1 relations: the second time around chapter 7 equivalence classes

1

Relations: The Second Time Around

Chapter 7

Equivalence Classes

2

7.1 Relations Revisited: Properties of Relations

Ex. 7.3 Consider a finite state machine M=(S,I,O,v,w).(a) For s1,s2 in S, define s1Rs2 if v(s1,x)=s2 for some x in I. Relation R establishes the first level of reachability.(b) The second level of reachability. s1Rs2 if v(s1,x1x2)=s2

for some x1x2 in I2. For the general reachability relation we have v(s1,y)=s2 for some y in I*.(c) Given s1,s2 in S, the relation 1-equivalence, which is denoted by s1E1s2, is defined when w(s1,x)=w(s2,x) for all x in I. This idea can be extended to states being k-equivalence, where s1Eks2 if w(s1,y)=w(s2,y) for all y in Ik.If two states are k-equivalent for all k in Z+, then they are calledequivalent.

3


Def. 7.2 A relation R on a set A is called reflexive if for all x in A, (x,x) is in R.

Ex. 7.4 For = {1,2,3,4}, a relation R will be reflexiveif and only if R {(1,1), (2,2), (3,3), (4,4)}. R = {( , )|, , } is reflexive on .

Ex. 7.5 If | |= , | |= There are relations on .How many of these are reflexive?We must include {( The remaining pairscan be either included or excluded from the relation. Therefore,

there are reflexive relations.

2

i

A A Ax y

x y A x y A

A n A A n A

a a a A n n

n

i i

n n

.

, )| }.

2

2

2

2

2

4


Def. 7.3 Relation R on set is call symmetric if ( , ) R( , ) R, for all , .

A x yy x x y A

Ex. 7.6 With A={1,2,3}, we have:(a) R1={(1,2),(2,1),(1,3),(3,1)}, symmetric, but not reflexive;(b) R2={(1,1),(2,2),(3,3),(2,3)}, reflexive, but not symmetric;(c) R3={(1,1),(2,2),(3,3)}, R4={(1,1),(2,2),(3,3),(2,3),(3,2)}, both reflexive and symmetric;(d) R5={(1,1),(2,3),(3,3)}, neither reflexive nor symmetric.

5


To count the symmetric relations on = { write as where and

contains 1

2 subsets of

the form {( Therefore, there are

symmetric relations on . And reflexive and symmetric relations on .

Def. 7.4 For a set , a relation R on is called if forall , , , ( ,

11 2

2

A a a aA A A A A a a i n A

a a i j n i j A n n

a a a a

AA

A A transitivex y z A x y

ni i

i j

i j j in n n

n n

, , , },, {( , | }

{( , )| , , }. ( )

, ), ( , )}.( )

( )

21 2

2

1

2

1

2

1

1

2 2

2

2

2

), ( , ) R ( , ) R.y z x z

6


Ex. 7.8 Define the relation R on the set Z+ by aRb if a exactlydivides b. Then R is transitive, reflexive, but not symmetric (2R6,but not 6R2)

Ex. 7.9 Consider the relation R on the set Z where R when0. Then R is reflexive, symmetric, but not transitive.

(3R0,0R - 7, but not 3R - 7)

Def. 7.5 Consider a relation R on a set , R is called if for all , , ( R and R ) = .Ex. 7.11 the subset relation: R if , reflexive, transitive,antisymmetric, but not symmetric.Ex. 7.12 = {1,2,3}, R = {(1,2), (2,1), (2,3)} is neither symmetricnor antisymmetric. R = {(1,1), (2,2)} is both symmetric and antisymmetric.

a bab

A antisymmetrica b A a b b a a b

A B A B

A

7


To count the antisymmetric relations on = { write as where and

contains 1

2 subsets of

the form {( There are 3 choices for this kind ofsubsets: select one or none of them. Therefore, there are

antisymmetric relations on . And reflexive and antisymmetric relations on .

11 2

2

A a a aA A A A A a a i n A

a a i j n i j A n n

a a a a

AA

ni i

i j

i j j i

n n n n n

, , , },, {( , | }

{( , )| , , }. ( )

, ), ( , )}.

( ) ( )

21 2

2

1

2

1

2

1

1

2 3 32 2

8


Def. 7.6 A relation R os a set A is called a partial order, or apartial ordering relation, if R is reflexive, antisymmetric, andtransitive. (It is called a total order if for any a,b in A, either a Rbor bRa).

