23.4. Triangulations of Convex Polygons

In question number 4 we are interested to find out the number of triangulations of a convex polygon. So, let T denotes the number of triangulations of a convex polygon with n + 2 sides and what is basically a triangulation: it is the process of dividing a convex polygon by non intersecting diagonals.

Triangulation is the process of dividing a polygon into smaller triangular regions using diagonals that do not intersect each other. For a convex polygon with n sides, if we denote T as the number of ways to do this, we note that a polygon with n + 2 sides can be triangulated into simpler parts through this method. Understanding triangulation is crucial in computational geometry as it aids in various applications like rendering graphics or analyzing shapes.

Imagine you have a flat sheet of paper shaped like a convex polygon. If you wanted to create more stability when applying pressure (for instance, by folding), you would make small triangles out of that polygon by folding lines that connect non-adjacent corners together, thus reducing the shape into simpler, stable parts.

So, now I want to find out a recurrence relation or want to find out the number of ways of triangulating a convex polygon with n + 2 sides so I will find that by formulating a recurrence equation and by showing that the solution for that reference equation is same as the nth Catalan number.

To express the number of triangulations of a convex polygon with n + 2 sides, we can use a recurrence relation which relates the number of triangulations of smaller polygons to the larger polygon. This is done by recognizing that if you fix one edge (say between vertex n + 1 and n + 2) and choose a third vertex as part of the triangle, the remaining vertices will form two smaller polygons, each of which can also be triangulated. This relationship leads us to the conclusion that the number of ways to triangulate an n + 2 sides polygon can also be described by the nth Catalan numbers, well-known in combinatorial mathematics.

Think of triangulating a large field (the polygon) where you need to set up tents for an event. By placing two tents along the boundary (the fixed edge) and using one of the remaining points (one vertex) as another tent, you break the large field into two smaller areas, each needing its own arrangement of tents. Thus, counting the ways for the whole area involves counting the arrangements for both smaller ones, just like in triangulation of polygons.

Now to solve my problem of triangulating a convex polygon with n + 2 sides into smaller problems I consider an arbitrary edge, so for simplicity I take that arbitrary edge to be the edge or the side v and v , namely the side with end points v and v .

In order to formulate the problem of triangulating a convex polygon, we start by fixing an arbitrary edge—the line connecting two vertices. From this fixed line, we can select any other vertex from the remaining vertices to form a triangle. Each triangle formed allows us to additionally triangulate the two resulting smaller polygons independently, leading to a summation over all possible selections—it’s a matter of choosing the right vertex to connect with.

Consider a triangular pizza slice. If you start by keeping the edge of the pizza slice fixed while selecting a topping to add your pinwheel sandwich to the top, you can visualize how this processes “splits” the remaining toppings to focus on—to either fill the left or right side of the triangle accordingly.

But now my k can range from 1 to n, my k could be vertex number v , my k could be vertex number v and so on. So, if I take the summation over k being equal to 1 to n then I get the total number of triangulations for n + 2 sided convex polygon.

The final step in deriving the total number of triangulations involves summing the results from all possible vertex choices (denoted by k) for the triangle we are forming. Each combination gives us unique triangulations, and since k can take multiple values from 1 to n (each representing a specific configuration), the summation of all possible combinations will yield the total triangulations for the polygon.

Picture a group of friends making a chain of paper airplanes, with one airplane for each vertex of a polygon. Each friend (or vertex) can have multiple configurations for how they can sit next to each other to create triangles. Summing all those configurations gives you not just one way to form them but how many ways you can arrange the entire group.

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