# How many glue flaps are needed for each shape's net?

So I was creating some dice from printed nets, and I noticed it doesn't matter where you put the glue flaps(1) around the net(2) of a d6(cube) you'll always end up with 7 flaps if the net can produce a valid cube: This also happens for d20(icosahedron), for which there can be many nets with different shapes, But still with the same number of flaps(11):  and d12(dodecahedron) which one would think should have a simpler net because the smaller number of faces, actually had even more flaps than d20 which made it much harder to craft: So this made me curious, is there a set number of flaps needed for a certain shape or this is a coincidence that I found nets with the same number of flaps?
if there is a certain number, is there a formula for it? and if there isn't, how can I make sure the net I make is the most efficient one available (there are other factors like the number of creases needed and having a least one flapless face for that, but we only talk about the number of the flaps for now).

Any additional information about nets and creatable shapes is appreciated. Internet seems void of any information on any net other than the net of a normal cube.

(1) A flap is used to glue two faces of the shape which aren't already connected in the net and every edge between two such faces needs a flap on one of the faces it connects. you can search for a video of a cube being made from the net to fully understand the flap's usage.
(2) Opened 3d shape on the 2d plane that can be folded to make the 3d shape.

• Not sure why you say there is only one net for a cube. $11$ nets are pictured here. Is there some restriction you're making that they aren't? – saulspatz Sep 9 '19 at 14:39
• @saulspatz My bad. I never saw the other nets anywhere before so I took that for granted without any additional research. Thanks for mentioning. edited – mpower Sep 13 '19 at 21:16

Let's consider the cube. A cube has $$12$$ edges. In the net, there are $$5$$ edges that join two faces, three horizontal and two vertical. The remaining $$7$$ edges have to be joined by glue flaps. In the icosahedron, we count $$19$$ edges joining two faces of the net, so the remaining $$11$$ edges of the icosahedron's $$30$$ edges will need glue flaps.
To see that the number of glue flaps is constant, think of the faces in the net as vertices in a graph, and two vertices as adjacent if they have a common edge in the net. This will be a tree. If we had a cycle, then there would be some point completely surrounded by faces in the net. That is, the sum of the angles of the faces meeting at that point would be $$360^\circ$$, which is impossible. Since the number of edges in a tree is one less than the number of vertices, we see that the number of edges in the net is one less than the number of faces in the polyhedron, and the number of glues flaps is $$e-(f-1)$$ where $$e$$ is the number of edges, and $$f$$ the number of faces, of the polyhedron.