6 the pumping lemma for regular languages corollary

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6.6 The Pumping Lemma for Regular Languages Corollary 6.6.2. Let G be the state diagram of a DFA with k states, and let p be a path of length k or more. The path p can be decomposed into sub paths (some of them can be empty) q , r , and s , where p = qrs , | qr | k and r is a cycle . Theorem 6.6.3 ( Pumping Lemma for Regular Languages ) Let L be a regular language that is accepted by a DFA M with k states. Let Z L with | Z | k , and let Z = uvw such that | uv | k , | v | > 0, and uv i w L , 2200 i 0
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18 6.6 The Pumping Lemma for Regular Languages Theorem . If a regular language contains strings of the form x n y n for arbitrarily large integers n , then it must contain strings of the form x m y n , where m n Proof : Suppose M is a DFA . 220d . L ( M ) contains x n y n , n 0. Then, 5 k N such that k is larger than the number of states in M and x k y k L ( M ). Since there are more symbols in x k than there are states in M , the process of accepting x k y k will result in some state in M being traversed more than once before any y s in the string are reached, i.e., in reading some of the x s a circular path will be traversed. If j is the number of x ’s read while traversing this path, then M can accept the string x k+j y k by traversing the path an extra time. Hence, 5 l = k + j k . 220d . x l y k L ( M ), a contradiction.
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  • Winter '12
  • DennisNg
  • Formal language, Regular expression, Regular language, Nondeterministic finite state machine

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