What is the maximum number of edges in a planar bipartite graph of order 2n that has n vertices from each side?
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Perhaps this may be of some use. – wormram Apr 27 '20 at 13:11
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Its not, because its not planar over there. But thanks. – dave Apr 27 '20 at 13:16
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In every planar graph we have (Euler's formula):
$$V - E + F = 2$$
Where $V$, $E$, $F$ is the number of vertices, edges and faces respectively. Since the bipartite graph has no odd length cycles, the smallest cycle (and thus face) has $4$ edges. The sum of numbers of edges of each face equals twice the number of edges, so:
$$4F \le 2E$$
Which in conclusion with previous equality yields:
$$E \le 2V - 4$$
And in case of a bipartite graph having $n$ vertices on each side we have:
$$E \le 4n - 4$$
This can be achieved by an easy construction - picture $n$ vertices in a row numbered $1a$, $2a$, ... $na$ from one set and the other (call them $1b$, $2b$, ..., $nb$) directly below them (like in the grid). Then connect $1a$ to $1b$, $2a$ to $2b$ and so on with straight lines ($n$ edges), then connect $1b$ to $2a$, $2b$ to $3a$ and so on, again with straight lines ($n - 1$ edges) creating a "zigzag" pattern. Finally, connect $1a$ to all vertices on the other side by curved edges over of the grid ($n - 1$ edges because $1a$ and $1b$ are already connected) and connect $nb$ to all vertices on the other side by curved edges under the grid ($n - 2$ edges because $nb$ is already connected with $na$ and $1a$).
Bartek
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What is the construction? I tried to find it but couldn't please see my other question... – dave Apr 27 '20 at 14:24
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You also wrote in your proof "no even length cycles", you meant odd right? – dave Apr 27 '20 at 14:33
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@dave yes, for $n = 1$ we can achieve more than $0$ edges by connecting the only two vertices with $1$ edge. That is because our only "face" in this case is quite strange. – Bartek Apr 27 '20 at 16:50
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There is a mistake I believe at the end of your construction: you meant that nb is connected to 1a and na. Isn't it? – dave Apr 27 '20 at 18:14
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