Can a graph be both Eulerian and Hamiltonian?

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.

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What is an example of a graph which is both Eulerian but not Hamiltonian?

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.

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Which graph is neither Hamiltonian and Eulerian?

Some graphs lack both an Eulerian and a Hamiltonian cycle such as Star Graph. Star Graph contains no cycle.

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Is it possible for a graph to have a Hamiltonian cycle but not an Eulerian 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.

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Can a graph have both an Eulerian circuit and an Eulerian path simultaneously?

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.

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Euler and Hamiltonian Graph Problem - Graph Theory - Discrete Mathematics

21 related questions found

Is it true that every Eulerian graph is also a Hamiltonian graph?

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.

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Can a graph be both Euler path and circuit?

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.

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Is every Euler cycle a Hamiltonian cycle?

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.

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Can a graph have a Hamilton path but no Hamilton circuit?

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.

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Does Euler cycle imply Hamiltonian cycle?

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.

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Does every graph have a Hamiltonian circuit?

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.

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Does every graph have a Hamiltonian path?

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.

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Is a graph Hamiltonian if and only if its closure is Hamiltonian?

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.

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How do you determine whether the graph is Eulerian or Hamiltonian?

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.

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Can there exist a graph that is both Eulerian and is bipartite?

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.

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Is a graph always an Euler path if it has both odd and even vertices?

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.

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How do you determine whether the graph has a Hamilton circuit or 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.

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What are the rules for 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.

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How do you prove a graph does not have a Hamiltonian circuit?

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.

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Can a graph have more than one Hamiltonian cycle?

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.

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Does a Hamiltonian cycle have to visit every edge?

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.

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Can a graph that only has one vertex with an odd degree have an Euler path but not an Euler circuit?

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).

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Does a graph have an Euler circuit if it is connected and every vertex has even degree?

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.

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Can a connected graph have a Euler circuit if it has 4 even vertices and 4 odd vertices?

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.

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What graphs are not Hamiltonian?

A nonhamiltonian graph is a graph that is not Hamiltonian. All disconnected graphs are therefore nonhamiltoinian, as are acylic graphs.

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