– Euler circuit is in P, but
Hamiltonian circuit is NP-complete
. – Shortest path between two points is computable in O(1112), but longest path is NP- complete.
Is Euler’s circuit problem in NP?
A graph is called Eulerian if it has an Eulerian Cycle and called Semi-Eulerian if it has an Eulerian Path.
The problem seems similar to Hamiltonian Path which is NP complete problem for a general graph
. Fortunately, we can find whether a given graph has a Eulerian Path or not in polynomial time.
Is Eulerian a path NP?
Euler path is in NP
. Proof. Consider any path in a graph G. Simply check every edge; the path is a solution if and only if every edge is in the path.
How do you find the Eulerian cycle?
To find the Euler path (not a cycle), let’s do this: if and are two vertices of odd degree,then just add an edge ( V 1 , V 2 ) , in the resulting graph we find the Euler cycle (it will obviously exist), and then remove the “fictitious” edge ( V 1 , V 2 ) from the answer.
How do you determine if a graph has an Euler cycle?
Thus for a graph to have an Euler circuit,
all vertices must have even degree
. The converse is also true: if all the vertices of a graph have even degree, then the graph has an Euler circuit, and if there are exactly two vertices with odd degree, the graph has an Euler path.
How do you know if a graph is Eulerian?
An Euler circuit always starts and ends at the same vertex. A connected graph G is an Euler graph if and only if all vertices of G are of even degree, and
a connected graph G is Eulerian if and only if its edge set can be decomposed into cycles
.
Is halting problem NP-complete?
– If we had a polynomial time algorithm for the halting problem, then we could solve the satisfiability problem in polynomial time using A and X as input to the algorithm for the halting problem . – Hence the halting problem is an NP-hard problem which is not in NP. – So
it is not NP-complete
.
Which of the following is NP-complete problem?
Explanation:
Hamiltonian circuit, bin packing, partition problems
are NP complete problems. Halting problem is an undecidable problem.
Is a 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 circuit an Euler path?
An Euler path is a path that uses every edge of a graph exactly once.
An Euler circuit is a circuit that uses every edge of a graph exactly once
.
Which complete graph is Eulerian?
Odd Order
Complete Graph is Eulerian.
Is every Euler circuit an Euler path?
An Euler circuit is a circuit that travels through every edge of a graph once and only once. Like all circuits, an Euler circuit must begin and end at the same vertex. Note that
every Euler circuit is an Euler path
, but not every Euler path is an Euler circuit. Some graphs have no Euler paths.
How do you find the Euler path in a directed graph?
A directed graph has an Eulerian path if and only if the following conditions are satisfied: At most one vertex in the graph has out-degree = 1 + in-degree , and at most one vertex in the graph has in-degree = 1 + out-degree . All the remaining vertices have in-degree == out-degree .
What is complete graph in discrete mathematics?
In the mathematical field of graph theory, a complete graph is
a simple undirected graph in which every pair of distinct vertices is connected by a unique edge
. A complete digraph is a directed graph in which every pair of distinct vertices is connected by a pair of unique edges (one in each direction).
How do you get Euler trail?
What is Euler Graph Theorem?
Theorem:
An Eulerian trail exists in a connected graph if and only if there are either no odd vertices or two odd vertices
. For the case of no odd vertices, the path can begin at any vertex and will end there; for the case of two odd vertices, the path must begin at one odd vertex and end at the other.
Does this graph have an Euler circuit?
What is Euler path example?
One example of an Euler circuit for this graph is
A, E, A, B, C, B, E, C, D, E, F, D, F, A
. This is a circuit that travels over every edge once and only once and starts and ends in the same place.
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 scheduling NP-complete?
So, the scheduling, where the output is the optimal set of schedule, would be NP-hard, but
not NP-complete
.
Is Sudoku NP-complete?
Introduction.
The generalised Sudoku problem is an NP-complete problem
which, effectively, requests a Latin square that satisfies some additional constraints. In addition to the standard requirement that each row and column of the Latin square contains each symbol precisely once, Sudoku also demands block constraints.
What is NP-complete with example?
In terms of solving a NP problem,
the run-time is not polynomial
. It would be something like O(n!) or something larger. However, this class of problems are given a specific solution, and checking the solution would have a polynomial run-time. For example, the Sudoku game.