TITLE: Planar networks
QUESTION [5 upvotes]: Let $(b_1,...,b_n)$ be a sequence of positive integers. Define a $(b_1,...,b_n)$-network as the following graph $G$:

$V(G)=V_1\cup...\cup V_n$
$V_i\cap V_j=\emptyset$ ($1\leq i<j\leq n$)
$|V_i|=b_i$ ($1\leq i\leq n$)
$E(G)=\{\{u,v\}:u\in V_i,v\in V_{i+1} \text{ for some }i<n\}$

Example of the $(3,5,3)$-network:

It is straightforward that the condition $b_i\leq 2$ or $b_{i+1}\leq 2$ ($\text{for all }i<n$) is necessary for a $(b_1,...,b_n)$-network to be planar. Is it true that this is also the sufficient condition? If not, what the sufficient condition could be?

REPLY [2 votes]: The answer to the first question is no due to the following theorem which gives the necessary and sufficient condition.
Theorem. Let $G$ be a $(b_1,...,b_n)$-network, then:

If $n=1$, then $G$ is planar
If $n=2$, then $G$ is planar iff $b_1\leq2\vee b_2\leq2$
If $n\geq3$, then $G$ is planar iff $\forall(1<i<n):b_i>2\rightarrow b_{i-1}=b_{i+1}=1$

Proof. The first case is obvious. In the second case, if the condition doesn't hold, then $G$ contains $K_{3,3}$ which is not planar. Otherwise, $G$ is a subgraph of a planar graph $K_{2,x}$ $(x\geq1)$. Consider the case $n\geq3$. Suppose that the condition doesn't hold. Then for some $1<i<n$ we have $b_i>2$ and at least one of $b_{i-1}$ or $b_{i+1}$ is greater than $2$. Again, this implies that $G$ contains $K_{3,3}$. Finally, let's prove by induction that the last condition is sufficient for $G$ to be planar.
Base case. Suppose $n=3$ and the condition holds. The cases when $b_2=1$ or $b_1=b_3=1$ are straightforward. Now, if $b_2=2$, then $G$ is just $K_{2,b_1+b_3}$ which is planar.
Induction step. Suppose $G$ is a $(b_1,...,b_{n+1})$-network and the condition holds for $n$ ($n\geq3$). Then the condition $b_i>2\text{ }(1<i<n+1)\rightarrow b_{i-1}=b_{i+1}=1$ implies the $(b_1,...,b_n)$-network $G'$ is planar. There are two cases:

$b_{n+1}\geq2$. We have $b_n\leq2$. If $b_n=1$, then it's obvious how to use $G'$ to draw $G$ on a plane. Assume $b_n=2$. Then we have $b_{n-1}\leq2$. If $b_{n-1}=2$, consider the sets $V_{n-1}=\{u_1,u_2\}$ and $V_{n}=\{u_3,u_4\}$ from the definition. When drawn on a plane, $G'$ contains a face $F$ that is the cycle $u_1-u_3-u_2-u_4-u_1$. We can draw $G$ planarly putting $b_{n+1}$ vertices inside of $F$. If $b_{n-1}=1$, then both vertices from $V_n$ in $G'$ have the degree $1$, and it's straightforward to draw $G$ on a plane.

$b_{n+1}=1$. If $b_n>2$, then $b_{n-1}=1$ and all the vertices from $V_n$ in $G'$ have the degree $1$, so $G$ can be drawn on a plane. The case $b_n\leq2$ is analogous to the previous step.


$\blacksquare$
The following picture demonstrates why $(2,3,2)$-network can not be planar: