= Solution
Rename the old halt state $q_H$ as a nonhalting state $q_F$, including every occurrence of $q_H$ in the original program, and introduce a new halt state $q_H$. Attach the following output-flipping routine:
$$
\begin{array}{c|l}
q_F&\mapsto ?_0(0,q_{0},q_{1})\\
q_{0}&\mapsto -(0,q_{+1},q_{+1})\\
q_{1}&\mapsto -(0,q_{+0},q_{+0})\\
q_{+1}&\mapsto +_1(0,q_H)\\
q_{+0}&\mapsto +_0(0,q_H)\\
q_H&\mapsto ?_\varepsilon(0,q_H,q_H).
\end{array}
$$
Because the original machine computes a characteristic function, it reaches $q_F$ with register $0$ containing exactly $0$ or $1$. The routine tests that symbol, removes it, and appends its complement. It therefore halts with $1-\chi_X(w)=\chi_{\mathbb B\setminus X}(w)$.
Back to article page
