Announcements ICS 6B � Regrades for Quiz #3 and Homeworks #4 & 5 are due Thursday Boolean Algebra & Logic Lecture Notes for Summer Quarter, 2008 Michele Rousseau Set 7 – Ch. 8.4, 8.5, 8.6 2 Lecture Set 7 - Chpts 8.4, 8.5, 8.6 Grades Overall Grades & Quiz #3… Quiz #3 � Max: 100% 90-100 � Min: 42% 80-89 � Median: 83% 70 79 70-79 60-69 Overall less � Max: 99% than 50 � Min 45% 0 1 2 3 4 5 6 7 � Median: 84% Overall Quiz #3 Lecture Set 7 - Chpts 8.4, 8.5, 8.6 3 Lecture Set 7 - Chpts 8.4, 8.5, 8.6 4 Today’s Lecture � Chapter 8 �8.4, 8.5, 8.6� ● Closures of Relations �8.4� Chapter 8: Section 8.4 ● Equivalence Relations �8.5� ● Partial Orderings �8.6� Closures of Relations Lecture Set 7 - Chpts 8.4, 8.5, 8.6 5 1
Reflexive Closure Closure of Relations Example Let R be a relation on set A. A��1,2,3� Let P be a property (reflexive, symmetric, etc.) The closure of R with respect to the property P R���1,1�,�1,2�,�1,3�� is the smallest relation containing R which has P�“being reflexive” this property. R is not reflexive, b/c its missing �2,2�, �3,3� � In other words, add the minimum number of pairs to obtain property P. property P. The smallest reflexive relation containing R is The smallest reflexive relation containing R is � Note: This may not be possible. S���1,1�,�1,2�,�1,3�,�2,2�,�3,3�� Example: This is the reflexive closure of R & it’s the intersection A��1,2,3,4�, R���1,1�,�1,3�,�14�� of all of the reflexive relations that contain R P is being “irreflexive” Any relation on A which is reflexive and contains R � If the closure S of R w.r.t. P exists, must include: ● Then the relations S is the intersection of all the �1,1�,�1,2�,�1,3� and �1,1�, �2,2�, �3,3� relations R which satisfy property p. R The diagonal pairs in AxA 7 8 Lecture Set 7 - Chpts 8.4, 8.5, 8.6 Symmetric Closure Reflexive Closure (2) Example A��1,2,3� � Let R be a relation on set A. R���1,1�,�1,2�,�1,3�� � Then the reflexive closure of R always P�“being symmetric” exists: we just need to add all the R is not symmetric , b/c it’s missing �2,1�, �3,1� elements of the form �a,a� with a � A. The smallest symmetric relation containing R is The smallest symmetric relation containing R is � In other words the “diagonal � in AxA” S���1,1�,�1,2�,�1,3�,�2,1�,�3,1�� Note: we are adding R -1 This is the symmetric closure of R. Theorem: If R is a relation on A, denote by � ={(a,a): a � A} Generalized: the diagonal in AxA. Then the reflexive closure of R If R is a relation on A. Then the symmetric closure of R exists and is equal to exists and is equal to S reflexive = R � � S sym = R � R -1 y Lecture Set 7 - Chpts 8.4, 8.5, 8.6 9 10 Irreflexive, AntiSymmetric & Symmetric Closure (2) Asymetric Closures Example Assume P � “being irreflexive” A � �1,2,3,4� A��1,2,3,4�, R���1,1�,�1,3�,�14�� R � ��1,3�,�2,2�, �2,4�, �3,3�, �3,4�, �4,3�� � Shows that if R is not irreflexive we can’t make it R ‐1 ���3,1�, �2,2�, �4,2�, �3,3�, �4,3�, �3,4�� irreflexive. Then Then � Thus the irreflexive closure of R does not exist R � R ‐1 ���1,3�,�3,1�,�2,2�,�2,4�,�4,2�,�3,3�,�3,4�,�4,3�� � When R is irreflexive � the irreflexive closure of R exists – it is R itself. This is the symmetric closure of R � The relation R then is the smallest irreflexive R � R ‐1 is the smallest symm‐relation containing R, relation containing R basically we are adding �3,1� & �4,2� which is � This also applies to: what R needed to become symmetric � Antisymmetric & Asymetric closures. Lecture Set 7 - Chpts 8.4, 8.5, 8.6 11 12 2
