Re: Infinite knight's tour?



Mike Williams wrote:
> Wasn't it Prai Jei who wrote:
> >
> >Here's one that's just come to mind. Is it possible to have a knight's tour
> >on an infinite board that consists of two opposite quarters, i.e. two
> >quarter-infinite boards that touch at their finite corners?
> >
> ><fixed pitch>
> >
> > | | | | | | | |
> > |---|---|---|---|---|---|---|--
> > | | | | | | | |
> > |---|---|---|---|---|---|---|--
> > | | | | | | | |
> > |---|---|---|---|---|---|---|--
> > | B | | | | | | |
> > |---|---|---|---|---|---|---|--
> > | C | B | | | | | |
> >-----------------------------------------------------------------
> > | | | | | | | D | A |
> >--|---|---|---|---|---|---|---|---|
> > | | | | | | | | D |
> >--|---|---|---|---|---|---|---|---|
> > | | | | | | | | |
> >--|---|---|---|---|---|---|---|---|
> > | | | | | | | | |
> >--|---|---|---|---|---|---|---|---|
> > | | | | | | | | |
> >--|---|---|---|---|---|---|---|---|
> > | | | | | | | | |
> >
> ></fixed pitch>
> >
> >At first I thought this must be impossible, since the knight can only - and
> >must - pass twice through the origin.
>
> I see that the knight can pass through the origin at most twice, but I
> don't see why it can't just pass once.
>
> For any pair of finite boards for which a closed knight's tour is
> possible on each board, it's easy to generate an open knight's tour on
> the pair of boards that only crosses the origin once.

It depends on what we call a knight's tour.

If it is a bijection f: (integers) - > (board positions), such that
f(n) is always a knights move away from f(n+1), then there need only be
one traverse of the origin.

If it is a bijection f: (naturals) -> (board positions), (i.e., if we
state that "the knight starts at sqaue whatever"), then the origin must
be crossed an infinite number of times. Suppose it crosses from the
initial quadrant the first time at move n_1; then there remain an
infinite number of moves to completed on the first quadrant. So the
knight must cross again at some other move, n_2, leaving an infinite
number of unmarked squares in the other quadrant. If there is a "last
crossing" there will still remain an infinite number of unvisited
squares on one or the other other quadrant.

Cheers - Chas


Cheers - Chas

.



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