GATE CS

Context-free languages

Closure properties and the pumping lemma

Context-free languages are closed under some operations (union, concatenation, star, reversal, homomorphism, substitution, inverse homomorphism, intersection with a regular language) and not under others (intersection, complement). The pumping lemma for CFLs is the standard tool for showing a language is not context-free. In GATE questions you use these closure facts, the decidability facts and the pumping argument to place a given language in the right class: regular, deterministic context-free, context-free or not context-free.

8 previous-year questions · 6 solved · 5 practice questions

Conventions and closure properties

Σ is the input alphabet, ε the empty string, |w| the length of w, and $\overline{L}$ the complement of L with respect to $Σ^{*}$ . CFL means context-free language. DCFL means a language accepted by a deterministic pushdown automaton (DPDA) by final state.

For nondeterministic PDAs, acceptance by final state and acceptance by empty stack give the same class, the CFLs. For DPDAs they differ: a DPDA accepting by empty stack accepts only prefix-free languages (no string in the language is a proper prefix of another), so DCFLs are defined by acceptance by final state.

Closure means: if the operands are in the class, the result is also in the class. To show closure, build a grammar or automaton for the result. To show non-closure, give one counterexample.

Operation	CFL	DCFL	Regular
Union	Closed	Not closed	Closed
Concatenation	Closed	Not closed	Closed
Kleene star	Closed	Not closed	Closed
Reversal	Closed	Not closed	Closed
Homomorphism	Closed	Not closed	Closed
Substitution (each symbol replaced by a language of the same class)	Closed	Not closed	Closed
Inverse homomorphism	Closed	Closed	Closed
Intersection	Not closed	Not closed	Closed
Complement	Not closed	Closed	Closed
Intersection with a regular language	Closed	Closed	Closed
Union with a regular language	Closed	Closed	Closed
Difference $L - R$ (L in the class, R regular)	Closed	Closed	Closed
Difference $R - L$ (R regular, L in the class)	Not closed	Closed	Closed

Grammar constructions for CFLs. If $S_{1}$ and $S_{2}$ are the start symbols (with disjoint variables) and S is a new start symbol, then union uses $S \to S_{1} ∣ S_{2}$ , concatenation uses $S \to S_{1} S_{2}$ , star uses $S \to S_{1} S ∣ ε$ , and reversal reverses the right side of every production.

Why intersection and complement fail

Take $L_{1} = {a^{n} b^{n} c^{m} ∣ n, m \geq 0}$ and $L_{2} = {a^{n} b^{m} c^{m} ∣ n, m \geq 0}$ . Both are context-free (indeed deterministic context-free). Their intersection is ${a^{n} b^{n} c^{n} ∣ n \geq 0}$ , which is not context-free. So CFLs are not closed under intersection.

Complement fails by De Morgan. $L_{1} \cap L_{2} = \overline{\overline{L_{1}} \cup \overline{L_{2}}}$ . If CFLs were closed under complement, then since they are closed under union, they would be closed under intersection, which is false. So CFLs are not closed under complement.

INTERSECTION WITH A REGULAR LANGUAGE

L CFL, R regular \Rightarrow L \cap R is a CFL

Proved by running a PDA and a DFA together; the same holds for DCFLs. Hence $L - R = L \cap \overline{R}$ is a CFL, but $R - L$ need not be.

EXAM TIP

Use intersection with a regular language as a tool to show a language is not context-free. If $L \cap a^{*} b^{*} c^{*}$ is ${a^{n} b^{n} c^{n}}$ , then L cannot be context-free, because otherwise the intersection would be context-free.

EXAM TIP

A CFL whose complement is not a CFL exists. Over ${a, b, c}$ , the complement of ${a^{n} b^{n} c^{n} ∣ n \geq 0}$ (with respect to ${a, b, c}^{*}$ ) is context-free: it is the union of the strings not in $a^{*} b^{*} c^{*}$ and the strings $a^{i} b^{j} c^{k}$ with $i \neq = j$ or $j \neq = k$ . But its complement, ${a^{n} b^{n} c^{n}}$ , is not context-free.

