MATH 501: Discrete Mathematics

Transcription

1 January 5, 016 : Final exam Model Solutions Instructions. Please read carefully before proceeding. (a) The duration of this exam is 180 minutes. (b) Non-programmable calculators are allowed. (c) No books or other aids are permitted for this test. (d) This exam booklet contains a total of 11 pages, including this one. Page 1/11

2 Exercise 1 First-order Logic (8 Marks) (a) State whether the following statements are TRUE or FALSE. Justify your answer with a concise explanation. (a) x(at(x, GUC) Smart(x)) means Everyone at the GUC is smart. False (b) The formula x y(x = y) is valid. True (c) Any propositional logic formula can be translated into an equivalent formula using only the negation symbol. False (d) The and the are the only two connective that can represent all the other connectives of propositional logic. False (e) Two simple statements joined by a connective to form a compound statement are know as a disjunction. False (b) Choose the correct answer. Justify your answer with a concise explanation. (a) Which of the following formulas is a formalization of the sentence: Any person can fool some of the people all of the time, all of the people some of the time but not all of the people all of the time. i. x(person(x) y t(person(y) Fools(x, y, t)) t y(person(y) Fools(x, y, t)) t y(person(y) Fools(x, y, t))) ii. x(person(x) t y(person(y) Fools(x, y, t)) y t(person(y) Fools(x, y, t)) t y(person(y) Fools(x, y, t))) iii. x(person(x) y t(person(y) Fools(x, y, t)) t y(person(y) Fools(x, y, t)) t y(person(y) Fools(x, y, t))) i. Final exam Page /11

3 (b) It is not the case that the car is red and the truck is blue. Let proposition p = the car is red and q = the truck is blue. Which formula represents the above statement? i. p q ii. (p q) iii. (p q) iv. p q iii. (c) Propositional logic uses propositions and connectives to express i. Statements ii. The relationships between subject and predicate iii. Truth values iv. Statements and the relationships between statements iv. Final exam Page 3/11

4 Exercise First-order Logic (0 Marks) (a) Given the following predicate symbols: (10 Marks) P(x,y) which holds iff x is the parent of y. S(x,y) which holds iff x is the sibling of y. M(x) which holds iff x is a male. In the First-order language introduced in the lecture, express the following family relations: (a) x is the sibling of y (without using the S(x,y) predicate) z(p(z, x) P(z, y)) (b) x is the great-grandparent of y w, v(p(x, w) P(w, v) P(v, y)) (c) x is the nephew of y (nephew is the son of one s sister or brother) z(p(z, x) S(y, z) M(x)) (d) x is the first cousin of y (child of your parent s siblings) w, v(s(v, w) P(v, x) P(w, y)) (e) x is the second cousin of y (child of parent s cousin) w, v, q, r(s(v, w) P(v, q) P(w, r) P(q, x) P(r, y)) Final exam Page 4/11

5 (b) Given the following KB: (10 Marks) Anyone whom Mary loves is a football star. x(loves(mary, x) Star(x)) Any student who does not pass does not play football. x((student(x) Pass(x)) Play(x)) John is a student. Student(john) Any student who doesn t study doesn t pass x((student(x) Study(x)) Pass(x)) Anyone who doesn t play football isn t a football star x( Play(x) Star(x)) Convert the set of sentences into clausal normal form then apply resolution to prove: If John does not study, then Mary does not love John. Study(john) Loves(mary, john) Convert the sentences into clausal normal form C1: Loves(mary, x) Star(x) C: Student(x) Pass(x) Play(x) C3: Student(john) C4: Student(x) study(x) Pass(x) C5: Play(x) Star(x) Add negation of the goal C6: Study(john) C7: Loves(mary, john) Apply resolution to derive C8 (C7RC1): Star(john) C9 (C8RC5): Play(john) C10 (CRC3): Pass(john) Play(john) C11 (C9RC10): Pass(john) C1 (C3RC4): Study(john) Pass(john) C13 (C11RC1): Study(john) C14 (C6RC13): Final exam Page 5/11

6 Exercise 3 Proofs (1 Marks) Prove the following statements: (a) The difference of any rational number and any irrational number is irrational. (4 Marks) Let m be a rational number such that m = x for integers x, y 0, and n be an irrational number. y Assume m n is rational, then m n = q r for integerq, r 0, then x y n = q r Thus, n = x y q xr qy =, then n is rational (Contradiction!!) r yr (b) For any positive integer n (8 Marks) n ( 1) i i = ( 1)n n(n + 1) i=1 Base case (n = 1): Hypothesis (assume for positive integer k): ( 1) 1 1 = 1 = ( 1)1 1 k ( 1) i i = ( 1)k k(k + 1) i=1 Induction step: k+1 ( k ) ( 1) i i = ( 1) i i + ( 1) (k+1) (k + 1) i=1 i=1 = ( 1)k k(k + 1) + ( 1) (k+1) (k + 1) = ( 1)k k(k + 1) + ( 1) (k+1) (k + 1) = ( 1)k (k + 1) = ( 1)k (k + 1) = ( 1)k (k + 1) (k + ( 1)(k + 1)) ( k ) ( 1)(k + ) = ( 1)(k+1) (k + 1)(k + ) Final exam Page 6/11

