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**Chapter 6**Relations**Contents**1. Relations and their properties 2. n-ary relations and their applications 3. Representing relations 4. Closure of relations 5. Equivalence relations 6. Partial orderings**6.1 Relations and their properties**Def. 1: A, B: two sets. A (binary) relation from A to B is a subset of AxB. (i.e., any subset of {(x,y) | x Î A and y Î B} ) Notes: • If R Í AxB, we usually use a R b (or not aRb) to denote that (a,b) Î R (or (a,b) Ï R). • Binary relations can be generalized to n-ary ( n > 0) relations Ex1: A = the set of all students; B = the set of all courses let R = {(a,b) | student a is enrolled in course b} Í AxB is a binary relation from A to B. Ex2: A = the set of all cities; B = the set of all nations. Let R = {(x,y) | city x is located in nation y}. ==> (New York, USA) Î R and Tokyo R Japan, ...**Relations on a set**Ex3: A = {1,2,3,4}; B={a,b} ==> {(0,a),(0,b),(1,a),(2,b)} is a relation from A to B. ==> Graphical representation = ? ==> table (or matrix) representation = ? Def. 2: A relation R on (or over) a set A is a subset of AxA (= A2). Ex4: A={1,2,3,4}; Let R = {(a,b) Î A2 | a divides b} ==> R= ? ; graphical and table representation = ? Ex 5: Let R1,R2,...,R6 be relations on integers Z defined as R1= {(a,b) | a £ b); R2 = {(a,b) | a > b }; R3 = {(a,b) | a = b or a = -b}; R4={(a,b) | a = b}; R5={(a,b) | a = b+1}; R6={(A,B) | A + B < 4} ==> Which relations contain the pairs: (1,1), (1,2),(2,1),(1,-1),(2,2)?**Examples of relations**Ex 6: |A| = n; |B| = m ==> What is the number of relations from A to B ? (i.e., let Rel(A,B)= {R | R is a relation from A to B}, ==> |rel(A,B)| = ? ) Sol: 1. |AxB| = ____. 2. Since Rel(A,B) = 2AxB. Hence |rel(A,B)| = ___ Ex6': |Rel(A,A)| = #relations on A = ____**Properties of (binary) relations on a set**Def 3,4,5: R: a binary relation on a set A; • R is reflexive if aRa for every a Î A. • R is irrelexive if (a,a) Ï R for each a Î A. • R is symmetric if for every a,b ÎA, (a,b) Î R => (b,a) ÎR. • R is asymmetric if for every a,b Î A, (a,b) Î R => (b,a) Ï R. • R is antisymmetric if for every a,b Î A, if (a,b) Î R and (b,a) Î R, then a = b. • R is transitive if for every a,b,c Î A, if aRb and bRc then aRc. Def: • R is a partial ordering(p.o.) iff it is ref. sym. and transitive. • R is a strict partial ordering(s.p.o) if it is irref. and trans. • R is a preordering iff it is ref. and transitive. • R is an equivalence relation iff it is ref. sym. and transitive.**Example:**• Let B = {1,2,3,4}; • R1= {(1,1),(1,2),(2,1),(2,2),(3,4),(4,1),(4,4)} • R2={(1,1),(1,2),(2,1)} • R3={(1,1),(1,2),(1,4),(2,1),(2,2),(3,3),(4,1),(4,4)} • R4={(2,1),(3,1),(3,2),(4,1),(4,2),(4,3)} • R5={((1,1),(1,2),(1,3),(1,4),(2,2),(2,3),(2,4),(3,3),(3,4),(4,4)} • R6={(3,4)} Ex 7: Which of these relations are ref. ? Ex 7': Which of these relations are irref. ? Note: 1. using graphical representations may help determining properties of relations. 2. There are relations which is neither ref. nor irref. Ex 8: Is the "divides" relation on integers reflexive ? Ex 9: which of the relations on Ex5 are ref. ?