Deterministic CFLs

DCFLs sit strictly between regular languages and CFLs: $Regular \subset DCFL \subset CFL$ . ${a^{n} b^{n}}$ is a DCFL that is not regular. ${w w^{R} ∣ w \in {a, b}^{*}}$ is a CFL that is not a DCFL; informally, a DPDA cannot know where the middle is.

DCFLs are closed under complement. This is the key difference from general CFLs.
DCFLs are closed under intersection with a regular language, union with a regular language, difference with a regular language (either order) and inverse homomorphism.
DCFLs are not closed under union, intersection, concatenation, Kleene star, reversal or homomorphism.
Standard example of union failing: ${a^{i} b^{j} c^{k} ∣ i = j or j = k}$ is a union of two DCFLs, is a CFL, and is not a DCFL.
Every DCFL has an unambiguous grammar, so an inherently ambiguous CFL is never a DCFL.
The converse fails: ${w w^{R}}$ has the unambiguous grammar $S \to a S a ∣ b S b ∣ ε$ but is not a DCFL.

EXAM TIP

If a language is a DCFL then its complement is a DCFL, hence a CFL. So if you can show that $\overline{L}$ is not context-free, then L is not a DCFL (it may still be a CFL if you are told so).

Decision properties

Here the language is given by a context-free grammar G. A property is decidable if some algorithm always halts with the correct yes/no answer.

Question for CFGs	Decidable?	Idea
Is $L (G) = \emptyset$ ?	Yes	Mark variables that derive a terminal string; check whether the start symbol is marked
Is $L (G)$ finite?	Yes	Remove useless symbols, ε- and unit productions; draw an edge A → B when B appears on the right of an A-production; L(G) is finite if and only if this graph has no cycle
Is $w \in L (G)$ ?	Yes	CYK algorithm on a Chomsky normal form grammar, time $O (∣ w ∣^{3})$ for a fixed grammar
Is $L (G) = Σ^{*}$ (universality)?	No	Undecidable
Is $L (G_{1}) = L (G_{2})$ (equivalence)?	No	Undecidable
Is $L (G_{1}) \cap L (G_{2}) = \emptyset$ ?	No	Undecidable
Is G ambiguous?	No	Undecidable
Is $L (G)$ regular?	No	Undecidable
Is $L (G)$ deterministic context-free?	No	Undecidable
Is $L (G) \subseteq L (R)$ for R regular?	Yes	Test $L (G) \cap \overline{R} = \emptyset$
Is $L (R) \subseteq L (G)$ for R regular?	No	With $R = Σ^{*}$ this is universality

EXAM TIP

Universality and equivalence are decidable for regular languages and for DCFLs given by DPDAs (for universality, complement the DPDA and test emptiness; equivalence of DPDAs is decidable, a deep result). Inclusion $L (D_{1}) \subseteq L (D_{2})$ and emptiness of $L (D_{1}) \cap L (D_{2})$ stay undecidable even for DPDAs. All of these are undecidable for general CFGs. Check which class the question gives you.

The pumping lemma for CFLs

PUMPING LEMMA FOR CONTEXT-FREE LANGUAGES

L CFL \Rightarrow \exists p \geq 1 \forall z \in L with ∣ z ∣ \geq p \exists u, v, w, x, y : z = uv w x y, ∣ v w x ∣ \leq p, ∣ v x ∣ \geq 1, u v^{i} w x^{i} y \in L for all i \geq 0

A necessary condition only: it can prove a language is not context-free, never that it is. If L is generated by a Chomsky normal form grammar with k variables, $p = 2^{k}$ works.

Compare the regular pumping lemma, where $z = x y z^{'}$ with $∣ x y ∣ \leq p$ and $∣ y ∣ \geq 1$ . For CFLs the string splits into five parts, v and x are pumped together (the same number of times), and the window $v w x$ is short. The usual consequence is that two positions of z that are p or more apart cannot both lie inside $v w x$ .

