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The constraint is not on the number of neighbors directly. If there is just one node (say, A) with 4 neighbors (say B1, B2, B3, B4), we can color A red and B1,
by meuk 7y ago
The constraint is not on the number of neighbors directly. If there is just one node (say, A) with 4 neighbors (say B1, B2, B3, B4), we can color A red and B1, B2, B3, B4 blue. It only becomes a problem when all nodes are interconnected (i.e. the graph is complete). So, the result does imply that graphs which are not colorable with 4 colors are not planar. In particular, the complete graph (where all nodes of the graph are connected) with 4 nodes is not planar.
- thaumasiotes 7y agoThe complete graph with 4 nodes is planar, and obviously can be colored with only 4 colors -- that gives every node a unique color. The complete graph with 5 nodes is what isn't planar. (That, and the complete bipartite graph with 3 nodes on each side.)
- deleted 7y ago[deleted]