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Answer : In your second image, the floating disconnected vertex is throwing off your total. Euler's Theorem only applies to connected graphs -- otherwise you could arbitrarily add as many isolated vertices as you want and make F − E + V F-E+V F − E + V come out to any arbitrary whole number greater than or equal to 2 2 2 . The last face is the entire plane that isn’t in the other faces you mentioned, which they called the outer, infinitely large region. In the case of a single point, it is the entire plane. A face can be thought of as a region of space such that you ca go anywhere in the region without having to cross over a line. Also, Euler's Theorem only applies to connected graphs. As an example, here is a planar graph representing a cube. You'll notice that the left image tells you to remove the bottom, but it should say that the area that isn’t labeled "side" or "top" is the bottom of the cube.

Answer : Any rigid framework, hence all regular polygons, can be converted to a triangle-free equivalent. Simply chaining copies of the 12 12 12 -vertex triangle-free braced square shown in the question (which I discovered) along the two collinear edges gives a rigid line segment of arbitrary whole number length without triangles: Then any triangular grid can be mimicked without triangles as follows (all straight fuchsia edges are made with the graph chaining construction above, all black edges are single sticks): For example, to brace the hexagon without triangles: However, the above hexagon bracing is quite big. Another approach to triangle-free bracing is the virtual edge : in any embedding of the cubical graph with one edge removed, the distance between the two degree- 2 2 2 vertices (incident to the missing edge) must always be 1 1 1 . This leads to the following triangle-free rigid regular hexagon in 16 16 16 vertices and 29 29 29 edges (Shibuya commit proof): Th...

Answer : UTU = Universally throughput unlimited There exist UTU balancers for 3, 4, and 5 belts, and likely for any number of belts . Examples are given in this Jupyter notebook, along with a Python implementation of an iterative algorithm for computing the flow for any set of belts and splitters . I will quote some of the notebook here for those who don't wish to follow the link. I refer to splitters as junctions, since that's how they really function. Each belt is composed of unit length segments, referred to in the game as tiles. In the model, we imagine each belt flowing from left to right along coordinate direction x x x and each junction uniting two belts at the left edge of a given tile. Note that some balancers used in practice contain loops and could not be represented by our model without an infinite length belt. The state of a belt tile is represented by a density 0 ≤ ρ ≤ 1 0 \le \rho \le 1 0 ≤ ρ ≤ 1 and a velocity 0 ≤ v ≤ 1 0 \le v \le 1 0 ≤ v ≤ 1 . T...