The chromatic number is 4 because the graph contains a K4 (D, E, F, and G.)
Theorem. Any planar graph is 4-colorable.
The theorem was proved by exhaustive checking of over 1900 cases. A computer program was used to do it, which makes the proof difficult for other people to understand. People are still looking for a "human" proof.
The chromatic number is 2 because the graph is made up of paths and contains no cycles.
The chromatic number is 2. No, the graph is not planar. This graph is named K3,3 and is an example of a "bipartite" graph. The vertices of a bipartite graph can be divided into two groups in such a way that there are no edges between any two vertices that are in the same group. In the graph above, all the edges are between a vertex in the top group and a vertex in the bottom group. Thus all the vertices in the top group can be colored one color, and all the other vertices can be colored a different color.
A path has chromatic number 2, an even cycle has chromatic number 2, an odd cycle has chromatic number 3, K3 has chromatic number 3, and K6 has chromatic number 6.
Either every node must have even degree or there may be exactly two odd nodes.
There is an Euler trail in a K3 because every vertex has degree 2, but there is no Euler trail in a K4.
Yes, there is. Start at one of the odd nodes and go to the other.
Edges have been added to the graph so that there are only two odd nodes in the whole graph. Three of the nodes in the K4 have been given extra edges to make them even.
2, 8, 14, 16, 17, 19, 35, 42, 49, 50, 59, 67, 77, 86, 91
One, in both cases. Every node in a complete graph is adjacent to every other node.