Definitions Given a language L , we define the following two-player pumping games. In both games, the winner is the last player to make a legal move. So if they finish the game, N wins.

Regular Pumping Game for L: 1. R chooses an integer p 0.

2. N chooses a string s L such that |s| p.

3. R chooses strings x, y, z such that s = xyz, |xy| p, and |y| > 0.

4. N chooses an integer i 0 such that xyiz L.

Pumping Claim 1: If L is regular, then R has a winning strategy. (Chapter 1 pumping lemma.)

Context-Free Pumping Game for L: 1. C chooses an integer p 0.

2. N chooses a string s L such that |s| p.

3. C chooses strings u, v, x, y, z such that s = uvxyz, |vxy| p, and |vy| > 0.

4. N chooses an integer i 0 such that uvixyiz L. Pumping Claim 2: If L is context-free, then C has a winning strategy. (Chapter 2 pumping lemma.) Roughly speaking: N tries to pick a a unpumpable string s, while R (or C) tries to find a pumpable substring (or pair of substrings) in s. We often use the above pumping claims in the contrapositive direction: If N has a winning strategy for the regular game, then L is not regular.

If N has a winning strategy for the context-free game, then L is not context-free.

question:

Consider this modified version of the regular pumping game:

1. R chooses an integer p 0.

2. N chooses a string s L such that |s| p.

3. R chooses strings x, y, z such that s = xyz, |xz| p, and |y| > 0.

4. N chooses an integer i 0 such that xyi z 6 L.

(The difference is that R is renamed R, and we require |xz| p instead of |xy| p.) If L is regular, must R have a winning strategy? Either prove this is true (by modification of the book argument, pages 7879), or give a counterexample (a specific regular L, and a winning strategy for N).