[Freeciv-Dev] Re: [RFC] Tech costs and AI problem

[Gregory Berkolaiko <gberkolaiko@xxxxxxxxxxx>]

 --- Raimar Falke <hawk@xxxxxxxxxxxxxxxxxxxxxxx> wrote: 
[...]
>     k = pplayer->ai.num_unknown_techs[unit_types[i].tech_requirement];
>     ...
>     l = k * (k + pplayer->research.techs_researched) *
> game.researchcost /
>        (game.year > 0 ? 2 : 4); /* cost (shield equiv) of gaining these
> techs */
> 
> K is the number of techs which have to researched till the goal tech
> (unit_types[i].tech_requirement) _can_ be researched/is reachable.
> 
> Note that currently
> 
> int total_bulbs_required(struct player *pplayer)
> {
>   int timemod = (game.year > 0) ? 2 : 1;
> 
>   return timemod * pplayer->research.techs_researched *
> game.researchcost;
> }
> 
> So I assume that the condition in the AI code can wants to mimic this.
> 
> I want to compute l from a call to total_bulbs_required_for_goal
> (which is the new method for this).
> 
> If we assume that
>   s=pplayer->research.techs_researched,
>   rc=game.researchcost and 
>   tbrfg=the result from the total_bulbs_required_for_goal 
> than the old method will compute:
> 
>  l_old = (k*k + k*s) * rc

you mean l_old = (k*k + k*s) * rc / 4 ?
factors of 2 are my weak point :(

> the new method calculates correctly:
>  tbrfg = (0.5*k*k + (s - 0.5)*k) * rc
> 
> So for large k's a good approximation would be "l_new = tbrfg / 2" but
> this
> fails for small ks:
>  l_old(k=1) = (s + 1)*rc
>  l_new(k=1) = (s*rc)/2
> 
>  l_old(k=2) = (2*s + 4)*rc
>  l_new(k=2) = ((s + 1)*rc)/2
> 
> I would have no problems if these numbers wouldn't get mixed with
> build_costs and the number of cities.
> 
> So any advise from the "AI-people"?

Prediction: Ross will say "don't worry too much about it, it's just some
fuzzy weighting".  And he will be right :)

I agree with you that the right formula is
  tbrfg = (0.5*k*k + (s - 0.5)*k) * rc
althoug I prefer the traditional way
  tbrfg = (s*k + k*(k-1)/2) * rc  
both because it's traditional and because it's obvious that the result is
integer.

I expect variable "l" to be useful only when s is big and when k is small
(because the tech tree is comparatevely wide and because AI shouldn't
pretend it is standing on the shoulders of giants).  So if you want to
keep your approximation as close as possible to the wrong one, aim to get
the s prefactor right (sort out the factors of 2 too).

BTW, in "AI-people", are quotation marks expressing sarcasm? ;)

G.

P.S. I reread my email and found that it's full of pretentious rubbish. 
The message in clear English is "useful ranges are: k is of order 1 (1, 2
or 3), s is of order 10 (20, 30 or more)"

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