Circuit satisfiability problem: Given a Boolean combinatorial circuit composed…

Correct selection and reasoning:

Selected true statements:

• The circuit satisfiability problem belongs to class NP: This is true because a proposed input assignment serves as a certificate and evaluating the output of a combinational Boolean circuit on that assignment can be done in time polynomial in the size of the circuit.

• The circuit satisfiability problem is at least as hard as any language in NP (NP-hard): This is true because any nondeterministic polynomial-time computation can be encoded as a Boolean circuit that accepts exactly those inputs the machine would accept. Therefore every problem in NP can be reduced to circuit satisfiability.

• The circuit satisfiability problem is NP-complete: Since it is both in NP and NP-hard, it meets the definition of NP-completeness.

Why the other statements are false or unsupported:

• A stated formula about the circuit size being Θ(K 2 +1) is not generally correct as given: circuit size depends on the encoding and on what K denotes; the provided expression is unsupported and not a standard property of circuit satisfiability.

• The claim that 'if P ≠ NP this situation would contradict NP-completeness' is incorrectly worded and false: NP-completeness is a statement about relative difficulty inside NP. If P = NP then all NP-complete problems would be in P; P ≠ NP does not contradict a problem being NP-complete.

Final conclusion: The correct true statements are the membership in NP, the NP-hardness, and therefore NP-completeness. The combination that lists those three statements is the correct answer.