There is a difference. In the first one, the agents have a slight difference in their source code. In the second one, the source code of the two agents is identical.
If you’re claiming that TDT does not pay attention to such differences, then we only have a definitional dispute, and by your definition, an agent programmed the way I described would not be TDT. But I can’t think of anything about the standard descriptions of TDT that would indicate such a restriction. It is certainly not the “whole point” of TDT.
For now, I’m going to call the thing you’re telling me TDT is “TDT1”, and I’m going to call the agent architecture I was describing “TDT2″. I’m not sure if this is good terminology, so let me know if you’d rather call them something else.
Anyway, consider the four programs Alice1, Bob1, Alice2, and Bob2. Alice1 and Bob1 are implementations of TDT1, and are identical except for having a different identifier in the comments (and this difference changes nothing). Alice2 and Bob2 are implementations of TDT2, and are identical except for having a different identifier in the comments.
Consider the Newcomb’s problem variant with the first pair of agents (Alice1 and Bob1). Alice1 is facing the standard Newcomb’s problem, so she one-boxes and gets $1,000,000. As far as Bob1 can tell, he also faces the standard Newcomb’s problem (there is a difference, but he ignores it), so he one-boxes and gets $1,000,000.
Now consider the same problem, but with all instances of Alice1 replaced with Alice2, and all instances of Bob1 replaced with Bob2. Alice2 still faces the standard Newcomb’s problem, so she one-boxes and gets $1,000,000. But Bob2 two-boxes and gets $1,001,000.
The problem seems pretty fair; it doesn’t specifically reference either TDT1 or TDT2 in an attempt to discriminate. However, when we replace the TDT1 agents with TDT2 agents, one of them does better and neither of them does worse, which seems to indicate a pretty serious deficiency in TDT1.
Either TDT decides if something is identical based on it’s actions, in which case I am right, or it’s source code, in which case you are wrong, because such an agent would not cooperate in the Prisoner’s Dilemma.
They decide using the source code. I already explained why this results in them cooperating in the Prisoner’s Dilemma.
In the architecture I’ve been envisioning, Alice and Bob can classify other agents as “identical to me in both algorithm and implementation” or “identical to me in algorithm, with differing implementation”, or one of many other categories. For each of the two categories I named, they would assume that an agent in that category will make the same decision as they would when presented with the same problem (so they would both be subcategories of “functionally identical”). In both situations, each agent classifies the other as identical in algorithm and differing in implementation.
In the prisoners’ dilemma, each agent is facing the same problem, that is, “I’m playing a prisoner’s dilemma with another agent that is identical to me in algorithm but differing in implementation”. So they treat their decisions as linked.
Wait! I think I get it! In a Prisoner’s Dilemma, both agents are facing another agent, whereas in Newcomb’s Problem, Alice is facing an infinite chain of herself, whereas Bob is facing an infinite chain of someone else. It’s like the “favorite number” example in the followup post.
Because if Bob does not ignore the implementation difference, he ends up with more money in the Newcomb’s problem variant.
But there is no difference between “Bob looking at Alice looking at Bob” and “Alice looking at Alice looking at Alice”. That’s the whole point of TDT.
There is a difference. In the first one, the agents have a slight difference in their source code. In the second one, the source code of the two agents is identical.
If you’re claiming that TDT does not pay attention to such differences, then we only have a definitional dispute, and by your definition, an agent programmed the way I described would not be TDT. But I can’t think of anything about the standard descriptions of TDT that would indicate such a restriction. It is certainly not the “whole point” of TDT.
For now, I’m going to call the thing you’re telling me TDT is “TDT1”, and I’m going to call the agent architecture I was describing “TDT2″. I’m not sure if this is good terminology, so let me know if you’d rather call them something else.
Anyway, consider the four programs Alice1, Bob1, Alice2, and Bob2. Alice1 and Bob1 are implementations of TDT1, and are identical except for having a different identifier in the comments (and this difference changes nothing). Alice2 and Bob2 are implementations of TDT2, and are identical except for having a different identifier in the comments.
Consider the Newcomb’s problem variant with the first pair of agents (Alice1 and Bob1). Alice1 is facing the standard Newcomb’s problem, so she one-boxes and gets $1,000,000. As far as Bob1 can tell, he also faces the standard Newcomb’s problem (there is a difference, but he ignores it), so he one-boxes and gets $1,000,000.
Now consider the same problem, but with all instances of Alice1 replaced with Alice2, and all instances of Bob1 replaced with Bob2. Alice2 still faces the standard Newcomb’s problem, so she one-boxes and gets $1,000,000. But Bob2 two-boxes and gets $1,001,000.
The problem seems pretty fair; it doesn’t specifically reference either TDT1 or TDT2 in an attempt to discriminate. However, when we replace the TDT1 agents with TDT2 agents, one of them does better and neither of them does worse, which seems to indicate a pretty serious deficiency in TDT1.
Either TDT decides if something is identical based on it’s actions, in which case I am right, or it’s source code, in which case you are wrong, because such an agent would not cooperate in the Prisoner’s Dilemma.
They decide using the source code. I already explained why this results in them cooperating in the Prisoner’s Dilemma.
Wait! I think I get it! In a Prisoner’s Dilemma, both agents are facing another agent, whereas in Newcomb’s Problem, Alice is facing an infinite chain of herself, whereas Bob is facing an infinite chain of someone else. It’s like the “favorite number” example in the followup post.
Yes.
Well that took embarrassingly long.