# CmSc 365 Theory of Computation

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```					CmSc 365 Theory of Computation Homework 01 SOLUTION 1. Problem 1.1.3 (a), p.9: Prove the equality A  (B  C) = (A  B)  ( A  C) (a) Prove A  (B  C) = (A  B)  (A  C) 1. We shall show that A  (B  C)  (A  B)  (A  C) Let x  A (B  C) Two cases are to be considered: a. x  A By def of union and (1) we have: x  A  B By def of union and (1) we have: x  A  C From (2) and (3) => x  (A  B)  (A  C) b. x  (B  C) By def of intersection and (5) => x  B By def of union and (6) => x  A B By def of intersection and (5) => x  C By def of union and (8) => x  A C From (7) and (9) => x  (A  B)  (A  C) From (4) and (10): A  (B  C)  (A  B)  (A  C) 2. We shall show that (A  B)  (A  C)  A  (B  C) Let x  (A  B)  (A  C) By def of intersection we have x  (A  B) x  (A  C) Let x  A By def of union and (15) => x  A  (B  C) Consider now the case when x  A Then from (13) x  B From (14) x  C (12)

(1) (2) (3) (4) (5) (6) (7) (8) (9) (10)

(11)

(13) (14) (15) (16)

(17) (18)

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From (18) and (17) x  B  C From (19) and def of union x  A  (B  C) From (20) (A  B)  (A  C)  A  (B  C) From (21) and (11) we have (A  B)  (A  C) = A  (B  C)

(19) (20)

(21)

2. Problem 1.2.2, p.13: Let R = {(a,b),(a,c),(c,d),(a,a),(b,a)}. What is R  R, the composition of R with itself? What is R-1, the inverse of R? Is R, R R, or R-1 a function? R is not a function R  R = {(a,a),(a,d),(a,b),(a,c),(b,b),(b,c),(b,a)} R-1 = {(b,a),(c,a),(d,c),(a,a),(a,b)}

not a function not a function

In all relations there is an element that has two images in the relation, and therefore they are not functions.

3. Problem 1.3.6, p.19. Let R  A x A be a binary relation as defined below. In which cases R is a partial order? A total order? Prove your answers by examining the properties of the relations a. A = the positive integers; (a,b)  R iff b is divisible by a Week partial order: reflexive, anti-symmetrical, transitive a. reflexivity: Let a is any positive integer. a is divisible by a (property of positive integers) therefore (a,a)  R b. anti-symmetrical Let (a,b)  R and a b a is divisible by b, therefore b is not divisible by a, therefore (b,a)  R c. transitive Let (a,b)  R and (b,c)  R. a is divisible by b, and b is divisible by c. Therefore a is divisible by c (property of positive integers) Therefore (a,c)  R

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Therefore R is partial order. It is not a total order, because there are elements of A that are not in the relation R, e.g. (3,7) and (7,3) are not in R b. A = N; (a,b)  R iff b = a or b = a+1 Not a partial order because it is not transitive. Let (a,b)  R and (b,c)  R, and let b = a+1, and c = b+1 Consider (a,c): c = b + 1 = a + 2, therefore (a,c)  R c. A is the set of all English words; (a,b)  R iff a is the same as b or occurs more frequently than b in our textbook. Partial order Reflexive: (a,a)  R by the definition of R Anti-symmetric: Let (a,b)  R, and a b. Therefore a occurs more frequently than b, therefore b does not occur more frequently than a, therefore (b,a) is not in the relation. Transitivity: Let (a,b)  R and (b,c)  R A occurs more frequently than b, and b occurs more frequently than c, therefore a occurs more frequently than c, therefore (a,c)  R Not a total order, because different words that occur with same frequency cannot be ordered.

4. Problem 1.3.8 (a), p.20 Prove that if S is any collection of sets, then RS = {(A,B): A, B  S and A  B} is a partial order Reflexivity: Let A S By definition of subsets, A  A, Therefore (A,A)  RS Anti-symmetric: Let (A,B) RS and A B By def of Rs, A B. Since A  B, B  A, therefore (B,A)  RS

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Transitivity: Let (A,B) RS and (B,C) RS A B , and B  C Therefore A  C (by definition of subsets) Therefore (A,C) RS 5. Prove that the set of all finite strings of the letter a is countable. Hint: show how the elements can be ordered, i.e. find a bijection that maps elements to N Let w be a finite string consisting of the letter “a”. Consider the function F(w) = k, where k is the number of the letters in the string. This function is a bijection to the set N, therefore the set of all finite strings of the letter “a” is countable.

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