bonus - l in to at most l ! classes. For 1 i ( l !), let z...

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Solution to Bonus Problem May 1, 2009 Proof. Given a CFG L over single alphabet a , we will show it is also a regular language. Suppose the pumping length of langauge L is l ; then for any string z L of length larger than l , we know we can write into uvwxy such that uv i wx i y is also in the language for every i . Notice that uv i +1 wx i +1 y is just adding za i ( | v | + | x | ) . Also Notice that | vwx | ≤ l , therefore ( | v | + | x | ) is divisible by ( l !). Therefore, we know the following property of language L: for any z L , if | z | > l , then z ( a l ! ) * is also in L . Then we can divide the string in L with length larger than
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Unformatted text preview: l in to at most l ! classes. For 1 i ( l !), let z i be the shortest string in L such that | z i | l, and | z i | l i mod ( l !) . Notice that for some i , there might not exists some z i that satisfy above equation. We use S to denote the set that contains all the i such that z i exists. Now we can write L by the following regular expression, L = { x | | x | l,x L } i S z i ( a ( q !) ) * . 1...
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