**Examples**Ex10. Which of these relations are symmetric, antisymmetric ? Ex10': Which are asymmetric ? • What is the graphical characteristic of sym., asym. and antisym. relations ? • What is the relation among symmetric, antisymmetric and asymmetric properties ? Ex 12: Is the "divides" relation on integers sym. ? Is it antisym. or asym. ? Ex13: Which of these relations in Ex 7 are transitive. ? Ex13': Which of the relations inEx 7 are preordering, p.o., s.p.o. or equivalence? Ex15: is the "divides" relation transitive ?**Counting the number of relations**• |A| = n. 1. # relations on A = ? (2^(n^2)) 2. # reflexive relations on A = ? (2 ^ (n^2 - n)) 3. # irreflexive relation on A = ? (= sol of 2, why?) 4. #irref and symmetric relation on A = ? (2^(n(n-1)/2) ) 5. #symmetric relation on A = ? (2^(n(n+1)/2)) 6. #asymmetric relations on A = ? (3^(n(n-1)/2)) 7. #antisymmetric relations on A = ? (3^(n(n-1)/2) x 2^n) 8. #transitive relations on A = ? (hard!) 9. #equivalence relations on A = ? (hard!) 10. #strict partial ordering on A = ? (hard!) 11. #preordering on A = ? (hard!) 12. #partial ordering on A = ? (hard!)**Combining relations**• Relations are sets: • ==> All set operations can be applied to relations • Ex: R1; R2: relations on A; • ==> R1 U R2; R1 ⋂ R2; R1 \ R2 and ~R are defined as usual. • [Binary ] relation compositions: R Í AxB ; S Í BxC; ==> R·S (the composition of R and S) Í AXC is the set: {(x,z) | $y Î B s.t. xRy ∧ ySz} Note: The textbook use S·R (less popular) instead of R·S to denote the composition.**1**0 1 2 2 1 3 2 3 4 R S C A B Examples of relation compositions Ex19: A={1,2,3}; B={1,2,3,4}; C={0,1,2} R ÍAXB = {(1,1),(1,4),(2,3),(3,1),(3,4)} S Í BXC = {(1,0),(2,0),(3,1),(3,2),(4,1)} => R·S= ? Fact: Relation composition is associative (i.e., R·(S·T) = (R·S)·T ) Hence we can write RST w/t parentheses.**The power of relations**Def.7 : R: a relation on A (i.e., a subset of AxA); n : nonnegative integers. The relation Rn is defined inductively as follows: • 1. R0 = IA =def {(x,x) | x ∈ A}. • 2. Rk+1 = RRk = {(x,z) | $y Î R s.t. (x,y) ÎR /\ (y,z) ÎRk }. • Hence R2 = RR; R3=RR2 = R(RR) = (RR)R = R2 R = RRR . • in general we have Rn = RRRR...R. (n R’s) Ex: show that If a, b 0 then RaRb = Ra+b. Pf : by induction on a. • basis: a = 0: => RaRb = IARb = • {(x,z) | $y (x,y) Î IA /\ (y,z) Î Rb } = • {(x,x) | (x,x) Î IA /\ (x,z) Î Rb} = {(x,z) | (x,z) Î Rb } = Rb. • Ind. step : assume RaRb= Ra+b => • Ra+1Rb = (RRa)Rb = R(RaRb) = RRa+b = Ra+1+b. Corollary: RRk = Rk R.**Example**Ex20: A= {1,2,3,4}; R={(1,1),(2,1),(3,2),(4,3)} is relation on A. => R0, R1, R2, R3, R4,R5, ...= ____? (use graphical representation) Lemma: If Rn = Rn+1 then Rk = Rk-1 for all k > n. Ex: If R5 = R6 => R10 = R6R4 = R5R4 = R9 Pf: left as an exercise. Corollary: If Rn = Rn+1 ==> Rk = Rn for all k > n. Pf: Rk = Rk-1 = R(k-1)-1 = Rk-2 = ...Rn+1 = Rn. Theorem 1: Relation R on a set A is transitive iff RnÍ R for all n > 0. Pf: (=>) : trivial (by induction on n). Basis: n = 1 (ok). Ind. step: for any (x,z) Î Rn+1 => $z s.t. (x,y) Î Rn /\ (y,z) Î R. By ind. hyp. both (x,y) and (y,z) Î R. By tran. of R, (x,z) Î R. (<=) : for any x,y,z, (x,y) Î R /\ (y,z) Î R => (x,z) Î R2 => (x,y) Î R. QED Corollary: R is transitive iff R2 Í R. (why ?)**6.2 n-ary relations**Def. 1: A1,A2,...,An : n sets. 1. An n-ary relation R on A1,A2,...,An is any subset of A1xA2x...xAn = {(x1,...,xn) | x1Î A1,...,xnÎ An }. 