I am having troubles with this question. Seems non-trivial as I see it. The tightest lower-bound I could reach was "at least n/12" number of nodes in the independent set, with the assumption that all nodes have degree at most 11. Can someone help me with a line of reasoning?
2026-03-27 00:05:56.1774569956
Argue that in a planar graph on n nodes there is an independent set of at least n/8 nodes, each of degree at most 11.
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The above attempt is not the right approach. One can imagine a solution as follows.
We have that in a planar graph G, there are at least n/2 nodes with a degree at most 11. This can be proved using contradiction. Assume there are > n/2 nodes with degree >=12. Then, we have the sum of degrees for the nodes in the graph to be 6n. This contradicts the result that the sum of degrees in a planar graph is at most 6n-12.
Next, we let A to be a set of size n/2 with each degree at most 11. Let G' be the sub-graph induced by these nodes. Following the four colour theorem, we know that we can color sub-graph G' with 4 colours. We also know that the colour with the maximum number of nodes will have at least |A|/4 number of nodes by pigeon hole principle. Further, nodes of a single colour in a colouring scheme would be independent by definition. Hence, we have an independent set with at least n/8 nodes, each with degree at most 11.