Chapter 12: Context-Free Languages and Pushdown Automata

1. Chapter 12: Context-Free Languages and Pushdown Automata Section 12.4 Context-Free Language Topics James L. Hein - Discrete Structures, Logic, and Computability

2. Remove Λ-productions Algorithm. Remove Λ-productions from grammars for langauges without Λ. 1. Find nonterminals that derive Λ. 2. For each production A → w construct all productions A → w’ where w’ is obtained from w by removing one or more occurrences of the nonterminals from Step 1. 3. Combine the original productions with those of step 2 and eliminate any Λ-productions. James L. Hein - Discrete Structures, Logic, and Computability

3. Example Remove Λ-productions from the grammar S → ABc A → aA | Λ B → bB | Λ. Solution. • Step 1: The nonterminals A and B derive Λ. • Step 2: From the production S → ABc we construct S → Bc | Ac | c. From the production A → aA we construct A → a. From the production B → bB we construct B → b. • Step 3: S → ABc | Bc | Ac | c A → aA | a B → bB | b. James L. Hein - Discrete Structures, Logic, and Computability

4. Quiz Remove Λ -productions from S → ABc | Ab | c A → ABa | Λ B → Bbc | Λ. Solution. S → ABc | Ab | c| Bc | Ac | b A → ABa | Ba | Aa | a B → Bbc | bc. James L. Hein - Discrete Structures, Logic, and Computability

5. Chomsky Normal Form Productions have one of the following forms • A → b (b a terminal) • A → BC • S → Λ (if Λ is in the language). Advantages: Parse trees are binary, which are easy to represent. Any string of length n > 0 can be derived in 2n – 1 steps. James L. Hein - Discrete Structures, Logic, and Computability

6. Transform context-free grammar to Chomsky normal form Algorithm. Transform context-free grammar to Chomsky normal form 1. Remove A → Λ (if A ≠ S) by previous algorithm. (If S → Λ is removed, add it back.) 2. Remove unit productions (i.e., A → B): If A → B or A + B, then construct productions A → w where B → w is not a unit production. Now remove all unit productions. 3. For each production whose right side has two or more symbols, replace all occurrences of each terminal a with a new nonterminal A and also add the new production A → a. 4. Replace each production B → C1…Cn with n > 2 with B → C1D where D → C2 …Cn. Repeat this step until all right sides have length two. James L. Hein - Discrete Structures, Logic, and Computability

7. Example Construct a Chomsky normal form for the grammar S → aSb | D D → Dc | Λ. Solution. Step 1: S → aSb | ab | D | Λ D → Dc | c. Step 2: S → aSb | ab | Dc | c | Λ D → Dc | c. Step 3: S → ASB | AB | DC | c | Λ D → DC | c A → a B → b C → c. Step 4: Replace S → ASB by S → AE and E → SB. James L. Hein - Discrete Structures, Logic, and Computability

8. Quiz Construct a Chomsky normal form for the grammar S → aSbb | T | Λ T → cT | d. Solution. Step 1: No change in Λ -productions. Step 2: Remove unit production S → T to obtain S → aSbb | cT | d | Λ T → cT | d. Step 3: Transform right sides of length at least two into strings of nonterminals. S → ASBB | CT | d | Λ T → CT | d A → a B → b C → c. Step 4: Transform right sides into strings of length at most two. S → AD | CT | d | Λ D → SE E→ BB T → CT | d A → a B → b C → c. James L. Hein - Discrete Structures, Logic, and Computability

9. Greibach Normal Form Productions have one of the following forms A → b (b a terminal) A → bD1…Dk S → Λ (if Λ is in the language). Advantage: Any string of length n > 0 can be derived in n steps. James L. Hein - Discrete Structures, Logic, and Computability

10. Transform context-free grammar to Greibach normal form Algorithm (idea). Transform context-free grammar to Greibach normal form. 1. Remove all left-recursion. 2. Remove Λ−productions. (If S → Λ is removed, add it back.) 3. Make substitutions to transform the grammar into the proper form. James L. Hein - Discrete Structures, Logic, and Computability

11. Example Put the following grammar into Greibach normal form. S → AB | Ac | d A → aA | a B→ Ab | c. Solution: Steps 1 and 2 are unnecessary for this grammar. Step 3: Replace A in S → AB | Ac | d with aA | a to obtain S → aAB | aB | aAc | ac | d. Replace A in B→ Ab | c with aA | a to obtain B→ aAb | ab | c. Add the new productions C → c and D → b to obtain the proper form: S → aAB | aB | aAC | aC | d A → aA | a B→ aAD | aD | c C → c D → b. James L. Hein - Discrete Structures, Logic, and Computability

12. Quiz Put the following grammar into Greibach normal form. S → BaS | B B→ cSd | a. Solution: Steps 1 and 2 are unnecessary for this grammar. Step 3: Replace B in S → BaS | B with cSd | a to obtain: S → cSdaS | aaS | cSd | a. Add the new productions A → a and D → d to obtain the proper form: S → cSDAS | aAS | cSD | a A → a D → d B→ cSD | a (Not needed in this example). James L. Hein - Discrete Structures, Logic, and Computability