Assume L is context-free and let p be its pumping length.
Choose a string $z \in L$ with $∣ z ∣ \geq p$ , built from p, so that the argument works for every p.
Consider every split $z = uv w x y$ with $∣ v w x ∣ \leq p$ and $∣ v x ∣ \geq 1$ . Do not pick the split; the opponent does.
For each split (grouped into a few cases by where $v w x$ lies), find an i such that $u v^{i} w x^{i} y \in / L$ . Usually i is 0 or 2; pumping down ( $i = 0$ ) is allowed and often needed.
Since every split fails, the assumption is false, and L is not context-free.

Language	Not context-free? String to choose (or reason)
${a^{n} b^{n} c^{n}}$	Yes; $z = a^{p} b^{p} c^{p}$
${w w ∣ w \in {a, b}^{*}}$	Yes; $z = a^{p} b^{p} a^{p} b^{p}$
${a^{i} b^{j} c^{k} ∣ i < j < k}$	Yes; $z = a^{p} b^{p + 1} c^{p + 2}$
${a^{n^{2}}}$ , ${a^{2^{n}}}$ , ${a^{q} ∣ q prime}$	Yes; every CFL over a one-letter alphabet is regular, and these are not regular
${a^{n} b^{n} c^{m}}$ , ${a^{n} b^{m} c^{n}}$ , ${w w^{R}}$	No; these are context-free

EXAM TIP

The lemma gives only a necessary condition. A language that satisfies it may still fail to be context-free, so you can never use it to show a language is context-free.

Worked examples

WORKED EXAMPLE

Show that $L = {a^{n} b^{n} c^{2 n} ∣ n \geq 0}$ is not context-free.

Solution

Assume L is context-free with pumping length p, and take $z = a^{p} b^{p} c^{2 p}$ , so $∣ z ∣ \geq p$ .
Let $z = uv w x y$ with $∣ v w x ∣ \leq p$ and $∣ v x ∣ \geq 1$ . The window $v w x$ is too short to contain both an a and a c, since the p b's lie between them.
Case 1: $v x$ contains an a. Then it contains no c. Pumping with $i = 2$ gives at least $p + 1$ a's but still $2 p$ c's, while membership needs the number of c's to be twice the number of a's, which is at least $2 p + 2$ . So the pumped string is not in L.
Case 2: $v x$ contains no a. Then the number of a's stays p, so membership needs exactly p b's and $2 p$ c's. But $v x$ contains a b or a c, so pumping with $i = 2$ raises one of these counts. The pumped string is not in L.
Every split fails, so L is not context-free.

Answer: L is not context-free.

WORKED EXAMPLE

Classify $L = {w \in {a, b, c}^{*} ∣ n_{a} (w) = n_{b} (w) = n_{c} (w)}$ , where $n_{x} (w)$ is the number of occurrences of x in w.

Solution

Intersect L with $a^{*} b^{*} c^{*}$ (a regular language).
A string in $a^{*} b^{*} c^{*}$ with equal numbers of a, b and c has the form $a^{n} b^{n} c^{n}$ , so $L \cap a^{*} b^{*} c^{*} = {a^{n} b^{n} c^{n}}$ .
This intersection is not context-free (pumping lemma).
CFLs are closed under intersection with a regular language, so if L were context-free then the intersection would be. It is not, so L is not context-free.

Answer: L is not context-free.

Classifying a language

Language	Regular	DCFL	CFL
$(a + b)^{*} abb$	Yes	Yes	Yes
${a^{n} b^{n}}$	No	Yes	Yes
${w c w^{R} ∣ w \in {a, b}^{*}}$	No	Yes	Yes
${w w^{R}}$	No	No	Yes
${a^{i} b^{j} c^{k} ∣ i = j or j = k}$	No	No	Yes
${w w}$	No	No	No
${a^{n} b^{n} c^{n}}$	No	No	No

These are rules of thumb; confirm each with a construction or a proof.
First ask whether a finite amount of memory suffices (regular). If it does, say so.
If one counter or stack comparison matches a pattern read left to right ( $a^{n} b^{n}$ , a marker in the middle), it is usually a DCFL.
If the machine must guess the middle or choose between two comparisons (palindromes, 'i = j or j = k'), it is a CFL and usually not a DCFL.
If two comparisons must hold at once and they cross or share a block ( $a^{n} b^{n} c^{n}$ , $a^{n} b^{m} c^{n} d^{m}$ ), or a string must be compared with a copy of itself ( $w w$ ), it is not context-free. Nested or side-by-side comparisons ( $a^{n} b^{m} c^{m} d^{n}$ , $a^{n} b^{n} c^{m} d^{m}$ ) are still context-free.
Confirm with closure: intersect with a regular language to reduce the question to a standard one.