2. A1,..., An are the domains of R 3. n: the degree (or arity) of R. 4. 1-ary (0-ary, 2-ary, 3-ary) relations are called unary (nullary, binary, ternary) relations. Ex: If R Í Z3 (= ZxZxZ) =def {(x,y,z) | x < y < z, x,y,z ∈Z}. Then (1,2,3) Î R but (2,4,3) Ï R.**Relations and databases**• Database: • In a database, lots of information about the real world are extracted and stored. • Since the database is a "representation" of the world, if the real world changes, the database should change over time accordingly • Issues in database theory (data model): • How to represent various information (or data) in a database? • (i.e., can we have a unifying method by which we can always (or in most cases) easily organize or arrange all data to be stored in the database ). • Any such method is called a data model.**The relational data model**• A database DB (at a time instant, say, time t) consists of a pair <Dom,Rel> where • Dom = {A1,...,An} is a set of sets, each Ai is called a domain of the database. • Rel = {R1,...,Rm} where each Ri Aj1xAj2x...xAjki is a ki-ary relations on domain Aj1,Aj2,...,Ajki. • Each Ki-tuple (a1,a2,...,aki) Ri is called a record (or row) of Ri. • Each Ajs ( 1 <= s < = ki) is called a field (or a column) of the relation Ri (and the record). • Each relation Ri can be represented by a table (w/t duplicate rows) with |Ri| rows and Ki columns.**Example relations**Ex: Student records S in a school may be represented by 4-tuples of the form: (student-name, register-id, major, GPA ) Let A = the set of all student [names] B = the set of all student identifier numbers. C = the set of all departments [names] D = the set of all possible GPA numbers ([0,5]). => DB = <{A,B,C,D}, {S}> where S must be a subset of AxBxCxD. A possible value of S may be represented by the table: NAME ID MAJOR GPA ------------------------------------------------------- Allen 23 CS 3 Bob 15 math 4.1 Chen 34 CS 4.0 ...**Operations on n-ary relations**• A domain Bj of an n-ary relation R (in database) is called a primary key iff it can be used to uniquely identify records in R at all time. • (in other words, If R is an n-ary relation on domain B1,B2,...,Bn, then for any time t, R(t) must be a subset of B1xB2...xBn. • Bj is a primary key iff for any time t and for any (a1,a2,...,an) and (b1,..,bn) Î R(t), if aj = bj then (a1,...an) = (b1,...,bn). • Note: R(t) changes over time. • Operations on n-ary relations • Selection (or restriction) --- find subset of R with a certain property. • Projections. • Joins --- generalization of binary composition • and other usual set operations (U, /, intersection,...)**relation operations**• Restriction (selection): P(x1,...,xn): a statement (or property) on n-tuples (x1,...,xn). R: a relation on A1xA2...,xAn. ==> R|P ( find tuples of R with property P) is the set {(a1,...,an) | (a1,...,an) Î R and P(a1,...,an) is true }. Ex: Let P(x1,x2,x3) = " x1 < x2 < x3 or x3 = 2 " and R = { (1,2,3), (1,3,2), (2,1,3), ==> R|p = {(1,2,3), (1,3,2), (3,1,2) }. (2,3,1), (3,1,2), (3,2,1) }**Projections**R : a relation on A1,A2,...,An w= j1j2...jm is a sequence with 0< js< n+1 for all s = 1,..,m. ==> Pw(R) : the projection of R on Aj1Aj2...Ajm