Problem Session 5 (CFGs) Talk about the building blocks of CFGs: S 0S 1S ε - everything. S 0S0 1S1 A - waw R. S 0S0 0S1 1S0 1S1 A - xay, where x = y.

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1 CSE2001, Fall Problem Session 5 (CFGs) Talk about the building blocks of CFGs: S 0S 1S ε - everything. S 0S0 1S1 A - waw R. S 0S0 0S1 1S0 1S1 A - xay, where x = y. S 00S1 A - xay, where x = 2 y. Problem 1 (Textbook 2.6d) Design a CFG for the following language L: L = {x 1 #x 2 #...#x k k 1, x i {a, b}, and for some i j, x i = x R j } This strings in L can be viewed as follows:...#...#w#...#w R #...#... Let A be a non-terminal that generates {a, b, #}, i.e. A aa ba #A ε Then, the part w#...#w R can be generated as follows B aba bbb # #A# And then, the CFG that generates L is as follows: S A#B#A B#A A#B B B aba bbb # #A# A aa ba #A ε

2 CSE2001, Fall Problem 2 Design a CFG for the language L = {a i b j c k i + j = k}, and prove its correctness. The strings a i b j c k in the language L can be viewed as a i b j c j c i, and so the following grammar G generates L: S asc A A bac ε Now, we need to prove that it is correct, i.e. L = L(G). As usual, there re two directions to handle. Claim 1 If w L(G), then w L. Proof: If w L(G), then there exist a derivation S w. The proof is by induction on the length of the derivation, d. But first, we need to prove something about A: Claim 1.1 If A w, then w = b i c i for some i 0. Proof: Again, by induction on the length of the derivation, d. Base case: the shortest derivation from A is A ε. So, the base case is for d = 1. And since ε = b 0 c 0, we are done. I.H.: assume that every derivation of length d = n generates a string of the form b i c i for some i 0. I.S.: prove that every derivation of length d = n + 1 generates a string of the form b i c i for some i 0. Let A w be a derivation of length d = n + 1. Since d > 1, the first rule of the derivation must be A bac. Thus, we have A bac bwc, where w is derived from A in n steps. By induction hypothesis, w = b i c i for some i, and so A bac bb i c i c = b i+1 c i+1. Now, back to the proof of Claim 1. Remember, we re doing induction on the length of the derivation S w. Base case: since both with S on the left have a non-terminal symbol, we conclude that there are no derivations of length 1. The only derivation of length 2 is S A ε (check this). Since ε = a 0 b 0 c 0, and = 0, we conclude that ε L. I.H.: assume that every derivation of length d = n generates a string w L. I.S.: prove that every derivation of length d = n + 1 generates a string w L. Let S w be a derivation of length d = n + 1. Since there are only two rules with S on the left-hand, we have two choices:

3 CSE2001, Fall If S asc awc, then by inductive hypothesis, w L, i.e. w = a i b j c k and i + j = k. But then, awc L, because awc = aa i b j c k c = a i+1 b j c k+1, and i j = k + 1, since i + j = k. If S A w, then by Claim 1.1, w = b j c j L. Now, we re proving the opposite direction: Claim 2 If w L, then w L(G). Proof: Again, we want to do induction, but it cannot be induction on the length of w, because w L doesn t imply the every string of length w + 1 is in L. So, we do induction on the number of a s, for example (or number of b s or c s). To cover the base case we will need the following claim about A: Claim 2.1 If w = b i c i for some i 0, then A w. Proof: induction on i. Base case: i = 0, then w = b 0 c 0 = ε, and A ε. I.H.: assume that for w = b i c i, A w. I.S.: prove that A b i+1 c i+1. By I.H., A b i c i, and so A bac bb i c i c = b i+1 c i+1. So, back to the proof of Claim 2: induction on i - the number of a s. Base case: i = 0. Then w = b j c k, and since i + j = k, we have j = k, and so w = b k c k. By Claim 2.1, A w, and so S A w. I.H.: assume that for any string w = a i b j c k, where i + j = k, S w. I.S.: prove that for any string w = a i+1 b p c m, where i p = m, S w. Note that w = aa i b p c m 1 c, and since i p = m, we have i + p = m 1, and so, by induction hypothesis, S a i b p c m 1. But then, S asc aa i b p c m 1 c = w. This completes the proof of Claim 2. And so, from Claims 1 and 2, we conclude that L = L(G).

4 CSE2001, Fall Problem 3 Design a CFG for the language L = {a i b j c k d m i + j = k + m}. This is similar to the previous question, except we have to consider two cases: if j k, then a i b j c k d m can be viewed as a m a k j b j c j c k j d m. if j k, then a i b j c k d m can be viewed as a i b j k b k c k d j k d i. With this in mind, the grammar is constructed as follows: A will generate b j c j, as follows: A bac ε. B will generate a k j b j c j c k j, as follows: B abc A. B will generate b j k b k c k d j k, as follows: B bb d A. And finally, S will generate string in L as follows: S asd B B. So, the resulting grammar is: S asd B B B abc A B bb d A A bac ε Problem 4 (Textbook, 2.22) Design a CFG for the language L = {x#y x, y {0, 1}, x y}. the trick here is to realize that x y implies one of the following: either 1. either x is a proper prefix of y, or 2. y is a proper prefix of x, or 3. there is at least one position in x and y in which they are differ. Let s deal with these, case by case: 1. If x is a proper prefix of y, the strings will look like w#wz, where z is non-empty string over {0, 1}. Note, that we can t easily generate w#wz, but instead we can generate w 1 #w 2 z with w 1 = w 2, which will cover the case of w#wz, and some of the strings in the case 3 (but that s ok, all it means is that we have more that one derivation for some strings, i.e. the grammar will be ambigous).

5 CSE2001, Fall So, the rules for this case are: A 1 BC B 0B0 0B1 1B0 1B1 # C 0C 1C If y is a proper prefix of x, the strings will look like wz#w, where z is non-empty string over {0, 1}. So, by similar reasoning to case 1 we get A 2 0A 2 0 0A 2 1 1A 2 0 1A 2 1 C# 3. Here, we have to make sure that the strings on the left and on the right of # differ in at least one position. So, the strings will look like either or w 1 0 y # w 2 1 z w 1 1 y # w 2 0 z where w 1 = w 2, and y and z can be anything. So, the rules for this case are: A 3 D1E D 0E D 0D0 0D1 1D0 1D1 0E# D 0D 0 0D 1 1D 0 1D 1 1E# E C ε So, putting it all together, the CFG for L will be S A 1 A 2 A 3 A 1 BC B 0B0 0B1 1B0 1B1 # A 2 0A 2 0 0A 2 1 1A 2 0 1A 2 1 C# A 3 D1E D 0E D 0D0 0D1 1D0 1D1 0E# D 0D 0 0D 1 1D 0 1D 1 1E# C 0C 1C 0 1 E C ε