Question

Let \(\Sigma = \{a, b, c, d\}\) and let \(L = \{a^i b^j c^k d^l \mid i, j, k, l \geq 0\}\). 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. \(i+j = k+l\)

Hint:

Check whether each numeric constraint can be verified with a single stack acting as one running counter, or whether it forces two independent counts to be remembered at the same time.

Answer 1

The Correct Option is A, C, D

Step 1: Understand what the question is really asking.
The base set \(L=\{a^ib^jc^kd^l \mid i,j,k,l\geq 0\}\) written on its own, with no extra restriction linking \(i,j,k,l\), is just all strings of the form \(a^*b^*c^*d^*\), which is a regular language. Each option below adds an EXTRA numeric constraint tying the four exponents together, and we must check, for each such constraint, whether the resulting restricted language can still be recognised by a pushdown automaton, that is, a machine with a single stack.

Step 2: Recall the tool for this kind of check.
A pushdown automaton has exactly one stack, so it can compare at most one running count against another running count, as long as the two counts do not have to be remembered independently at the same time. If a constraint forces the machine to remember two separate counts simultaneously and match each against a different, non adjacent, block later, that constraint typically breaks context-freeness. This is the same idea used to show that crossing style constraints such as those behind \(a^nb^nc^nd^n\) are not context-free, while nested, parenthesis style constraints are.

Step 3: Check option (A), \(i+k=j+l\).
Read the string in order \(a^ib^jc^kd^l\) and use a single stack as an up down counter. Push one symbol for every \(a\) and every \(c\), which counts as adding, and pop one symbol for every \(b\) and every \(d\), which counts as subtracting. Because the a's and c's always occur before the pops from the b's and d's are needed to finish, the stack height after reading the whole string is exactly \((i+k)-(j+l)\). Accept only if the stack is empty at the end. This single counter design is a legal pushdown automaton, so the language for constraint (A) is context-free.

Step 4: Check option (B), \(i=k\) and \(j=l\).
Now the requirement links the FIRST block to the THIRD block (\(i=k\)) and, separately, the SECOND block to the FOURTH block (\(j=l\)). To verify \(i=k\), the machine must still remember \(i\) while it is busy reading the b's, which need their own memory of \(j\) to compare against \(l\) later. A single stack cannot hold two independent, growing counts and pop them back in this criss cross order, because whichever count is pushed on top blocks access to the count underneath. This is the same obstruction that makes languages like \(a^nb^mc^nd^m\) provably not context-free by the pumping lemma: pumping any one block of a string like \(a^pb^pc^pd^p\) with a large \(p\) breaks either the first-third match or the second-fourth match. So the constraint in (B) does NOT give a context-free language.

Step 5: Check option (C), \(i=l\) and \(j=k\).
Here the outer blocks match each other (\(i=l\)) and the inner blocks match each other (\(j=k\)), and the blocks are already arranged in the nested order \(a^i\,b^j\,c^k\,d^l\). This is exactly the shape of a balanced bracket language: every \(a\) is an opening bracket matched by a \(d\) at the very end, and every \(b\) is an opening bracket matched by a \(c\) soon after. A simple grammar generates exactly this language:
\[ S \rightarrow aSd \mid T, \qquad T \rightarrow bTc \mid \varepsilon \]
Since a grammar exists, constraint (C) is context-free.

Step 6: Check option (D), \(i+j=k+l\).
This has the same structure as option (A): the first two blocks \(a^ib^j\) form one running total and the last two blocks \(c^kd^l\) form the other, and the split between adding and subtracting happens at a single fixed point in the string, right after the b's finish and the c's begin. Push one symbol for every \(a\) and every \(b\), then pop one symbol for every \(c\) and every \(d\), and accept when the stack is empty at the end. This is again a legal single counter pushdown automaton, so constraint (D) is context-free.

Step 7: Final answer.
Constraints (A), (C), and (D) each keep the language context-free; constraint (B) creates a crossing dependency that cannot be checked with a single stack.
\[ \boxed{(A),\ (C),\ (D)} \]