Prove whether or not language
$L$ ={$a^pb^q : p ≥ 100$ and $q ≥ 100$ are fixed integer values, and $i ≥ 0$} is regular.
I'm not sure how to prove this.
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What is a^pb^q? is it $a^pb^q$ or $a^{pb^q}$? – S.C.B. Mar 02 '16 at 03:08
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is the first one (a^p) (b^q) sorry – user142743 Mar 02 '16 at 03:10
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1Where does $i$ come into it? – MJD Mar 02 '16 at 03:17
2 Answers
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If $p$ and $q$ are fixed values, then the language just has one string, and is clearly a regular language.
If $p$ and $q$ are not fixed values, but arbitrary values, the question is slightly more interesting. A language is regular iff there exists a corresponding regular expression to it. The following is a regular expression corresponding to $L$
$$ (\underbrace{aaa \ldots a}_\text{100 times})a^* (\underbrace{bbb \ldots b}_\text{100 times})b^*$$
eepperly16
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Prove whether or not language $L$ ={$a^pb^q : p ≥ 100$ and $q ≥ 100$ are fixed integer values, and $i ≥ 0$} is regular.
Ignoring $i$, this seems to be a regular language. Here is a NFA for $L = \{ a^p b^q \mid p\ge 3, q\ge 3 \}$, which is easy to modify for a lower bound of $100$ instead of $3$.
mvw
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