Computers and Technology, 19.10.2021 05:10 kiki4832
Using Pumping Lemma (slides 30-35 of the notes ‘Regular Languages & Finite Automata-IV’) one can show the language L= {a^n b^n | n ϵ N } is not regular (We need this property in the notes ‘Context-free Languages and Pushdown Automata I). This is done by way of contradiction. We assume L is regular. Since L is infinite, Pumping Lemma applies. We then consider the string s=a^m b^m where m is the number of states in the DFA that recognizes L. Since the length of s is bigger than m, by Pumping Lemma, there exists strings x, y and z such that s=xyz, y≠Λ, |xy|≤2m and xy^k z∈L for all k∈N. If |xy|<2m then the first repeated state on the acceptance path cannot be a final state. Is this statement true, why or why not?
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Using Pumping Lemma (slides 30-35 of the notes ‘Regular Languages & Finite Automata-IV’) one can...
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