Let $\Sigma = \{a, b, c, d\}$ and let $L = \{a^i b^j c^k d^\ell \mid i, j, k, \ell \ge 0\}$. Which of the following constraints ensure(s) that the language $L$ is context-free?

A language recognized by a Pushdown Automaton (PDA) is context-free.

• Option (A): Constraint $i + k = j + \ell$ can be rewritten as $(i - j) + (k - \ell) = 0$. A PDA can push all $a$'s onto the stack, pop for $b$'s (or push remaining excess $b$'s if $j > i$), then process $c$'s and $d$'s similarly to verify that the total net sum is zero. Hence, this language is context-free.

• Option (B): $L = \{a^i b^j c^i d^j \mid i, j \ge 0\}$ contains cross-serial dependencies (matching $a$ with $c$, and $b$ with $d$). Using the Pumping Lemma for Context-Free Languages with the string $z = a^n b^n c^n d^n$, it is easily shown that this language is not context-free.

• Option (C): $L = \{a^i b^j c^j d^i \mid i, j \ge 0\}$ has nested dependencies. A PDA can push all $a$'s, push all $b$'s, match and pop the $b$'s using $c$'s, and then match and pop the $a$'s using $d$'s. Thus, it is context-free.

• Option (D): $L = \{a^i b^j c^k d^\ell \mid i + j = k + \ell\}$. A PDA can push all symbols in the first group ($a$'s and $b$'s, total $i + j$) onto the stack, and then pop one symbol for each $c$ and $d$ (total $k + \ell$). Thus, this language is context-free.

Therefore, options A, C, and D are correct.