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Unformatted text preview: ’(iUU Lllt‘lll (ll Lin." tum. U‘y DCLLULES vv .— YY V 1? SECTION 8.5 Equivalence Relations 2. _a) This is an equi falence relation by Exercise 9 (:f(:.r) is .r’s age). b) This is an equivalence relation by xercise 9 (:f(.z') is its parents). c) This is not an (jiquivalence relation, since it need not be transitive. (We assume that biol<:>gical parentage: is at issue here, so it is possible for A to be the child of li' and X. B to be the child of X and Y, and C to be the child of Y and Z. Then A is related to B, and B is related to C', but A is not related to C'.) d) This is not an equivalence relation since it is clearly not transitive. e) Again, just as in part (C), this is not transitive. One relation is that a and b are related if they were born in the same US. state (with "not in a state of the US.” counting as one state). Here the equivalence classes are the nonempty sets of students from each state. 2 Another example is for a to be related to b if (z and l) have lived the same number of complete decades. The . difference be :vveen as age and bs age. Where “age" means the whole number of years since birth. as of the first day of class. For each '2' : 0.1.... , 9. there is the equivalence class (if it is nonempty) of those students , Whose age, ends with the digit i. One way to partition the clas s would be by level. At many schools. classes have three-digit numbers. the , first digit of which is approximately the level of the course, so that courses numbered 100-195) are taken by freshman, ZOUWZ‘JQ by sophomores, and so on. Formally, two classes are related if their nmnbers . same digit in the lmndreds column; the equivalence classes are the set of all MiG-level classes, the ,. QUU-level cla. ses, and so on. A second example would focus on department. Tvvo classes are e<p1ivalcnt i’ they are offered by the same department: for example, MATH 15-1 is equivalent to MATH 372. but not W EGR 141. The equivalence classes are the, sets of classes offered by each department (the set of math classes the set of engineering classes, and so on). A third—and more egocentric classification would be to have em? equivalence class be the set of classes that you have completed successfully and the other equivalence class to be all the other classes. Formally, two classes are equivalent if they have the same answer to the q uesiioll "Have I completed this class successfully?” Recall (Definition 4 in Section 2.4) that two sets have the same cardinality if there is a bijection (one-town? and onto function) from one set to the other. We must show that R is reflexive, symmetric, and transitive- Every set has the same cardinality as itself because of the identity function. If f is a bijection from S 1'0 r then f—1 is a bijection from T to S, so R is synnnetric. Finally, if f is a bijection from S to T and {I is a bijection from T to U , then 9 o f is a biject-ion from T to U, so B is transitive (see Exercise '3” in Section 2.33) Equivalence Relations 219 The equivalence class of {1.2 3} is the set of all three element sets of 1eal numbers. including su<h éets as {4, 25. 1948} and {6.711 v’2}. Similarly. [Z] is the set of all infinite countable sets of real numbers (see Section 2.4). such as the set of natural 111,1111bers. the set of rational numbers. and the set of the prime numbe us but not including the set {1.2. 3(} (its too small) or the set of all real numbers (it’s too big). See Section 2.4 {0}; more on countable sets. The function that sends each .17 E A to its equivalence class [1‘] is (b1 ionsl such a funttion This follows f1o111 Exe1cise911he1e f is the function that takes a bit string of length n > 3 to its last 11 - 3 his fOHOWS from Exercise 9‘ Where f is the function that takes a string of uppercase and 1011’81'case English l6 ers and Changes all the lower case letters to their Uppercase equivalents (and leaves the Uppercase letters unchanged). in follows from Exercise 9. Where f is the function from the set of pairs of positive integers to the set of ive rational numbers that takes ((1.11) to (1/11. since clearly (Id 2 be if and only if 11/11 = (7/4, ‘9. [1). (c. (1)) R then (1(1— — be. 11l11( 11 also means that (711—- — do. so ((c.d).)(t1. bl) E R; this tells us that R is . Final lv. 1 ((0.6). (c. (ll) E R and (led) (e. f)) E I? then ad :2 be and cf be gives ucdf : (mic. and since all these 11111111 If1ve want an explicit proof. we, can argue as follows. For reflexivitnl (0. l1).