Quote:
Original post by InnocuousFox
*facepalm*
Yeah, a little (it does seem like he's uncomfortable with the idea of applying Bayes' rule, but in honesty it's also enough of a pain in the ass that I haven't sat down to figure it out myself either) but... how
do you solve an extensive-form Bayesian game with signaling? In general; forget practical "but it'll take 1000 years" arguments. I mean, I don't know how you'd do it; do you?
I think this is something we'd do well to figure out.
See e.g.
Link 1Link 2Link 3...etc.
I had assumed that things generalized from games of perfect information to games of imperfect information in the same way that MDPs generalize to POMDPs -- especially since solving Markov Games is almost the same as solving MDPs. That's where I'd been going with my "this is the gamestate" post. But now I'm really not so sure that's enough.
I've studied a little game theory, but not enough to help me answer this question.
We should figure this out.
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Unrelated to above, but useful. This is a Chebyshev (not Bayes) estimator for the number of remaining pieces of each type.
Suppose that there were 4 kinds of pieces (it's easy to generalize to the full game, but I only feel like writing 4), with initial counts N1,...,N4.
Now suppose you know that
- pieces of rank 1 have made k1 kills (including equal-value annihilations)
- ...
- pieces of rank 4 have made k4 kills ("")
and you want to estimate the values
- 'n1,' the number of pieces of rank 1 remaining
- ...
- 'n4,' the number of pieces of rank 4 remaining
Well, we know the following:
[These inequalities have been edited.]
1.) The numbers of remaining pieces are within bounds:
0 <= n1 <= N10 <= n2 <= N20 <= n3 <= N30 <= n4 <= N4
2.) At least as many of your enemy's low-rank units have been killed as your low-rank units have gotten kills.
k1 + k2 + k3 + k4 <= (N1 + N2 + N3 + N4) - (n1 + n2 + n3 + n4)k1 + k2 + k3 <= (N1 + N2 + N3 ) - (n1 + n2 + n3 )k1 + k2 <= (N1 + N2 ) - (n1 + n2 )k1 <= (N1 ) - (n1 )
3.) Your high-rank units are responsible for at least the high-rank kills:
k1 + k2 + k3 + k4 >= (N1 + N2 + N3 + N4) - (n1 + n2 + n3 + n4) k2 + k3 + k4 >= ( N2 + N3 + N4) - ( n2 + n3 + n4) k3 + k4 >= ( N3 + N4) - ( n3 + n4) k4 >= ( N4) - ( n4)
This system of inequalities defines a bounded, convex set in R^4. One estimate for how many pieces remain is the "center" of this set; this is called the Chebyshev center. It can be found in the following way.
Letting n=(n1,n2,n3,n4), write the above inequalities and equations as
A n - B <= 0
Aeq n - Beq = 0
for appropriate matrix A and vector B. Then the Chebyshev center is found by adjoining one more scalar variable, 'z', to the system of inequalities and solving
min zs.t. z >= A n - B Aeq n - Beq = 0
with the minimization taking place over (z,n).
If the optimal point is (zbar, nbar), then the Chebyshev center is nbar; this gives you an estimate for your opponent's troop numbers.
This minimization problem is a linear program, and can be solved by standard LP solvers.
This is suboptimal compared to Bayes' Rule, but still useful, I think.
[Edited by - Emergent on November 6, 2010 2:29:46 PM]