# lect9_1 - Homework#3 is posted(due on June 26 Todays topics...

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1 Today’s topics: Associative and distributive properties of composition Powers of a relation Inverse relation Homework #3 is posted (due on June 26)

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2 Theorem. Let R A × B , S B × C and T C × D denote three binary relations. Then relation compositions satisfy the following associative law: ( R S ) T = R ( S T ) Properties of composition R S A C B D T R S S T
3 Theorem. Let R A × B , S B × C and T C × D denote three binary relations. Then relation compositions satisfy the following associative law: ( R S ) T = R ( S T ) Proof . First note that both sides define a relation from A to D . We also note the following: ( R S ) T = {( a , d ) | a A and d D and 5 c C [( a , c ) R S cTd ]} ( by the definition of composition T ) = {( a , d ) |… ( 5 c C, 5 b B ) [( aRb bSc ) cTd ]} ( by definition of R S ) = {( a , d ) |… ( 5 c C, 5 b B ) [ aRb ( bSc cTd )]} ( by associative property of ) = {( a , d ) |… ( 5 b B ) [ aRb ( b , d ) S T ]} ( by definition of S T ) = R ( S T ) ( by the definition of composition R )

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4 Associative property of composition means, that no ambiguity arises if we simply write R S T . We can also define powers of a relation R A × A recursively as R 1 =R and R n +1 = R n R ( n 1) Definition. A path in a relation R is a sequence a 0 , … , a k with k 0 such that ( a i , a i +1 ) R for every i<k . We call k the length of the path. 1 7 3 2 A a 0 a 1 a 2 a 3 1, 5, 7, 6 – a path of length 3
5 a 1 a 0 a 2 a 3 R (parent-of) a 1 R R = R 2 (grandparent-of) a 0 a 2 a 3 R R R = R 3 a 1 a 0 a 2 a 3 R 4 = a 1 a 0 a 2 a 3 Powers of a relation

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6 Theorem . ( a , b ) R n iff there is a path of length n in R .
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lect9_1 - Homework#3 is posted(due on June 26 Todays topics...

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