CMPS 260 Spring 2024
HW #5: Context-free Grammars/Languages; Chomsky Normal Form
Sample Solutions

1. Let G₁ be the following context-free grammar:

S	⟶	aS  |  aHb	(1) (2)
H	⟶	aaHb  |  a	(3) (4)

(a) Offer a precise description of L(G₁), the language generated by G₁.

Solution: L(G₁) = { a^kbⁿ  |  n≥1, k≥2n }

(b) Provide some evidence for the claim that G₁ is unambiguous. To do so, choose a "generic" member of L(G₁), describe its derivation from S, and argue that there is no other way to derive it from S.

Solution: By observing the productions of the grammar, it is clear, with respect to any derivation from the start symbol S, that

• Initially, only productions (1) and (2) are applicable, because S occurs in the sentential form but H does not.
• Once production (2) has been applied, H occurs in the sentential form but S no longer does, making only productions (3) and (4) applicable.
• Once production (4) has been applied, a terminal string has been produced and thus no productions are applicable.

Thus, every derivation of a terminal string from the start symbol S must have this form, for some m≥0 and n≥1:

• m applications of production (1), yielding a^mS, followed by
• one application of production (2), yielding a^maHb, followed by
• n−1 applications of production (3), yielding a^ma(aa)^n-1Hb^n-1b, followed by
• one application of production (4), yielding a^ma(aa)^n-1ab^n-1b, which is the same as a^m+2nbⁿ.

Summarizing, the derivation is

The ⟹^(j,k) notation indicates k applications of production (j).

If, rather than applying (1) m times and (3) n−1 times, we instead applied (1) m' times and (3) n'−1 times, where either m'≠m or n'≠n (or both), the terminal string produced would be x' = a^m'+2n'b^n', which is distinct from x = a^m+2nbⁿ. (Reason: If n'≠n, then the number of occurrences of b in x and x' differ. Meanwhile, if n'=n but m'≠m, then the number of occurrences of a in x and x' differ.)

It follows that for any string generated by G₁, there is a unique derivation by which to produce it. Hence, G₁ is unambiguous.

(c) Present an alternative CFG G'₁ that generates the same language as G₁ but whose only nonterminal symbol is S. Demonstrate that your grammar is ambiguous.

Solution: A grammar that generates L(G₁) but has only one nonterminal symbol is as follows:

To produce a^m+2nbⁿ, where m≥0 and n≥1, one could apply production (1) m times and production (2) n−1 times, in any order, followed by an application of (3). For any values of m and n satisfying m>0 ∧ n>1, there are multiple derivations of that string!

Note: A context-free grammar is ambiguous iff there are distinct derivation trees both yielding the same terminal string. For every grammar, there is a one-to-one correspondence between its derivation trees and its leftmost derivations. For linear grammars, such as those described in this problem (including the solution to part (c)), every derivation is a leftmost derivation. Hence, for linear grammars, the question of ambiguity reduces to the question of whether any string has distinct derivations.

2. Let G₂ be the following context-free grammar:

S	⟶	HS  |  K	(1) (2)
H	⟶	aHb  |  c	(3) (4)
K	⟶	cK  |  c	(5) (6)

Give a precise description of L(G₂), the language generated by G₂.

Solution:

The star/asterisk superscript means, in effect, zero or more occurrences, just as in regular expressions.

3. Present a context-free grammar G₃ that generates the language { a^j b^k c^2k b^j  |  j≥1, k≥0 }

Small Bonus: Present a context-free grammar G'₃ that generates the language { a^j b^k c^2k d^* b^j  |  j≥1, k≥0 }.

The star/asterisk superscript means, in effect, zero or more occurrences, just as in regular expressions.

Solution:

S ⟶ aSb  |  aHb   (1) (2) H ⟶ bHcc  |  λ   (3) (4)	S ⟶ aSb  |  aHDb   (1) (2) H ⟶ bHcc  |  λ   (3) (4) D ⟶ dD  |  λ   (5) (6)
G3	G'3 (bonus)

4. Present a context-free grammar G₄ that generates the language { a^j b^j+k a^{k  |  j≥1, k≥0 }}

Small Bonus: Present a context-free grammar G'₄ that generates the language { a^j c^* b^j+k a^k  |  j≥1, k≥0 }

The star/asterisk superscript means, in effect, zero or more occurrences, just as in regular expressions.

