An "O" Solution

(for December 94 "O" Teaser)

The last issue of ONB Notes in December 1994 quickly followed on the heels of the Annual General Meeting, which was well attended, or was it? The warm-up question from the "O" Teaser wondered how many people attended the meeting if every person shook hands exactly once with every other person present and there were 55 handshakes. The answer is 11. Although checking the AGM minutes we see that there were actually 12 people in attendance (we did that just to keep the attendees honest in figuring this one out).

The second part of the "O" Teaser provided a network of passable paths within a park (essentially two connected pentagons) for which a control was placed at each intersection. The question was to determine the shortest path of travel for the individual who must vet the controls and run each path to ensure that they are passable. The answer is 2592 m. Since the park contains 10 odd nodes, a complete tour will require 5 separate journeys. This means that 4 back-trackings will have to be made. As many of these 4 as possible should be the short inner corridors, but experiment shows that at least 2 back-trackings must be made along outer corridors. The total tour will therefore be 10 inner walks plus 2 inner back-trackings of 100 m each, plus 10 outer walks plus 2 outer back-trackings of 116 m each. Total 2592. Such a vetting route could be commenced from any of the eight outside corners, an example being:

36-KR-35-KR-37-35-34-37-34-MN-35-36-MN-32-33-32-31-33-30-31-30-BL-33-MN-BL.

(Now if only we always had 4 months to decide on the best route choice...)


This page is maintained by: Benjamin Lee
Last updated: 1995-10-12