( (1. l1) ) E B because (1 vi) = ll-(I. dc. Multiplving these ll ers are nonzmo. 1“e lime (If l—1€r._ so (l (1, l1). ‘ u.f))_ 7—” 1?; Ms us that l? is transitive. This follows from Exercise 9.11'here the turn tion 7‘ f1 om the set of pol1nomials to the set of polfx noniials is e operator that takes the dcnimthe u ti111es—~i.e.. f of a function g is the function g ‘7“. The best way to nil; about this is that any relation defined by a statement of the form “a and b are equivalent if they have ‘ one Whatever" is an equivalence relation. Here “Whatever" is “11‘“ derivative"; in the general situation _. e1c1se 9. ‘11‘hatev1e‘" is “function value under f." The third derivative of .zr" is 2417. Since the third llcu‘ivative of a polynomial of degree 2 or less ‘ (l. the 7 Tnonnals of the term .1“ -:~ (1.17~ —.‘~ 132.17 + c have the same third derivative. Thus these are the functions in the e. etpiivalence class as f . follows from Exercise 9. Where the function f from the set of people to the set of Heb—traversing behaviors 11g at the given particular \Veb page takes the person to the behavior that person exhibited. Fecal to observe Whether the relation is reflexive (there is a loop at each vertex). symmetric (every edge “tippears is accompanied by its antiparallel mate—an edge involving the same two vertices but pointing Opposite direction). and transitive (paths of length 2 are accompanied by the path of length 1~i.e.. between the same two verticcs in the same direction). 'We see that this relation is an equivalence on. satisfving all three properties. The equivalence classes are {(1. d} and {11.0}. 11s is not an equivalence relation. since it is not symmetric. is an equivalence relation; one equivalence class consists of the first and third elements. and the other ,s of the second and fourth elements. “3 is an equivalence relation; one equivalence class consists of the first. second. and third elements. and ‘ consists of the fourth element. art (a) and part (c) are equivalence relations. In part (a) each element is in an equivalence class by In part (c) the elements 1 and 2 are in one equivalence class. and O and 3 are each in their own ice class. 220 28. 30. 32. 341. 36. 38. tn [0 44. Chapter 8 Relat 1-0”; Only part (a) and part (d) are equivalence relations. In part (a) there is one equivalence class for each n E Z. and it contains all those functions Whose value at 1 is n. In part (Cl) there really is no good way to describe the ecnlivalence classes. For one thing. the set of equivalence classes is uncountable. For each function f : Z —: Z. there is the equivalence class consisting of all those functions 9 for which there is a constant C such that gtn) = f(n) + C for all n E Z. a) all the strings whose first three bits are 010 b) all the strings Whose first three bits are 101 C) all the strings Whose first three hits are 111 d) all the strings Whose first three bits are 010 Since two strings are related if they agree beyond their first 3 hits, the equivalence class of a bit string .’I‘_z/;-r" where .r. y. and z are hits. and t is a bit string, is the set of all hit strings of the form .ri’y’z’t. Where .1", 1/. and 3’ are any bits. a) the set of all bit strings of length 3 (take t = ,\ in the formulation given above) b) the set of all bit strings of length 4 that end with a 1 C) the set of all hit strings of length 5 that end 11 cl) the set of all bit strings of length ‘1? that end 10101 a) Since this string has length less than 5. its equivalence class consists only of itself. b) This is similar to part (a): ilUlllRa : {1011}. 6) Since this string has length 5. its equivalence class consists of all strings that start 11111, d) This is similar to part (C): [010101011135 = {01010.9 I .5 is any bit string }. In each case. the equivalence class of 4 is the set of all integers congruent to 4., modulo 172. j a) {4+2n l 11E Z} = {....—2,0.2,4,...} 1)) {4+3}: ] n E Z} = {....~2.1.4.7....} c){4~:—611§12E Z} : {....——2,J.,10,16....} cl) {4+8n i n E Z} = {....—4.4.12.2()....} in each case we need to allow all strings that agree with the given string if we ignore the case in which the letters occur. 21) {NC}. N0. 110. no} 1)) {YES YES. 1'65. its. 1/ES. yEs. 1/65. yes} c) {HELP HELp.HE/P,HE112.H6LP.HeLp. .HGZP.]1TGZ]).