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[Freeciv-Dev] Re: Comments on CMA 2.6
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[Freeciv-Dev] Re: Comments on CMA 2.6

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To: Christian Knoke <ChrisK@xxxxxxxx>
Cc: freeciv-dev <freeciv-dev@xxxxxxxxxxx>
Subject: [Freeciv-Dev] Re: Comments on CMA 2.6
From: Raimar Falke <hawk@xxxxxxxxxxxxxxxxxxxxxxx>
Date: Tue, 23 Oct 2001 18:12:51 +0200
Reply-to: rf13@xxxxxxxxxxxxxxxxxxxxxx

On Tue, Oct 23, 2001 at 02:50:04PM +0200, Christian Knoke wrote:
> Am Dienstag, 23. Oktober 2001 14:15 schrieb Raimar Falke:
> > On Tue, Oct 23, 2001 at 01:41:28PM +0200, Christian Knoke wrote:
> > > Am Montag, 22. Oktober 2001 23:28 schrieb Raimar Falke:
> > > > On Mon, Oct 22, 2001 at 10:51:45PM +0200, Christian Knoke wrote:
> > > > > For that case, I suggested a 0 (zero) weight. Everything
> > > > > considered as not important can have a 0.
> > > >
> > > > Technically the core CMA needs non-zero weights. However the
> > > > interface doesn't have to use these weights. It may map them to
> > > > other using a different scale of example.
> > >
> > > The CMA core should have the ability of ignoring a sort of
> > > production at all. E.g. in my games a food surplus is often wasted
> > > *at* *all*. Is a 0.01 weight possible?
> >
> > Yes. Just multiply all weights by 100. So you have core weights
> > (which are ints and always >1) and floating point interface weights.
> 
> Not the interface, the core. Is the core able to ignore e.g. food
> at all in the maximize calculation?

No why should it? Which one should CMA use if you have two
combinations which only differ in food. It chooses the one with the
bigger food value. If you want to use a stat only as such a tiebreaker
multiply all other stats with 100 or 1000.

> > > > So what about the surplus (the left sliders)?
> > >
> > > Am Sonntag, 21. Oktober 2001 19:39 schrieb Daniel L Speyer:
> > > > On Sun, 21 Oct 2001, Christian Knoke wrote:
> > > > > > > Why not: (negative, 0, positive) three possibilities, where
> > > > > > > positive is just (+1).
> > > > > >
> > > > > > No. You may need +2 or +3 to finish your unit in n turns. So
> > > > > > a slider is still needed.
> > > > >
> > > > > Yes, but really really seldom. You'd rather buy. Too much
> > > > > hassle with the agent. In high production phases, you would
> > > > > need to set +30 e.g.
> > > >
> > > > I do this sort of thing very often.  If I'm using potentially
> > > > high-prod cities where I cant afford to buy, or if I want enough
> > > > armor or cruisers for a major assault some time in the next ten
> > > > turns.  I would definitely want CMA to support this sort of
> > > > strategy.
> > >
> > > So if Daniel and others really need this - it means changing CMA's
> > > setting for a single turn - why not take the needed shields from
> > > the city info itself? That means that the CMA will pass control the
> > > very round after, but I think that is intended. So:
> > >
> > > -------------------------------------------------------------------
> > >-
> > >
> > > Food:   X allow loss   X no loss   X surplus
> > >
> > > Prod:   X allow loss   X no loss   X finish current
> >
> > So "finish current"=shield_stock-cost?!
> 
> I don't know. "finish current" == The CMA shall guarantee that the
> city produces at least so many shields that it can finish the current
> production in the next round.

And what happens if the city can produce this many shields?

> > > Gold:   X allow loss   X no loss
> > >
> > > ...
> > >
> > > -------------------------------------------------------------------
> > >-
> > >
> > > For the right side sliders I fancy a slider with values n of
> > > (0, 1, 2, 3, 4, 5) mapped according weight = e ^ (( n - 1) / 2) to
> > > (0, 1, 1.6, 2.7, 4.5, 7.4).
> >
> > Why?
> 
> According to my previuos posts: 0 is to ignore a stat, 1 is the base
> factor for other stats, 7.4 is a good approximation to the maximum
> ratio probably ever needed to weight a very important stat over other
> stats (e.g. production vs. gold when a bomber is being produced), and
> 1.6, 2.7, and 4.5 are the steps inbetween, giving a total of 6 steps
> which gives reasonable control and is still handy. pfhhh

Ok. But why e ^ (( n - 1) / 2) and not just plain linear? Do you have
any deeper knowledge? Or do you just feel that this is better?

        Raimar

-- 
 email: rf13@xxxxxxxxxxxxxxxxx
 1 + 1 = 3, for large values of 1


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