is a m-ary relation on Aj1Aj2...Ajm with value: {(xj1,...,xjm) | (x1,...,xn) Î R }. Ex: R= {(1,2,3), (1,3,3), (2,3,1), (2,2,2)} => P1(R) = {1,1,2,2} = {1,2} P2(R) = {2,3} P31(R) = {(3,1),(3,1),(1,2),(2,2)} = {(3,1),(1,2),(2,2)} P313(R) = {(3,1,1), (1,2,1),(2,2,2)}**Joins (combining two relations into one)**• R : A1xA2x...xAm-pxAm-p+1x...xAm; S : B1 xB2x..Bpx Bp+1x ..xBm; p ≤ min(m,n). ==> Jp(R,S) = {(a1,...,am-p, c1,c2,...,cp, bp+1,..., bn) | $ (a1,...,am-p, c1,c2,...,cp ) Î R /\ (c1,c2,...,cp, bp+1,..., bn) Î S } Ex: R={(1,2,3), (2,2,3), (3,3,3) (6,1,2)}; S={(2,3,4), (3,3,3), (2,2,2)} ==> J2(R,S) = ? J1(R,S) = ? J3(S,R) = ?**A**B A->B 1 1 a a 2 2 b b 3 3 6.3 Relation representation • Three possible ways to represent a finite relation • a list (or array) of ordered pairs. • (0-1) matrix • digraph (directed graph) Ex: A={1,2,3}, B={a,b}, R = {(2,a),(3,a),(3,b)}. 1. list representation: ((2,1),(3,1),(3,2)) possibly ordered by fields. 2. Matrix (table): 3. Digraph: A\B a b 1 0 0 or 2 1 0 3 1 1**Formalization of relation representation**• A={a1,a2,..,am}; B = {b1,...,bn}; R: a binary relation from A to B. MR (the matrix representation of R) = (mij) mxn where mij = 1 if (ai,bj) ∈ R and = 0 if (ai,bj) ∉ R. Ex: A={a1,a2,a3}; B = {b1,b2,...,b5} If R1={(a1,b1),(a1,b2),(a3,b2),(a3,b5)} ==> MR1 = ? 01000 If MR2= 10110 ==> R2 = ? 10101 Note: 1. If R ∈ AxA with |A| = n, ==> MR is a nxn 0-1 matrix. 2. R is ref. (resp. irref. ) iff mii = 1 (0) for every 0<i < n+1. 3. R is symmetric (asym) iff for all i, j mij = mji (mij ≠ mji) 4. R is tansitive iff for all i,j,k if mij=mjk=1 => mik = 1.**0-1 matrix operation**• Boolean operation on {0,1} • x ∨ y = max(x,y) • x ∧ y = min(x,y) • ~x = 1-x • Boolean operation on 0-1 matrices: • (M1 \/ M2) ij = M1ij \/ M2ij for all i,j. --- pairwise or • (M1 /\ M2) ij = M1ij /\ M2ij for all i,j. --- pairwise and • (~M1) ij = 1- M1ij for all i,j. -- complement. • (M1x M2)ij = \/k=1,n M1ik /\ M2kj. --- product • M0 = I; Mn+1 = Mn x M. --- power of 0-1 matrix.**Matrix representation for relation operations**• R1, R2: two relations from A to B. 1. MR1 UR2 = MR1 ∨ MR2. 2. MR1ÇR2 = MR1 ∧ MR2 3. M~R1 = ~MR1. • R: AXB; S: BxC with A = {a1,...,am}, B = {b1,...,bn} and C= {c1,..,cp}. ==> RS = {(ai,ck) | $bj ∈ B s.t. (ai,bj) R and (bj,ck) S } Hence (MRS)ik = 1 iff there is 1 j n s.t. (aj,bj) R and (bj,ck) S iff there is 1 j n s.t. (MR)ij = 1 /\ (MS)jk = 1 iff \/j=1,n (MR)ij /\ (MS)jk = 1 • Hence MRS = MR x MS. ==> MRk = (MR)k for any k 0.**Examples:**Ex3: MR = [110; 111; 011] ==> Is R ref, symmetric or antisymmetric ? Ex4: R1,R2: two relation on A. MR1= [101;100;010]; MR2 = [101;011;100]. ==> MR1UR2 = ? MR1Ç R2 = ? M~R1 = ? Ex5: MR = [101;110;000]; MS = [010; 001;101] => MRS = ? Ex6: MR = [010;011;100] ==> MR2 = (MR)2 = ?**b**a d c Representing relations using digraphs • Def 1. A diagraph (or directed graph) is a pair (V,E) where • V is a set, each member is called a node or a vertex. • E is a set of ordered pairs of nodes. (i.e. E Í V2), each member of which is called an edge. • If (a,b) E is an edge, we say a is the initial (or starting) vertex and b the terminal (or ending) vertex of the edge. • If (a,a) is an edge, we say it is a loop. Ex7: In Fig 3, we have G=(V,E) where V= {a,b,c,d} and E={ (a,b),(a,d),(b,b)(b,d), (c,a),(c,b),(d,b)}. Note: E in fact is a relation on V. Fig 3: (p 377)**Representing relations by digraphs (cont'd)**• A = {a1,...,am}, B = {b1,...,bn} R: a binary relation from A to B ==> we can use the digraph G=(V,E) to represent R, where • V= A U B. • E = R. Ex: A = {1,2,3} ; B = {2, 3,4} R = {(x,y) | x > y} ⊆ AXB. ==> R = ? The digraph GR for R = ({1,2,3,4}, R). ==> What is its graphical representation ?