13. Properties of Context-Free Languages When we know some properties of context-free languages they can help us argue, BWOC, that certain languages are not context-free. James L. Hein - Discrete Structures, Logic, and Computability

14. The Pumping Lemma If L is an infinite context-free language, then any grammar for L must be recursive, so there must be derivations of the following form where u, v, w, x, and y are terminal strings. S ➾+ uNy N ➾+ vNx (where v and x are not both Λ) N ➾+ w. These derivations lead to derivations like S ➾+ uNy ➾+ uvNxy ➾+ uv2Nx2y ➾+ uvkNxky ➾+ uvkwxky ∊L for all k ∊N. This is the basis for the Pumping Lemma: There is an integer m > 0 such that if z ∊L and | z | ≥ m, then z has the form z = uvwxy where 1 ≤ | vx | ≤ | vwx | ≤ m and uvkwxky ∊L for all k ∊N. Note: The number m depends on the grammar as we’ll see in the following example. James L. Hein - Discrete Structures, Logic, and Computability

15. Example Suppose we have the following grammar for {Λ, bbc} ⋃ {abcnd | n ∊N}. S → aNd | bbc | Λ N→ Nc | b. Here are a few derivations: S ➾aNd ➾abd S ➾aNd ➾aNcd ➾abcd S ➾aNd ➾aNcd ➾aNccd ➾abccd S ➾+ abckd for any k in N. For this grammar m = 4 can be used in the pumping lemma because any derivation of a string z with | z | ≥ 4 must use the nonterminal N. For example, if | z | = 8 and z = abcccccd, then the pumping lemma factors z = abcccccd = uvwxy where 1 ≤ | vx | ≤ | vwx | ≤ 4 and uvkwxky ∊L for all k ∊N. In this case let u = a, v = Λ, w = b, x = c, and y = ccccd. James L. Hein - Discrete Structures, Logic, and Computability

16. Example The language L = {anbncn+k | k, n ∊N} is not context-free. Proof: Assume, BWOC, that L is context-free. L is infinite, so pumping lemma applies. Choose z = ambmcm where m is the positive integer from the lemma. Then z = ambmcm = uvwxy where 1 ≤ | vx | ≤ | vwx | ≤ m and uvkwxky ∊L for all k ∊N. Observe neither v nor x can contain distinct letters. For example, if v = …a…b…, then v2 = …a…b……a…b…, which can’t appear as a substring of any string in L. So v and x must be strings of repeated occurrences of a single letter. Now since | vwx | ≤ m, there are two possible places in ambmcmwhere v and x must occur: (1) v and x occur in ambm. (2) v and x occur in bmcm. But we obtain the following contradictions because v and x are not both Λ. (1) Let k = 2 to obtain uv2wx2y = am+ibm+jcm, where i > 0 or j > 0. So uv2wx2y ∉ L (2) Let k = 0 to obtain uwy = ambm-icm–j, where i > 0 or j > 0. So we have uwy ∉ L. These contradictions imply that L is not context-free. QED. James L. Hein - Discrete Structures, Logic, and Computability

17. Example/Quiz Prove that the language L = {ss | s ∊ {a, b}*} is not context-free. Proof: Assume, BWOC, that L is context-free. L is infinite, so pumping lemma applies. Choose z = ambmambmwhere m is the positive integer from the lemma. Then z = ambmambm = uvwxy where 1 ≤ | vx | ≤ | vwx | ≤ m and uvkwxky ∊L for all k ∊N. Now since | vwx | ≤ m, there are three possible places in ambmambmwhere v and x must occur: (1) v and x occur in ambm(on the left of z). (2) v and x occur in bmam(in the center of z). (3) v and x occur in ambm(on the right of z). Notice that v and x can consist only of repetitions a single letter. For example, in case (1) suppose v = aibjfor some i > 0 and j > 0 and x = bnfor some n ≥ 0. Then, letting k = 0, we would obtain uwy = am–ibm–j–nambm, which cannot be in L. The argument is similar for the other cases. So v and x must consist only of repetitions of a single letter. James L. Hein - Discrete Structures, Logic, and Computability

18. Example/Quiz Cont’d We need to find a contradiction in each of the three cases. We’ll do it by using k = 0. This tells us that uwy ∊L. But we obtain the following contradictions because v and x are not both Λ. (1) uwy = am–ibm–jambmwhere either i > 0 or j > 0 So uwy ∉ L, (2) uwy = ambm–iam–jbmwhere either i > 0 or j > 0. So uwy ∉ L. (3) uwy = ambmam–ibm–jwhere either i > 0 or j > 0. So uwy ∉ L. Therefore L is not context-free. QED. Remark: Be careful that the choice of z is not in a context-free sublanguage of L. For example, if we chose z = (ab)m(ab)min the preceding example, we would not get any contradictions. James L. Hein - Discrete Structures, Logic, and Computability

19. The End of Chapter 12 - 4 James L. Hein - Discrete Structures, Logic, and Computability