Proof of the Pumping Theorem for Regular Languages

1. Proof of the Pumping Theorem for Regular Languages Richard Beigel CIS Temple University

2. The Pumping Theorem for Regular Languages If L is regular then N z such that z  L and |z|  N u,v,w such that z = uvw , |uv|  N, and |v| > 0 i [uviw  L]

3. Proof • Assume L is regular • Then there is a (standardized) DFR P that recognizes L (no EOF or NOOP) • Let be N be the number of control states in P • Let z  L and |z|  N • Consider P’s accepting computation on input z • Let q0, q1, …, qn be the sequence of control states in that computation. Then n N.

4. Proof • Then there is a standardized DFR P that recognizes L (no EOF or NOOP) • Let be N be the number of control states in P • Let z  L and |z|  N • Consider P’s accepting computation on input z • Let q0, q1, …, qn be the sequence of control states in that computation. Then n N. • By the pigeonhole principle qj=qk for some j<k

5. Proof • Let be N be the number of control states in P • Let z  L and |z|  N • Consider P’s accepting computation on input z • Let q0, q1, …, qn be the sequence of control states in that computation. Then n N. • By the pigeonhole principle qj=qk for some j<k • Let u be the string scanned between q0 and qj

6. Proof • Let z  L and |z|  N • Consider P’s accepting computation on input z • Let q0, q1, …, qn be the sequence of control states in that computation. Then n N. • By the pigeonhole principle qj=qk for some j<k • Let u be the string scanned between q0 and qj • Let v be the string scanned between qj and qk

7. Proof • Consider P’s accepting computation on input z • Let q0, q1, …, qn be the sequence of control states in that computation. Then n N. • By the pigeonhole principle qj=qk for some j<k • Let u be the string scanned between q0 and qj • Let v be the string scanned between qj and qk • Let w be the string scanned between qk and qn

8. Proof • Let q0, q1, …, qn be the sequence of control states in that computation. Then n N. • By the pigeonhole principle qj=qk for some j<k • Let u be the string scanned between q0 and qj • Let v be the string scanned between qj and qk • Let w be the string scanned between qk and qn • Then uvw = z, |uv|  N, |v|  1, • and P accepts uviw for all i  0

9. Proof • By the pigeonhole principle qj=qk for some j<k • Let u be the string scanned between q0 and qj • Let v be the string scanned between qj and qk • Let w be the string scanned between qk and qn • Then uvw = z, |uv|  N, |v|  1, • and P accepts uviw for all i  0 • Therefore uviwL for all i  0, completing the proof of the Pumping Theorem

10. qj = qk Picture-proof that uv*w  L v u w q0 qj qn

11. Example: L = {an2 : n 0} • Assume L is regular • Let N be given by the Pumping Theorem • Let z = aN2 • Let u, v, w be given by the Pumping Theorem • Then v = ak where 0 <k  N • Let i = 2 • Then uviw = uv2w = uvvw = aN2+ k • Since N2 < N2 + k N2 + N < N2 + 2N + 1 = (N + 1)2, N2 + k is not a square, so uviw = aN2+ k L • This contradicts the Pumping Theorem, so L is not regular

12. Example: L = {ambn : mn} • Assume L is regular • Let N be given by the Pumping Theorem • Let z = aN bN • Let u, v, w be given by the Pumping Theorem • Then v = ak where 0 <k  N • Let i = 2 • Then uviw = uvvw = aN+k bN • Since k > 0, N+k> N, so uviw = aN+k bN L • This contradicts the Pumping Theorem, so L is not regular

13. Example: L = {ambn : mn} • Assume L is regular • Let N be given by the Pumping Theorem • Let z = aN bN • Let u, v, w be given by the Pumping Theorem • Then v = ak where 0 <k  N • Let i = 0 • Then uviw = uw = aN-k bN • Since k > 0, N-k< N, so uviw = aN-k bN L • This contradicts the Pumping Theorem, so L is not regular