As the particles pass through it, they knock the silicon's electrons around in ways that can be used to track those particles' point of origin (or "vertex").
A Hamilton circuit is one that passes through each point exactly once but does not, in general, cover all the edges; actually, it covers only two of the three edges that intersect at each vertex.
This remarkable result derives from two preliminary theorems: (a) the areas of all triangles on the same base, whose third vertex lies anywhere on an indefinitely extended line parallel to the base, are equal; and (b) the area of a triangle is half that of any parallelogram (including any rectangle) with the same base and height.
The central vertex represents the identity.
Consider any nonroot vertex v.
Place a guard on each blue vertex.
If the backward-cone relation were chosen, then simultaneous events would be those seen simultaneously by an observer at the cone's vertex.
So a red edge from a vertex \(\rv\) representing \(\rs\) in \(S_{3}\) ends at a vertex representing \(\rs\rr\) and a black edge from \(\rv\) ends at a vertex representing \(\rs\rf\).
(Paths here are assumed not to contain any vertex more than once).
Each vertex v represents the word consisting of the letters encountered along the path from the root to v. The concatenation of u and v is the vertex arrived at by taking the subtree whose root is the vertex u, noticing that this tree is isomorphic to the full tree, and locating v in this subtree as though it were the full tree.
Furthermore instead of a tree we have the upper right quadrant of the plane, that is, N², rotated 135 degrees clockwise, with every vertex v at the top of a diamond whose other vertices are v0 and v1 at the next level down, and the identified pair v01 = v10 below both.
