To prove that a problem P is PSPACE-complete, you need to do two things:
- Show that P can be computed in PSPACE.
- Show that every problem Q∈PSPACE reduces to P, in the sense that there exists a polytime function f such that f(x)∈P iff x∈Q.
They show that the problem of solving a level in "Super Mario Brothers" is as hard as the hardest problems in the "complexity class" PSPACE, meaning that it's even more complex than the traveling-salesman problem, or the problem of factoring large numbers, or any of the other hard problems belonging to the better-known
1 Answer. Assuming we use polytime reductions in both, all PSPACE-hard problems are NP-hard, which follows directly from the definition and from the easy fact NP⊆PSPACE: a languages L is PSPACE-hard if all languages in PSPACE polytime-reduce to L.
The complexity classes L, NL, P, NP and PSPACE are closed under log-space reduction.
Explanation: The complement of all the problems in PSPACE are also in PSPACE, meaning co-PSPACE= PSPACE.
Since, PSPACE is closed under reductions and NP is contained in PSPACE, then we have that NP = PSPACE. The P versus NP problem is a major unsolved problem in computer science. This problem was introduced in 1971 by Stephen Cook [1]. It is considered by many to be the most important open problem in the field [2].
In fact NP is a subset of PSPACE. Proving this involves taking any problem in NP and showing how you can use the verifying algorithm to answer the question, using only polynomial space. Note that any poly-time algorithm can only use poly-space, since the algorithm can only use one new memory space per time step.
Traveling Salesman Optimization(TSP-OPT) is a NP-hard problem and Traveling Salesman Search(TSP) is NP-complete. However, TSP-OPT can be reduced to TSP since if TSP can be solved in polynomial time, then so can TSP-OPT(1).
Theorem 1. D-HAM-PATH is NP-Complete. containing both literals can be removed with φ without affecting its satisfiability. The reduction uses the following steps.
13. How many steps are required to prove that a decision problem is NP complete? Explanation: First, the problem should be NP. Next, it should be proved that every problem in NP is reducible to the problem in question in polynomial time.
It is easy to prove that the halting problem is NP-hard but not NP-complete. There are also NP-hard problems that are neither NP-complete nor Undecidable. For instance, the language of true quantified Boolean formulas is decidable in polynomial space, but not in non-deterministic polynomial time (unless NP = PSPACE).
We can solve Y in polynomial time: reduce it to X. Therefore, every problem in NP has a polytime algorithm and P = NP. then X is NP-complete. In other words, we can prove a new problem is NP-complete by reducing some other NP-complete problem to it.
A Hamiltonian cycle in a graph is a cycle that passes through every vertex in the graph exactly once. This is known as the P versus NP problem. In my research, we focus on finding Hamiltonian Cycles in a specific group of graphs: cubic planar.
To reduce from 3SAT, create a “gadget” for each variable and a “gadget” for each clause, and connect them up somehow. Recall that input to Subset sum problem is set A = {a1 ,a2 ,,am} of integers and target t. The question is whether there is A ⊆ A such that elements in A sum to t.
NP-complete problem, any of a class of computational problems for which no efficient solution algorithm has been found. Many significant computer-science problems belong to this class—e.g., the traveling salesman problem, satisfiability problems, and graph-covering problems.
A problem is NP-hard if an algorithm for solving it can be translated into one for solving any NP-problem (nondeterministic polynomial time) problem. NP-hard therefore means "at least as hard as any NP-problem," although it might, in fact, be harder.
- Euler circuit is in P, but Hamiltonian circuit is NP-complete.
Depth first search and backtracking can also help to check whether a Hamiltonian path exists in a graph or not. Simply apply depth first search starting from every vertex v and do labeling of all the vertices. All the vertices are labelled as either "IN STACK" or "NOT IN STACK".
Replace each vertex v with a cycle Cv of your favourite length--as long as it is no larger than polynomial in H that is, and if uv is an edge in H then put a complete bipartite graph between the vertices of Cu and Cv. Then the resulting graph G has a hamiltonian cycle iff H does.
The Hamiltonian cycle problem is a special case of the travelling salesman problem, obtained by setting the distance between two cities to one if they are adjacent and two otherwise, and verifying that the total distance travelled is equal to n (if so, the route is a Hamiltonian circuit; if there is no Hamiltonian
A graph is Hamiltonian-connected if for every pair of vertices there is a Hamiltonian path between the two vertices. A Hamiltonian cycle, Hamiltonian circuit, vertex tour or graph cycle is a cycle that visits each vertex exactly once.
A Hamiltonian path, also called a Hamilton path, is a graph path between two vertices of a graph that visits each vertex exactly once. If a Hamiltonian path exists whose endpoints are adjacent, then the resulting graph cycle is called a Hamiltonian cycle (or Hamiltonian cycle). has no Hamiltonian paths.
A dodecahedron ( a regular solid figure with twelve equal pentagonal faces) has a Hamiltonian cycle. A Hamiltonian cycle is a closed loop on a graph where every node (vertex) is visited exactly once.
permutations of the non-fixed vertices, and half of those are the reverse of another, so there are (n-1)!/2 distinct Hamiltonian cycles in the complete graph of n vertices.
In an undirected graph, the Hamiltonian path is a path, that visits each vertex exactly once, and the Hamiltonian cycle or circuit is a Hamiltonian path, that there is an edge from the last vertex to the first vertex.
The modern definitions of NP-Completeness use the Karp reduction. For a reduction from Hamiltonian Cycle to Path. Given a graph G of which we need to find Hamiltonian Cycle, for a single edge e={u,v} add new vertices u′ and v′ such that u′ is connected only to u and v′ is connected only to v to give a new graph Ge.
