Re: [this_is_spam] Re: computer-go: Ko

["Chris Fant" <Fantius@xxxxxxxxxxx>]

Obvious correction below:


> I don't think that is a counterexample.  Let's use some letters:
>
>  Assume that this position has been reached without captures:
>  ......
>  ..OX..
>  .O.OX.
>  ..OX..
>  ......
>
>  Black takes the ko by playing at 'a'.  Since that was the first stone to
> play on 'a', this is a new board.
>
>  ......
>  ..OX..
>  .Oa.X.
>  ..OX..
>  ......
>
> White takes the ko back by playing at 'b'.  Since that was not the the
first
> stone to play on 'b'
> , we can't be sure this is a new board.
> ......
>  ..OX..
>  .O.bX.
>  ..OX..
>  ......
>
>
> ----- Original Message -----
> From: "Gunnar Farnebäck" <gunnar@xxxxxxxxxxxxxxxxx>
> To: <computer-go@xxxxxxxxxxxxxxxxx>
> Sent: Wednesday, June 04, 2003 12:33 AM
> Subject: Re: computer-go: Ko
>
>
> > Peter Drake wrote:
> > > Is the following statement true?
> > >
> > > If a stone played at a point is the first stone to ever occupy that
> > > point, no subsequent full-board position can be identical to a
> > > preceding position.
> >
> > Definitely not, it doesn't even hold for a basic ko. I.e., assume that
> > this position has been reached without captures:
> >
> > ......
> > ..OX..
> > .O.OX.
> > ..OX..
> > ......
> >
> > Black takes the ko, playing the first stone ever at that vertex
> >
> > ......
> > ..OX..
> > .OX.X.
> > ..OX..
> > ......
> >
> > If O immediately recaptures we get a counterexample.
> >
> > /Gunnar Farnebäck
> >
> >
>
>
>