**6.4 closure of relations**• R: a relation on A • P(-): a property (or statement or predicate) about relations. eg: 1. P(R) = " R is reflexive" or 2. P(R) = " R is symmetric and transitive" or 3. P(R) = " |R| is not finite" etc. Def: P: a property about relations on A; The operation CLp: 2AxA -> 2AxA is defined to be the function s.t. for each each relation R on A, CLP(R) is the smallest of all relations on A which includes R and satisfies the property P. (i.e., CLp(R) is the relation on A s.t. • 1. R Í CLP(R) --- including R • 2. P(CLP(R)) is true --- satisfying P • 3. For any relation S on A, if it satisfies (1) and (2), • then CLP(R) Í S. --- smallest • CLp(R) is called the P closure of R.**Properties about closure**Ex: A = {1,2,3}; R={(1,1),(1,2),(2,1),(3,2)} Let P(S) means " S is reflexive" ==> CLp(R) (reflexive closure of R) = ? (exist ? unique ?) 1. Let TT = { T | T ÍA2 /\ T is ref. and R Í T}. Note: by def. every T ∈ TT satisfies item (1) and(2) of the def. of P-closure. ==> TT = ? 2. Which is the smallest in TT ? (= R U IA !) • Theorem: CLP(R), if existing, is unique. pf: Let S and T be any two P-closures of R. ==> By (3) S ÍT and T Í S, hence S = T. 3. Problem: Does CLP(R) always exist ? (no!) Does the irreflexive closure of R exist in the example ?**closure properties**Def: P: a property about relations P is said to be increasing if for each relation R, there exists a relation S satisfying P and including R. P is said to be closed under intersection(or conjunctive) if for any set TT of relations, if all R Î TT satisfies P, then so is their intersection Ç TT. Ex: The following properties are increasing and closed under intersection: • 1. reflexivity; 2. symmetry; 3. transitivity. • Since A2 (the largest relation on A) includes R • and satisfies 1,2,3. • Theorem: If P is increasing and closed under intersection, then CLp(R) always exist for all relations R (on a universe set A).**Example:**• Ex: property increasing closed under intersection • ref. yes yes • symmetric yes yes • transitivity yes yes • irref. no(why?) yes(why?) • asymmetric ? ? • antisymmetric ? ? • equivalence ? ? • partial order ? ? • total ordering ? ? • ref or symmetric yes no (why?) Fact: If P is not increasing, then CLp may not exist. If P is not conjunctive,then CLp may not exist.**reflexive and symmetric closures of relations**Theorem 1': R: a binary relation on A, then 1. reflexive closure of R (Rref) = R U IA. 2. symmetric closure of R (Rsym) = R U R-1. Ex: if MR = [100,101,000], then 1. IA = ? 2. MRref = ? 3. MR-1 = ? 4. MRsym = ?. pf: Let Tref = {S Í R | S is a relation on A and is ref.} ==> 1. (1) RUIAÎ Tref (2) S Î Tref ==> SÍ R U IA. Hence R U IA = CLref(R). 2. Similar to 1.