Assuming F(x,y,D) is a function, and we can evaluate it in polynomial time with input x, y and D.

Consider the problem P1: With D as input, computes $(x^*,y^*)=argmax_{(x,y)}F(x,y|D)$ where x and y are two sets of variables.

Another problem P2: With y and D as input, computes $x^*=argmax_xF(x|y,D)$.

If we know P2 is NP-hard, can we infer the hardness of P1? Is P1 also NP-hard?

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    $\begingroup$ Please explain your notation. The statement of P2 is unclear, what does $x\mid y$ mean? $\endgroup$ – Jan Johannsen Oct 5 '16 at 7:53
  • $\begingroup$ @JanJohannsen Sorry for the abused notation, F(x|y,D) means "given y and D as input, F as a function of x only"; similarly, F(x, y | D) means "given D as input, F as a function of x and y". $\endgroup$ – user2789928 Oct 5 '16 at 16:37

No, you cannot infer hardness of P1. (And your question looks suspiciously close to homework.) Consider the special case where

  • $D$ is an undirected graph $G=(V,E)$
  • $x$ is a subset $E_x\subseteq E$
  • $y$ is a non-negative integer

The function value $F(x,y,D)$

  • takes the value $2$, whenever $y\ge1$;
  • takes the value $1$, if $y=0$ and if $E_x$ induces a Hamiltonian cycle in $G$;
  • takes the value $0$, if $y=0$ and if $E_x$ does not induce a Hamiltonian cycle in $G$.

Fact. The problem P1 of maximizing $F$ for a given graph $G$ is polynomially solvable.

(Just pick $E_x=\emptyset$ and $y=1$ to reach an objective value of $2$.)

Fact. The problem P2 of maximizing $F$ for a given graph $G$ and a given number $y$ is NP-hard.

(By fixing $y=0$, the problem becomes the NP-hard Hamilton cycle problem.)

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  • $\begingroup$ Many thanks to your explanation! But it is not for homework:) $\endgroup$ – user2789928 Oct 5 '16 at 19:02

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