EXAM TIP

${w w}$ is not context-free but its complement over ${a, b}$ is context-free. Do not assume that a language and its complement are in the same class unless the class is closed under complement (regular, DCFL).

PREVIOUS-YEAR QUESTIONS

8 questions from the CS papers

Try each question first, then open the answer. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE CSE 2026 Set 2 · GATE CSE 2025 Set 1 · GATE CSE 2025 Set 2 · GATE CSE 2022 · GATE CSE 2020 · GATE CSE 2018

GATE 2026 CS2 Q.48

GATE 2026 · CS2Q.48 · 2 marks · MSQ

Let $Σ = {a, b, c, d}$ and let $L = {a^{i} b^{j} c^{k} d^{ℓ} ∣ i, j, k, ℓ \geq 0}$ .

Which of the following constraints ensure(s) that the language $L$ is context-free?

(A)
$i + k = j + ℓ$
(B)
$i = k$ and $j = ℓ$
(C)
$i = ℓ$ and $j = k$
(D)
$i + j = k + ℓ$

Answer and solution

GATE 2026 CS2 Q.48: answer and solution

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GATE 2025 CS1 Q.45

GATE 2025 · CS1Q.45 · 2 marks · MCQ

Consider the following two languages over the alphabet ${a, b, c}$ , where $m$ and $n$ are natural numbers.

$L_{1} = {a^{m} b^{m} c^{m + n} ∣ m, n \geq 1}$
$L_{2} = {a^{m} b^{n} c^{m + n} ∣ m, n \geq 1}$

Which ONE of the following statements is CORRECT?

(A)
Both $L_{1}$ and $L_{2}$ are context-free languages.
(B)
$L_{1}$ is a context-free language but $L_{2}$ is not a context-free language.
(C)
$L_{1}$ is not a context-free language but $L_{2}$ is a context-free language.
(D)
Neither $L_{1}$ nor $L_{2}$ are context-free languages.

Official answer

Official answer: CSource: Official answer key, GATE 2025 CS1 (IIT Roorkee)

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GATE 2025 CS2 Q.30

GATE 2025 · CS2Q.30 · 1 mark · MSQ

Consider the two lists List I and List II given below:

List I (i) Context free languages (ii) Recursive languages (iii) Regular languages List II (a) Closed under union (b) Not closed under complementation (c) Closed under intersection

For matching of items in List I with those in List II, which of the following option(s) is/are CORRECT?

(A)
(i) – (a), (ii) – (b), and (iii) – (c)
(B)
(i) – (b), (ii) – (a), and (iii) – (c)
(C)
(i) – (b), (ii) – (c), and (iii) – (a)
(D)
(i) – (a), (ii) – (c), and (iii) – (b)

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GATE 2022 CS Q.47

GATE 2022 · CSQ.47 · 2 marks · MSQ

Consider the following languages:

$L_{1} = {a^{n} w a^{n} ∣ w \in {a, b}^{*}}$

$L_{2} = {w x w^{R} ∣ w, x \in {a, b}^{*}, ∣ w ∣, ∣ x ∣ > 0}$

Note that $w^{R}$ is the reversal of the string $w$ . Which of the following is/are TRUE?

(A)
$L_{1}$ and $L_{2}$ are regular.
(B)
$L_{1}$ and $L_{2}$ are context-free.
(C)
$L_{1}$ is regular and $L_{2}$ is context-free.
(D)
$L_{1}$ and $L_{2}$ are context-free but not regular.

Answer and solution

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GATE 2022 CS Q.48

GATE 2022 · CSQ.48 · 2 marks · MSQ

Consider the following languages:
$L_{1} = {w w ∣ w \in {a, b}^{*}}$
$L_{2} = {a^{n} b^{n} c^{m} ∣ m, n \geq 0}$
$L_{3} = {a^{m} b^{n} c^{n} ∣ m, n \geq 0}$
Which of the following statements is/are FALSE?

