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- Oct 10 2012 12:43 PM
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Posted 10 Oct 2012If possible, post the source where you found the information.
Posted 29 May 2012Okay I got it. Duh, a palindrome would have the same first and last letter. So string w with length(w) = n + 2 is equal to string aua, with length(u) = n and a being an element of the alphabet of language b. Thanks a lot.
Posted 29 May 2012Then I'm not sure what to do. Could you help me get started?
Posted 29 May 2012Okay, where that "length(" and then there is a face with sunglasses, that should be "length("
Dang it! It did it again. After the open parenthesis, there should be a 'b' and then a close parenthesis.
Posted 28 May 2012Okay, well here is an attempt I made (using the Basis and Inductive Hypothesis you provided in your post that was posted at 9:14 AM). I'm sorry if it is a little hard to follow:
Assume w is a string generated from B with length = n + 2. Since length(w) = n > 0, w = ua, with u being a string generated from B with length = n + 1 and a being from the alphabet of B. Since length(u) = n > 0, u = vb with v being a string generated from B with length = n and b being from the alphabet of B, or u = va. Basically, w = vba or w = vaa. Since the length(a) = length( = 1, then according to the Basis, a and b can be generated from A. Since length(v) = n, then according to the Inductive Hypothesis, v can be generated from A. Since w consists of strings that are generated from A, and are concatinated in a way such that w = w^R, w is a palindrome. Since w is a palindrome, awa is also a palindrome.
I know it looks pretty rough, but am I close to getting this right?
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