Sep 6, 2012
The thing about dead mathematical proofs is that it's practically always worth looting the corpse; sometimes you even find there's still a pulse! That appears to be the case with my recent post lamenting the demise of the TDT-like "Masquerade" algorithm. I think I've got a rigorous proof this time, but I'd like other opinions before I declare that the rumors of Masquerade's demise were greatly exaggerated...
To recap quickly, I've been trying to construct an algorithm that, given the payoff matrix of a game and the source code of its opponent, does some deductions and then outputs a move. I want this algorithm to do the commonsense right things (defect against both DefectBot and CooperateBot, and mutually cooperate against both FairBot and itself), and I want it to do so for simple and general reasons (that is, no gerrymandering of actions against particular opponents, and in particular no fair trying to "recognize itself", since there can be variants of any algorithm that are functionally identical but not provably so within either's powers of deduction). I'd also like it to be "un-exploitable" in a certain sense: it has a default move (which is one of its Nash equilibria), and no opponent can profit against the algorithm by forcing it below that default payoff. If the opponent does as well or better in expected value than it would by playing that Nash equilibrium, then so too does my algorithm.
The revised Masquerade algorithm does indeed have these properties.
In essence, there are two emendations that I needed: firstly, since some possible pairs of masks (like FairBot and AntiFairBot) can't knowingly settle on a fixed point, there's no way to determine what they do without a deductive capacity that strictly exceeds the weaker of them. That's a bad feature to have, so we'll just have to exclude potentially troublemaking masks from Masquerade's analysis. (In the special case of Prisoner's Dilemma I know that including DefectBot and FairBot will suffice; I've got what looks like a good solution in general, as well.)
The second emendation is that FairBot needs to alternate between trying harder to prove its opponent cooperates, and trying harder to prove its opponent defects. (There needs to be an asymmetry, like cooperation proofs going first, to guarantee that when FairBot plays against itself, it finds a Löbian proof of mutual cooperation rather than one of mutual defection.) The reason for this is so that when agents reason about masks, they should be able to find a proof of the mask's action without needing to exceed that mask's powers of deduction. Otherwise we get that arms race again.
This escalation of proof attempts can be represented in terms of proof limits (since there exists a constant C such that for N sufficiently large, a proof that "there are no proofs of X of length less than N" either exists with length less than C^N or not at all), but the simplest way to do this is with the formalism of PA+N. That is, PA is Peano Arithmetic; PA+1 is the formal system with the axioms of Peano Arithmetic and an extra axiom that PA is self-consistent (that is, if PA proves X, then PA does not prove not-X); PA+2 has those axioms and an extra one stating that PA+1 is self-consistent and so on. (Note that none of these formal systems know themselves to be self-consistent, and for good reason!) In every use, we'll assume that N is a fixed number (anything greater than 4 will work).
So without further ado, let's define our masks for the Prisoner's Dilemma:
for i in range(N):
if "X(FairBot)=C" is provable in PA+i:
if "X(FairBot)=D" is provable in PA+i:
Lemma 1: For any X, "DefectBot(X)=D" is provable in PA. (Obvious.)
Lemma 2: "FairBot(FairBot)=C" is provable in PA.
Proof: If we label the FairBots separately (after all, they might be logically equivalent but not know they are), we can quickly show that "if PA proves FairBot1(FairBot2)=C, then FairBot2(FairBot1)=C" is provable in PA (and the same with the subscripts reversed). This makes what we might term a Löbian cycle, which will come up a few more times in this post: if we have a formal system S, and "if S proves A, then B" and "if S proves B, then A" are theorems of S, then S indeed proves both A and B. (Proof reserved for a comment; hat tip to Eliezer and co.)
Lemma 3: "FairBot(DefectBot)=D" is provable in PA+1.
Proof: Note that this does not work in PA, even though PA can prove "DefectBot(FairBot)=D". Why not? Because PA can't prove that PA doesn't find a proof of "DefectBot(FairBot)=C" before it can look for a proof of defection! However, PA+1 can handle this just fine, because it proves "if PA proves DB(FB)=D, then PA does not prove DB(FB)=C".
Note the weird fact that it takes a stronger formal system to prove a result that's much more intuitive from outside the system!
