Read the post again...
The adversary then chooses the split up, that is, the part of the
string you've given him that loop
you can only choose the string 'w' (=n), the adversary can only _CHOOSE X,
Y and Z_ but ofcourse the adversary is restricted by the fact that he had
chosen a 'n' earlier and
You've made some very fundamental mistakes.
Pumping Lemma is used as an _adversarial_ argument. This has the
following implications,
1) _You_ as the one who is trying to prove that a language is NOT
regular will not be choosing `n'. The adversary chooses `n'.
2) Once the adversary has chosen `n',
He could quite easily have choosen x=0^n-2,
y=0^2, z=the rest.
But may choose it my way instead to prove me wrong. I took the roles
of both to get both the sides of the problem.
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