Planar bipartite graph I know that the number of edges for a face in a bipartite is at least 4. Clearly, we cannot have 3, because we will need to move back to the other side to have face. But, I'm having difficulty in imagining a face of a planar bipartite graph. Where can I find an example of it?
On the other hand, I have found the following equality and inequality, but I couldn't grasp exactly what they are saying, if someone can elaborate for me them, I would be glad.
$$2 |E| = \sum_{i \geq 4} i F_i$$ where $i$ is the number of edges of the face $F_i$ and $F_i$ is the number of faces with $i$ edges. Clearly this is somehow consequence of the handshaking lemma, but why multiplying number of edges of a face with the faces give us this result? Shouldn't we use the degree of each vertex for handshaking lemma?
The second inequality is this one
$$4 \sum_{i \geq 4} F_i \leq \sum_{i \geq 4} i F_i$$
$$4 |F| \leq 2|E|$$
I mean, the second part is form the above equation, but first part the 4 confuses me. Is it because we need at least 4 edges for a face in bipartite graph? If yes, why do we need it actually in this inequality? Second question, why does this inequality hold? A clarification would be highly welcomed.
 A: Here’s a simple example of a planar bipartite graph:
           a  
          / \  
         /   \  
        b--c--d  
         \   /  
          \ /  
           e

It has three vertices, $a,c$, and $e$, in one part and two, $b$ and $d$, in the other. It has three faces: one is the triangular region bounded by the cycle $abcd$, another is the triangular region bounded by the cycle $bcde$, and the third is the external face bounded by the cycle $abed$. Each of these faces has $4$ edges and $4$ vertices.
To see why $$2|E|=\sum_{i\ge 4}iF_i\;,$$ note that each edge of a planar graph is adjacent to two faces, one on each side of it. Suppose that the graph has faces $f_1,\ldots,f_n$, and face $f_k$ has $e_k$ edges for $k=1,\ldots,n$. Then $e_1+\ldots+e_n$, the sum of the numbers of edges of the faces, counts each edge twice, once for each of the two faces adjacent to it. Thus, $e_1+\ldots+e_n=2|E|$. But $e_1+\ldots+d_n$ is the same thing as $\sum_{i\ge 4}iF_i$. $F_i$ is the number of faces with $i$ edges, so $iF_i$ is the total number of edges for all of the faces with $i$ edges, and when we sum over $i$, we get the total number of edges of all of the faces of the graph. The degrees of the individual vertices don’t enter into it.
In the example above, $F_4=3$, and $F_i=0$ for $i\ne 4$, so $\sum_{i\ge 4}iF_i=4\cdot3=12=2\cdot6$, which is as it should be, since there are $6$ edges.
The see why the inequality
$$4\sum_{i\ge 4}F_i\le\sum_{i\ge 4}iF_i$$
holds, notice that if $i\ge 4$, then $4F_i\le iF_i$. Since we’re summing only over values of $i$ greater than or equal to $4$, we therefore have
$$\sum_{i\ge 4}4F_i\le\sum_{i\ge 4}iF_i\;:$$
each term on the left is less than or equal to the corresponding term on the right, so the sum on the left must be less than or equal to the sum on the right. And of course
$$4\sum_{i\ge 4}F_i=\sum_{i\ge 4}4F_i\;.$$
Finally, $\sum_{i\ge 4}F_i$ is simply $|F|$, the total number of faces, since every face has at least $4$ edges: we’re just adding up the number with $4$ edges, the number with $5$ edges, and so on. Thus, $4\sum_{i\ge 4}F_i=4|F|$. Combining this with the inequality and the original equation, we see that $4|F|\le 2|E|$ (and hence $2|F|\le|E|$: the number of edges of a planar bipartite graph is at least twice the number of faces).
In the example above $|F|=3$ and $|E|=6$, so in this case the inequality is actually an equality: the number of edges is exactly twice the number of faces.
