Solutions to Assignment 5

October 27, 2000

Exercise 1 (40 points) Give context-free grammars generating the following languages.

1. The set of strings over the alphabet Σ = {a, b} with twice as many a’s as b’s.

2. The complement of the language {anbn | n ≥ 0}.

3. {w#x | wR is a substring of x for w, x ∈ {0, 1}}.

4. {w | w = wR for w ∈ {0, 1}∗}.

Solution We present the context-free grammar rules for each case.

1.

S → SS | aaSb | abSa | baSa | ε

2.

S → T | aU | V bT → aT | Ta | bT | Tb | baU → aU |WV → V b |WW → aWb | ε

3.

S → S0 | S1S → TT → 0T0 | 1T1T → UU → U0 | U1 | #

4.

S → 0S0 | 1S1 | 0 | 1 | ε

Exercise 2 (15 points) Give a context-free grammar that generates the language.

A = {aibjck | i, j, k ≥ 0 and either i = j or j = k}

Is your grammar ambiguous? Why or why not?

1

Solution Let G = (V,Σ, S, R), where V = {S,U, V,X, Y }, Σ = {a, b, c} and R consists of thefollowing rules.

S → U | VU → aU | XV → V c | YX → bXc | εY → aY b | ε

The context-free grammar G is ambiguous. For instance, we have S ⇒ U ⇒ aU ⇒ abUc ⇒ abcand S ⇒ V ⇒ V c⇒ aY bc⇒ abc, which are two different derivations of abc.

Exercise 4 (35 points) Use the pumping lemma to show that the following languages are notcontext-free.

1. (15 pts) L1 = {0n#02n#03n | n ≥ 0}

2. (20 pts) L2 = {w#x | w is a substring of x for w, x ∈ {0, 1}}.

Solution Assume L1 is context-free. Let p be its pumping length. and s = 0p#02p#03p. By thepumping lemma for context-free languages, we have s = uvxyz satisfying the following.

• uvixyiz ∈ L1 for i = 0, 1, 2, . . ., and

• |vy| > 0, and

• |vxy| ≤ p.

We do a case analysis here.

• Either v or y contains #. Then uvvxyyz contains more than two #’s, which contradictsuvvxyyz ∈ L1.

• v = 0k1 and y = 0k2 for some k1, k2 ∈ N . We have k1 + k2 > 0 since |vy| > 0.

– vxy occurs before the second # in s. Then there are 3p+ k1 + k2 0’s in uvvxyyz beforethe second # and only 3p 0′s after, implying uvvxyyz 6∈ L1.

– vxy occurs after the first # in s. Then there are 5p + k1 + k2 0’s in uvvxyyz after thefirst # and only p 0′s before, implying uvvxyyz 6∈ L1.

Therefore, L1 is not context-free.Assume that L2 is context-free. Let p be the pumping length of L2 and s = 10p#10p. By the

pumping lemma for context-free languages, we have s = uvxyz satisfying the following.

• uvixyiz ∈ L1 for i = 0, 1, 2, . . ., and

• |vy| > 0, and

• |vxy| ≤ p.

We do a case analysis here.

• Either v or y contains #. Then uvvxyyz contains more than one #’s, which contradictsuvvxyyz ∈ L1.

2

• vxy occurs before #. Then uvvxyyz = w1#w2 for some w1, w2 such that w1 is longer thanw2. Therefore, w1 cannot be a substring of w2, contradicting uvvxyyz ∈ L2.

• vxy occurs after #. Then uxz = w1#w2 for some w1, w2 such that w1 is longer than w2.Therefore, w1 cannot be a substring of w2, contradicting uxz ∈ L2.

• x straddles #. We have two subcases.

– y begins with 1. Then uxz = w1#w2 for some w1, w2 such that w1 begins with 1 and w2

consists of 0’s. Therefore, w1 cannot be a substring of w2, contradicting uxz ∈ L2.

– y = ε. Then uvvxyyz = w1#w2 for some w1, w2 such that w1 is longer than w2. There-fore, w1 cannot be a substring of w2, contradicting uvvxyyz ∈ L2.

Therefore, L2 is not context-free.

3