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Let $L_1$ and $L_2$ be two decision problems.

Is there an algorithm deciding whether $L_1 \leq_P L_2$, that is, whether $L_1$ is reducible to $L_2$ in polynomial time?

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  • $\begingroup$ How are $L_1$ and $L_2$ given to the algorithm? $\endgroup$ – Yuval Filmus Dec 4 '18 at 4:54
  • $\begingroup$ Elaborate the possible permutations of, decidable, undecidable languages anf Polynomial time and Non-deterministic polynomial time algirithms. $\endgroup$ – CHETAN RAJPUT Dec 4 '18 at 6:03
  • $\begingroup$ I don't see how this answers my question. At any rate, once you pose your question in a well-defined way, the answer will likely be that the problem is undecidable. $\endgroup$ – Yuval Filmus Dec 4 '18 at 6:04
  • $\begingroup$ I guess then i didn't get your question, if you can eleborate more , i can add it. $\endgroup$ – CHETAN RAJPUT Dec 4 '18 at 7:17
  • $\begingroup$ The input to an algorithm has to be finite. Decision problems are infinite objects. You can't given a decision problem as an input. Therefore I find it hard to understand your question. $\endgroup$ – Yuval Filmus Dec 4 '18 at 7:25
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I'm going to assume that your question is whether the language $$L = \{\langle M_1\rangle; \langle M_2\rangle\mid L(M_1)\leq_\mathrm{p} L(M_2)\}$$ is decidable, where $\langle M\rangle$ is the description of Turing machine $M$ and $\leq_\mathrm{p}$ denotes polynomial-time many-one reducibility.

This language is undecidable.

The only language that is many-one reducible to $\Sigma^*$ is $\Sigma^*$ itself. This is because a many-one reduction from $X$ to $\Sigma^*$ must, by definition, map all "no" instances of $X$ to "no" instances of $\Sigma^*$. However, $\Sigma^*$  has no "no" instances, so the reduction can only exist if $X$ also has no "no" instances, i.e., if $X=\Sigma^*$.

So, now let $Y$ be any Turing machine that accepts every input. The language $H=\{\langle M\rangle\mid \langle M\rangle;\langle Y\rangle\in L\}$ is clearly reducible to $L$. But $H$ is the language of Turing machines that accept every input, which is undecidable.

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  • $\begingroup$ A good explanation. May be I couldn't able to ask properly but expecting this kind of explanation. Thanx. $\endgroup$ – CHETAN RAJPUT Dec 4 '18 at 12:53
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Let $L_2$ be some easy to decide language. For example,

$$L_2 = \{ \omega : |\omega| \equiv 0 \mod 2 \}$$

Then reducing instances of $L_1$ to instances of $L_2$ is as hard as deciding $L_1$ itself, and your question boils down to:

Given a description of a TM halting on every input, can you simulate it in polynomial time?

This is undecidable, which you can prove by the same argument as for the Halting problem.

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