CPSC 121: Models of Computation Assignment #4, due Thursday, March 16 th,2017at16:00

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1 CPSC 121: Models of Computation Assignment #4, due Thursday, March 16 th,2017at16:00 [18] 1. Consider the following predicate logic statement: 9x 2 A, 8y 2 B,9z 2 C, P(x, y, z)! Q(x, y, z), where A, B, and C are sets and P and Q are predicates. There is not enough information about the domains and predicates to prove this theorem. Instead, lay out as much of the structure of a proof of this theorem as you can for each proof technique that we specified. Whenever you choose a specific value for a variable, specify what (if anything) this choice can depend on. Number each distinct step so it s easy for us to follow your proof structure (but we wouldn t normally write these numbers in an English-language proof like this). [7] a. Suppose that we are trying to prove this statement using a direct proof. [7] b. Suppose that we are trying to prove this statement using a proof by contrapositive. [4] c. Suppose that we are trying to prove this statement using a proof by contradiction. [6] 2. Alice has recently found a piece of paper that describes a new theorem. Unfortunately the piece of paper had gotten wet in the rain, and the ink had run o, so all that was readable is the following portion of a proof: Proof: Consider an unspecified real number m. Choose the real number n to be 1 m Consider two unspecified real numbers a and b. Assume that a b <n.... Therefore, f(a) f(b) < m. Based on this information, reconstruct the entire theorem that was being proved, and write it using predicate logic. [12] 3. Let a and b be positive integers. Suppose that d is the greatest common divisor of a and b, denoted gcd(a, b), Then d has the following two properties: 1. d is a common divisor of a and b. In other words, a is divisible by d and b is also divisible by d. 2. For all integers c, ifc is common divisor of both a and b, thenc is less than or equal to d.

2 Using the definition of the greatest common divisor, prove the two theorems below. [8] a. For any two positive integers a and b, b is divisible by a if and only if gcd(a, b) = a. Hint: To prove a biconditional, remember that it s equivalent to two conditionals. Prove each conditional separately! [4] b. For any two positive integers a and b, ifa>b, then gcd(a, b) = gcd(b, a b). Hint: One way to prove an equality is to show that the left hand side of the equation is less than or equal to the right hand side of the equation, and then to also prove that the right hand side of the equation is less than or equal to the left hand side of the equation. You will want to repeatedly apply the definition of the greatest common divisor in your proof. Translated into predicate logic, the theorem statement becomes 8a 2 Z +, 8b 2 Z +,a>b! gcd(a, b) = gcd(b, a b). We have completed the first half of the proof for you. Please fill in the second half of the proof. Proof: Consider two unspecified positive integers a and b. Assume that a>b. First, we will show that gcd(a, b) apple gcd(b, a b). We will show that gcd(a, b) is a common divisor of b and a b. By property 1 of gcd, gcd(a, b) dividesb. Now, we will show that gcd(a, b) dividesa b. By property 1 of gcd, gcd(a, b) divides both a and b. Let a = m gcd(a, b) and b = n gcd(a, b) wherem and n are integers. Then we have a b = m gcd(a, b) n gcd(a, b) = (m n) gcd(a, b). Since (m n) is an integer, gcd(a, b) dividesa b. Therefore, gcd(a, b) is a common divisor of b and a b. By property 2 of gcd, since a common divisor of b and a b must be less than or equal to the greatest common divisor of b and a b, gcd(a, b) is less than or equal to gcd(b, a b). Second, we will show that gcd(b, a b) apple gcd(a, b). Please complete this part of the proof. QED [16] 4. Recall that for any two functions f(n) and g(n), f(n) isino(g) if9c 2 R +, 9n 0 2 N, 8n 2 N,n n 0! f(n) apple cg(n). Using this definition, prove the following two theorems. [8] a. For any three functions f, g and h, prove that if f(n) isino(g) and g(n) is in O(h), then f(n) isino(h). Hint: If you assume that a statement such as 9x 2 D, P(x) is true, you cannot choose a specific value of D for which P (x) is true, unlike when proving an existential. Instead, you simply know that for some unspecified value x in D,

