Question:hard
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?
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?
Show 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.
Updated On: Jul 22, 2026
- \(i+k = j+l\)
- \(i=k\) and \(j=l\)
- \(i=l\) and \(j=k\)
- \(i+j = k+l\)
Show Solution
The Correct Option is A, C, D
Solution and Explanation
Step 1: What context-free means here.
A language is context-free if some pushdown automaton, a finite control plus one stack, accepts it, or equivalently if some context-free grammar generates it. We have four candidate extra rules on $i,j,k,l$ for strings of the form $a^ib^jc^kd^l$, and we test each one separately.
Step 2: Test $i+k=j+l$, option A, with a stack machine.
Build a machine that pushes a marker for each $a$ and each $c$, and pops a marker for each $b$ and each $d$, reading the string left to right in the fixed order a's, then b's, then c's, then d's. Since the letters that add to the stack and the letters that remove from it never need to be tracked at the same time as a second, independent quantity, the stack empties exactly when $i+k=j+l$. This works, so option A is context-free.
Step 3: Test $i+j=k+l$, option D, the same way.
Push for a's and b's, pop for c's and d's. Again only one running total is ever needed at a time, so this is context-free by the same one stack argument as option A.
Step 4: Test $i=l$ and $j=k$, option C, with a grammar.
Write $S \to aSd \mid T$ and $T \to bTc \mid \varepsilon$. Expanding $S$ a number of times matches every leading $a$ with a trailing $d$, giving $i=l$, and expanding $T$ matches every $b$ with a following $c$, giving $j=k$. Both matchings are properly nested inside each other, not crossed, so an ordinary context-free grammar builds the language directly. Option C is context-free.
Step 5: Test $i=k$ and $j=l$, option B, with the pumping lemma.
Suppose this language were context-free with pumping length $p$. Take the string $a^pb^pc^pd^p$, which satisfies $i=k=p$ and $j=l=p$. The pumping lemma lets us split it as $uvwxy$ with $|vwx|\leq p$ and $|vx|\geq 1$, then pump $v,x$ together. Because $|vwx|\leq p$, the piece $vwx$ can touch at most two of the four letter blocks, it cannot reach across three or four blocks at once. Pumping up then increases the count of at most two of the four letters and leaves the other two unchanged, which always destroys either $i=k$ or $j=l$, since those two equalities pair up NON adjacent blocks, block one with block three, block two with block four. So no valid split exists, and the language is not context-free. Option B fails.
Step 6: Collect the result.
Only the constraints that let a single stack track one running total, A and D, or that produce a properly nested bracket structure, C, survive. The crossing constraint in B does not.
\[ \boxed{\text{Options A, C, D are context-free; option B is not}} \]
A language is context-free if some pushdown automaton, a finite control plus one stack, accepts it, or equivalently if some context-free grammar generates it. We have four candidate extra rules on $i,j,k,l$ for strings of the form $a^ib^jc^kd^l$, and we test each one separately.
Step 2: Test $i+k=j+l$, option A, with a stack machine.
Build a machine that pushes a marker for each $a$ and each $c$, and pops a marker for each $b$ and each $d$, reading the string left to right in the fixed order a's, then b's, then c's, then d's. Since the letters that add to the stack and the letters that remove from it never need to be tracked at the same time as a second, independent quantity, the stack empties exactly when $i+k=j+l$. This works, so option A is context-free.
Step 3: Test $i+j=k+l$, option D, the same way.
Push for a's and b's, pop for c's and d's. Again only one running total is ever needed at a time, so this is context-free by the same one stack argument as option A.
Step 4: Test $i=l$ and $j=k$, option C, with a grammar.
Write $S \to aSd \mid T$ and $T \to bTc \mid \varepsilon$. Expanding $S$ a number of times matches every leading $a$ with a trailing $d$, giving $i=l$, and expanding $T$ matches every $b$ with a following $c$, giving $j=k$. Both matchings are properly nested inside each other, not crossed, so an ordinary context-free grammar builds the language directly. Option C is context-free.
Step 5: Test $i=k$ and $j=l$, option B, with the pumping lemma.
Suppose this language were context-free with pumping length $p$. Take the string $a^pb^pc^pd^p$, which satisfies $i=k=p$ and $j=l=p$. The pumping lemma lets us split it as $uvwxy$ with $|vwx|\leq p$ and $|vx|\geq 1$, then pump $v,x$ together. Because $|vwx|\leq p$, the piece $vwx$ can touch at most two of the four letter blocks, it cannot reach across three or four blocks at once. Pumping up then increases the count of at most two of the four letters and leaves the other two unchanged, which always destroys either $i=k$ or $j=l$, since those two equalities pair up NON adjacent blocks, block one with block three, block two with block four. So no valid split exists, and the language is not context-free. Option B fails.
Step 6: Collect the result.
Only the constraints that let a single stack track one running total, A and D, or that produce a properly nested bracket structure, C, survive. The crossing constraint in B does not.
\[ \boxed{\text{Options A, C, D are context-free; option B is not}} \]
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