I sh'd've numbered the knights. The problem was (sh'd've been) How many moves needed to swap Si with si for i = 1,2,...,n. The 8 half moves you mention swap the knights, but leave them in a different order. R. On Sat, 16 Oct 2010, Maximilian Hasler wrote:

I'm sorry to bother you, but I don't see why the 3x3 case is not possible.

The following 8 half-moves seem to do it: 1. a1-c2 b3-a1 2. c1-b3 c3-a2 3. b1-c3 a3-b1 4. c2-a3 a2-c1

see https://docs.google.com/present/view?id=d34bxwj_519c2s3jcr9 or enclosed PDF for illustration.

Unless I'm wrong, at least n=3k seems as simple as n=3 (same moves k times). (Maybe there's a shorter solution, I did not search for.)

Maximilian

On Fri, Oct 15, 2010 at 3:53 PM, Richard Guy <rkg@cpsc.ucalgary.ca> wrote:

...

n pairs of knights on a 3 by n chessboard. I.e., to get from

S  S  S  ...  S          s  s  s  ...  s -  -  -       -    to    -  -  -       - s  s  s  ...  s          S  S  S       S

where  s,  S  are chess knights of opposite colors.  Not possible for  n = 1, 2, 3.

n = 4: 32 moves are necessary and sufficient. n = 5, 6, 7, ... are Rikki-Tikki-Tavi questions