I understand that if $A$ is NP-hard, and $A \leq_{T} B$, then $B$ is NP-hard. Is the converse also true? So if $B$ is NP-hard, and $A \leq_{T} B$, then $A$ is NP-hard? Why (not)?
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No. Take for example $A = \emptyset$ and let $B$ be any NP-hard problem. Clearly $A \le_T B$ but $A$ is not NP-hard.
Steven
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That makes sense. Thank you! – meikyu Oct 24 '22 at 12:15
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The Turing reduction $A\leqslant_T B$ can be explained as: "If I have a black box solving $B$, can I use it to solve $A$?".
Since such a definition does not have any constraint of time complexity, any computable set is reducible to any other. That means that even if $A$ is a problem $\mathsf{NP}$-complete, then $\emptyset\leqslant_T A$ (but $A$ is not $\mathsf{NP}$-hard). Note that the converse is also true: $A\leqslant_T \emptyset$, so your first affirmation is wrong.
If you find the last reduction odd, you can add a condition on complexity, for example "I cannot use the black box solving $B$ more than a polynomial number of times".
Even if you limit the number of uses to a polynomial number of times, your first affirmation is not necessarily true, because if $A$ is any problem, then $A\leqslant_T^p \text{co}A$. If your affirmation was true, that would prove $\mathsf{NP} = \text{co}\mathsf{NP}$.
See here for some details.
Nathaniel
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