"Tricky" questions on graph theory

Question

So , I am revising graph theory. I need to gain some help/ feedback for those, because at least to me they are tricky. They are supposed to be answered quickly , because they come from a tight timed - exam.( so I guess they come with either some strong theory requirements or smart "insight")

- There is a graph with 1871 vertices that is Eulerian and bipartite.True or false?

To be bipartite, there should be no cycle of odd length. If there is an Euler circle, it can be partioned to a set of disjoint circles ( and we need all of those to be of even length) . So , if we add up everything we should have an even number of edges so $\exists k$ such that $|Ε|=2k \rightarrow \frac{\sum_{i}^n deg(v_i)}{2}=2k \rightarrow \sum_{i}^n deg(v_i)=4k$. But I don't use anywhere the number of vertices . I end up nowhere..

What's the max value of edges for a simple (no - parallel edges), undirected graph with $n \geq 10$ that it is Eulerian and has at least two different Hamilton circles?

Inclass ,we proved that if n is even in order to be Eulerian it has at most $n(n-2)$ edges and if n is odd it has at most $\frac{n(n-1)}{2}$ .We basically know , that there is no bridge , eith cut vertex here . Hence, every edge belongs to a circle. It is Eulerian , hence every vertex has an even number degree. Adding edges , to a graph with Hamilton circles could do no harm into it's Hamiltonian property so it only impacts the Eulerian property. So we need to examine , how a graph with two Hamilton circles can be Eulerian. A Hamilton circle , does not repeat vertices so neither edges. If these 2 Hamilton circles are completely disjoint (each one is constructed by n edges) then the Euler one has at least 2n edges

Note: There was an extra question in the second one :Find the minimum number of edges for a simple (no - parallel edges), undirected graph with $n \geq 10$ that it is Eulerian and has at least two different Hamilton circles?

Answer 1

• There is a graph with 1871 vertices that is Eulerian and bipartite.True or false?

This is trying to fool you with the odd number of vertices, which means the 2 parts of the bipartition can't have equal size. But this is only a problem for Hamilton cycles, not Euler cycles.

This picture shows that you can create a bipartite Eulerian Graph for any odd number ( $\ge 7$) of vertices:

It doesn't matter that the regular red pattern ends on the same side as it started, you can just add 2 more (green) edges to get the Euler cycle.

• What's the max value of edges for a simple (no - parallel edges), undirected graph with $n\ge 10$ that it is Eulerian and has at least two different Hamilton circles?

For odd $n$ that's easy, it's the complete $K_n$ with ${n \choose 2} = \frac{n(n-1)}2$ edges.

For even $n$, at least one possible edge per vertex can't be in the graph (to make the degree of each vertex even). That means you need to remove at least $\frac{n}2$ edges from the complete graph. If you do this by pairing the vertices, you get a complete graph minus a perfect matching. That graph is Eulerian, and (because $n$ is big enough) still has 2 different Hamiltonian cycles). So the answer here is ${n \choose 2} -\frac{n}2= \frac{n(n-2)}2$

ADDED: Why do the graphs for the second example have at least two Hamiltonian cycles? Because they are for odd $n$ exactly, and for even $n$ almost the complete graph $K_n$! With a given set of vertices, the more edges you have the better (as you don't need to, but can use them for the Hamiltonian cycle). Those graphs have lots and lots of Hamiltonian cycles, 2 is just a very easy lower bound.

Let's look at the case of even $n$:

The construction above means all edges between points are in the graph, except the red "ladder rungs". I've just select 2 cycles in green on each side of the ladder and a few more edges (blue and purple).

You get one Hamiltonian if you start anywhere on the left side of the ladder, go up until you reach the first blue edge, then cross over to the right side with that blue edge, then take the downward "outer" green edge and go up again on the right side until you meet the other blue edge, cross back on it to the left side, take the left "outer" green edge to the bottom and go up until you get to your initial starting point.

You get a different Hamiltonian if you do this procedure again, but use the purple edges to cross from left to right and back.

It should be clear that there are so many way you can select the left and right points and choose how they are ordered in the green cycles that "just 2" Hamiltonian cycles is already a vast understatement for the minimum $n$ of 10.