I have some problems when I prove "For a planar graph $G$ and $\deg(v) = 3$ for any vertex $v$, there is a face with at most $5$ edges". I want to prove with contradiction.
Suppose that every face has more than $5$ edges.
Then $2e > 5r$ ($e$ is the number of edges and $r$ is the number of faces) because each edge occurs on the boundary of a face exactly twice.
Also, I can get $2e = 3v$. (for every edges has $2$ vertices)
So we have $3v > 5r$.
Then, with Euler's formular: $r = e – v + 1$, and $2e = 3r, 3v > 5r$ we can get $r > 20$.
But I cannot lead to a contradiction with the steps above.
Did I miss some points or my idea is not ok? Thanks.
Prove that a planar graph with all vertices degree $3$ must has a face with at most $5$ edges.
graph theoryplanar-graphs
Related Solutions
The discharge method works here. (You can forget this proof, but try to remember the method. It can be very powerful. It belongs in your toolbox next to the pigeonhole principle and the principle of induction).
Assume $G$ is a counterexample with a minimum number of vertices ($n$). We may also assume that we have maximized the number of edges ($e$). So $G$ has minimum degree at least three and $d(u)+d(v)>13$ for every edge $uv$. The edge maximization implies that if there are any non-adjacent $u$ and $v$ with $d(u)+d(v)>13$, then $G+uv$ is not planar.
Now we assign a charge to each vertex: a vertex $v$ gets charge $6-d(v)$. Applying the formula $e\leq 3n-6$ (equality only holds for triangulations), we find the total charge $\sum_v(6-d(v))=6n-2e\geq6n-(6n-12)=12$.
Now we discharge: a vertex $v$ with a positive charge distributes its charge evenly among its neighbours. We do this for all vertices simultaneously.
Since no charge is lost, there must still be vertices with a positive charge.
Let $v$ be a vertex with a positive charge.
Case 1: $d(v)\leq 8$. Since a vertex can only obtain a positive charge from a neighboring vertex $u$ with degree at most 5, we find an edge $uv$ with $d(u)+d(v)\leq 13$. Contradiction.
Case 2: $d(v)=9$. $v$ has no neighbours of degree 3 or 4, so its positive charge can only be obtained from neighbours of degree 5 and each of them contributes a charge of $\frac 15$ (they started with charge 1 and sent an equal part, i.e. $\frac 15$ to each of their neighbours). Since $v$ started out with charge $-3$ it would need 16 neighbours to end up with a positive charge. Contradiction.
Case 3: $d(v)=10$. As case 2: no neighbours of degree 3, and each neighbour of degree 4 or 5 contributes at most charge $\frac 12$. Since $v$ started with charge $-4$, it needs at least 9 neighbours of degree 4 or 5. Let $v_1,\ldots,v_{10}$ be the neighbours of $v$, arranged clockwise. Without loss of generality we may assume that $v_{10}$ is the only vertex that (possibly) has a degree larger than 5. Since $v_1$ and $v_2$ cannot be adjacent (since $d(v_1)+d(v_2)\leq10$), the face containing $v,v_1,v_2$ contains at least one other neighbour $u$ of $v_1$. The degree of $u$ must be at least 9 (or $d(v_1)+d(u)\leq13$). If $u$ is not adjacent to $v$, we can add edge $uv$, contradicting the edge-maximality. So $u$ is another neighbor of $v$, which implies $u=v_{10}$. We have shown that $v_1$ is adjacent to $v_{10}$. Exactly the same argument shows that $v_2,\ldots,v_9$ are adjacent to $v_{10}$. Now consider the adjacent "4-gons" $vv_2uv_1$ and $vv_2uv_3$. Since $d(v_2)\geq4$, $v_2$ must have neighbours that are inside one of these 4-gons, and not neighbours of $v$. Without loss of generality we may assume that $v_2$ has clockwise neighbours $w_0=v,w_1,\ldots,w_t,w_{t+1}=v_2$ where $t\geq1$ and $w_1,\ldots,w_t$ are inside $vv_2uv_1$. Since $d(w_1)\geq3$ we can now add edge $vw_1$ and again contradict edge-maximality.
Case 4: $d(v)\geq11$. Each neighbour of degree <6 contributes at most 1. $v$ starts with charge $6-d(v)$, so it needs at least $d(v)-5$ neighbours of degree less than 6 to end up with a positive charge. Since $d(v)-5>\frac{d(v)}2$ we again must find two consecutive neighbours of low degree and, like in case 3, these must either be adjacent or expose a high-degree neighbour that can be connected to $v$ without breaking planarity. This final contradiction finishes the proof.
For the counterexample we can repeat the same arguments and find that there must be at least 12 vertices of degree 10, each of them surrounded by 5 vertices of degree 3.
This immediately brings up the idea of the icosahedron and indeed, if you take the icosahedron, and add a vertex in the center of each triangle face and connect it to the three corners, you get a very symmetric planar graph with 20 vertices of degree 3 and 12 of degree 10. No two vertices of degree 3 are adjacent, so this is our desired counterexample.
Edit to add: a stellated icosahedron:
Let $\mathcal F$ be the set of faces. Clearly $2E=\sum_{f\in \mathcal F} s(f)$ where $s(f)$ is the number of edges in the face $f$ (this is because every edge is adjacent two exactly two faces and every face is adjacent to $|f|$ edges).
in this particular case we get that $2E=\sum_{f\in \mathcal F} k= kF\iff F=\frac{2E}{k}$.
Now use euler's formula: $V-E+\frac{2E}{k}=2\iff \frac{2-k}{k}E=2-V\iff E=\frac{k(V-2)}{k-2}$
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Best Answer
If $G = (V, E)$ is a planar graph with $|E|\geq g$ and no cycle of length $< g$, then $$e \leq \frac{g}{g − 2}(v − 2)$$. This is a standard generalisation of the result $e \leq 3v-6$ which can be proved easily.
In this problem $g=6$, so we get $e \leq \frac{3}{2}(v-2)$, but since all vertices have degree $3$, number of edges is exactly equal to $\frac{3v}{2}$, hence a contradiction. Hence the graph must have a face with at most 5 edges.