Let $\Sigma=\{a,b,c,d\}$ and $L=\{a^i b^j c^k d^{\ell}\mid i,j,k,\ell\geq0\}$.
Which of the following constraints ensure(s) that the language $L$ is context-free?

Question

A grammar is given:
$E \rightarrow E + T \mid T$
$T \rightarrow T * F \mid F$
$F \rightarrow (E) \mid id$
For the input: $id + id * id$, what is the correct interpretation?

Answer 1

Every string of $L=\{a^ib^jc^kd^l : i,j,k,l\ge 0\}$ has the letters in the strict order a...a b...b c...c d...d. To test each option, try to write a context-free grammar (CFG) for it directly; if a CFG exists, the option is correct.

Option A ($i+k=j+l$): Rewrite as $(i+k)-(j+l)=0$. Treat a and c as +1 tokens and b and d as -1 tokens; since they appear in the fixed order a-block, b-block, c-block, d-block, a single counter (stack) can be incremented on every a or c and decremented on every b or d, switching stack symbols nondeterministically whenever the running total would go negative. The string is valid iff the counter returns to zero, so a PDA exists. Context-free.

Option B ($i=k$ and $j=l$): This restricts $L$ to $\{a^nb^mc^nd^m\}$. The outer pair (a,c) must match in count AND, separately, the inner pair (b,d) must match in count, but none of the equal blocks sit next to each other. A CFG would need two independent nested counters running at once, which one stack cannot support; the standard pumping-lemma split across the non-adjacent blocks breaks either $i=k$ or $j=l$ for long strings. Not context-free.

Option C ($i=l$ and $j=k$): Substituting gives strings of the form $a^i b^j c^j d^i$, which is nested (outer a/d pair, inner b/c pair). Grammar: $S \to aSd \mid A$ and $A \to bAc \mid \epsilon$. This is a textbook Dyck-style CFG, so this option is context-free.

Option D ($i+j=k+l$): The combined length of the first two blocks must equal the combined length of the last two blocks. Grammar: $S \to (a\mid b)\,S\,(c\mid d) \mid \epsilon$ pairs one symbol from $\{a,b\}$ with one from $\{c,d\}$ at each derivation step, generating exactly the strings with $i+j=k+l$. Context-free.

So the constraints that preserve context-freeness are A, C and D; only B fails because it demands two simultaneous non-adjacent count matches, which needs more than one stack.

Final answer: A, C, D

Let \(\Sigma=\{a,b,c,d\}\) and \(L=\{a^i b^j c^k d^{\ell}\mid i,j,k,\ell\geq0\}\).
Which of the following constraints ensure(s) that the language \(L\) is context-free?

Show Hint

The Correct Option is A, C, D

Solution and Explanation

Top Questions on Context Free Languages and Pushdown Automata

Questions Asked in GATE CS exam

Let \(\Sigma=\{a,b,c,d\}\) and \(L=\{a^i b^j c^k d^{\ell}\mid i,j,k,\ell\geq0\}\).Which of the following constraints ensure(s) that the language \(L\) is context-free?