Let's split up the long definition of Masquerade with a helper function.
for i in range(N):
for j,k in [C,D]:
if "Y(X)=j and X(Y)=k" is provable in PA+i:
This function, given two agents, simply searches (using increasingly strong formal systems) for a proof of what they do against each other; once it finds one, it halts and returns that result, and if it doesn't find one in time, it announces its failure. We will incorporate this into the definition of Masquerade. (The other function call here is to the payoff matrix, to retrieve the utility U(j,k) of a certain pair of moves.)
utility = -∞
my_move = null
their_move = null
for Y in [DefectBot, FairBot]:
if Search(X,Y) != False:
(j,k) = Search(X,Y)
if U(j,k) > utility:
my_move = j
their_move = k
utility = U(j,k)
if utility > U(D,D):
for i in range(N):
if "X(Masquerade) = their_move" is provable in PA+i:
A few things to note: first, this agent actually halts and outputs an action against any opponent... but for it to provably do so in a system below PA+N, Search(X,Y) has to find proofs quickly. This is the reason that my prior attempt didn't work- it had to wait at one point for the old FairBot to run out of time/power before concluding what it did, and that made it impossible for the old FairBot to know what the old Masquerade did. But with the new and improved agents, we get to ground in a fixed number of steps.
For brevity, I'll label DefectBot, FairBot, and Masquerade as DB, FB, and M, respectively.
Lemma 4: "Search(DB,DB)=(D,D)" is provable in PA+1. (Follows from Lemma 1; note that it needs to use PA+1 in order to rule out finding proofs of other action-pairs.)
Lemma 5: "Search(FB,DB)=Search(DB,FB)=(D,D)" is provable in PA+2. (Follows from Lemma 3.)
Lemma 6: "Search(FB,FB)=(C,C)" is provable in PA. (Follows from Lemma 2; since (C,C) is the first one tried, we don't even need to go up to PA+1.)
Lemma 7: "Masquerade(DB)=D" is provable in PA+3.
Proof: Lemmas 4 and 5, plus the fact that PA+3 knows the consistency of PA+2. There is no sanity-check step, since utility=U(D,D) here.
Lemma 8: "Masquerade(FB)=C" and "FB(Masquerade)=C" are provable in PA+3.
Proof: Lemmas 5 and 6, and the consistency of PA+2, imply that when Masquerade arrives at the sanity-check stage, it has the variables set as utility=U(C,C), my_move=C and their_move=C. Thus PA+3 can prove that "if 'FB(M)=C' is provable in PA+3, then M(FB)=C". And of course, "if 'M(FB)=C' is provable in PA+3, then FB(M)=C" is provable in PA+3, since again PA+3 can prove that PA through PA+2 won't have found proofs of contrary conclusions before it gets around to trying to find cooperation in PA+3. Therefore we have the desired Löbian cycle!
Theorem: "Masquerade(Masquerade)=C" is provable in PA+4.
Proof: Lemmas 7 and 8, and the consistency of PA+3, allow PA+4 to prove that when each Masquerade arrives at the sanity-check stage, it has set utility=U(C,C), my_move=C and their_move=C. Thus we achieve the Löbian cycle, and find proofs of mutual cooperation!
Well, assuming that I haven't made a mistake in one of my proofs, I'm going to run the same proof for my generalization: given a payoff matrix in general, Masquerade enumerates all of the constant strategies and all of the "mutually beneficial deals" of the FairBot form (that is, masks that hold out the "stick" of a particular Nash equilibrium and the "carrot" of another spot on the payoff matrix which is superior to the "stick" for both players). Then it alternates (at escalating PA+n levels) between trying to prove the various good deals that the opponent could agree to. There are interesting complexities here (and an idea of what bargaining problems might involve).
Secondly, I want to see if there's a good way of stating the general problem that Masquerade solves, something better than "it agrees with commonsense decision theory". The analogy here (and I know it's a fatuous one, but bear with me) is that I've come up with a Universal Turing Machine but not yet the Church-Turing Thesis. And that's unsatisfying.
But before anything else... I want to be really sure that I haven't made a critical error somewhere, especially given my false start (and false halt) in the past. So if you spot a lacuna, let me know!