(A)
$L_{1}$ is not context-free but $L_{2}$ and $L_{3}$ are deterministic context-free.
(B)
Neither $L_{1}$ nor $L_{2}$ is context-free.
(C)
$L_{2}$ , $L_{3}$ and $L_{2} \cap L_{3}$ all are context-free.
(D)
Neither $L_{1}$ nor its complement is context-free.

Answer and solution

GATE 2022 CS Q.48: answer and solution

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GATE 2020 CS Q.10

GATE 2020 · CSQ.10 · 1 mark · MCQ

Consider the language $L = {a^{n} ∣ n \geq 0} \cup {a^{n} b^{n} ∣ n \geq 0}$ and the following statements.

I. $L$ is deterministic context-free.

II. $L$ is context-free but not deterministic context-free.

III. $L$ is not LL( $k$ ) for any $k$ .

Which of the above statements is/are TRUE?

(A)
I only
(B)
II only
(C)
I and III only
(D)
III only

Official answer

Official answer: CSource: Official answer key, GATE 2020 CS (IIT Delhi)

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GATE 2020 CS Q.32

GATE 2020 · CSQ.32 · 2 marks · MCQ

Consider the following languages.
$L_{1} = {w x y x ∣ w, x, y \in (0 + 1)^{+}}$
$L_{2} = {x y ∣ x, y \in (a + b)^{*}, ∣ x ∣ = ∣ y ∣, x \neq = y}$
Which one of the following is TRUE?

(A)
$L_{1}$ is regular and $L_{2}$ is context-free.
(B)
$L_{1}$ is context-free but not regular and $L_{2}$ is context-free.
(C)
Neither $L_{1}$ nor $L_{2}$ is context-free.
(D)
$L_{1}$ is context-free but $L_{2}$ is not context-free.

Answer and solution

GATE 2020 CS Q.32: answer and solution

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GATE 2018 CS Q.35

GATE 2018 · CSQ.35 · 2 marks · MCQ

Consider the following languages:
I. ${a^{m} b^{n} c^{p} d^{q} ∣ m + p = n + q, where m, n, p, q \geq 0}$
II. ${a^{m} b^{n} c^{p} d^{q} ∣ m = n and p = q, where m, n, p, q \geq 0}$
III. ${a^{m} b^{n} c^{p} d^{q} ∣ m = n = p and p \neq = q, where m, n, p, q \geq 0}$
IV. ${a^{m} b^{n} c^{p} d^{q} ∣ mn = p + q, where m, n, p, q \geq 0}$
Which of the languages above are context-free?

(A)
I and IV only
(B)
I and II only
(C)
II and III only
(D)
II and IV only

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OTHER EXAMS

ISRO: 1 question

From ISRO's Scientist/Engineer written test: ISRO CS 2023. They are not counted in the GATE numbers on this page.

ISRO CS 2023 Q.59

ISRO 2023 · CSQ.59 · 1 mark · MCQ

Which of the following statements is FALSE:

(A)
The intersection of a regular language and a context-free language is context-free
(B)
The intersection of a regular language and context-free language is regular
(C)
The union of two context-free languages is context-free
(D)
The union of two regular languages is regular

Answer and solution

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PRACTICE

5 questions to test yourself

PRACTICE 11 mark · MCQ

Which of the following languages over ${a, b, c}$ is NOT context-free?

