Modal SAT: Self Cooperation

scott-garrabrant

Modal SAT: Self Cooperation

post by Scott Garrabrant · 2015-08-10T05:48:33.000Z · LW · GW · 4 comments

  Theorem: If there exists a modal agent M such that Ci cooperates with M for each Ci∈C and such that M defects against Di for each Di∈D, then there exists an M′ which satisfies the above properties and cooperates with itself.
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Post 2 in in Modal SAT series. In this post, we show that SC Modal SAT is equivalent to Modal SAT.

For this, we need just need to prove the following theorem:

Theorem: If there exists a modal agent $M$ such that $C_{i}$ cooperates with $M$ for each $C_{i} \in C$ and such that $M$ defects against $D_{i}$ for each $D_{i} \in D$ , then there exists an $M^{'}$ which satisfies the above properties and cooperates with itself.

Proof: Let $n$ be greater than the total number of boxes in agents in $C$ and $D$ . Consider the agents $X_{n}$ and $X_{n + 1}$ defined by $X_{i} (B) = C \leftrightarrow \neg □^{i} ⊥$ .

We define $M^{'}$ by

$M^{'} (B) = C$ if $\neg □^{n + 1} ⊥$ and $□ (\neg □^{n} ⊥ \to B (C o o p e r a t e B o t) = C)$ and $\neg □ (\neg □^{n - 1} ⊥ \to B (C o o p e r a t e B o t) = C)$
$M^{'} (B) = D$ if $\neg □^{n + 2} ⊥$ and $□ (\neg □^{n + 1} ⊥ \to B (C o o p e r a t e B o t) = C)$ and $\neg □ (\neg □^{n} ⊥ \to B (C o o p e r a t e B o t) = C)$
$M^{'} (B) = C$ if $\neg □^{n + 3} ⊥$ and $□ (\neg □^{n + 1} ⊥ \to B (X_{n}) = C)$ and $□ (\neg □^{n + 2} ⊥ \to B (X_{n + 1}) = D)$
$M^{'} (B) = M (B)$ otherwise

Rule 1 says that $M^{'}$ cooperates with $X_{n}$ , rule 2 says that $M^{'}$ defects against $X_{n + 1}$ , and rule 3 says that $M^{'}$ cooperates with anyone who (provably assuming $\neg □^{n + 1} ⊥$ ) cooperates with $X_{n}$ and (provably assuming $\neg □^{n + 2} ⊥$ )defects against $X_{n + 1}$ . Thus, $M^{'}$ cooperates with itself.

For any bot $B$ with fewer than $n$ boxes, the conditions of 1, 2, and 3 are all false. For 1 and 2, this is because the actions of such bots against CoopearteBot stabilize by the time you assume $\neg □^{n - 1} ⊥$ . For 3, this is because these bots cannot distinguish between $X_{n}$ and $X_{n + 1}$ .

Therefore $M^{'}$ behaves the same as $M$ on all inputs with fewer than $n$ boxes, so $M^{'}$ cooperates with every bot in $C$ and defects against every bot in $D$ .

$□$

Note that I was lazy here, and took way more longer to cooperate with myself than I had to. In principle, if there are only $k$ bots that I need to consider (including bots I need to consider because they are referenced by bots I care about), then regardless of how many boxes are in each bot, It should be possible to achieve self cooperation within $2 {log}_{2} k$ worlds of the Kripke frame. That is, ${log}_{2} k$ worlds to identify a single bot that is distinguishable from all other bots, and another ${log}_{2} k$ worlds to ensure that the actions of $M^{'}$ differ on that bot from all other bots, so that $M^{'}$ can identify itself without changing its behavior against any other bot.

EDIT: Actually, I think $log k$ + a small constant should suffice, but it does not matter much.

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comment by orthonormal · 2015-08-10T00:05:48.000Z · LW(p) · GW(p)

I'm confused; does your definition of $M^{'}$ imply that it evaluates those conditionals in order? If so, consider the example of $C_{1} = □ (T h e m = D)$ and $M =$ DefectBot. The $M^{'}$ you construct will cooperate in world 0, and thus it will not get the cooperation of $C_{1}$ , while $M$ does. What am I missing?

Replies from: Scott Garrabrant

↑ comment by Scott Garrabrant · 2015-08-10T05:41:38.000Z · LW(p) · GW(p)

Your confusion was justified. It was wrong before. I think it is fixed now.

The conditionals are checked in order. As it is written now, none of the conditionals except the last one should trigger until world $n + 1$ .

The first only triggers against $X_{n}$ , the second only triggers against $X_{n + 1}$ and the third only triggers against $M^{'}$ .

Replies from: agilecaveman

↑ comment by agilecaveman · 2015-08-11T06:01:46.000Z · LW(p) · GW(p)

I am also confused. How does this do against EABot, aka C1=□(Them(Them)=D) and M = DefectBot. Is the number of boxes not well defined in this case?

Replies from: Scott Garrabrant

↑ comment by Scott Garrabrant · 2015-08-11T07:31:48.000Z · LW(p) · GW(p)

So according to the original Modal Combat framework, EABot is not a Modal Agent. The bots are not allowed to simulate Them(Them).

Modal SAT: Self Cooperation

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