$L= \{ xy \mid x,y \in \{a,b\}^* \}$ is not a regular language. But would it make a difference if we added another constraint that $|x|=|y|$. Would this enforce the condition of finiteness on the language?
Regularity of the language
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regular-language
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4Both $\{xy \mid x,y \in \{a,b\}^*\}$ and $\{xy \mid x,y \in \{a,b\}^*, |x|=|y|\}$ are regular languages. Please clarify your question. – 2012-06-17
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0Oh am I wrong in stating that {xy|x,y E {a,b}*} is not regular? Please correct me and give me a reasoning for the same. Also, would it be different from {ww|w E{a,b}*} – 2012-06-17
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1$L=\{xy \mid x,y \in \{a,b\}^*\}$ means exactly the same as $L=\{a,b\}^*$. Are you sure this is the language you mean? – 2012-06-17
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0Yes of course these are different. In order to write a string as $ww$ it must be a smaller string exactly repeated. But every string can be written as $xy$ for some $x$ and $y$. Can you imagine any string that can't? – 2012-06-17
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0yes I mean to evaluate L= {xy| x,y E {a,b}* and |x|=|y|} . If you mean so say that it equals {a,b}* then it is regular as there can be a DFA to represent it. However, if the length is infinite then it cannot. So does |x|=|y| state that the length is finite. or irrespective of any condition {a,b}* is regular? – 2012-06-17
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0Yes, $|x|=|y|$ implicitly means that the lengths are finite. But actually, $\{a,b\}^*$ only consists of finite-length strings. There is no such thing as an infinite string in a language. – 2012-06-17
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0I suppose that by now it's moot, but you've gotten four good answers to this question and you evidently know how to accept an answer, so why not accept one of these? If I were you, I'd pick Brian's, if for no other reason than he's really close to breaking into the 100K reputation club. – 2012-11-18