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:

Let L = {a i b j c kd l : i=0 or j=k=l}. Consider the context-free pumping game for L.

(a). State a winning strategy for C.

(b). See Ogdens Lemma on wikipedia. State a game version of it. (That is, state a modified version of our context-free puming game, which Ogdens Lemma says C can win, if L is context-free.)

(c). Describe a winning strategy for N in your game. (Remark: as a consequence, L is not context-free, so this example shows the converse of Pumping Claim 2 is false.)