**6**2 1 3 5 4 Transitive closure of relations • R: a binary relation on A ==> What is the transitive closure of R (i.e., the smallest transitive relation S on A including R (i.e., if (a,b) and (b,c) Î S, then (a,c) is also Î S)). Ex: If R = {(1,2)(2,6)(6,1)(1,3)(3,4)(4,5)} ==> GR = Fig 1. ==> MR = ? ==> GS = ? ==> MS = ? ==> S = ? • The reflexive and transitive closure of R is the smallest reflexive and transitive relation on A including R.**Path**Def. 1: G = (A,R): a digraph; a, b : two vertices in A. 1. A path of length n (n 0) from a to b is a sequence x0,x1,...,xn of vertices s.t. • x0 = a (starting point) and xn = b (ending point) and • (xi, xi+1) Î R for all i = 0,..,n-1. 2. If x0 = xn and n > 0 ==> the path is called a cycle ( or circuit). Ex: If R = {(1,2),(2,1),(2,3)(3,4)(4,2)} ==> "1,2,3,4,5" is a path of length ? from ? to ?.**Correspondence b/t relation and digraph**Theorem 1: |A|= m; R is binary relation on A; (Hence G=(A,R) is a digraph) ==> for all n ³ 0, (x,y) Î Rn iff $ a path of length n from x to y. pf: by induction on n. Basis: n = 0 (and R0 = IA = {(x,x) | x Î A}) $ path of length 0 from x to y iff x=y iff (x,y) Î IA. Ind. case: n = k+1. $ a path of length n from x to y <=> (by def) $ a path of length k from x to z and $ a path of length 1 from z to y for some z A <=> (by ind. hyp.) $ z Î A s.t. (x,z) Î Rk /\ (z,y) Î R <=> (by def) (x,y) Rn. QED**correspondence b/t relation and digrqph**Def. 2: R: a binary relation on A (hence (A,R) is a digraph). R+ =def {(x,y) $ a path of length > 0 from x to y in (A,R)} (i.e., R+ = Uk=1,¥ Rk) R* =def {(x,y) | there is a path of length ³ 0 from x to y} =Uk=0,¥ Rk = R+ U {(x,x) | x in A} = R+ U IA. Theorem 2: 1. R+ is the transitive closure of R. 2. R* is the reflexive and transitive closure of R. Note: If R encodes the property that " (a,b) R iff b can be reached from a by 1 step" then 1. (a,b) ∈ Rk means "b can be reached from a by n steps." 2. (a,b) ∈ R* means " b is reachable from a" or means " b is connected to a".**Proof of Theorem 2:**pf: (1) left to the students. (2). 1. R ÍR* ; 2. R* is ref. and transitive. 3. If T is another relation with property 1 & 2, then R* Í T. 1. R Í R* and R* is ref.: trivial, since R* = IA U R U R2 U ... 2. R* is transitive: (x,y) Î R* and (y,z) Î R* => $m and n ³ 0 s.t. (x,y) Î Rm and (y,z) Î Rn => (x,z)Î Rm+n => (x,z) Î R*. 3. Let T be any ref. and tran. relation including R. we show by induction on n that Rn ÍT for all n 0. Hence R* = R0 U R UR2U... Í T. case 1: n = 0 : R0 = IA Í T (since T is reflexive). case 2 : n = 1 : R Í T (since T includes R) case 3: n = k+1 > 1: by ind.hyp., RJ Í T for all J < n. ==> if (x,y) Î Rn => $z s.t. (x,z) Î Rk and (z,y) Î R => (by ind. hyp.) (x,z)Î T and (z,y)Î T => (by tran. of T) (x,y)Î T.**Some lemmas**Lem 1: |A|= n > 0, R: a relation on A. 1. If there is a path of length m > n from a to b, then there is a path of length 0 < p < m from a to b. 2. If there is a path of length > 0 from a to b then there is a path of length 0 < p n from a to b. pf: 1. Let a=x0, x1,x2,..., xm-1,xm=b be the path of length m from a to b. Since m > n, by pigeonhole principle, there is 1 i<j m s.t. xi = xj. ==> The path x0,x1,..,xi, xj+1,...,xm is a path from a to b with length m + (j-i) < m. 2. Let a0 be any path from a to b. if | a0 | ≤ n, we are done. O/w, by 1. we can