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Questions about decision problems that can be solved on nondeterministic Turing machines in time polynomial in the length of the input.

Note that you got it wrong in the question. What you have is that if $Y \leq^p_m X$ and $Y$ is NP-hard, then $X$ is also NP-hard. This is true by definition, unconditional on any other lower bounds … known for $X$ or $Y$ (though, as Ryan points out, such lower bounds could be very interesting). If $X$ is in NP and has a lower bound on its computational complexity that is super-polynomial, it would mean that $P \neq NP$. …
answered Apr 27 '17 by Pontus
As I understand your question, yes, this is possible. As an example, consider the (infinitely large) subset of SAT that contains all satisfiable formulae with no negations. This subset is trivially in …
answered Mar 20 '17 by Pontus
This problem is essentially just a rephrasing of the classic NP-complete problem PARTITION. If you can partition a set $S$ of natural numbers into two partitions $S_1$ and $S_2$ with equal sum, then $\sum_{s \in S_1} s + \sum_{s \in S_2} -s = 0$ and vice versa. …
answered Apr 20 '17 by Pontus
All directed graphs can be edge-partitioned into two subgraphs that are acyclic and therefore triangle free. Let $\prec$ be any total ordering of the vertices. For each edge $(u, v) \in A$, put it in …
answered Oct 22 '18 by Pontus