Assignment #2 COMP 3200 Spring 2012 Prof. Stucki 1) Construct deterministic finite automata accepting each of the following languages. In (a)-(c) the alphabet is = {0,1}. In (d)-(e) the alphabet is ASCII and you can specify transitions with character classes. a) {w w begins with a 1 and ends with a 0} b) {w w contains at least 3 1s} c) {w w has length at least 3 and its third symbol is a 0} d) {w w is an integer literal} e) {w w is a Java comment} 2) Now construct non-deterministic finite automata for each of the languages in (1). 3) Give state transition diagrams of NFAs recognizing the following languages. In all cases, the alphabet is {0, 1}. (a) (b) (c) (d) (e) {w w ends with 00} with three states {w w contains an even number of 0s, or exactly two 1s} with six states {w w contains the either the substring 010 or the substring 100} with six states {0} with two states {1, 00, 111} with five states
Homework #2 Solution C SC 381 Spring 2008 Prof. Stucki 1) 1.5.2, p. 29 Show by induction that n 4 4n 2 is divisible by 3 for all n 0. Base: n = 0 0 4 4(0) 2 = 0, which is certainly divisible by 3. IH: Assume that 3 k 4 4k 2 for 0 k n IS: Show that 3 (n+1) 4 4(n+1) 2 (n+1) 4 4(n+1) 2 = (n 4 + 4n 3 + 6n 2 + 4n + 1) (4n 2 + 8n + 4) = (n 4 4n 2 ) + (4n 3 + 6n 2 + 4n + 1 8n 4) = (n 4 4n 2 ) + (4n 3 + 6n 2 4n 3) = (n 4 4n 2 ) + 3(n 3 + 2n 2 n 1) + n 3 n = (n 4 4n 2 ) + 3(n 3 + 2n 2 n 1) + (n 1) n (n + 1) In this last expression, the IH tells us the first term is divisible by 3, the middle term is explicitly divisible by 3, and the last term, as the product of three consecutive numbers, must have a factor of 3. 2) 1.5.3, p. 29 What is wrong with the purported proof (in textbook) that all horses are the same color? The problem is that the asserted base case of one horse is not the correct one. Consider the case when n + 1 = 2. Call the horses in the set Bucephalus and Mister Ed. When you discard Bucephalus, all the remaining horses, meaning {Mister Ed}, have the same color by the inductive hypothesis. Call this color c 1. Put Bucephalus back into the set and discard another. The only possibility is Mister Ed. This time all the remaining horses, referring to {Bucephalus}, have the same color. Call this color c 2. So Bucephalus and Mister Ed have the same color as the ones that were not discarded either time, which we shall call c 3. That is c 1 = c 2 = c 3. But when n + 1 = 2, there are no horses in the set that were never discarded, so there are no horses of color c 3 to which we can apply the is the same color as relation to conclude c 1 = c 3 or c 2 = c 3, and no meaningful fixed value for c 3. Since the induction fails to show that if the claim holds for 1 horse, then the claim holds for two horses, the entire proof fails. Notice that if a proper base case were established, then the proof would show that all horses have the same color if any set of two horses were all the same color, then any set of three would be, etc. 3) 1.5.6, p. 29 Show that in any group of at least two people there are at least two persons that have the same number of acquaintances within the group. (Use the pigeonhole principle.) The pigeonhole primciple says that there is no one-to-one function from a set A of pigeons to a set B of holes, whenever A > B. So to demonstrate that at least two people in our group ( pigeons ) know the same number of people within the group ( fit into the same pigeonhole ), we must show that the set P of people has a greater cardinality than the set R of possible number of people that a person could know. We know that P = n. For this group of n people, we assume that knowing is not reflexive (we could also have assumed that knowing is reflexive; only a few minor details of the proof would be changed). So a person p can know anywhere from 0 to n 1 people in the group (i.e., can know no one, or can know everyone, or anywhere in between). Consider the set of numbers of people the p could know. That set R is {0, 1,, n 1}, and clearly has n elements. If there is a person p who knows everyone in the group, then there cannot also be a person r who knows nobody else. Either p and r know each other (and thus r knows someone), or they do not (and thus p does not know everyone). Thus R can contain either 0 or n 1, but not both, so in no case can R have a cardinality of more than n 1. As P = n, this demonstrates that R < P, so there is no one-to-one function from R to P, so by the pigeonhole principle at least two people must know the same number of people within the group.
4) 1.5.7, p. 29 Suppose we try to prove, by an argument exactly parallel to the proof of Theorem 1.5.2, that the set of all finite subsets of is uncountable. What goes, wrong? When we take the complement of a finite subset of, the set we obtain will be infinite. Thus the new diagonal set is not a finite subset of. As such it did not need to be included in our enumeration, so its absence from that enumeration poses no problem. 5) 1.5.8, p. 29 Give examples to show that the intersection of two countably infinite sets can be either finite or countably infinite, and that the intersection of two uncountable sets can be finite, countably infinite, or uncountable. The intersection of two countably infinite sets can be either finite or countably infinite: The intersection of the odd naturals with the even naturals is the empty set, which is finite. The intersection of the prime integers with the odd integers is the odd primes, which is countably infinite. On the other hand, the intersection of two uncountably infinite sets can be finite, countably infinite, or uncountably infinite: The intersection of 2 with the set of all real numbers is empty, and thus finite. The intersection of 2 with {2 2 } is, which is countably infinite. The intersection of 2 with itself is uncountable. Likewise, the intersection of the interval [0. 6) with the interval (5, 10] is the interval (5, 6), which is uncountable. 6) 1.6.1, p. 40 Are the following sets closed under the following operations? If not, what are the respective closures? a) The odd integers under multiplication. Yes, the product of any two odd integers is an odd integer (Can you prove this?). b) The positive integers under division. No, +, the positive rational numbers, is the closure of the positive integers under division. c) The negative integers under subtraction. No, the closure of the negative integers under subtraction is, the set of all integers. d) The negative integers under multiplication. No, the closure is the set of all non-zero integers. e) The odd integers under division. No, the closure is the set of all rational with only odd numerators and denominators.
7) 1.6.2, p. 40 What is the reflexive transitive closure R * of the relation R = {(a, b), (a, c), (a, d), (d, c), (d, e)}? Draw a directed graph representing R *. a b e d c 8) 1.6.3, p. 41 Is the transitive closure of the symmetric closure of a binary relation necessarily reflexive? Prove it or give a counterexample. No. Let R be the empty relation on any non-empty set S. R is its own symmetric closure, and its own transitive closure, but remains non-reflexive. 9) 1.6.5, p. 41 Give an example of a binary relation that is not reflexive but has a transitive closure that is reflexive. The relation {(0, 1), (1, 0)} on the set {0, 1} is not reflexive, but its transitive closure is the relation {(0, 0), (0, 1), (1, 0), (1, 1)}.