construct a path a1 with length | a1 | <| a0 |. if | a1| ≤ n, then we are done. O/w, repeat the same process, we can obtain a sequence of path a0, a1, a2,... from a to b with | a0 | > | a1 | > ... But since | a0 | is finite, we will eventually find a path ak with length n . QED**Corollaries**lemma: If a ¹ b and there is a path of length m n, then there is a path of length 0 < p < n from a to b. pf: similar to the previous one. Corollary: |A| = n. 1. R+ = Uk= 1, n Rk. 2. R* = Uk=0,n Rk. Pf: (1.) Since ∀m > n, RmÍ RUR2U...URn. (x,y) Î Rm <=> $ a path from x to y of length m > n => $ a path from x to y of length 0 < p n <=> (x,y) Î Rp <=> (x,y) Î RUR2U...URn. (2.) Similar to 1. Corollary: R* = Uk=0,n-1 Rk. Pf: Left as an exercise. Hint: RnÍ IA U Uk=1,n-1 Rk. (x,y) ∈ Rn ==> (x=y /\ (x,y) Î IA) or (xy /\ (x,y) Î Uk=1,n-1 Rk). Theorem: 1. MR+ = MR \/ MR2 \/...\/MRn. 2. MR* = MR0 \/ MR \/ MR2 \/...\/MRn-1.**Algorithm for transitive closure**• TR(MR: a 0-1 nxn matrix) • 1. A := MR; • 2. B := A • 3. for i = 2 to n do • 3.1 A := A x MR// A = MRi. • 3.2 B := B \/ A // B = Uk=1,i MRk. • 4. return(B) // B = MR+. • Exercise: Give an alg. for finding MR*. • Complexity: • each execution of step 3.1 takes time O(n3) • Total execution time = O(n4). • ==> more efficient algorithm ( O(n3) ) is possible!!**Warshall's algorithm**• Applying dynamic programming technique • running time O(n3). • R: a relation on A = {v1,v2,...,vn}. • note: (MR)ij = 1 iff (vi,vj) R. • W0,W1,...,Wn: a sequence of nxn 0-1 matrices, where • (WK)ij = 1 ( 0 £ k £ n ) iff there is a path of length >0 from vi to vj s.t. all interior vertices of this path are in the set {v1,...,vk}. • (i.e., $ a path vi,u1,u2,...,us,vj (s³ 0) s.t.{u1,...,us}Í {v1,...,vk}. Facts: 1. W0 = MR. [ since (vi,vj) Î R <=> $ a path of length 1 from vi to vj (w/o interior vertex) <=> (i,j) Î W0. ] 2. Wn = MR+. [ Since (vi,vj) Î R+ <=> $ a path of length > 0 from vi to vj <=> $ a path of length > 0 from vi to vj with interior vertices among the set {v1,v2,...,vn} <=> (i,j)Î Wn. ]**Warshall's algorithm (cont'd)**Fafct 3: Wk+1 can be defined in terms of Wk. Lem 2: (Wk+1)ij = (Wk)ij \/ ((Wk)i(k+1) /\ (Wk)(k+1)j ) pf: (Wk+1)ij = 1 <=> (by def.) $ a path a=vi,u1,...,us, vj from vi to vj with {u1,...,us} Í{v1,...,vk+1}. <=> two possible cases: case 1: no ui (i= 1,..,s) is vk+1 (i.e., vk+1Ï {u1,...,us}). ==> a is also a path from vi to vj with interior vertices from {v1, ...,vk} ==> (Wk)ij = 1.**Warshall's algorithm (cont'd)**case 2:$ a path a = vi,u1,...,us, vj from vi to vj with some ui = vk+1. (i.e., vk+1 ∈ {u1,...,us}). Now let IDX = {i | ui = vk+1} and let p = min(IDX); q = max(IDX). Note that p will equal to q iff |IDX| = 1. Now the path b = v1,u1,...,up-1,up,uq+1,...,us,vj is a path from vi to vjcontaining only one vk+1 = up. ==> v1,u1,..,up ia a path from v1 to vk+1 and up,u q+1,..,vj is a path from vk+1 to vj ==> (Wk)i(k+1) = (Wk)(k+1)j = 1. <=> (Wk)ij \/ ((Wk)i(k+1) /\ (Wk)(k+1)j = 1. QED**Warshall's algorithm (cont'd)**EX: Let MR = [0001,1010,1001,0010]. find W0,W1,W2,W3,W4. The Warshall's algorithm: Procedure WS1(MR: nxn 0-1 matrix) 1. W0 := MR; 2. for k := 1 to n do // compute Wk 3. for i := 1 to n do 4. for j = 1 to n do // compute (Wk)ij 5. Wk[i,j] := Wk-1[i,j] \/ (Wk-1[i,k]/\Wk-1[k,j] ); 6. Return(Wn) // Wn = MR+. analysis: 1. The algorithm takes time O(n3). 