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Figure 1
Example of a one-boundary virtual Cartesian point distance calculation. Only the VCP operation is shown, no reflections. Both points p1 and p2 are Selling reduced. The image is the three-dimensional, all-negative octant of the three [{\bf S^6}] axes, s1, s3 and s5; the reduction is done along the s1 axis, and s3 and s5 are the two scalars that will be interchanged. The points are shown above or below the s3/s5 plane, with their projections onto that plane marked with a +. To compute the minimal distance between p1 and p2, begin by computing the Euclidean distance between the two. The s1 reduction transforms p1 into [p_1'], but the metric changes when going from negative s1 to positive s1, so the simple Euclidean distance may not be minimal. To generate p1(VCP) to which the distance may be shorter, project p1 onto the s3/s5 plane, transform that projected point, and subtract s1 from that point. The distance from p1 to p1(VCP) can now be used to decide whether it is shorter than the p1p2 Euclidean distance. The best distance for this case is the shorter of the distances between p1 and p2 as opposed to the distance between p1(VCP) and p2.

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