Proof of the Pumping Theorem for Regular Languages

1. Proof of the Pumping Theorem for Regular Languages Richard Beigel EECS University of Illinois at Chicago

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