}2ELP. liELerElP. liEZ]),]ZGLP./26Lp. lzelPchlp} 5 . a) By our observation in the solution to Exercise 16. the equivalence class of (,1: 2) is the set of all pairs ((1.1)) such that the fraction (1/!) equals 1/2. b) Again by our observation, the equivalence classes are the positive rational numbers. (Indeed. this is the way one can rigorously define ivhat a rational number is. and this is Why fractions are so difficult for children to understand.) 3| . a) This is a partition. since it satisfies the definition. b) This is not a partition. since the subsets are not disjoint. c) This is a partition since it satisfies the definition. (1) This is not a partition. since the union of the subsets leaves out 0. ”t a) This is clearly a partition. . b) This is not a partition. since 0 is in neither set. c) This is a partition by the division algorithm. (1) This is a partition, since the second set mentioned is the set of all number between —100 and 100. inClUSlW' e) The first two sets are not disjoint (4 is in both). so this is not a partition. n 335 Equivalence Relations 221 ; for This is a partition. smce it satisfies the definition. )d )This is a partition. since it satisfies the definition. 11111 ) This is not a partition. since the intervals are not. disjoint (they share endpoints). 7n ) This is not a partition. since the union of the subsets leaves out the integers. ) This is a partition, since it satisfies the definition. This is a partition, since it satisfies the definition. Each equivalence class consists of all real numbers with fixed fractional part. each case. we need to list all the pairs we can Where both coordinates are chosen from the same subset. We 'nr rould proceed in an organized fashion, listing all the pairs corresponding to each part of the partition. “‘8‘ l , (11.11). (01b), (ltd)- (17.17) )1 (c. C). (0- (0- ’(1- Cl- ((1- d)- (9.9). (of): (€117). (f1 6)- if f): (f: .0)» (11.5). (91f), (91.61)} b) {{a a),(b. b). (c. c). (c. d). (d. c).( d d).( e e) .(e. f).( .e).( f. f).(g .g)} } {(1111) (a. b) ((1.5).)((1(.1[)}(b.a)).(b.b). (17.57). (170') (cf (1). (c. b). (c. c)( c. d). ((1.61).(Cl,b).((l,€).((fl,d). 11.132 fit 6 9M Hf a) (g f) 111.111.1171 ) {(0, (1H1?!) (a 1))(1i :17): c (1).(c.c). \c.c),(c,g),(e.a).( ).(e. e) ).(e g) (g. (1) (q. c). )(g. e)( g. g). ,6) ((7 d). (1.17). (11 (1). fl} need to show 1hat every equivalence class consisting of people living in the same county (or parish) and me state is contained in an equivalence class of all people living in the same state. This is clear. The {trivalence class of all people living in county c in state 5 is a subset of the set of people living in state s e are asked to show that ev e13 equivalence class for R1 is a subset of some equivalence c las: s‘for P3. Let R} be an arbitraiv equivalence class for R1. We claim that [gr/jg“: ; [1/];33; proving this claim finishes the roof. To show that one set is a subset of another set. we choose an arbitrary bit string .17 in the first set and 1m that it is also an element of the second set In th is1 case since I} E ill 1;. . we know that 1/ is equivalent :1: under R1. that is. that either 2/ : 1 or 1/ and 1 aiee each at leastJ l.7its long and agree on their fiistJ whi ts. Becaus strings that are at least 4 bits long and agree on their firstJ '. bits perforce are at least 3 bits ng and agree on their first 3 bits. we know that either 1/ = .1,‘ or y and .z‘ are each at least 3 bits long and glee. on their first 3 bits. This 111eans that y is equivalent to .1,‘ under R53, that is. that 1/ E [1:] 133. {he , suppose that R1 2 R3. We must short that P1 is a refinement of P3. Let MR1 be an equivalence I ass in P1 . “7e, must show that MR1 is contained in an equivalence class in P3. In fact. me Will show that _7ai J ;_ lale' To this end. let (7 E MR! . Then ((1.1)) E R1 ; R3. Therefore 1) E [algw as desired. Convers , suppose that P1 is a 1cfi11e111e11t of P3. Since a E icing. the definition of Hiefinement tomes this ‘ « lab? for all u E .4. This means that for all h E .4 we have ((1.1)) E R1 —-:- (11.)17) ER3; in otl1e1 \1 ords 3r 1-1111 __ This need not be an equivalence relation. since it need not be transitive. isSince the intersection of reflexive. svnnnetric. and transitive relations also have these properties (see 0 1181). the inteisection of equivalence 1elations is an equivalence relation _ This will never be an equivalence relation 011 a nonempty set. since it is not reflexive. QS C‘Xerr is very similar to Exercise 59. and the reader should look at the solution there for details. AS in Exercise 59. the motions of the bracelet form a dihedral group. in this case consisting of six motions: tations of 00.1200. and 2h )0 and three reflections. each keeping one bead fixed and interchanging the othei mi} The composition of anv tno of these ope1ations is again one of these operations. The 00 rotation plays ‘3 1016 of the identity. which says that the relation is reflexive. Each operation has an inverse (reflections are 222 60. 