Homework 11 Solutions W11

Homework 11 Solutions W11 - EECS 203: Homework 11 Solutions...

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Unformatted text preview: EECS 203: Homework 11 Solutions Section 8.5 1. (E) 14 In order to show that R is an equivalence relation, we need to show that it is reflexive, symmetric and transitive. It is easy to see that R is reflexive since x has the same n th character as x for any string x . It is also straightforward that R is symmetric since if the n th characters in x and y are the same letter, then n th characters in y and x are the same letter as well. Finally, if x and y have the same n th character and y and z have the same n th character, then the n th characters of x and z are same. Thus, R is transitive. 2. (E) 24ab a) It is not an equivalence relation since it is not symmetric. b) It is an equivalence relation. 3. (M) 54 Lets first show that if R 1 R 2 then P 1 is a refinement of P 2 . If R 1 R 2 and ( s,x ) R 1 then ( s,x ) R 2 . Hence, it follows that, for any s S , [ s ] R 1 = { x S | ( s,x ) R 1 } { x S | ( s,x ) R 2 } = [ s ] R 2 . By the definition of refinement, P 1 is a refinement of P 2 if [ s ] R 1 [ s ] R 2 for all s S . Hence we conclude that if R 1 R 2 then P 1 is a refinement of P 2 . Next lets prove the converse. By the definition of refinement, if P 1 is a refinement of P 2 then we have [ s ] R 1 [ s ] R 2 for all s S , which implies { x S | ( s,x ) R 1 } { x S | ( s,x ) R 2 } for all s S . Note that this further results in { ( s,x ) | ( s,x ) R 1 ,x S } { ( s,x ) | ( s,x ) R 2 ,x S } for all s S , and R 1 = uniondisplay s S { ( s,x ) | ( s,x ) R 1 ,x S } uniondisplay s S { ( s,x...
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Homework 11 Solutions W11 - EECS 203: Homework 11 Solutions...

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