Question

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?

• A. 𝑖+ 𝑘 = 𝑗 + ℓ
• B. 𝑖= 𝑘 and 𝑗 = ℓ
• C. 𝑖 = ℓ and 𝑗 = 𝑘
• D. 𝑖+ 𝑗 = 𝑘 + ℓ

Hint:

Check whether each constraint can be enforced by a single PDA stack: constraints linking adjacent blocks (sums of consecutive blocks) or nested outer-inner blocks stay context-free, but two independent equalities between non-adjacent same-position blocks, such as i=k and j=l together, need two counters at once and break context-freeness.

Answer 1

The Correct Option is A, C, D

The language is \(L = \{a^i b^j c^k d^l : i,j,k,l \ge 0\}\), where every string of \(L\) has the fixed block order a's, then b's, then c's, then d's, over \(\Sigma = \{a,b,c,d\}\). We must check, for each extra constraint on \(i,j,k,l\), whether the resulting language can still be recognized by a pushdown automaton (single stack).

Step 1: Option A - \(i+k=j+l\).
Because the letters always occur in the fixed order \(a^i b^j c^k d^l\), we can build a PDA with one stack symbol \(X\): push \(X\) while reading each \(a\) and each \(c\), and pop \(X\) while reading each \(b\) and each \(d\), switching nondeterministically to a second symbol \(Y\) for excess b's if \(j>i\), and popping \(Y\) first when c's arrive. The string is accepted exactly when the stack is empty after the last \(d\), which happens iff \(i+k-j-l=0\), i.e. \(i+k=j+l\). A single stack suffices, so this constraint keeps \(L\) context-free.

Step 2: Option B - \(i=k\) and \(j=l\).
This forces the language \(\{a^i b^j c^i d^j\}\), where the first block must match the third block AND, independently, the second block must match the fourth block, while the equal blocks are never adjacent to each other. A single stack can preserve at most one such correlation cleanly across the string; keeping the count of a's alive to compare with c's while also keeping the count of b's alive to compare with d's forces two independent tallies at once, which a single LIFO stack cannot do. The pumping lemma for context-free languages confirms this: pumping any sufficiently long string breaks either \(i=k\) or \(j=l\). Hence this constraint does NOT keep \(L\) context-free.

Step 3: Option C - \(i=l\) and \(j=k\).
With \(j=k\), the middle two blocks become \(b^jc^j\), and with \(i=l\) the outer blocks become \(a^i \dots d^i\). So the string has the nested form \(a^i(b^jc^j)d^i\). This is generated by the context-free grammar \[S \rightarrow aSd \mid A, \qquad A \rightarrow bAc \mid \varepsilon\] which is a standard nested-matching (Dyck-like) grammar. Since a CFG generates it, this constraint keeps \(L\) context-free.

Step 4: Option D - \(i+j=k+l\).
Here the first two (adjacent) blocks \(a^ib^j\) together must have the same length as the last two (adjacent) blocks \(c^kd^l\). A single stack can push one symbol for every \(a\) or \(b\) read, then pop one symbol for every \(c\) or \(d\) read; the string is accepted iff the stack empties exactly at the end, which holds iff \(i+j=k+l\). Equivalently, the grammar \[S \rightarrow (a\mid b)\,S\,(c\mid d) \mid \varepsilon\] generates precisely this sub-language of \(L\), so it is context-free.

Conclusion: Options A, C and D each admit a single-stack (or CFG) construction, so they preserve context-freeness, while option B requires two independent, non-adjacent counter matches that cannot be realized with one stack.

\[\boxed{\text{Correct options: A, C, D}}\]