使用抽引引理证明正则表达式不是正则语言

Question

Ok, I know that this isn't a programming question but it is a computing question so it is relevant.

Basically, how can I use the pumping lemma to prove that this language is not regular?

{w in {0,1}* | if the length of w is odd then the middle symbol is 0}

Please answer this as simple as possible as whilst I know about models of computation, I am relatively new to it.

Thank you very much in advance!

Answer 1

According to the pumping lemma, if that language is regular then there must exist a number p such that for all strings longer than p in the language, we can decompose that string into x + y + z , where each of x , y , and z are strings and | y | >= 1, | x + y | <= p , and x + ( y * i ) + z is in the language for all non-negative integers i .

Now observe that for every non-negative integer i , the string " 1 " * i + " 0 " + " 1 " * i is in the language. (That is, the string of i 1 s followed by a single 0 and then i more 1 s)

Specifically, the string S consisting of p 1 s followed by a 0 and then p more 1 s is in the language. Since this string has length 2 p + 1, this string is long enough that it can be broken into three strings x , y , and z as in the pumping lemma. Since | x + y | <= p , it must be that x and y are all 1 s, and the only 0 character in S is in z . Now consider the string S' = x + y + y + y + z . Since we added 2*| y | characters to it, S' must also have an odd length. But we added some number of 1 characters to the left of the only 0 in S , and didn't add any 1 characters to the right of the 0 . So S' doesn't have a 0 as its middle character, and therefore S' isn't in the language.

Therefore, we've shown that the language can't be pumped as the pumping lemma requires. Therefore, the language is not regular.