[Jprogramming] integers on a pentagon puzzle

[Devon McCormick]

Oh, I see.  I was assuming in the second step that you could make X the new
Y but
in fact its adjacency to a fourth vertex does propagate the changes around
the pentagon.
I was thinking I could keep the other two numbers unchanged.

I'll have to think about this some more.

On 5/25/06, Miodrag Milenkovic <miodrag.milenkovic at gmail.com> wrote:
>
> sorry, haven't even noticed the last mistake. you can't do what you
> did in the last step, you can only apply a transformation centered on
> a negative number. in the last step you are starting with y = 1.
>
> even though one transformation involves only three adjacent numbers,
> the total number of vertices does matter.
>
> On 5/25/06, Devon McCormick <devonmcc at gmail.com> wrote:
> > I only looked at three numbers because that's all you have to consider
> for a
> > given
> > transformation.  Just use the other 2 numbers to make the total sum > 0,
> e.g.
> > the
> > other 2 numbers are 1 & 2.
> >
> > On 5/25/06, Miodrag Milenkovic <miodrag.milenkovic at gmail.com> wrote:
> > >
> > > hmmm, not corresponding to the problem as stated on two counts:
> > >
> > > 1. there are only three numbers
> > > 2. the sum is not > 0
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> >
> >
> >
> > --
> > Devon McCormick
> > ^me^ at acm.
> > org is my
> > preferred e-mail
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> >
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>



-- 
Devon McCormick
^me^ at acm.
org is my
preferred e-mail