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3 votes
3 votes

is the language WXWR is regular? can any one provide the proof?

2 Answers

Best answer
12 votes
12 votes

WXWis regular on {a,b}  . X can be anything

so we can rewrite this language as " starting bit and ending bit are same " which is a regular language .  

Here, If alphabet set = (a,b)+,

Then , Language accepted by the finite automata,  L = a(a+b)+a + b(a+b)+b .

see this : http://stackoverflow.com/questions/14521300/why-l-wxwr-w-x-belongs-to-a-b-is-a-regular-language

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5 votes
5 votes
i assume W,X belongs to {a,b}* i will always intentionally put W=null and X=string you provide me now X belongs to {a,b}* so the whole language is regular..

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