66. 68. . No. Here is a counterexample. Start with {(1.2) Chapter 8 Relate)... their own inverses. the O0 rotation is its own inverse. and the 1200 and 211)0 rotations are inverses 01081111 other); this proves symmetry. And transitivitv follows from the group table. b) The equivalence classes are the indistinguishable bracelets. If we denote a bracelet by the colors Of 115 beads. then these classes can be described as RRR. WW'W. BBB, RRW. RRB. WWR, XVXVB. BBR, BB“: and RW B. Note that once we specify the colors. then every two bracelets with those colors are equivalent; This would not be the case if there were four or more beads. however. For example, in a 1—bead bracelet with two reds and two whites. the bracelet in which the red beads are adjacent is not equivalent to the one in which they are. not. a) In Exercise 25 of Section 3.2. we shoved that f(.z ) is @(g(.zr)) if and only if f( r) is 0(g (1')) and (/( :11 ‘3 O(f(.r)). ‘To show that R is reflexive. we need to $1 row that f(.1:) is O( f(.1)) 11 hich is clear bv taking C':1_ and I; = 1 in the definition. Synnnetrjv is immediate from the definition. since if fit) is O(g(.r )) )and {/1171 is O(f(:z,')). then g(.r) is O(f(.’zt)_) and f(.r) is O(g(.r)). Finally. transitivity follows innnediatelv £10111th transitive of the "is big‘O of” relation, which was proved in Exercise 17 of Section 13.2. b) This is the class of all functions that asymptotically (i.e.. as 71 «2 oc) grow just as fast as a multiple of f(n) = 722. So. for example. functions such as g(n) 2 5112 +log n. or 51(11) 2 (123 — 17)’(1U()n +11) ”) belong: to this class. but g(n) = 12101 does not (it grows too fast). and 9(a) : nil/log 71 does not (it gross too slowly). Another way to express this class is to say that it is the set of all functions g such that there exist constants positive C71 and C3 such that the ratio f (11) /g(n) always lies between C1 and Cg. . We will count partitions instead. since equivalence relations are in one—to—one correspondence with partitions. Without loss of generality let the set be {1. 2. 2.3. 1}. There is 1 partition in which all the elements are in the same set namelv {{1. '2 13. 1}}. There are 1 partitions in which the sizes of the sets are 1 and :3. 11211111111' {{1}. {2.3. 1}} and hree more like it. There are ‘3 partitions in which the sizes oi the sets are ‘2 and 2 namely, {1. ‘2}. {$3. 1}} and two more lil {e it. There are (11 partitions in which the sizes of the sets 1111.1 and 1. namely {{1. 2}. {13}. {1}} and five more like it. Finallv. there is 1 partition in which all the elements are in separate sets This gives a total of 1:3. To actuall} list the15 relations 11 ould be tedious. (3. 2)} on the set {1.2 .53}. Its transitive (711,151.11?) is it— sclf. The reflexive closure of that) is {(1 1). (1. 2). (‘2. ‘2). (3.2).( (1.3.1. )3} The symmetric closure oi that is {(1-1)311.2) (2 1). ( .2) (2. 3).( (1.3 3)}. The result is not transitive; for example. (1.3) is 1.. Therefore this is not an equix alehce )relation. we end up with the original partition P. We will develop this recurrence relation in the context of partitions of the set {1. 2 ..... n} . Note that 112111 Z 1 ~ since there is only one way to partition the empty set (namely. into the empty collection of subseisi. l‘i'I‘ warm-up. we also note that 11(1) 2 1, since {{1}} is the only partition of {1}; that 11(2) 2 ‘2. sincc 1“ partition {1. ‘2} eithei as {{1. 2}} or as {{1}. {2}}: and that 11(13 )— — 5 .since there are the followine par {{1-2 3}} {{1 21-13-11 {{1 3} {2}} {12. 3} {1}} {111- {2} {3}} N01110112111'1ti011i1-‘2 decid e how man) other elements oftl 1is set will go into the same subset as 11 goes into. Cal i this 111111 and note that i can take anv value from 0 through 11 — 1. Once we have determined j. we can spw paitition bv deciding 011 the subset of j elements from {1'2 ....... n — 1} that 11111 go into the same still-491 as n (and this can be done 111 C (n — 1.j) ways). and then we need to decide how to partition the remaining 11 — 1 — j elements (and this can be done in 11(11 -— j — 1) ways). The given recurrence relation now teller.» ...
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