For a string $w$, we define $w^R$ to be the reverse of $w$. For example, if $w = 01101$ then $w^R = 10110$. Which of the following languages is/are context-free?

Let us analyze each language:

• Option A: $\{w x w^R x^R \mid w, x \in \{0, 1\}^*\}$ This language requires matching two independent reversed patterns that are interleaved (cross-serial dependency). A single pushdown automaton cannot verify both $w$ against $w^R$ and $x$ against $x^R$ simultaneously. Hence, this language is not context-free.

• Option B: $\{w w^R x x^R \mid w, x \in \{0, 1\}^*\}$ This language is the concatenation of two CFLs: $L_1 = \{w w^R \mid w \in \{0, 1\}^*\}$ and $L_2 = \{x x^R \mid x \in \{0, 1\}^*\}$, where both are standard context-free languages (even-length palindromes). Since CFLs are closed under concatenation, $L_1 L_2$ is context-free.

• Option C: $\{w x w^R \mid w, x \in \{0, 1\}^*\}$ By choosing $w = \epsilon$, any string $s \in \{0, 1\}^*$ can be written as $\epsilon s \epsilon^R = s$. Thus, the language generates all strings over $\{0, 1\}^*$, i.e., $L = \{0, 1\}^*$, which is a regular language. Every regular language is also context-free.

• Option D: $\{w x x^R w^R \mid w, x \in \{0, 1\}^*\}$ Let $y = w x$. Then $y^R = (w x)^R = x^R w^R$. Thus, any string of this form is $y y^R$ for $y = w x \in \{0, 1\}^*$. Choosing $x = \epsilon$ gives all even-length palindromes $\{w w^R \mid w \in \{0, 1\}^*\}$. Therefore, the language is exactly the set of even-length palindromes, which is context-free.

Thus, the correct options are B, C, and D.