[Saber Cherry]

Originally posted by LaFollet:
I wasn't particularly asking for the big picture of values, just the ones in the test case. (But I'd have to say that way #2 sounds very interesting. ) Cause the program I made came out with 4c 3d with a score of 110 (113 for the units and -3 for resources), and when I pluged in data from a game I'm playing now all it made was Black Hunters.
(which is why I'm now trying to force it to make a combined arms aproach)

Oooh, a working solver already!

I think that the best way to achieve a "combined arms bonus" is like this:

Value of 1st unit of a type=(base value)*f1
Value of 2nd unit of a type=(base value)*f2
Value of 3rd unit of a type=(base value)*f3

...etc, where f1, f2, f3...fn are decreasing. For example, if a "militia" had a base value of 10, and f1=2, f2=1.5, f3=1.33, f4=1.25, etc:

Value of queue:
1 militia = 10*2 = 20
2 militia = 20+10*1.5 = 35
3 militia = 35+10*1.33 = 48
4 militia = 48+10*1.25 = 61

This would result in a much more balanced queue, and the AI would be less likely to flood the battlefield with a single type of valuable unit that could be countered with a simple tactic.

A few examples of fn (the factor for the value of the nth unit of a type) are:

0) Currently, of course, fn=1 for all n.

1) fn = (n+x)/n
That's the example shown above, where x=1: fn = 2/1, 3/2, 4/3, 5/4... 1

2) fn = 2 if (n = 1), else 1
: 2, 1, 1, 1, 1... 1

3) fn = 1+(11-n)/10, minimum 1.
: 2, 1.9, 1.8, 1.7, 1.6, 1.5 ... 1.1, 1, 1... 1

Of course, there are other ways to achieve combined arms bonuses. But this one would be especially easy to add to a "brute force approach".

What do you think, should combined arms be added to the problem statement? If so, I would favor solution 1, fn=(n+x)/1, where "x" is specified in the specific case to solve. That would allow "x" to be specified in a problem instance as "0", or "no combined arms bonus".