Is this a weaker or stronger form of the halting problem
The standard proof that the halting problem $L$ is undecidable also gives an efficient algorithm for constructing an instance on which a given Turing machine $H$ fails to solve the halting problem.
For any Turing machine $H$, let $M_H$ be a Turing machine implementing the following algorithm: &quot;On input $\langle P \rangle$ where $P$ is a Turing machine, run $H(\langle P, P \rangle)$. If it outputs $1$, run forever, otherwise halt.&quot;
By design, $H(\langle M_H, M_H \rangle)$ fails. Either it runs forever, or else it halts and gives the wrong answer to the question of whether $\langle M_H, M_H \rangle \in L$. Given $\langle H \rangle$, one can efficiently construct $\langle M_H, M_H \rangle$.