Allan, The seventh move shd be c2a3, not b2a3. I sh'd've numbered the knights, S1,S2,S3 and s1,s2,s3. The problem was (meant to be) swap Si with si for i = 1, 2, ...., n. R. On Fri, 15 Oct 2010, Allan Wechsler wrote:
On the 3 by 3 board: a1c2, c3a2, b1c3, b3a1, c1b3, a3b1, b2a3, a2c1. Why is this not a solution for n=3? I think I remember seeing this in a Martin Gardner column: it uses the fact that knights moves link up the outer eight cells in a cyclic graph.
On Fri, Oct 15, 2010 at 3:53 PM, Richard Guy <rkg@cpsc.ucalgary.ca> wrote:
I don't believe that the following sequence is in OEIS.
The (least) number of moves needed to swap n pairs of knights on a 3 by n chessboard.
I.e., the number of chess knight moves needed 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 on p.250 (R187) of The Inquisitive Problem Solver.
R.
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