3 P (x) is true. (You don t know which value, and you only know that there is at least one such value. There may be more!) [8] b. Diane Ross, a computer scientist, has written an excellent algorithm whose execution requires 7n 3 +2n +6 steps, where n is the size the input. Prove that the number of steps of the algorithm 7n 3 +2n +6isNOT in O(n 2 ). Hint: You may want to negate the definition, and then prove the negation directly. [8] 5. Prove this theorem using a proof by contrapositive: For any integer y, ify 5 + 2y 3 +8y 3y 4 +3y 2 + 4, then y 0. [8] 6. Prove this theorem using a proof by contradiction: For any integers a, b, and c, if a 2 + b 2 = c 2,thena is even or b is even. Hint: Try to show that c 2 is divisible by 4 and not divisible by 4 at the same time. [8] 7. Convert the following DFA to a sequential circuit. You must write down a truth table and the corresponding propositions to justify the design of your sequential circuit. Make sure that your sequential circuit has an output, labeled accept, which is true when the DFA is in an accepting state. If you use the approach taught in lecture, then your sequential circuit should have 3 sub-circuits which produce the next states for the 3 states of the DFA. Please clearly label which subcircuit corresponds to which starting state of the DFA.

4 The remaining questions are for additional practice only. You should not hand them in. The solutions are included below. 1. For every non-negative integer x, bx/2c + dx/2e = x. Additional information: You will find the definitions of the floor and ceiling functions in your textbook. In Epp s fourth edition, the floor and ceiling functions are defined on page 191. In Epp s third edition, they are defined on page 165. These functions are defined on pages 149 (143) in Rosen s seventh (sixth) edition. You should break the proof down into two cases. Solution : Translated into predicate logic, this is: 8x 2 N, bx/2c + dx/2e = x. Musings and scratch work (this is not part of the proof, and you don t need to include it with your answer; it s just a way in which you might have come up with the solution we ll give): the floors and ceiling functions don t do anything when x/2 is already an integer, that is when x is even. If x/2 is not an integer, when x is odd, then they will round up or down. So maybe we can divide the proof into two cases: one when x is even, and one when x is odd? Proof: Consider an arbitrary non-negative integer x. If x is even, then x/2 is a non-negative integer, and so bx/2c = x/2 and dx/2e = x/2. Therefore bx/2c + dx/2e = x/2+x/2 =x. If x is odd, then x =2y + 1 for some non-negative integer y. Hence bx/2c = by+0.5c = y and dx/2e = dy+0.5e = y+1. Therefore bx/2c+dx/2e = y+y+1 = 2y +1=x. Since every non-negative integer is even or odd, this completes the proof of the theorem. QED 2. Prove that for every integer n, theproductn(n 2 1)(n + 2) is divisible by 4. Solution : Consider an unspecified integer n. Since n 2 1=(n 1)(n + 1), the product n(n 2 1)(n + 2) is equal to (n 1)n(n + 1)(n + 2). Since n 1, n, n + 1 and n + 2 are four consecutive integers, at least one of them is divisible by 4. Therefore so is their product. QED Note: the product is in fact divisible by 8. Can you see why?

5 3. Given an arbitrary positive integer n, we can find n consecutive positive integers that are all composite. Additional information: a number is composite if it is greater than 1, and is not prime. Hint: use a trick similar to one we used in class. Solution : Translated into predicate logic, this is: 8n 2 Z + 9x 2 Z + 8i 2 Z +,i apple n! Prime(x+i). Another acceptable and simpler, but slightly less formal, formulation would be 8n 2 Z + 9x 2 Z +, Prime(x+1)^ Prime(x+2)^ ^ Prime(x+n). Musings and scratch work: the only trick we used in class (up to the day where the assignment was posted) was using n! when we proved that there is always a prime number larger than any given integer. Now, n! is clearly not prime (as long as n 3). What about n! + 1? Hmmm... I don t know. But n! + 2 will be even, and not prime. And n! + 3? Well n! is divisible by 3, so n! + 3 is also divisible by 3. And this will be true until n! + n. But that s only n 1 consecutive composite integers, since we don t know if n!+n + 1 is composite or not. So what can we do? Use (n + 1)! instead! Proof: Consider an unspecified positive integer n. Choose x =(n + 1)!. We claim that the integers x + 2, x + 3,... x +(n + 1) are all composite. Indeed, let i be an integer between 2 and n + 1. Because x =(n + 1)! is divisible by i, x + i is also divisible by i, and hence x + i is not prime. QED 4. Prove that for every real number c we pick, there will be some positive integer n such that 4 n >c3 n. Hint: the inside back cover of your textbook contains a list of properties of exponents and logarithms. A web search for logarithm reference will also turn up many alternative resources. Solution : This theorem contains a universal quantifier, with an existential quantifier nested inside it. Hence, we start by considering an unspecified real number c. The case where c apple 0 is trivial, since then c3 n apple 0 < 4 n. So assume that c is positive. Now, we need to show how to choose a value of n for which 4 n >c3 n. Scratch work (not part of the answer as such, but you might have scribbled that on the back of the page): we want 4 n >c3 n.thatwillbesatisfiedaslongas4 n /3 n >c, which means as long as (4/3) n >c,thatis,aslongasn>log (4/3) c. Back to the proof: pick any integer n larger than log (4/3) c. For instance, 1+dlog (4/3) ce would do (recall that dxe is the smallest integer that is greater than or equal to x). Then (4/3) n > (4/3) log (4/3) c, which means that (4/3) n >cand hence 4 n /3 n >cwhich is the same as writing that 4 n >c3 n, as desired. QED 5. Prove that for every three non-zero integers a, b and c, at least one of the three products ab, ac, bc is positive. Hint: use a proof by contradiction.