Examples of partial order: , , , ,

Ex. 7.15. Define the relation R on the set Z+ by aRb if a exactlydivides b. R is a partial order.

Def. 7.7 An equivalence relation R on a set A is a relation that isreflexive, symmetric, and transitive.

Examples: aRb if a mod n=b mod n

The equality relation {(ai,ai)|ai in A} is both a partial order andan equivalence relation.

9

7.2 Computer Recognition: Zero-One Matrices and Directed Graphs

Def. 7.8 If , , and are sets with R and Rthen the composite relation R R is a relation from to defined by R R and there are exists

with ( , ) R R

1 21 2

1 21 2

A B C A B B CA C

x z x A z Cy B x y y z

,

{( , )| , ,, ( , ) }.

(Note the different ordering with function composition.)

Ex. 7.17 A={1,2,3,4}, B={w,x,y,z}, and C={5,6,7}. ConsiderR1={(1,x),(2,x),(3,y),(3,z)}, a relation from A to B, and R2={(w,5),(x,6)}, a relation from B to C. Then R1 R2={(1,6),(2,6)}is a relation from A to C.

Theorem 7.1 Let A, B, C, and D be sets with RR and R Then R R R( R R R

12 3 1 2 3

1 2 3

A BB C C D

,, . ( )

) .

10


Def. 7.9 Given a set and a relation R on , we define the R recursively by (a) R R; and (b) for Z R

R R

Ex. 7.19 If = {1,2,3,4} and R = {(1,2), (1,3), (2,4), (3,2)}, thenR R and for 4, R

Def. 7.10 - (Use boolean operation, 1 +1 =1)

Ex. 7.20

1

is a 3 4 (0,1) - matrix.

1 + +1

2 3

A A powerof n

An

zero one matrix

n

n

n

,.

{( , ), ( , ), ( , )}, {( , )}, .

1 4 1 2 3 4 1 4

0 0 1

0 1 0 1

1 0 0 0

11


Ex. 7.21 A={1,2,3,4}, B={w,x,y,z}, and C={5,6,7}. R1={(1,x),(2,x),(3,y),(3,z)}, a relation from A to B, and R2={(w,5),(x,6)}, a relation from B to C.

relation matrixM M

w

x

y

zw x y z

M M M

( ) , ( )

( ) ( ) (

R R

5 6 7

R R R

1 2

1 2

1

2

3

4

0 1 0 0

0 1 0 0

0 0 1 1

0 0 0 0

1 0 0

0 1 0

0 0 0

0 0 0

0 1 0 0

0 1 0 0

0 0 1 1

0 0 0 0

1 0 0

0 1 0

0 0 0

0 0 0

0 1 0

0 1 0

0 0 0

0 0 0

1 2)R

12


Let A be a set with |A|=n and R a relation on A. If M(R) is therelation matrix for R, then(a) M(R)=0 (the matrix of all 0's) if and only if R is empty(b) M(R)=1 (the matrix of all 1's) if and only if R=A A(c) M(Rm)=[M(R)}m, for m in Z+.

Def. 7.11 Let = ( , = ( be two (0,1) -

matrices. We say that precedes, or is less than, , and wewrite , if for all , .

Ex. 7.23 With =1

0

0

0

1

1 and =

1

0

0

1

1

1 we have .

In fact, there are eight (0,1) - matrices for which .

E e F f m n

E FE F e f i m j n

E F E F

G E G

ij m n ij m n

ij ij

) )

,

,

1 1

13


Def. 7.12 For Z is the (0,1) - matrix

where if =

if Def 7.13 Let = ( be a (0,1) - matrix. The transpose of

, written , is the matrix ( where

for all 1 , .

Ex. 7.24 =

+

ij

tr * *

tr

n I n n

i j

i jA a

A A a a a

i m j n

A A

n ij n n

ij m n

ji n m ji ij

, ( )

,

,)

) ,

, .

1

0

10 1

0 0

1 1

0

1

0

0

1

1

14


Theorem 7.2 Given a set with | |= , and a relation R on ,let denote the relation matrix for R. Then(a) R is reflexive if and only if

(b) R is symmetric if and only if =

(c) R is transitive if and only if =

(d) R is antisymmetric if and only if

tr

2

tr

A A n AM

I M

M M

M M M M

M M I

n

n

.

.

.

