Re: Helix, Ring, and Bracelet

On Feb 6, 8:09 pm, Bill Taylor <w.tay...@[EMAIL PROTECTED]
> wrote:
> "markste...@[EMAIL PROTECTED]
" <markste...@[EMAIL PROTECTED]
> wrote:
> > Winning conditions for Black:
> >   1. Form a ring
> >   2. Form an r-helix
> >   3. Form an l-helix and a bracelet
>
> > Winning conditions for White:
> >   1. Form a bracelet
> >   2. Form an l-helix
> >   3. Form an r-helix and a ring
>
> Quite so.   As I outlined some days ago with my "NZ" post!
>
> With considerably more accuracy, and examples shown.
>
> -- Bill of Board-games

No, Bill.  The problem that Torben identified in my earlier attempt
applies to your game as well.  I'll translate it for you in a way that
applies specifically to your game.

In your game N wins if he forms an N-S loop, and Z wins if he forms an
E-W loop.  In the case that neither player achieves his loop goal and
instead there are two helices, N wins if there is a negative gradient
and Z wins if there is a positive gradient.

The problem is that if Z makes an N-S loop *and* a helix with a
negative gradient, he hasn't achieved either of his goals but he
prevents N from achieving either of his.  Now (picking up where Torben
left off) if N makes a negative gradient line connecting N to S (not a
loop), Z will be unable to form either an E-W loop or a positive
gradient helix.  Now neither player can win.

Try to keep up with the discussion.

Oh and one more thing.  And I quote:

"OC it would be possible to play a similar game on a hexagonal-gridded
torus, but it would not be able to have isom****phically tasked winning
criteria - so it could never be a "fair" game, as is quadric NZ."

My new game, which is different from your game, and which doesn't have
the problem that Torben identified, is played on a hexagonally
tessellated grid. *And* my game has the property that a filled board
produces exactly one winner.

Any questions?