Proof that the language is not regular (Pumping Lemma)

Asked Feb 23 '18 at 15:47

Active Feb 23 '18 at 17:06

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I have to prove that the following language is not regular:

$$\{ x | x = 10^{2n} + 10^n + 1, n ≥ 1\}$$

I am trying to prove it using Pumping Lemma, however, when I expand the expression I have both addition and multiplication.

I assume that the pumping length is three ($p=3$) and I work on the following word of this language:

$$\{ 10^6 + 10^3 + 1\}$$ $$\{ 1000000 + 1000 +1\}$$

edited Feb 23 '18 at 17:06

Yuval Filmus

asked Feb 23 '18 at 15:47

Pantelis

2

You cannot "assume that he pumping length is three". Your proof has to work for any pumping length. – Yuval Filmus Feb 23 '18 at 17:06
2

How does the input look like? Is it over the alphabet ${1,0,+}$? – Yuval Filmus Feb 23 '18 at 17:06
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I don't understand what the language is. What's the alphabet? Is the language literally strings such as "100+10+1", "10000+100+1", "1000000+1000+1", etc? Or is it encodings of the integers of the form $10^{2n}+10^n+1$? If so, in what base? – David Richerby Feb 23 '18 at 20:37
Please state a question, in the title if possible. – reinierpost Feb 24 '18 at 20:34
I think the language is over {0,1} and it consists of the strings {111,10101,1001001,...}, which is clearly not regular – Maiaux Feb 25 '18 at 10:35
Tip: note that, in base 10, all strings in the language are palindromes. – Guilherme Rito Mar 01 '18 at 18:12
Possible duplicate of How to prove that a language is not context-free? – Ran G. Mar 03 '18 at 07:53

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