.( , )0 0 0 1 1 0 0 1 1 1

15

Proof of (c) Let If ( , ), ( , ) R,

1 1 1 Hence ( , ) R.

Conversely, if R is transitive and is the relation matrix for R,

let be the entry in row and column of with

For to be 1 in there must exist at least one where in . This happens only if ( , ), ( , ) R. With

R transitive, it then follows that ( , ) R.

2

2

2

M M x y y z

M x z

M

s x z M s

s M y Am m M x y y z

x z

xy yz xz

xz xz

xz

xy yz

.

.

, .

,

1

1

So and2

m

M Mxz

1

.


16


Def. 7.14 Directed Graphs G=(V,E) (Digraph)

Ex. 7.25

1 2

4 3 5isolated vertex (node)

a loop V={1,2,3,4,5}E={(1,1),(1,2),(1,4),(3,2)}

1 is adjacent to 2.2 is adjacent from 1.

Undirected graphs: no direction in edges

17


Ex. 7.26

(s1) b:=3;(s2) c:=b+2;(s3) a:=1;(s4) d:=a*b+5;(s5) e:=d-1;(s6) f:=7;(s7) e:=c+d;(s8) g:=b*f;

Construct a directed graph G=(V,E), whereV={s1, s2, s3, s4, s5, s6, s7, s8} and (si, sj) in Eif si must be executed before sj.

s3 s1 s6

s4 s2 s8

s5 s7

precedence graph

precedence constraint scheduling3 processors: 3 time units2 processors: 4 time units

In general, n tasks,m processors: NP-completem=2: polynomialm=3: open problem

18


Ex. 7.27 relations and digraphs

A={1,2,3,4}, R={(1,1),(1,2),(2,3),(3,2),(3,3),(3,4),(4,2)}

1

2

34

directed graphrepresentation

1

2

34

associatedundirected graph

a connected graph: a pathexists between any two vertices

cycle: a closed path (thestarting and ending verticesare the same)

path: no repeated vertex

Def. 7.15 Strongly connected digrapha directed path exists between any two vertices

The above graph is not strongly connected. (no directed path from 3 to 1)

19


Ex. 7.28 Components

1 2

3 4

two components

1 2

3 4

one component

Ex. 7.29 Complete Graphs: Kn

K1 K2 K3 K4

K5

(n vertices with n(n-1)/2 edges)

20


directed graphs relations

adjacency matrices relation matrices

Ex. 7.30 If R is a relation on finite set A, then R is reflexive if and only if its directed graph contains a loop at each vertex.

Ex. 7.31 A relation R on a finite set A is symmetric if and onlyif its directed graph contains only loops and undirected edges.

Ex. 7.31 A relation R on a finite set A is transitive if and only if inits directed graph if there is a path from x to y, (x,y) is an edge.A relation R on a finite set A is antisymmetric if and only if inits directed graph there are no undirected edges aside from loops.

21


Ex. 7.33 A relation on a finite set A is an equivalence relationif and only if its associated (undirected) graph is one complete graph augmented by loops at every vertex or consists of the disjoint union of complete graphs augmented by loops at every vertex.

reflexive: loop on each vertexsymmetric: undirected edgetransitive: disjoint union of complete graphs

22

7.3 Partial Orders: Hasse Diagrams

natural counting: Nx+5=2 : Z2x+3=4 : Qx2-2=0 : Rx2+1=0 : C

Something was lost when we wentfrom R to C. We have lost the abilityto "order" the elements in C.

Let A be a set with R a relation on A. The pair (A,R) is calleda partially ordered set, or poset, if relation R is a partial order.

Ex. 7.34 Let A be the courses offered at a college. Define R on Aby xRy if x,y are the same course (reflexive) or if x is a prerequisitefor y. Then R makes A into a poset.

23


Ex. 7.36 PERT (Performance Evaluation and Review Technique)

J1 J6

J3

J2J5

J4

J7

Find each job's earliest start time and latest start time.Those jobs which earliness equals to lateness are critical.All critical jobs form a critical path.

A poset

24


not partial order

1 21

2

3

not antisymmetric not transitive or not antisymmetric

Ex. 7.37 Hasse diagram

1

2 3

4

1

2 3

4

a partialorder

corresponding Hasse diagram

Read bottom up.reflexivity andtransitive linksare not shown.