In terms of a Digraph In terms of a Matrix � To find the reflexive closure � To find the reflexive closure ● add loops. ● Put 1’s on the diagonal. � To find the symmetric closure � To find the symmetric closure ● Take the transpose M T of the connection ● add arcs in the opposite direction. matrix M R � To find the transitive closure ‐ if there is a path from a to b Note: This relation is denoted R T or R c and ● add a direct arc from a to b. and called the converse of R Note: Reflexive and Symmetric closures are easy Transitive can be complicated 13 14 Lecture Set 7 - Chpts 8.4, 8.5, 8.6 Lecture Set 7 - Chpts 8.4, 8.5, 8.6 Transitive Closure t(R) Paths � This is a little more difficult � A path of length n in a diagram G is the c •Because (a,b) and (b,c) the sequence of edges: b transitive closure must contain (a,c) ● �x 0 , x 1 � �x 1 , x 2 �…�x n‐1 , x n � •Similarly it must contain (b,d) ● The terminal vertex of the previous arc a d matches the initial vertex of the following arc t h th i iti l t f th f ll i � The edges �a,c� and �b,d� seem to be the least amount of � If x 0 � x n the path is called a cycle or a circuit . edges that need to be added in order to make R transitive This is similarly true for relations c •This is not Transitive – b because of (a,c), (b,d) – we need to add (a,d) Now it is transitive – it may take a d several iterations so t(R)=R ���a,c�,�b,d�,�a,d�� Lecture Set 7 - Chpts 8.4, 8.5, 8.6 15 Lecture Set 7 - Chpts 8.4, 8.5, 8.6 16 Theorem: Let R be a relation on set A. From the induction Hypothesis There is a path of length n from a to b iff �a,b� � R n �a,x�� R Proof: �by induction� � Basis: And since �x,b� is a path of length n An arc from a to b is a path of length 1 which is in R 1 �R. �x,b�� R n Hence the assertion is true for n�1 c x if �a,x�� R � Induction Hypothesis: yp and �x,b�� R n , Assume the assertion is true for n. then �a,b�� R n �R� R n � 1 Show it is true for n�1 a b � There is a path of length n�1 from a to b iff Q.E.D � quod erat demonstrandum there is an x � A such that �“that which was to have been demonstrated”� there is a path of length 1 from a to x and a path of length n from x to b. Lecture Set 7 - Chpts 8.4, 8.5, 8.6 17 Lecture Set 7 - Chpts 8.4, 8.5, 8.6 18 3
Proof Transitivity Closure (2) Theorem: R * is the transitive closure of R. Proof: Theorem: Let R be a relation on set A. We must show that R * is transitive The connectivity relation or the star closure Suppose �a,b�� R * and �b,c� � R * ∞ is the relation R * = � R n n�1 Show �a,c� � R * � R * is the union of all powers of R ● By definition of R * , �a,b� � R m for some m and �b,c� � R n for some n � Notice that R * contains the ordered set �a,b� if ● Then �a,c� � R n R m � R m�n which is contained there is a path from a to b in R * . � t�R� is the smallest transitive relation containing R Hence R * must be transitive � R is transitive iff R n is contained in R for all n � Notice that R * contains R ● Because R�R 1 � R n �R * 19 20 Lecture Set 7 - Chpts 8.4, 8.5, 8.6 So R * is a transitive relation containing R In fact we only have to consider paths of n or less By definition the transitive closure of R, t�R�, is the Theorem: If |A|� n , then any path of length � n must smallest transitive relation containing R. contain a cycle Proof: To prove this lets suppose S is any transitive relation If we write down a list of more than n vertices that contains R representing a path in R, some vertex must appear at We must show S contains R * to show R * is the R * t R * i th least twice in the list �by the Pigeon Hole Principle�. least twice in the list �by the Pigeon Hole Principle� W t h S t i h smallest such relation. Thus R k for k � n doesn’t contain any arcs that don’t R � S, so R * � S 2 � S since S is transitive already appear in the first n powers of R. There fore R n � S n � S for all n. Hence S must contain R * since it must also contain the union of all powers of R. Q.E.D Lecture Set 7 - Chpts 8.4, 8.5, 8.6 21 Lecture Set 7 - Chpts 8.4, 8.5, 8.6 22 3 Methods to construct Corollaries R* = R � R 2 � . . . � R n Corollary: If | A | � n, then 1. Digraphs t�R� � R* � R �R 2 � . . . � R n 2. Binary Matrices 3. Warshall’s Algorithm �detailed in book� g � � Corollary: We can find the connection matrix of t�R� by computing the join of the first n powers of the connection matrix of R. Lecture Set 7 - Chpts 8.4, 8.5, 8.6 23 Lecture Set 7 - Chpts 8.4, 8.5, 8.6 24 4
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