Corollary: The precise no-hiding conclusion

If $\mathcal{N}$ is constant, $B$ exactly recovers the code. If $\mathcal{C}$ is constant, $R$ exactly recovers it. Both complementary outputs cannot independently recover the entire unknown code when $d\ge 2$.

Proof. Interchange $R$ and $B$ in Theorem thm:exact. An exactly recoverable output has constant complement, which cannot itself recover two distinct inputs.