I am trying to wrap my head around the following proof

  • Choose some language A $\in$ $TIME(n^3)$ \ $TIME(n)$ (the existence of such a language is guaranteer by the hierarchy theorem)
  • Let B = {1}
  • Note that $A \leqslant_{p} B, B \in TIME(n)$ but $A \not\in TIME(n^{3})$

The way I understand is that A is in $TIME(n^{3})$ but not in $TIME(n)$. I assume the main point in the proof is the polynomial reduction to B, but can we say that it is more or less than $n^{3}$?

I guess I would understand that in case the reduction can be completed in $O(n)$ then A can actually be solved in $TIME(n)$ which indeed is a contradiction..

  • $\begingroup$ A cannot be solved in time $O(n)$ by hypothesis. The proof is not a proof contradiction, it is exhibiting two languages $A, B$ such that $B \in TIME(n)$ but $A \not \in TIME(n)$, yet $A \le_p B$, which means that $TIME(n)$ is not closed under poly-time reductions. The time spent to compute the reduction does not matter (as long as it is polynomial, which can be guaranteed to be the case since $A$ can be solved in polynomial time by hypothesis). Also, you have a typo in the last item, it should be $A \not\in TIME(n)$. $\endgroup$
    – Steven
    Nov 18 at 21:47
  • $\begingroup$ Ahh I see. But why does it not matter what is the time of the reduction? Oh yes the typo now makes sense, thank you! $\endgroup$
    – Meki21
    Nov 18 at 23:59


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