25


Ex. 7.38

{1,2,3}

{1,2} {1,3} {2,3}

{1} {2} {3}

subset relation exactly division relation

1

2

4

8

2 3 5 72 3 5 7 11

6

12385

35

equalityrelation

26


Def. 7.16 If (A,R) is a poset, we say that A is totally ordered iffor all x,y in A either xRy or yRx. In this case R is called a totalorder.

For example, <,> are total order for N,Z,Q,R. But partiallyordered in C.

But can we list the elements for a partially ordered set insome way?

sorting for a totally ordered set

27


topological sorting for a partially ordered set

A

C

B E

G F D

Hasse diagram fora set of tasks

How to execute the tasks one at a timesuch that the partial order is not violated?

For example, BEACGFD, EBACFGD, ...

28


topological sorting sequence (linear extension)

a b

c d

a^bc^d: 2 jumps

a^bd^c: 2 jumps

b^ac^d: 2 jumps

b^a^d^c: 3 jumps

bd^ ac: 1 jumps

Find a linear extensionwith minimum jumps.

(an NP-complete problem)

29


Def. 7.17 If ( , R) is a poset, then an element of is calleda of if for all , ( , ) R.( R = ) An element is called a minimal element of

if whenever and , then ( , ) R ( R = ).

Ex 7.42 With R the "less than or equal to" relation on the set Z, (Z, ) is a poset with neither a maximal nor a minimal element.The poset (N, ), however, has a minimal element 0 but nomaximal element.

A x Aimal element A a A a x x a

x a x a y AA b A b y b y b y b y

max

30


{1,2} {1,3} {2,3}

{1} {2} {3}

1 min

2

4

8 max

2 3 5 72 3 5 7 11

6

12385

35

Ex.7.43

{1,2,3}max

min

max and min

min

max

unique max and min

31


Theorem 7.3 If ( , R) is a poset and is finite, then has botha maximal and a minimal element.Proof: Let . If there is no element where and

R , then is maximal. Otherwise there is an element with and R If no element , satisfies

R , then is maximal. Otherwise we can find so that while R and R Continuing in this

manner, since is finite, we get to an element with( R for any where

11 1

1 2 22 2

3 1 2 2 3

A A A

a A a A a aa a a a A

a a a a a A a aa a a a Aa a a a a a a a

A a Aa a a A a a

nn

12

2 13

3 2 1

. ,

, .

, ) n na, so is maximal.The proof for minimal element is similar.

In topological sorting, each time we find a maximal element,or each time we find a minimal element.

32


Def 7.18 If ( , R) is a poset, then an element is calleda element if R for all . Element is called a

element if R for all .

Ex. 7.44 Let = {1,2,3} and R be the subset relation.(a) With = ( ), the poset ( , ) has as a least elementand as a greatest element.(b) For = the collection of nonempty subsets of , the poset ( , ) has as a greatest element. There is no leastelement.(c) For = the collection of proper subsets of , the poset ( , ) has as a least element. There is no greatestelement

A x Aleast x a a A y A

greatest a y a A

UA P U A

UB U

B U

C UC

33


Theorem 7.4 If the poset (A,R) has a greatest (least) element,then that element is unique.Proof: Suppose that x,y in A and that both are greatest elements.Then (x,y) and (y,x) are both in R. As R is antisymmetric, it follows that x=y. The proof for the least element is similar.

Def. 7.19 Let ( , R) be a poset with . An element iscalled a ( ) for if R ( R ) for all .An element ' is called a , , (

, ) if it is a lower bound (upper bound) of andfor all other lower bounds (upper bounds) " of we have

"R ' ( 'R ").

A B A x Alower upper bound B x b b x b B

x A greatest lower bound leastupper bound B

x Bx x x x

glblub

34


Ex. 7.46 Let U={1,2,3,4}, with A=P(U), let R be the subsetrelation on A. If B={{1},{2},{1,2}}, then {1,2}, {1,2,3},{1,2,4}, and {1,2,3,4} are all upper bounds for B(in A), whereas {1,2} is a least upper bound (in B). Meanwhile, a greatest lower bound for B is the empty set, which is not in B.

Ex. 7.47 Let R be the "less than or equal to" relation for the poset(A,R). (a) If A=R and B=[0,1] or [1,0) or (0,1] or (0,1), then B has glb 0 and lub 1. (b) If A=R, B={q in Q|q2<2}. Then B has

2 as a lub and - 2 as glb, and neither of these is in B.

(c) A=Q, with B as in part (b). Here B has no lub or glb.

