The 3-SAT problem can be reduced to both the graph coloring and the directed ham
ID: 646527 • Letter: T
Question
The 3-SAT problem can be reduced to both the graph coloring and the directed hamiltonian cycle problem, but is there any chain of reductions which reduce directed hamiltonian cycle to graph coloring in polynomial time?
A:
Let Q?NP and Q??NP-hard. Then, by definition, Q is (many-one) reducible to Q? in polynomial time.
The exact chain of the reductions will depend on the NP-hardness proof of Q?. Typically, it is proven by a chain of reductions starting from SAT and ending with Q? and then using the Cook-Levin theorem. So the chain of reductions will be a reduction from Q to SAT followed by the chain of reductions from SAT to Q?.
There is usually a more direct reduction for specific problems (without using Cook-Levin), since it is usually easy to find a propositional formula directly expressing the required property (with no reference to TMs). For example, in the case of Directed Hamiltionian Path (DHP) and Graph Coloring (GC), you can reduce:
DHP to SAT,
SAT to 3SAT,
3SAT to GC.
Explanation / Answer
Let Q?NP and Q??NP-hard. Then, by definition, Q is (many-one) reducible to Q? in polynomial time.
The exact chain of the reductions will depend on the NP-hardness proof of Q?. Typically, it is proven by a chain of reductions starting from SAT and ending with Q? and then using the Cook-Levin theorem. So the chain of reductions will be a reduction from Q to SAT followed by the chain of reductions from SAT to Q?.
There is usually a more direct reduction for specific problems (without using Cook-Levin), since it is usually easy to find a propositional formula directly expressing the required property (with no reference to TMs). For example, in the case of Directed Hamiltionian Path (DHP) and Graph Coloring (GC), you can reduce:
DHP to SAT,
SAT to 3SAT,
3SAT to GC.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.