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   RegEx: reverse(L)
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   Author  Topic: RegEx: reverse(L)  (Read 2572 times)
Dudidu
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RegEx: reverse(L)  
« on: Oct 27th, 2003, 9:22am »
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Let L be a regular language. Show that the following language is also regular.  
 
reverse(L) = { xy : yx [in] L}
 
Hint: You can look at the RegEx:half(L) thread in the CS section for hints and an example (e.g. my solution) how these kind of problems can be formally solved.
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towr
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Re: RegEx: reverse(L)  
« Reply #1 on: Oct 27th, 2003, 9:45am »
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I'd call it rotation(L), rather than reverse(L).. (since it isn't really reversing much)
And I think that's allready somewhere in this forum..
 
[e]of course William has his own ideas, and he called it Cycle(L)[/e]
« Last Edit: Oct 27th, 2003, 9:48am by towr » IP Logged

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Dudidu
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Re: RegEx: reverse(L)  
« Reply #2 on: Oct 28th, 2003, 12:02am »
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on Oct 27th, 2003, 9:45am, towr wrote:
And I think that's allready somewhere in this forum..

towr you are right ! (I'm so embarassed Embarassed recycling problems)
So, Lets change it a little bit... Let L be a regular language. Show that the following language is also regular.  
 
reverse(L) = { w : wR [in]  L}
 
* If w = x1x2...xn then wR = xnxn-1...x1.
 
Quote:
I'd call it rotation(L), rather than reverse(L).. (since it isn't really reversing much)
towr, now you can see that un-intensionally I choose the right name for the thread Wink.
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