Re: Classic game LUNAR and G=1E-03

["Michael J. Mahon" <mjmahon@aol.com>]

Nama wrote:
> On May 6, 12:45 am, -scooter- <scooter....@gmail.com> wrote:
> > > Ok...that makes sense.
> > > So one last question before I can say this port is finished:
> > > According to the document here (http://www.atariarchives.org/
> > > basicgames/showpage.php?page=106):
> > > (Q-Q*Q/2-Q^3/3-Q^4/4-Q^5/5)
> > > can be replaced by the notation
> > > -Q*(1+Q*(1/2+Q*(1/3+Q*(1/4+Q/5))))
> > > If that is the case, then what would be the replacement for:
> > > (Q/2+Q^2/6+Q^3/12+Q^4/20+Q^5/30)
> > Look up "Horner's Method" -- the quick and dirty explanation is that
> > Horner's is a linear time evaluation method for polynomials. It allows
> > you to avoid computing the exponents. However, there is a bug in that
> > expression -- it shouldn't start with -Q.
> >
> > Q*(1-Q*(1/2-Q*(1/3-Q*(1/4-Q/5))))
> > = Q-Q^2*(1/2-Q*(1/3-Q*(1/4-Q/5)))
> > = Q-Q^2/2-Q^3*(1/3-Q*(1/4-Q/5))
> > = Q-Q^2/2-Q^3/3-Q^4*(1/4-Q/5)
> > = Q-Q^2/2-Q^3/3-Q^4/4-Q^5/5
> >
> > BTW: Horner's method still shows up frequently on the Computer Science
> > Graduate Record Exam here in the US. And it's a favorite interview
> > torture question of mine, along with the XOR hack.
> >
> > -scooter
>
> So I should replace:
>
> -Q*(1+Q*(1/2+Q*(1/3+Q*(1/4+Q/5))))
>
> with
>
> Q*(1-Q*(1/2-Q*(1/3-Q*(1/4-Q/5))))

I think the leading minus sign is correct for this use of the
function.  Certainly the program will function differently if it is
removed, and I presume that it once worked.  ;-)

And both series computations can be replaced with simple LOG functions.
See other replies in this thread by myself and John.

-michael

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