(A) NP-complete = NP
(B) NP-complete P =
(C) NP-hard = NP
(D) P = NP-complete
R is NP-complete
R is NP-hard
Q is NP-complete
Q is NP-hard
There is no polynomial time algorithm for X.
If X can be solved deterministically in polynomial time, then P = NP.
If X is NP-hard, then it is NP-complete.
X may be undecidable.
The problem 3-SAT and 2-SAT are
both in P
both NP complete
NP-complete and in P respectively
undecidable and NP-complete respectively
The Boolean satisfiability problem (SAT) is a decision problem, whose instance is a Boolean expression written using only AND, OR, NOT, variables, and parentheses.
The problem is: given the expression, is there some assignment of TRUE and FALSE values to the variables that will make the entire expression true?
A formula of propositional logic is said to be satisfactory if logical values can be assigned to its variables to make the formula true. 3-SAT and 2-SAT are special cases of k-satisfiability (k-SAT) or simply satisfiability (SAT) when each clause contains exactly k = 3 and k = 2 literals respectively. 2-SAT is P while 3-SAT is NP-Complete.
1, 2 and 3
1 and 3
2 and 3
1 and 2
If we want to prove that a problem X is NP-Hard, we take a known NP-Hard problem Y and reduce Y to X
The first problem that was proved as NP-complete was the circuit satisfiability problem.
NP-complete is a subset of NP Hard
All of the above
None of the above
1. The problem of determining whether there exists a cycle in an undirected graph is in P. 2. The problem of determining whether there exists a cycle in an undirected graph is in NP. 3. If a problem A is NP-Complete, there exists a non-deterministic polynomial time algorithm to solve A.
1, 2 and 3
1 and 2 only
2 and 3 only
1 and 3 only
solvable in polynomial time by reduction to directed graph reachability.
solvable in constant time since any input instance is satisfiable.
NP-hard, but not NP-complete.
Both DHAM3 and SHAM3 are NP-hard
SHAM3 is NP-hard, but DHAM3 is not
DHAM3 is NP-hard, but SHAM3 is not
Neither DHAM3 nor SHAM3 is NP-hard