2. The algorithm require memory space O(n3). ==> can be improved to O(n2) by collapsing all Wk into one W. (and become the final Warshall's algorithm shown in the book). (why? since Wk[i,k]=Wk-1[i,k] /\ Wk[k,j]=Wk-1[k,j] )**6.5 Equivalence relations**Def. 1: R: a relation on a set A. • R is an equivalence relation on A iff it is ref. symmetric and transitive. • x and y in A are said to be equivalent (or R-equivalent) if (x,y) R. Ex1: R: a relation on the set of all bit strings {0,1}* defined by xRy iff |x| = |y|. ==> 1. R is an equivalence relation on {0,1}* ?. 2. 00110 and 11100 are R-equivalent since both have the same length. Ex 2: R: a relation on integers s.t. xRy iff x=y or x = -y. ==> R is an equivalence relation ? Ex3: R: a relation on real numbers s.t. xRy iff x-y is an integer. ==> R is an equivalence relation ?**More equivalence relations**Ex5: [congruence modulo m](共餘) m: a positive integer > 1. ºm: a binary relation on Z s.t. x ºm y iff m divides (x-y). ==> ºm is an equivalence relation on Z. pf: Equivalence classes: Def 2: R: an equivalence relation on A; a : any element of A. ==> the equivalence class of a w.r.t. R, denoted [a]R, is the set of all elements of A equivalent to a, i.e., [a]R = { b | aRb ∧ b Î A}. • If b [a]R, then say b is a representative of the class [a]R.**Equivalence classes**Ex 5: R: a binary relation on Z s.t. xRy iff x = y or x = -y. => 1. R is an equivalence relation. 2. For each x in Z, [x]R = {x, -x}. 3. e.g., [3]R = { y Î Z | 3 = y or 3 = -y} = {3, -3}. Ex6: R Z2 s.t. xRy iff x º y (mod 4). => [0]R = {y ÎZ: 4 | (0-y) } = {..., -8, -4, 0, 4, 8 ,...} [1]R = {y ÎZ: 4 | (1-y) } = {..., -7, -3, 1, 5, 9 ,...}, [2]R = {y ÎZ: 4 | (2-y) } = {..., -6, -2, 2, 6, 10 ,...}, [3]R ={y ÎZ: 4 | (3-y) } = {..., -5, -1, 3, 7,11,...} [4]R = {y ÎZ: 4 | (0-4) } = {..., -8, -4, 0, 4, 8 ,...}= [0]R. [5]R = [1]R, [6]R = [2]R,... => 1. [i]R = [i+4k]R, where i and k are any integers. 2. There are only 4 equivalence classes w.r.t. R. i.e., {[a]R | a Î Z} = {[0]R, [1]R, [2]R, [3]R}= { {...,-4,0,4,..}, {...,-3,1,5,..}, {...,-2,2,6,...}, {...,-1,3,7,...}}.**Congruence relation**3. the class [a]R is called a congruence class (modulo m). 4. For any integer x, xÎ [y]R iff xRy. Quotient set: A: a set; R: an equivalence relation on A. The quotient of A w.r.t R, denoted A/R, is the set of all R-equivalence classes of A. (i.e., A/R = {[x]R | x Î A} ). Ex: The set Z/º4 = {[0]4,[1]4,[2]4,[3]4}. in general we have Z/ºn = {[0]n,[1]n,...,[n-1]n}. Ex: |A| = n; R: an equivalence relation on A s.t. |[a]R| = k for all a Î A. ==> | A/R | = ___**x**x1 y y1 Equivalence classes and partitions Lemma 1: R: an equivalence relation on A. Then for any x,y Î A, 1. x R y iff [x]R = [y]R iff [x]RÇ [y]R ¹Æ. 2. not x R y iff [x]R ¹ [y]R iff [x]RÇ [y]R = Æ. 3. Every class [x] is not empty and every x belongs to exactly one class. pf: 1. (=>:) x R y => [x] = {b | xRb} = {b | yRb} = [y] => since [x]=[y] and x Î [x], [x]RÇ [y]R ¹Æ . 1. (<=:) let z in [x]RÇ [y]R ¹Æ. => [x] = {b | xRb} = {b | zRb} = {b |yRb} = [z]. => y Î [x] => xRy. 2,3: direct from 1. not xRy xRy A

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