2 10 (C(10,0) + C(10,1) + C(10,2) + C(10,3)) = ( ) = 848 EC-4. How many bit strings of length 10 have an equal number of 1's and 0's? 10

11 To be equal #-1s = #-0s = 5 So question is, how many bit strings have exactly 5 1's? C(10,5) = 252 EC-5. What is the coefficient of x 5 y 8 in (x + y) 13? C(13,5) = 1287 EC-6. What is the coefficient of x 7 in (x + 1) 11? C(11,7) = 330 EC-7. What is the coefficient of y 3 x 5 in (3y + 2x) 8? (3y 3 )(2x 5 )*C(8,3) = *8*7*6/3*2*1 = 27*32*56 = EC-8. What is the coefficient of x 9 in (2 x) 19? *C(19,9) = *92378 = -94,595,072 EC-9. Prove or disprove: If a bc, with a,b & c positive integers, then either a b or a c. DISPROOF:: by counterexample: a = 15, b =9, c = 5, bc = but 15 9 & 15 5 OR: a = 6, b =2, c = 9, bc = but 6 2 & 6 9 EC-10. Prove by mathematical induction: n! > 2 n, for all n > c, where c is some positive integer. To Prove: n! > 2 n, for all n > 3. PROOF:: Prove base case: 4! = 24 > 16 = 2 4 Inductive step: Assume: k! > 2 k, for all k > 3. To prove: (k+1)!> 2 k+1. Proof: (k+1)! = (k+1)*k! > (k+1)* 2 k > 2*2 k = 2 k+1 EC-11. Prove by mathematical induction: The sum of the first n even nonzero integers is n(n+1). In other words, prove: n i=1 i = n(n+1) 11

12 PROOF:: Prove base case: To prove: 1 i=1 i = n(n+1) Proof: 1 i=1 i = 2*1 = 2 If n =1, then n(n+1) = 1*(1+1) = 2 Inductive step: Assume: k i=1 i = k(k+1) To prove: k+1 i=1 i = (k+1)[(k+1) + 1] Proof: k+1 i=1 i = k i=1 i + 2(k+1) = k(k+1) + 2(k+1) = (k+1)(k+2) = (k+1)[(k+1) + 1] EC-12. Prove by mathematical induction: n! > 25 n, for all n > c, where c is some positive integer. To Prove: n! > 25 n, for all n > 65. PROOF:: Prove base case: 65! = 8.2 x > 7.3 x = Inductive step: Assume: k! > 25 k, for all k > 65. To prove: (k+1)!> 25 k+1. Proof: (k+1)! = (k+1)*k! > (k+1)* 25 k > 25*25 k = 25 k+1 EC-13. Prove: For all b > 0, c b such that n! > b n, for all n > c b. PROOF:: It is very likely that c b = ceil(2*e) is the correct value. But for a Q&D solution, c b = b b^2+b will work. First, we look at two examples: b = 3 b b^2+b = 3 12 (b 2 +b)! = 1*2*3*4*5*6*7*8*9*10*11*12 3*4*5 > 3 3 6*7*8 > = > > > 3 2 All together, (b 2 +b)! > 3 2*3+4*2 = 3 14 >

13 b = 5 b b^2+b = 5 30 (b 2 +b)! = 1*2*3*4*5 *.. *29*30 5*6*7*8*9 > *11*12*13*14 > *16*17*18*19 > *21*22*23*24 > = > > > > > 5 2 All together, (b 2 +b)! > 5 4*5+6*2 = 5 32 > 5 30 In general, b b^2+b represents b multiplied by itself b 2 + b times (b 2 +b)! is a number greater than b multiplied by itself (b-1)*b + 2*(b+1) = b 2 b +2b +2 = b 2 + b +2 times Since b 2 + b +2 > b 2 + b, (b 2 + b)! > b b^2+b 13