Reasoning behind the formula to determine the number of fused polygons

Question

I'm trying to find the reasoning behind the following formula to determine the number of fused polygons $N$, which I found in my Organic Chemistry textbook:

$$S-A+1=N$$

where $S$ is the number of edges, and $A$ is the number of vertices.

I verified the above formula for different combinations of polygons, and it worked flawlessly in all cases I considered. The following image shows two of the different configurations I considered and the application of the above formula giving the corresponding number of fused polygons:

It works irrespective of the number of common edges. Is this formula an empirical (just based on observations) or is there any good reason behind it? Further, are there any exceptions to this formula?

Edit:

Example to address the comments below the question:

$30-24+1=7$

Answer 1

If there are holes, then the formula does not hold as I've already commented.

Without holes, the formula holds.

In the following, let us consider fused polygons without holes.

The reason why the formula holds is that if you add a new hexagon to fused polygons, then the following relation $$(\text{the number of the added edges})-(\text{the number of the added vertices})=1$$ holds.

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One can prove that the formula holds by induction on the number of polygons.

Proof :

If we consider one hexagon, then $S=A=6$, so $S-A+1=N$ holds.

Suppose that $S_F-A_F+1=N$ holds in the fused polygons $F$.

• Case 1 : If you add one hexagon $H$ to $F$ where $H$ and $F$ has exactly one common edge, then in the new polygons $F'$, we have $$N_{F'}=N+1,\quad S_{F'}=S_F+5,\quad A_{F'}=A_F+4$$So, we see that $S_{F'}-A_{F'}+1=N_{F'}$ holds.

• Case 2 : If you add one hexagon $H$ to $F$ where $H$ and $F$ has exactly two common edges (note that the two edges have to be adjacent), then in the new polygons $F'$, we have $$N_{F'}=N+1,\quad S_{F'}=S_F+4,\quad A_{F'}=A_F+3$$So, we see that $S_{F'}-A_{F'}+1=N_{F'}$ holds.

• Case 3 : If you add one hexagon $H$ to $F$ where $H$ and $F$ has exactly three common edges (note that the three edges have to be adjacent), then in the new polygons $F'$, we have $$N_{F'}=N+1,\quad S_{F'}=S_F+3,\quad A_{F'}=A_F+2$$So, we see that $S_{F'}-A_{F'}+1=N_{F'}$ holds.

• Case 4 : If you add one hexagon $H$ to $F$ where $H$ and $F$ has exactly four common edges (note that the four edges have to be adjacent), then in the new polygons $F'$, we have $$N_{F'}=N+1,\quad S_{F'}=S_F+2,\quad A_{F'}=A_F+1$$So, we see that $S_{F'}-A_{F'}+1=N_{F'}$ holds.

• Case 5 : If you add one hexagon $H$ to $F$ where $H$ and $F$ has exactly five common edges, then in the new polygons $F'$, we have $$N_{F'}=N+1,\quad S_{F'}=S_F+1,\quad A_{F'}=A_F$$So, we see that $S_{F'}-A_{F'}+1=N_{F'}$ holds.

From the five cases above, the formula holds for $N+1$. $\quad\square$

Answer 2

You can prove this equation by induction on the number of edges: When you remove an edge, you reduce either the number of vertices or the number of (closed) polygons by one.

This is a specific case of the more general fact that the Euler characteristic is a topological constant. You can relate it to the more well-known case $V-E+F=2$, where $V$, $E$ and $F$ are the numbers of vertices, edges and faces of a polyhedron, respectively, by applying stereographic projection to map the plane to the unit sphere and noting that the infinite domain outside the polygons gets mapped to an additional face of the resulting polyhedron.