35


Theorem 7.5 If ( ,R) is a poset and , then has at mostone lub (glb).

Def 7.20 The poset ( ,R) is called a if for any ,the element lub{ , } and glb{ , } both exist in .

Ex. 7.48 For = N and , N, define R by . Then{ , } = { , }, { , } = { , }, and (N, )

is a lattice.

Ex. 7.49 ( ( ), ) is a lattice with lub{ , } = and{ , } = for , .

A B A B

A lattice x y Ax y x y A

A x y x y x yx y x y x y x y

P U S T S TS T S T S T U

lub max glb min

glb

36

7.4 Equivalence Relations and Partitions

Def. 7.21 Given a set and an index set , let foreach . Then { is a partition of if(a) = and (b) for all , where .

Each subset is called a or of the partition.

Ex. 7.51 = {1,2,3,4,5,6,7,8,9,10}, then each of the followingdetermines a partition of :(a) b) c)

11

A I A Ai I A A

A A A A i j I i j

A cell block

AA

A AA A AA i i

ii i I

ii I

i j

i

i

},

{ , , , , }, { , , , , }( { , , }, { , , , }, { , , }( { ,

1 2 3 4 5 6 7 8 9 101 2 3 4 6 7 9 5 8 10

5

22 3

}, .

[ , ).}

1 5

1

i

A i A i iA

ii i

Ex. 7.52 = R, and for each Z, let Then { is a partition of R.Z

37


Def. 7.22 Let R be an equivalence relation on a set . Forany , the equivalence class of , denoted is definedby R

Ex. 7.53 R if 4|( - ) for , Z. For this equivalence we find that 0 Z 1 Z 2 Z 3

Ax A x xx y A y x

x y x y x yk k

k kk k

,{ | }.

{ , , , , , , } { | }{ , , , , , , } { | }{ , , , , , , } { | }{ , , , , , , } {

8 4 0 4 8 47 31 5 9 4 16 2 2 6 10 4 25 1 3 7 11

4 3

1 2 3k k Z | }.

, , , }And { 0 provides a partition of Z.

38


Ex. 7.54 For , Z, R if Then R is an equivelencerelation. What can we say about the corresponding partition of

Z? In general, forn any Z Therefore,

Z = [ ]

Theorem 7.6 If R is an equivalence relation on a set , and , , then (a) [ ]; (b) R if and only if and

(c) [ ] = [ ] or [ ] [ ] = .

2

+

=

a b a b a b

n n n n n

n

Ax y A x x x y x y

x y x y

n

2

0

.

, { , }.

.

;

39


Ex. 7.58 If an equivalence relation R on A={1,2,3,4,5,6,7} inducesthe partition ? What is R?A { , } { } { , , } { }1 2 3 4 5 7 6

R = ({1,2} {1,2}) ({3} {3}) {(4,5,7} {4,5,7})

({6} {6}), |R|= 22

1 3 1 152 2 2 .

Theorem 7.7 If A is a set, then (a) any equivalence relation R on Ainduces a partition of A, and (b) any partition of A gives rise to anequivalence relation on A.

Theorem 7.8 For any set A, there is a one-to-one correspondencebetween the set of equivalence relations on A and the set ofpartitions of A.

40


Ex. 7.59 (a) If = {1,2,3,4,5,6}, how many relations on areequivalence relations? one - to - one correspondence between equivalence relations and partitions

( , )

(b) How many of the equivalence relations satisfy 1,2 4

1,2,4 are in the same partition. ( , )=

A A

S i

S i

i

i

6 203

4 15

1

6

1

4.

.

41

7.5 Finite State Machines: The Minimization Process

Given s1,s2 in S, the relation 1-equivalence, which is denoted by s1E1s2, is defined when w(s1,x)=w(s2,x) for all x in I. This idea can be extended to states being k-equivalence, where s1Eks2 if w(s1,y)=w(s2,y) for all y in Ik.

If two states are k-equivalent for all k in Z+, then they are calledequivalent, denoted by s1Es2. Hence our objective is to determinethe partition of S induced by the equivalence relation E and selectone state for each equivalence class.

42


observations:(1) If two states in a machine are not 2-equivalent, could they possibly be 3-equivalent?

No. If and are not 2 - equivalent, then there exists in

such that ( , ) ( So for any , ( = (

1 22

s s xy

I w s xy w s xy z Iw s xyz w s xy w v s xy z w s xy w v s xy z

w s xyz

1 2

1 1 1 2 2

2

, )., ) ( , ) ( ( , ), ) ( , ) ( ( , ), )

, ).

In general, to find states that are (k+1)-equivalent, we look atstates that are k-equivalent.

43


observations:(b) Suppose E We wish to determine whether EThat is does ( ( for all strings

Because E E w(sConsequently, ( ( if

( ( ( ( That is, if ( E

In general, for s

2 1 3

2 1 1

2

s s s sw s x x x w s x x x

x x x I s s s s x w s xw s x x x w s x x x

w v s x x x w v s x x xv s x v s x

1 2 2

1 1 2 3 2 1 2 3

1 2 33

1 2 1 2 1 2 1

1 1 2 3 2 1 2 3

1 1 2 3 2 1 2 3

1 1 2 1

. ?, ) , )

? , , ) ( , )., ) , )

, ), ) , ), )., ) ( , ).

1 2 1 +

1

, s we have E if and only if(i) E and (ii) ( E for all .

S s ss s v s x v s x x I

k

k k

1 2

2 1 2, ) ( , )

44


s1 s4 s3 0 1s2 s5 s2 1 0s3 s2 s4 0 0s4 s5 s3 0 0s5 s2 s5 1 0s6 s1 s6 1 0

Ex. 7.60

0 1 0 1v w

step 1: determine 1-equivalent states by examining outputs

P1: {s1},{s2, s5, s6},{ s3, s4}

input 0

s5 s2 s1 s2 s5

P2: {s1},{s2, s5},{ s6},{ s3, s4}

input 1

s2 s5 s4 s3

P3=P2, the process is complete with 4 states.

A

C

B

D

45


Could it be that P3=P2, but P3=P4?

Def. 7.23 If are any partitions of a set , then is called a refinement of and we write if everycell of is contained in a cell of When and

we write This occurs when at least one cellin is properly contained in a cell in

In the minimization process, because ( + 1) -equivalence implies - equivalence. So each successivepartition refines the preceding partition.

2 21

2 1 22 2

2 1

+

P P A PP P P

P P P PP P P P

P P

P P kk

k k

12 1

11 1

1

,, ,

..

.

46


Theorem 7.9 In applying the minimization process, if 1and and are partitions with then for any + 1.Proof: If not, let ( + 1) be the smallest subscript such that

Then so there exist with Ebut E But E E for all .And with we then find that E forall . So

+ + +

+ + 1+ 1

kP P P P P P

r kr k

P P P P s s S s ss s s s v s x v s x x I

P P v s x v s xx I

k k k k r r

r r r r rr r r

r r r

1 1 1

1 1 1 2 21 1 2 2 1 1 2

1 1 2

,

. , ,. ( , ) ( , )

, ( , ) ( , )s s P Pr r r1 + +E Consequently, 1 2 1. .

If there must be a smallest integer 0 such that

E but E That is, there is an such that( (but they will agree on the first outputs) is called a for and

1

1 1+

1 2

s s k

s s s s x Iw s x w s x kx distinguishing string s s

k kk

2

2 1 21

1 2

,

., ) ( , ).

.

47


How to find the minimal length distinguishing string?

s1 s4 s3 0 1s2 s5 s2 1 0s3 s2 s4 0 0s4 s5 s3 0 0s5 s2 s5 1 0s6 s1 s6 1 0

0 1 0 1v w

Ex. 7.61P1: {s1},{s2, s5, s6},{ s3, s4}

input 0

s5 s2 s1 s2 s5

P2: {s1},{s2, s5},{ s6},{ s3, s4}

input 1

s2 s5 s4 s3

A distinguishing string for s2 and s6 is 00 (or 01).

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s1 s4 s2 0 1s2 s5 s2 0 0s3 s4 s2 0 1s4 s3 s5 0 1s5 s2 s3 0 0

0 1 0 1v w P1: {s1 ,s3, s4},{s2, s5}

input 1s2 s2 s5 s2 s3

input 1s2 s2 s5

Ex. 7.62

P2: {s1 ,s3, s4},{s2},{ s5}

P3: {s1 ,s3},{ s4},{s2},{ s5}

distinguishing string for s1 and s4: 111

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Exercise: P318:10 P330:20 P340:26 P346:10 P356:14

1 relations: the second time around chapter 7 equivalence classes

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