To set the record clear: Yes. A Path can be both Eularian and Hamiltonian. A Hamiltonian path is a spanning path, and an Eularian path goes through each edge exactly once.
The complete bipartite graph K2,4 has an Eulerian circuit, but is non-Hamiltonian (in fact, it doesn't even contain a Hamiltonian path). Any Hamiltonian path would alternate colors (and there's not enough blue vertices). Since every vertex has even degree, the graph has an Eulerian circuit.
Some graphs lack both an Eulerian and a Hamiltonian cycle such as Star Graph. Star Graph contains no cycle.
Take a graph which is just a cycle on at least 4 vertices, then add an edge between one pair of vertices. Where you added the edge, you will have an odd degree, so the graph cannot have an Eulerian cycle. But the original cycle gives a Hamiltonian cycle.
With that definition, a graph with an Euler circuit can't have an Euler path. Other people say that an Euler path has no restriction on start and end vertices.
An Eulerian graph G (a connected graph in which every vertex has even degree) necessarily has an Euler tour, a closed walk passing through each edge of G exactly once. This tour corresponds to a Hamiltonian cycle in the line graph L(G), so the line graph of every Eulerian graph is Hamiltonian.
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. Other graphs have several Euler paths.
If the euler path ends at the same vertex from which is has started it is called as Euler cycle. A Hamiltonian path is a path that passes through every vertex exactly once (NOT every edge). Similarly if the hamilton path ends at the initial vertex from which it has started then it is known as Hamiltonian cycle.
A Hamilton Circuit is a Hamilton Path that begins and ends at the same vertex. *Unlike Euler Paths and Circuits, there is no trick to tell if a graph has a Hamilton Path or Circuit. A Complete Graph is a graph where every pair of vertices is joined by an edge.
A Hamiltonian cycle in a graph is a cycle that visits every vertex at least once, and an Eulerian cycle is a cycle that visits every edge once. In general graphs, the problem of finding a Hamiltonian cycle is NP-hard, while finding an Eulerian cycle is solvable in polynomial time.
We observe that not every graph is Hamiltonian; for instance, it is clear that a dis- connected graph cannot contain any Hamiltonian cycle/path. There are also connected graphs that are not Hamiltonian. For example, if a connected graph has a a vertex of degree one, then it cannot be Hamiltonian.
Not all graphs have a Hamilton circuit or path. There is no way to tell just by looking at a graph if it has a Hamilton circuit or path like you can with an Euler circuit or path. You must do trial and error to determine this. By the way if a graph has a Hamilton circuit then it has a Hamilton path.
A graph is Hamiltonian if it has a Hamiltonian cycle. Closure: The (Hamiltonian) closure of a graph G, denoted Cl(G), is the simple graph obtained from G by repeatedly adding edges joining pairs of nonadjacent vertices with degree sum at least |V (G)| until no such pair remains.
A graph that has a Hamiltonian circuit is called a Hamiltonian graph. For instance, the graph below has 20 nodes. The edges consist of both the red lines and the dotted black lines. : A graph is Eulerian if and only if each vertex has an even degree.
Can there exist a graph which is both eulerian and is bipartite? Explanation: If a graph is such that there exists a path which visits every edge atleast once, then it is said to be Eulerian. Taking an example of a square, the given question evaluates to yes.
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.
Hamilton Path A Hamilton path is a simple path that traverses every vertex in G exactly once. If G is a simple graph with n vertices with n ≥ 3 such that deg(u) + deg(v) ≥ n for every pair of nonadjacent vertices u and v in G, then G has a Hamilton circuit.
A Hamiltonian circuit is a circuit that visits every vertex once with no repeats. Being a circuit, it must start and end at the same vertex. A Hamiltonian path also visits every vertex once with no repeats, but does not have to start and end at the same vertex.
The most natural way to prove a graph isn't Hamiltonian is to do a case by case analysis of possible paths, showing it doesn't work. For instance, in lecture we outlined the proof that if you remove a vertex from the Icosian graph, than the result isn't Hamiltonian.
There are some useful conditions that imply the existence of a Hamilton cycle or path, which typically say in some form that there are many edges in the graph. An extreme example is the complete graph Kn: it has as many edges as any simple graph on n vertices can have, and it has many Hamilton cycles.
A Hamiltonian path in an undirected graph G = (V,E) is a path that goes through every vertex exactly once. A Hamiltonian cycle (or Hamiltonian tour) is a cycle that goes through every vertex exactly once.
Euler's Theorem 6.3. 1: If a graph has any vertices of odd degree, then it cannot have an Euler circuit. If a graph is connected and every vertex has an even degree, then it has at least one Euler circuit (usually more).
The question that should immediately spring to mind is this: if a graph is connected and the degree of every vertex is even, is there an Euler circuit? The answer is yes. Theorem 5.2. 2 If G is a connected graph, then G contains an Euler circuit if and only if every vertex has even degree.
If a graph G has an Euler circuit, then all of its vertices must be even vertices. Or, to put it another way, If the number of odd vertices in G is anything other than 0, then G cannot have an Euler circuit.
A nonhamiltonian graph is a graph that is not Hamiltonian. All disconnected graphs are therefore nonhamiltoinian, as are acylic graphs.