The "Good AI, Bad AI" Problem

[moonlord]

Why don't we need to substract one if he's dashed? I mean, even if he's chosen as a victim, he still can't commit suicide... :?

[PWrong]

because then we'd be subtracting one for him twice. Once for being the attacker, and again for being dashed.

[moonlord]

Aaaand, isn't that supposed to happen?

[PWrong]

Well he's only one person, and we only count him once, so we shouldn't subtract him twice.

[moonlord]

Then we need some other variables... Uhh... This didn't seem so difficult.

This is my initial approach (+ victim sharing). Suppose we have GEEE, or {4,3}, with k=0. There are m^(n-1) = 3^3 = 27 total situations after round one. Possible outcomes:

1. Nothing. Not possible, because there cannot be three evils and four casualties.
2. G - the three kill each other in a circular fashion. Possibilities: 2 (CW and CCW).
3. E - two evils kill each other, and the third kills a good. Possibilities: 3 = C[3,1] (who's not participating in the duel).
4. E - an evil kills another, who kills the third, who kills the good. Possibilities: 6 (three for "who's the first to shoot - aka who will survive" and two for "who's he going to shoot").
5. GG is not possible.
6. GE - two evils kill each other, and the third kills one of them. Possibilities: 6 (three for "who's not in the duel" and two for "who will he kill").
7. EE - one kills the good, and the other two kill him. Possibilities: 3 (who kills the good).
8. EE - two kill the good, and the third kills one of them. Possibilities: 6 (who doesn't kill the good --3-- and who does he kill --2--)
8. GGG is not possible.
9. GGE is not possible.
10. GEE is not possible, because the guy that dies has to kill someone else.
11. EEE - all three kill the good. Possibilities: 1.

They sum to 27 possibilities, so we've taken all into consideration. The utility of GEEE is therefore 17.75/27 = 79/108. However, taking into account this is the third different result I get for GEEE, I wouldn't wonder if it weren't correct.

[PWrong]

The utility of GEEE is therefore 17.75/27 = 79/108.

How can you know that without working out the utility of all the possible outcomes?

Then we need some other variables... Uhh... This didn't seem so difficult.

We could let all the undashed guys go first, then the remaining dashed guys. But then some people would get to attack twice.
Maybe we should see what happens when we allow suicide. Then we can just kill one person for each evil guy. That would make it a lot simpler. As long as it doesn't become completely trivial.

[PWrong]

Actually, it is trivial. The matrix look ok, but the best solution is always to kill noone.

[moonlord]

The utility of GEEE is therefore 17.75/27 = 79/108.

How can you know that without working out the utility of all the possible outcomes?

Well, that's what I've done. I just haven't written the calculations.

Let's allow suicide... for the moment.

[edit: posted after you posted your second post - word play]

[PWrong]

Here's something that might work. Suppose that being murdered is worse than being executed by the government, by a number 'a'. If a<0, then execution probably involves some torture. We don't get to decide how painful the execution is, (this government is pretty ridiculous).

So for each person we execute, we add 'a' to the utility. Obviously not being executed is better than being executed, so a<1. We're still allowing suicide here.

I'm still experimenting with this idea on mathematica, but it's looking good.

[moonlord]

a<0 doesn't imply torture, necessarily. Perhaps torture is good sometimes. I suppose a<0 if a good leader is executed and he shouldn't be.

By the way, did anything work out with this?

[PWrong]

I suppose a<0 if a good leader is executed and he shouldn't be.

Well, a is the measure of the difference in utility between being executed and murdered. If a good leader shouldn't be executed, he shouldn't be murdered either.

I worked out a consistent set of rules for mathematica, and I got a few matrices out.

row = m
column = k

Code: Select all

n=2:
1 a
0 a

n=3:

2    (1+a)    2a
f(a) (1/2+a)   2a
0      a      2a


where f(a) = (max[0,a] + max[1,a]) /2

I can make two conjectures.
1. In every entry, we get a continuous (piecewise defined) function of 'a'.
That is, the function never jumps instantly from one value to another.
We call this function f(a).

2. f(a) is differentiable everywhere outside the set [0,1]. That is, the slope of the graph might change any number of times between 0 and 1, but never outside this range.

[moonlord]

Erm, I don't understand your last post at all :?

[PWrong]

Sorry, I keep forgetting you don't know any calculus. I'll post the equations when I get home. (I'm at a friend's house and possibly not entirely sober).

What I mean is, the function is just an ordinary linear equation for x<0, and a different linear equation for x>1. Inside 0<x<1, it goes through several linear equations, all joined together.

[PWrong]

Here's an example for what I was talking about before:

U(6,1,2) works out to be this:

Code: Select all

7/3 + 2 a           a < 0
4/3 + 4 a       1 < a

7/3 +  8/3 a    0 < a < 1/2
13/6 +  3  a  1/2 < a < 2/3
11/6 + 7/2 a  2/3 < a < 1


Notice how the function is just a straight line for a<0 and a>1, and a bunch of connected straight lines when 0<a<1

[moonlord]

You know, I've been thinking about this "a" stuff. To me, it only seems to complicate the problem a lot. I'd first like to solve the problem with the old rules. Unfortunately, I've got virtually no free time to work on the problem. Maybe you do.

Offtopic: Mods and admins can post in the genius forum too .

[Nick]

Offtopic: Mods and admins can post in the genius forum too .

I will add that to the announcement at the top.