(A)
${a^{n} b^{m} c^{n} ∣ n, m \geq 0}$
(B)
${w w^{R} ∣ w \in {a, b}^{*}}$
(C)
${a^{i} b^{j} c^{k} ∣ i = j or j = k}$
(D)
${a^{i} b^{j} c^{k} ∣ i \leq j \leq k}$

Answer and solution

Answer: (D)

A: generated by $S \to a S c ∣ B$ , $B \to b B ∣ ε$ , so context-free.
B: generated by $S \to a S a ∣ b S b ∣ ε$ , so context-free.
C: union of ${a^{n} b^{n} c^{k}}$ and ${a^{i} b^{n} c^{n}}$ , both context-free, so context-free.
D: take $z = a^{p} b^{p} c^{p}$ . The window $v w x$ cannot contain both an a and a c.
Case 1: $v x$ contains an a (so no c). Pump up with $i = 2$ : the a count exceeds p while the c count stays p, violating $i \leq k$ .
Case 2: $v x$ contains a b but no a (it may also contain c's). Pump down with $i = 0$ : the b count drops below p while the a count stays p, violating $i \leq j$ .
Case 3: $v x$ contains only c's. Pump down with $i = 0$ : the c count drops below p while the b count stays p, violating $j \leq k$ .
Every split fails, so D is not context-free.

PRACTICE 22 marks · MSQ

Let $L_{1} = {a^{n} b^{n} c^{m} ∣ n, m \geq 0}$ and $L_{2} = {a^{n} b^{m} c^{m} ∣ n, m \geq 0}$ . Which statements are true?

(A)
$L_{1} \cap L_{2}$ is context-free
(B)
$L_{1} \cup L_{2}$ is context-free
(C)
$L_{1}$ and $L_{2}$ are both deterministic context-free languages
(D)
$L_{1} \cap L_{2}$ is not context-free

Answer and solution

Answer: (B), (C), (D)

$L_{1} \cap L_{2} = {a^{n} b^{n} c^{n}}$ , which is not context-free, so A is false and D is true.
The union of two CFLs is a CFL, so B is true.
A DPDA for $L_{1}$ pushes for a's, pops for b's, checks the bottom-of-stack marker, then reads any number of c's and accepts by final state. For $L_{2}$ it reads the a's without pushing, pushes for b's and pops for c's. Neither ever has to guess, so both are deterministic and C is true.

PRACTICE 32 marks · NAT

Let $L = {w \in {a, b}^{*} ∣ n_{a} (w) = n_{b} (w)}$ and let R be the regular language denoted by $a^{*} b^{*} a^{*}$ . How many strings of length exactly 6 belong to $L \cap R$ ? (Here $n_{x} (w)$ is the number of occurrences of x in w.)

Answer and solution

Answer: $4$

A string of length 6 in L has three a's and three b's.
A string in R has the form $a^{i} b^{j} a^{k}$ , so all its b's form one block: $j = 3$ .
Then $i + k = 3$ , giving $(i, k) = (0, 3), (1, 2), (2, 1), (3, 0)$ .
Each choice gives a distinct string (bbbaaa, abbbaa, aabbba, aaabbb), so the count is 4.

PRACTICE 42 marks · MCQ

Let L be a context-free language and R a regular language over the same alphabet. Which of the following languages is always context-free?

(A)
$\overline{L}$
(B)
$R - L$
(C)
$L - R$
(D)
$\overline{L \cup R}$

Answer and solution

Answer: (C)

$L - R = L \cap \overline{R}$ , and $\overline{R}$ is regular, so $L - R$ is context-free by closure under intersection with a regular language.
A: CFLs are not closed under complement.
B: $R - L = R \cap \overline{L}$ ; taking $R = Σ^{*}$ gives $\overline{L}$ , which need not be context-free.
D: $\overline{L \cup R} = \overline{L} \cap \overline{R}$ ; with $R = \emptyset$ this is $\overline{L}$ , again not always context-free.

PRACTICE 51 mark · MCQ

For context-free grammars $G$ , $G_{1}$ , $G_{2}$ and a string $w$ , which of the following problems is undecidable?

(A)
Is $L (G) = \emptyset$ ?
(B)
Is $L (G)$ finite?
(C)
Is $w \in L (G)$ ?
(D)
Is $L (G_{1}) = L (G_{2})$ ?

Answer and solution

Answer: (D)

Emptiness is decidable by marking variables that derive terminal strings.
Finiteness is decidable from the dependency graph after cleaning the grammar.
Membership is decidable with the CYK algorithm.
Equivalence of two context-free grammars is undecidable.

OFFICIAL GATE 2027 SYLLABUS WORDING

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GATE CS

context-free languages, pumping lemma

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