## Comment: Re:Conveyor belt problem... (Score 1) 60

Why? The only issue is the belt convexity. If the new pulley would pull the belt away from the intermediate pulley if it ran on the outside one can simply switch it to the inside. It cannot be pulled away on both sides. The point is that the run of the belt can be made topologically equivalent to a line drawn between all of the pulley centers in the configuration generated by a distance-ordered recursion from any starting point. All one has to prove is that for a trivial set of local geometries, one can always find a side of the pulley that maintains tension upon addition. The diameter of the pulley ends up being functionally irrelevant.

Perhaps I'm not being sufficiently clear. Take a configuration of N points/pulleys presumed to be spanned by a belt that was systematically structured by starting on the pulley closest to the center of the bounding circle. Add one pulley outside or on the bounding circle of the configuration (one can always reorder the problem to ensure that the N+1 pulley is outside or on this bounding circle). Pick the two (most distant from the center, if there are more than two) pulleys that bracket the new pulley inside rays drawn from the center of the circle through the pulley center. One can always loop in the new pulley in between the two thus selected, and one always does so without adding a loop that "occludes" a future distance-ordered addition. The insistence of maintaining rank order and radially ordered convexity as one proceeds suffices to ensure that one can always add a pulley to a suitably developed set of N pulleys.

Or, maybe I'm missing something, but when I draw sequences of points in this way there aren't really a lot of cases to consider on the addition. The convexity requirement eliminates, I think, your assertion of "distant points" for the nearest neighbors. They cannot be more distant than the diameter of the bounding N circle, and the construction ensures that one does not build a loop that twists around in some snaky way across angles. By posing the solution in this way, I simply avoid having to consider *going back* to do nonlocal rearrangements of some arbitrary looping selected from the (probably quite large) set of loopings that would work for any set of N pulleys. One really only needs to consider the two bracketing pulleys and the two next neighbors of those pulleys with the comfortable constraint that the next nearest neighbor has a belt that run at an angle of pi or less relative to the exterior of the radius of a/the bracketing pulley in question. Four permutations of possible side swap to consider, done, induction proven.