Corollary: The precise no-hiding conclusion

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

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