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Let $\Sigma$ be 0,1,2. I am interested in the language of all strings $w$ such that the sum of $w$ characters (where the char 0 is treated as the number 0, the char 1 as the number 1 etc) is divideable by 5. Obviously this is a regular language, and I can also find a simple linear grammer.

My idea for a regular expression was to check all the combinations where I can receive sum zero (in modulo 5) with the digits 1 and 2 and put 0* in between. But is there no better way? Something more elegant?

Edit: My expression is $0^* 1 0^* 1 0^* 1 0^* 1 0^* 1 0^* \cup 0^* 1 0^* 1 0^* 1 0^* 2 0^* \cup 0^* 1 0^* 1 0^* 2 0^* 1 0^* \cup 0^* 1 0^* 2 0^* 1 0^* 1 0^* \cup \ldots $ and so it goes on for all the possible combination. My question is if there is a way to simplify this - Or maybe a different approach?

Raphael
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Benicio Agüero
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1 Answers1

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Forget about the $0$, they are boring. Like you said, you can add $0^*$ basically everywhere and you are done.

Now for the $1,2$. Just try to order them in batches that total 5. Then you realised that this might be problematic in case at the count of 4 we read $2$ and overshoot by one. No problem, we remember this, and continue until we reach a proper multiple of five. Note however, that the overshoot just might repeat when we have $212$, or similar.

So here is how the regular expression is constructed: any multiple of five, or when we reach an overshoot, wait until we are back. Repeat.

First some abbreviations:

$E_5= \{11111,1112,1121,1211,2111,221,212,122\}$.

$E_4 = \{1111,112,121,211,22 \}$.

$E_{\mathrm{shoot}}= \{1112,212,122\}$.

Finally here is your expression:

$\Big ( E_5 \cup E_4\cdot 2 \cdot (E_{\mathrm{shoot}})^*\cdot E_4\Big )^* $.

Ah. Add $0^*$, lots of them.

Hendrik Jan
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