Question 1176889: An Eulerian Path is a graph traversal that starts at some vertex and traverses every edge exactly once, but ends at some other vertex. Using all five vertices given below, create and draw a simple graph (no loops and no multiple edges) with exactly 6 edges, containing an Eulerian Path (Reminder: Your graph cannot be an Eulerian Circuit, i.e. it cannot start and end at the same vertex).
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Is it possible to create the Eulerian Path from problem #6 in such a way that the graph is a tree? If yes, draw the tree below. If no, explain why not below (circle: YES or NO ):
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! **Creating the Eulerian Path Graph**
To create an Eulerian path, we need exactly two vertices with an odd degree (number of edges connected to them). The rest must have an even degree.
Here's one possible graph configuration:
```
A
/ \
B---C
/ \
D-------E
```
* **Degrees:**
* A: 2
* B: 3
* C: 3
* D: 2
* E: 2
* **Eulerian Path:** We can start at B and end at C (or vice versa).
* Example: B-A-C-E-D-B-C
**Is it Possible to Create an Eulerian Path as a Tree?**
* **NO.**
**Explanation:**
* A tree is a connected graph with no cycles.
* An Eulerian path requires exactly two vertices with odd degrees.
* In a tree, the sum of all vertex degrees must be 2(n-1) where n is the number of vertices. In this case, 2(5-1) = 8.
* To have an eulerian path, the sum of the vertex degrees must be an even number.
* To have an eulerian Path, two vertices must have an odd degree.
* To have a tree with 5 vertices, we must have 4 edges.
* The problem asked for 6 edges. Therefore, this graph can not be a tree.
* Also, in a tree, any two vertices are connected by exactly one path. If we have two vertices with odd degree, and the other vertices with an even degree, then those two vertices with odd degree must be connected by an edge. If we have 6 edges, it will be impossible to create a tree.
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