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Show that any PSPACE-hard language is also NP-hard

Short Answer

Expert verified

That if every NP-hard language is also PSPACE-hard, then PSPACE=NP.

Step by step solution

01

To NP-Hard and PSPACE Problems

The NP problems are a class of problems whose solutions are difficult to find but simple to prove, and which are solved in polynomial time by a Non-Deterministic Machine.

A problem is NP-hard if all problems in NP can be reduced to it in polynomial time, even if it isn't in NP.

PSPACE is the collection of all decision problems that a Turing machine can answer using a polynomial space in computational complexity theory.

02

Step-2:To explain the PSPACE Hardness

PSPACE hardness for a language L is defined as all A∈PSPACE,A≤LP. Because SAT is NP complete, it is also NP hard. As a result, SAT≤LP, and L s NP hard.

That if every NP-hard language is also PSPACE-hard, then PSPACE=NP.

If every NP-hard language is PSPACE-hard , then SAT is PSPACE-hard, and consequently every PSPACE language is polynomial-time reducible to SAT. then

PSPACE⊆NP because SAT∈NP and therefore PSPACE=NPbecause we know NP⊆PSPACE

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