6 Solution : We use a proof by contradiction. Suppose that none of the three products ab, ac and bc is positive. Because a, b and c are not zero, none of the three products is equal to 0, and so they must all be negative. The product of three negative values is negative, and hence (ab)(ac)(bc) < 0. However (ab)(ac)(bc) =a 2 b 2 c 2 =(abc) 2, and we know that (abc) 2 0. We have thus obtained a contradiction. Therefore at least one of the three products must be positive. QED 6. Suppose that you are given a group of n children, and that you want to divide them into two soccer teams. Unfortunately, there are pairs of children that do not like each other, and you need to make sure that two children who do not like one another do not end up on the same team. Prove that, if you can achieve this, then for every group of k children c 1,...,c k where c 1 does not like c 2, c 2 does not like c 3,...,c k 1 does not like c k, and c k does not like c 1,thenk must be even. Hint: use an indirect proof. Solution : We will not write a direct translation into predicate logic, because it would be rather ugly (or involve too many strange predicates such as CanMakeTeams). Musings and scratch work: The hint says to use an indirect proof, and there is an implication in the statement, so maybe we can try to use the contrapositive (it s simpler than a proof by contradiction)? So what will the teams look like if we have those children c 1,...,c k who can t stand one another as described? Yes, I see... Proof: We prove the contrapositive of the statement. That is, we show that if there is a group of k children c 1,...,c k where c 1 does not like c 2, c 2 does not like c 3,..., c k 1 does not like c k, and c k does not like c 1, and k is odd, then we will not be able to make the two teams. Suppose that we have this group of k children. Without loss of generality we assume that child c 1 is in team 1. This means that child c 2 is in team 2 (he/she can not be in the same team as c 1 ). Similarly, c 3 will be in team 1 and c 4 in team 2. Continuing with the same pattern, we get to child c k 1, who will be in team 2sincek 1 is even. But then we can not put child c k in team 1 (he/she doesn t like c 1 ), nor can we put c k in team 2 (he/she doesn t like c k 1 ). Hence there is no way to partition the group into two teams. QED 7. Can you draw a quadrilateral (4-sided figure) in which the sum of any two angles no matter which two you pick is less than 180? Explain how to do it, or prove that it can not be done. Hints: (1) the sum of the angles in a quadrilateral is always 360. (2) use a proof by contradiction. Solution : It can not be done. Let us call the angles of the quadrilateral a 1, a 2, a 3 and a 4. We know that the sum of the angles in a quadrilateral is 360. That is, a 1 + a 2 + a 3 + a 4 = 360

7 Let us now look at S =3a 1 +3a 2 +3a 3 +3a 4. On one hand S = (a 1 + a 2 )+(a 1 + a 3 )+(a 1 + a 4 )+(a 2 + a 3 )+(a 2 + a 4 )+(a 3 + a 4 ) < = 1080 Therefore S<1080. On the other hand S = 3(a 1 + a 2 + a 3 + a 4 )=3 360 which means that S = This is a contradiction, since S can not be both less than 1080 and equal to 1080 at the same time. QED