# Knight's Uncrossed Open and Polygons

In 1968, L. D. Yarbrough introduced a new variant on the classic problem of knight's tour. In addition to the rule that a knight touring a chessboard cannot visit the same cell twice, the knight is also not

permitted to cross its own path (the path consists of straight lines drawn between the center of the starting and ending square of every jump). The 6x6 tour found by Yarbrough can be improved from

16 to 17 jumps. It was discovered by a mathematician named Donal Knuth. This just proves that any Mathlete can improve on solutions by other mathematicians. So I challenged the Mathletes to improve on the former record for each Open and Closed Uncrossed Knight’s Tour for each grid size. If they improve on the old record, they get their faces on a Mathlete Dollar and win the 1,000,000 Mathlete Dollar.

What is the longest possible route without self-crossings for the chess knight to travel on the 3x3, 4x4, 5x5, 6x6, 8x8, and 10x10 board for an open and closed tour?

You will see on page one of the pdf, a chart with the records for each. For example, on the 8 x 8 board, the record for an open tour is 35 and for a closed tour is 32. My best on this board for a closed tour is 28. When you look at my solution, you see that it is a 28-sided polygon (straight sides, no crossing, and closed). This is called icosikaioctagon or you might call it octaicosagon. Each polygon has a name given it in Greek or Latin. The smallest number of sides a polygon can have is a triangle or trigon, next, a quadrilateral or tetragon, pentagon, hexagon, septagon (7) and so on. The children were fascinated by learning that September from septa is the 7th month, October the 8th, December the 10th, but …. Of course, we know that September is actually the 9th month, October, the 10th, and so on. This is because Julius Ceasar and Augustus Octavius, emperors of Rome wanted months named after themselves and added July and August moving the final four months back two.

They were able to find the names of 100 sided polygons by thinking of decade is to 10 years is to decagon (10 sides) as century is to 100 years is to 100 sides; then 1000 sides must be miligon. While they would be correct, it is actually called a chiligon.

I can’t wait to see their solutions.

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