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I have to write a regular expression that accepts any odd binary number not preceded by a 0. the best I can come up with is $1(0\cup1)^*1$, but that doesn't match just 1. The best it matches is 11.

Raphael
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Mirrana
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  • just add that single $1$ to your expression as an alternative to what you already have! – Hendrik Jan Apr 02 '13 at 23:38
  • I had thought of that, but I worried that it would not be the minimum possible. Thanks. – Mirrana Apr 02 '13 at 23:42
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    Minimal length is not always better than "elegant", or "easy to understand". A short alternative could be $1(0^1)^$. That one has nested stars, which is more complex in some respects. – Hendrik Jan Apr 02 '13 at 23:49
  • Obligatory hint: build an NFA and convert. Also, any question should contain, well, a question. – Raphael Apr 03 '13 at 07:15

1 Answers1

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First you might just add that single $1$ to your expression as an alternative to what you already have. One obtains $1 \cup 1(0\cup 1)^*1$.

An alternative is $1(0^*1)^*$. That one is shorter, but not necessarily better.

Actually that expression I obtained by starting with the FSA below, and realizing that the rightmost two states could be replaced by iterating $0^*1$.

enter image description here

Hendrik Jan
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