Solutions to In-­Class Problems Week 4, Mon

Massachusetts Institute of Technology 6.042J/18.062J, Fall ’05: Mathematics for Computer Science Prof. Albert R. Meyer and Prof. Ronitt Rubinfeld September 26 revised September 26, 2005, 1050 minutes Solutions to In­Class Problems Week 4, Mon. Problem 1. In each case, say whether or not R is a equivalence relation on A. If it is an equivalence relation, what are the equivalence classes and how many equivalence classes are there? (a) R ::= {(x, y) ∈ W × W | the words x and y start with the same letter} where W is the set of all words in the 2001 edition of the Oxford English dictionary. Solution. R is an equivalence relation since it is reflexive, symmetric, and transitive. The equiva­ lence class of x with respect to R is the set [x]R = the set of words y, such that y has the same first letter as x. There are 26 equivalence classes, one for each letter of the English alphabet. � (b) R ::= {(x, y) ∈ W × W | the words x and y have at least one letter in common}. Solution. S is reflexive and symmetric, but it is not transitive. Therefore, S is not an equivalence relation. For example, let w1 be the word “scream,” let w2 be the word “and,” and let w3 be the word “shout.” Then w3 Sw1 , and w1 Sw2 , but it is not the case that w3 Sw2 . � (c) R = {(x, y) ∈ W × W and the word x comes before the word y alphabetically}. Solution. R is not reflexive but it is transitive and antisymmetric. It is not an equivalence relation, but it is a partial order. � (d) R = {(x, y) ∈ R × R and |x| ≤ |y|}. Solution. R is reflexive and transitive. It is not symmetric. It is not antisymmetric either. As a counterexample, | − 3| ≤ |3|, |3| ≤ | − 3|, but 3 �= −3. � (e) R = {(x, y) ∈ B × B, where B is the set of all bit strings and x and y have the same number of 1s.} Copyright © 2005, Prof. Albert R. Meyer and Prof. Ronitt Rubinfeld. 2 Solutions to In­Class Problems Week 4, Mon. Solution. R is reflexive, symmetric and transitive, and therefore an equivalence relation. There is an equivalence class for each natural number corresponding to bit strings with that number of 1s. � Problem 2. False Claim. Suppose R is a relation on A. If R is symmetric and transitive, then R is reflexive. (a) Give a counter­example to the claim. Solution. The simplest counterexample is to let R be the empty relation on some nonempty set A. This R is vacuously symmetric and transitive, but obviously not reflexive. A slightly less trivial example is R ::= {(a, a), (a, b), (b, a), (b, b)} on the set A ::= {a, b, c}. It is not reflexive because (c, c) is not in R. � (b) Find the flaw in the following proof of the claim: False proof. Let x be an arbitrary element of A. Let y be any element of A such that xRy. Since R is symmetric, it follows that yRx. Then since xRy and yRx, we conclude by transitivity that xRx. Since x was arbitrary, we have shown that ∀x ∈ A (xRx), so R is reflexive. Solution. The flaw is assuming that y exists. It is possible that there is an x ∈ A that is not related by R to anything. No such R will be reflexive. Note that the theorem can be fixed: R restricted to its domain of definition is reflexive, and hence an equivalence relation. � Problem 3. Verify that each of the following relations is a partial order by describing a function, g, such that the relation is defined by g according to the Definition 4.2 in the Appendix. For each, is it a total order? (a) The relation, <, on R. Solutions to In­Class Problems Week 4, Mon. 3 Solution. Define g(r) ::= {t ∈ R | t < r}. It follows that r1 < r2 iff g(r1 ) ⊂ g(r2 ) so < satisfies the condition (3) on R that defines partial orders. Likewise, the relation, ≤, is a partial order because r1 ≤ r2 iff r1 < r2 , for all reals r1 = � r2 . � (b) The superset relation, ⊇, on P(B) for a set, B. � a2 ∈ P(B), Solution. Define g(a) ::= a ::= B − {a}, and note that for a1 = a1 ⊇ a2 iff a1 ⊃ a2 (since a1 �= a2 ) iff a1 ⊂ a2 (basic set theory) iff g(a1 ) ⊂ g(a2 ) (def of g). � (c) The “divides” relation on natural numbers. Solution. Let g(a) ::= the set of natural number that divide a. � Problem 4. Suppose you are given the description of an equivalence relation and want to cut down on the number of pairs that are stored without losing any information. For example, here are the pairs of an equivalence relation on a set of integers: G ::= {11, 33, 44, 55, 66, 77, 13, 31, 45, 54, 47, 74, 57, 75} . where for readability, we’ve written ”mk” to designate the pair (m, k). To start, we know that if we have mk, then we necessarily also have km, so there’s no need to keep both. This lets us cut down to: {11, 33, 44, 55, 66, 77, 13, 45, 47, 57} . Also, if we have 45 and 57, we don’t need 47, since that will necessarily be there (by transitivity), so we can further cut down to {11, 33, 44, 55, 66, 77, 13, 45, 57} 4 Solutions to In­Class Problems Week 4, Mon. In addition, as long as we keep some pair in which k appears, we don’t need the pair kk. This lets us cut down to the pairs {66, 13, 45, 57} . (1) These pairs are all that are needed to determine the entire original equivalence relation, G. More­ over, the set (1) of these pairs is minimal with this property; this means that if any pair was removed from the set, it wouldn’t determine the relation any more. (a) Describe another couple of minimal sets of pairs that determine the relation. Solution. {66, 31, 45, 57} , {66, 13, 47, 57} , and lots more . . . . � (b) Here are the pairs that are left after some unnecessary pairs have been removed from the description of an equivalence relation, E. What is the domain of E? What are the equivalence classes of E? 15, 20, 40, 57, 68, 79, 9a, bb, c3 Solution. domain of E = {0123456789abc} equivalence classes of E = {024} {1579a} {3c} {68} {b} . � (c) On a domain of n elements, what is the smallest number of pairs that could determine an equivalence relation? Solution. �n/2� by having all blocks of size 2, except for one block of size 1 if n is odd. � (d) Suppose you have an equivalence relation on a domain of size n with k equivalence classes, with no classes of just one element. Then every minimal set of pairs has the same size. What is that size? Explain. Solution. It takes n − k pairs. We’ll let you convince yourself that c − 1 is the minimum number of pairs to “connect up” all the elements and thereby determine an equivalence class of size c > 1. (We’ll prove this carefully in couple of weeks when we study tree structures.) So if there are k classes of sizes c1 , . . . , ck , you need (c1 − 1) + (c2 − 1) + · · · + (ck − 1) = (c1 + c2 + · · · + ck ) − k = n − k pairs. � Solutions to In­Class Problems Week 4, Mon. 5 Appendix Equivalence Definition 4.1. A binary relation, R, on a set, A, is an equivalence relation iff there is a function, f , with domain A, such that a1 R a2 iff f (a1 ) = f (a2 ) (2) for all a1 , a2 ∈ A. Theorem. A relation is an equivalence iff it is reflexive, symmetric and transitive. Partial Order Definition 4.2. A relation, R, on a set, A, is a partial order providing there is a function, g, from A to some collection of sets such that a1 R a2 iff g(a1 ) ⊂ g(a2 ), for all a1 = � a2 ∈ A. Theorem. A relation is a partial order iff it is transitive and antisymmetric. Relational Properties A binary relation, R, on a set, A, is • reflexive if aRa for every a ∈ A, • irreflexive if aRa holds for no a ∈ A, • symmetric if for every a, b ∈ A, aRb implies bRa, • antisymmetric if for every a = � b ∈ A, aRb implies ¬(bRa), • asymmetric if for every a, b ∈ A, aRb implies ¬(bRa), • transitive if for every a, b, c ∈ A, aRb and bRc implies aRc. (3)