7 Exercise 4 Summations (1 Marks) Consider the following recursive definition for function f: f(0) = 0 f(n + 1) = a f(n) + b where a and b are constants and a 1. Use the perturbation method to evaluate the summation: S n = Hint. Look at the difference between two adjacent values S n+1 S n = f(n + 1). Expand this difference and see if any relevant terms cancel out. f(k) f(n + 1) = f(1) + = b + = b + a = (a 1) f(k + 1) f(k+1) {}}{ (a f(k) + b) f(k) + f(k) + 0 k n = (a 1) S n + (n + 1) b b b f(k) f(k) f(k) Therefore: f(n + 1) = (a 1) S n + (n + 1) b thus S n = f(n + 1) (n + 1) b a 1 Final exam Page 7/11

8 Exercise 5 Counting (1 Marks) (a) In a race with 3 runners where 8 trophies will be given to the top 8 runners (the trophies are distinct: first place, second place, etc) (4 Marks) How many ways can this be done? N = How many ways can you do the above problem if a certain person must be one of the top 3 runners. N = Final exam Page 8/11

9 (b) The F-tree is an infinite tree such that at level k each node has k + 1 child nodes. For example in level 1, the root has children. Each of these has 3 children (see Figure 1). Find the number of nodes in level k. (4 Marks) Figure 1: F-tree The number of nodes at level k + 1is the product of the number of nodes in level k and the number of children per node in level k. Therefore, N(k + 1) = N(k) (k + 1) and N(1) = 1. This is exactly the factorial function. The number of nodes in level k is k!. (c) Let the days be numbered 1,,, 300. (4 Marks) On days of even number, you ll say you are sick. On days that are multiple of 3, you ll say you were stuck in traffic. On days that are multiple of 5, you ll say you got lost. How many days will you not come to the university? Let A i be the set of days that are divisble by i, in which you will not go to the university. First we compute the size of each set. N(A ) = 150, N(A 3 ) = 100, N(A 5 ) = 60 Then we have to find the size of the intersection of each pair of sets. N(A A 3 ) = 50, N(A 3 A 5 ) = 0, N(A A 5 ) = 30 Finally we compute the size of the intersection of A, A 3, and A 5 N(A A 3 A 5 ) = 10 N(A A 3 A 5 ) = N(A ) + N(A 3 ) + N(A 5 ) N(A A 3 ) N(A 3 A 5 ) N(A A 5 ) + N(A A 3 A 5 ) = = 0 Final exam Page 9/11

10 Exercise 6 Generating Functions (1 Marks) (a) Find a closed form for the generating function A(x) of the sequence a 0, a 1, a,... given through the recurrence: a 0 = 1, a n+1 = 5a n + ( 1) n. a n+1 = 5a n + ( 1) n a n+1 x n = 5a n x n + ( 1) n x n n 0 a n+1 x n = 5 n 0 a n x n + n 0 A(x) 1 x = 5A(x) x A(x) = 5xA(x) + (1 5x)A(x) = x 1 + x + 1 A(x) = x 1 + x + 1 ( 1) n x n x (1 + x)(1 5x) x (b) Find the coefficient [x n ]A(x) of the generating function A(x). Partial fraction expansion x = α (1 5x) + β (1 + x) x= 1 = 1 = α 6 + β 0 α = 1 6 x= 1 /5 = 1 5 = α 0 + β 6 β = A(x) = x x x A(x) = x x A(x) = 7 5 n x n 1 ( 1) n x n 6 6 A(x) = n 0 n 0 n 0 ( 7 6 5n 1 6 ( 1)n )x n a n = 7 5n ( 1) n 6 Final exam Page 10/11

11 Exercise 7 Chinese remainder theorem (1 Marks) Using the chinese remainder algorithm described in the lecture, find the smallest non-negative solution to the following simultaneous congruences. x (mod 3) x 3 (mod 7) x 4 (mod 11) We have N = = 31, N /3 = 77, N /7 = 33,and N /11 = 1. We solve the Bézout identities n s 1 77 = 1 (and e 1 = s 1 77) n 7 + s 33 = 1 (and e = s 33) n s 3 1 = 1 (and e 3 = s 3 1) using the extended Euclidean algorithm n 1 s 1 n 1 s n s n s n 3 s 3 n 3 s So e 1 = s 1 N/n 1 = ( 1) 77 = 77, e = s N/n = 3 33 = 99, and e 3 = s 3 N/n 3 = ( 1) 1 = 1. Thus x e 1 a 1 + e a + e 3 a 3 (mod N) ( 77) ( 1) 4 (mod 31) 59 (mod 31) We are looking for the smallest nonnegative x, so we take x = 59. Final exam Page 11/11