0

I am relatively new to the pumping lemma, and I have a problem here that I think I answered correctly, can anyone tell me if this works and if not why not

The problem: {www | w is {a,b}*}

My approach:

L = www

u* (v^k) * w has to be a subset of L

www

| | |

uvw

uvw = www

(u)(v^2)(w) = wwww

wwww is not part of the language www and therefore not regular

Edit: Well my according to my understanding the pumping lemma by taking the "test string" we are looking at and splitting it into a portion that stays the same follow by a portion that is repeatable and then lastly another portion that remains the same. In my "approach" I took the test string "www" and split it into u,v, and w each respectively holding a single "w", with v being the repeatable section and the other two being the ones that remain the same. I double the v section and end up with a resulting uvvw which translates to wwww which appears as if it is not a part of the language www. I have a good feeling that I am wrong because of the condition "w is {a,b}*" which I think includes the empty string, and since the empty string is viable in wwww and www my pumping lemma is faulty. I would just like to know what approach I would have to take to tackle such a problem, its just a practice problem

hyde
  • 2,525
  • 4
  • 27
  • 49
Guitardeon
  • 73
  • 1
  • 7

1 Answers1

0

I do not believe your answer works, because there is no way to be sure that wwww is not in the language.

For example, let |w| be a multiple 3 (i.e 3*k for some k) So your orignal string is length: |3k|+|3k|+|3k|=9*3k

So if you add another string length 3k. The length is now 12k, which is ALSO a multiple of 3.

Try something like: Let w = 100...001, where you have p zeros enclosed by 1s. Then no matter how you pump 10..0110..0110..01 u v w , you will be out of the language.

Mickey Sweatt
  • 167
  • 2
  • 10