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Is Pspace-complete?

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Sarah Oconnell

Published Feb 18, 2026

Is Pspace-complete?

In computational complexity theory, a decision problem is PSPACE-complete if it can be solved using an amount of memory that is polynomial in the input length (polynomial space) and if every other problem that can be solved in polynomial space can be transformed to it in polynomial time.

Beside this, is Hampath NP complete?

By HAMPATH we denote the following algorithmic problem: given a directed graph and two its vertices, s and t, find out whether there exists a Hamiltonian path from s to t. Theorem 1. HAMPATH is NP-complete.

Beside above, why is Hamiltonian cycle NP complete? Any Hamiltonian Path can be made into a Hamiltonian Circuit through a polynomial time reduction by simply adding one edge between the first and last point in the path. Therefore we have a reduction, which means that Hamiltonian Paths are in NP Hard, and therefore in NP Complete.

Just so, is Hamiltonian cycle NP complete?

A Hamiltonian path is a simple open path that contains each vertex in a graph exactly once. The Hamiltonian Path problem is the problem to determine whether a given graph contains a Hamiltonian path. Hamiltonian Cycle is NP-complete, so we may try to reduce this problem to Hamiltonian Path.

Is P equal to Pspace?

PSPACE is also equal to PCTC, problems solvable by classical computers using closed timelike curves, as well as to BQPCTC, problems solvable by quantum computers using closed timelike curves.

How do you prove a language is complete Pspace?

To prove that a problem P is PSPACE-complete, you need to do two things:
  1. Show that P can be computed in PSPACE.
  2. 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.

Is Pspace harder than NP?

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

Is NP hard in Pspace?

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.

Is Pspace downwards closed?

The complexity classes L, NL, P, NP and PSPACE are closed under log-space reduction.

Which are the complement of all the problems in Pspace complexity class?

Explanation: The complement of all the problems in PSPACE are also in PSPACE, meaning co-PSPACE= PSPACE.

Is Pspace an NP?

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].

Is NP subset of Pspace?

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.

Is Travelling salesman NP complete?

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).

Is Rudrata path NP complete?

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.

How many steps are required to prove that a decision problem is NP complete?

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.

Is halting problem NP hard?

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).

How do you prove a problem is NP-complete?

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.

Is Hamiltonian cycle in P?

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.

How can I reduce my 3SAT?

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.

What is NP-complete with example?

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.

What is meant by NP hard problem?

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.

Is Euler tour NP-complete?

- Euler circuit is in P, but Hamiltonian circuit is NP-complete.

How do you verify a Hamiltonian path?

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".

How do you know if a Hamiltonian cycle exists?

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.

Are Hamiltonian path problem and Travelling salesperson problem are similar?

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

Does a Hamiltonian path exist between every pair of nodes?

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.

Does a Hamiltonian path exist?

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.

What is Hamiltonian cycle with example?

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.

How many Hamiltonian cycles are in a complete graph?

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.

What is Hamiltonian cycle in DAA?

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.

How do you reduce Hamiltonian cycle to Hamiltonian path?

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.

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Is Pspace-complete?