Solution: The small bit of insight that makes this problem easy is to recognize that the language described is just the same as { a^jb^jb^ka^k  |  j≥1, k≥0 }. For the bonus, the language is { a^jc^*b^jb^ka^k  |  j≥1, k≥0 }

S ⟶ HK   (1) H ⟶ aHb  |  ab   (2) (3) K ⟶ bKa  |  λ   (4) (5)	S ⟶ HK   (1) H ⟶ aHb  |  aCb   (2) (3) K ⟶ bKa  |  λ   (4) (5) C ⟶ cC  |  λ   (6) (7)
G4	G'4 (bonus)

5. Consider the following Chomsky Normal Form grammar G₅:

S	⟶	HS  |  a	(1) (2)
H	⟶	KH  |  b	(3) (4)
K	⟶	KS  |  b	(5) (6)

For each of the strings bbaba and babab, use the CYK algorithm to determine whether it is generated by G₅.

Specifically, for each string, fill in the cells of the matrix pictured below so that the cell in row i and column j contains V_i,j = { X ∈ {S,H,K} : X ⟹⁺ w_i,j }, where w_i,j is the substring of w (i.e., the string of interest) beginning with its i-th symbol and ending with its j-th symbol (inclusive). (Your answers are expected to include not only the answer to the question "Is w ∈ L(G)?" but also the correctly filled in table.)

Recall that, for i satisfying 1≤i≤|w|, X ∈ V_i,i iff X → w_i,i is a production in the grammar. Meanwhile, for i and j satisfying 1≤i<j≤|w|, X ∈ V_i,j iff there exists k, where i≤k<j, and nonterminals Y and Z such that Y ∈ V_i,k, Z ∈ V_k+1,j, and X → YZ is a production in the grammar.

Solution:

1 2 3 4 5 +-------+-------+-------+-------+-------+ | H,K | H | S,K | H | S,K | 1 | (b) | (bb) | (bba) |(bbab) |(bbaba)| +-------+-------+-------+-------+-------+ | H,K | S,K | H | S,K | 2 | (b) | (ba) | (bab) |(baba) | +-------+-------+-------+-------+ | S | - | - | 3 | (a) | (ab) | (aba) | +-------+-------+-------+ | H,K | S,K | 4 | (b) | (ba) | +-------+-------+ | S | 5 | (a) | +-------+	1 2 3 4 5 +-------+-------+-------+-------+-------+ | H,K | S,K | H | K,S | H | 1 | (b) | (ba) | (bab) |(baba) |(babab)| +-------+-------+-------+-------+-------+ | S | - | - | - | 2 | (a) | (ab) | (aba) |(abab) | +-------+-------+-------+-------+ | H,K | S,K | H | 3 | (b) | (ba) | (bab) | +-------+-------+-------+ | S | - | 4 | (a) | (ab) | +-------+-------+ | H,K | 5 | (b) | +-------+
for bbaba	for babab

The (1,5) cell in the matrix for bbaba includes S (the start symbol) and thus bbaba ∈ L(G₅). However, the (1,5) cell in the matrix for babab does not include S and thus babab ∉ L(G₅).

6. Use the standard construction to obtain a Chomsky Normal Form grammar that is equivalent to (i.e., generates the same language as) the context-free grammar shown below. Present the resultant CNF grammar.

S	⟶	bHaH  |  c
H	⟶	SaH | baa

Solution:

S ⟶ b'VHaH  |  c VHaH ⟶ HVaH VaH ⟶ a'H H ⟶ SVaH | bVaa VaH ⟶ a'H Vaa ⟶ a'a' a' ⟶ a b' ⟶ b

7. A linear grammar is a CFG in which every production's right-hand side contains at most one nonterminal symbol. A right-linear grammar is a linear grammar in which any such occurrence of a nonterminal symbol must be the last (i.e., rightmost) symbol of that right-hand side.

(a) Describe a construction that, given a right-linear grammar G_R and a linear grammar G_Lin, produces a linear grammar G such that L(G) = L(G_R) · L(G_Lin).

Solution: Because nonterminal symbols can be renamed, without loss of generality we can assume that no nonterminal symbol appears in both grammars. We can also assume that S_R is the start symbol of G_R and S_Lin is the start symbol of G_Lin.

Take G to be the grammar whose start symbol is S_R and whose productions include all those in the two grammars, except that any production in G_R of the form A ⟶ x, where x is a terminal string, is replaced by A ⟶ xS_Lin.

(b) Apply your construction to these particular grammars and show the resulting linear grammar.

SR ⟶ aaK | ca H ⟶ cK | λ K ⟶ abSR | bH	SLin ⟶ caAdc | Ab A ⟶ dcSLind | a
GR	GLin

Solution: The constructed grammar (with start symbol S_R and with the right-hand sides of modified productions in red) is

SR ⟶ aaK | caSLin H ⟶ cK | SLin K ⟶ abSR | bH SLin ⟶ caAdc | Ab A ⟶ dcSLind | a