CS158 Section B Exam 1 Key

Transcription

1 CS158 Section B Exam 1 Key Name This is a closed-book exam. The only items not supplied that you are allowed to use are writing implements. You have 50 minutes to complete this exam. The total amount of points you can score is 61, but note that you only need 50 to score 100%. Good luck! 1. [8 pts] Use mathematical induction to prove that if n people stand in a line, where n is a positive integer, and if the first person in the line is a woman and the last person in line is a man, then somewhere in the line there is a woman directly in front of a man.

2 2. [8 pts] Show that the following program segment terminates with factorial = n! when n is a positive integer. i = 1 factorial = 1 while i < n do i = i + 1 factorial = factorial * i end while 3. [12 pts] Verify the correctness of the following program segment with the assertions shown. {y = 3} x = 2 z = x + y {z = 5} if y > 0 then z = z + 1 else z = 0 end if {z = 6}

3 4. [8 pts] Prove the following statement (try proof by contraposition): If 3n + 2 is odd, then n is odd. 5. [14 pts] Show whether from the hypotheses: It is not sunny this afternoon and it is colder than yesterday, We will go swimming only if it is sunny this afternoon, If we do not go swimming, then we will take a canoe trip, and If we take a canoe trip, then we will be home by sunset we can conclude we will be home by sunset or not.

4 6. [11 pts] Using predicate logic, prove that the following argument is valid. Use the predicate symbols shown. Argument: Some plants are flowers. All flowers smell sweet. Therefore, some plants smells sweet. P (x), F (x), S(x) Answer 1. Proof 1. Base case: n = 2, the first person is a woman and the last person is a man. 2. I.H.: Assume that when n = k(k 2), in the line there is a woman directly in front of a man. 3. Show that when n = k + 1, in the line there is a woman directly in front of a man. From k to k + 1 people, there is 1 new person added in the line. There are three cases. (a) the new person is before the woman that is directly in front of a man. In this case, the woman is still directly in front of the man. (b) the new person is after the man that is directly after a woman. In this case, the woman is still directly in front of the man. (c) the new person is added between the woman and the man who are next to each other. In this case, if the new person is a woman, she is directly in front of the man; if the new person is a man, he is directly after the woman. In both cases, there is still a woman directly in front of a man. Therefore, for any positive integer n (n 2), there is always a woman directly in front of a man in the line. 2. Proof: Let guess Q: factorial = i!. Prove by induction that Q is a loop invariant. 1. Base case: Before the first iteration of the loop, factorial = 1 and i = 1. Q( 0 ) holds. 2. I.H.: Assume that Q( k ): factorial k = i k! holds. 3. Show that Q( k+1 ): factorial k+1 = i k+1! holds. From the assignment statements in the loop, we have: i k+1 = i k + 1 and factorial k+1 = factorial k i k+1. Therefore, factorial k+1 = i k! i k+1 = i k! (i k + 1) = (i k + 1)! = i k+1! Therefore, Q is a loop invariant. At loop termination, factorial = i! and i = n. Hence, factorial = n!. 3. Proof: We can split the program segment into two halves. The first half contains the first two assignment statements and the second half contains the conditional statement. We first prove the first half. We work backwards from the postcondition using the assignment rule twice:

5 {2 + y = 5} or {y = 3} x = 2 {x + y = 5} z = x + y {z = 5} This agrees with the precondition given. Next we prove the second half using the conditional rule. We need to show: (1) {z = 5 y > 0} z = z + 1 {z = 6} and (2) {z = 5 y 0} z = 0 {z = 6} The first one is true by assignment rule. y > 0 adds nothing new to this conditional statement. The second one is true because the precondition is always false. Therefore, the whole program segment is correct. 4. Proof: Using proof by contraposition, let s prove: n is even 3n + 2 is even. if n is even, we can write n = 2k where k is an integer. Then 3n + 2 = 3(2k) + 2 = 6k + 2 = 2(3k + 1) where 3k + 1 is an integer. Therefore, 3n + 2 is an even integer. Hence, if 3n + 2 is odd, then n is odd. 5. S: it is sunny this afternoon C: it is colder than yesterday W: we will go swimming T: we ill take a canoe trip H: we will be home by sunset Argument in propositional wff is: (S C) (W S) (W T ) (T H) H Proof: 1. S C hyp 2. W S hyp 3. W T hyp 4. T H hyp 5. S 1 sim 6. W 2,5 mt 7. T 3,6 mp 8. H 4,7 mp Therefore, we can conclude that we will be home by sunset. 6. The argument is: ( x)[p (x) F (x)] ( x)[p (x) S(x)] A proof sequence is: 1. ( x)[p (x) F (x)] hyp 2. ( y)[f (y) S(y)] hyp 3. P (a) F (a) 1 ei 4. F (a) S(a) 2 ui 5. F (a) 3 sim 6. S(a) 4,5 mp 7. P (a) 3 sim 8. P (a) S(a) 6,7 con 9. ( x)[p (x) S(x)] 8 eg