Finite automaton for $\{0^{2n} : n \ge 0 \} \cup \{(100)^m : m \ge 0\}$

Asked Jan 27 '21 at 17:27

Active Jan 27 '21 at 17:41

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I have to construct an FA for the following language: $$ \{ w_1,w_2 \in \{a,b\}^* \mid w_1 = 0^{2n}, w_2 = (100)^m, \text{ for some } n,m \in \mathbb{N} \}. $$

Does anyone here have any experience with something like that? I'm totally lost how I should draw an FA for something like that. Would appreciate any kind of input or approach to this problem.

edited Jan 27 '21 at 17:41

Yuval Filmus

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asked Jan 27 '21 at 17:27

Prometheus

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This is a standard question. If you are completely lost, then there is no better recurse than reviewing class material. – Yuval Filmus Jan 27 '21 at 17:42
Mhm, so i don't have problems drawing DFA or NFAs, ist just that the 0^2n and w2 = (100)^m do give me problems. Would you like to help me out? – Prometheus Jan 27 '21 at 17:43
Can you construct an FA for just words of the first type or just words of the second type? When reading a word, can you tell which of the two types it belongs to? – Yuval Filmus Jan 27 '21 at 17:43
I am not planning to do this problem for you, if that's what you're after. – Yuval Filmus Jan 27 '21 at 17:43
So is w1 always 0? Is that correct? And w2 is 100^2 = 10000 or any other number depending on m. Is that the right direction? – Prometheus Jan 27 '21 at 17:47
No. These are powers of words. For example, $0^3 = 000$ and $(100)^2 = 100100$. – Yuval Filmus Jan 27 '21 at 17:52
There is a typo in your problem statement – the alphabet should be ${0,1}$ rather than ${a,b}$. – Yuval Filmus Jan 27 '21 at 17:53
Ah, okay, so w2 is identfied in the automate by any 100 sequence, and everything else is just w1. Okay, now that helps already. I'll try to draw it now – Prometheus Jan 27 '21 at 17:58
The regular expression would be in the case L = (00)∗·(100)∗, correct? – Prometheus Jan 27 '21 at 17:59
The description of the language is a bit weird. My interpretation is $(00)^+(100)^$. – Yuval Filmus Jan 27 '21 at 18:01
Okay, sounds good. I'll try it now. Thanks so far. – Prometheus Jan 27 '21 at 18:02
My guess was that the alphabet should have been { "0", "1", "," } and the language 0^2n "," (100)^m. – gnasher729 Jan 28 '21 at 00:05

Finite automaton for $\{0^{2n} : n \ge 0 \} \cup \{(100)^m : m \ge 0\}$

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