Good point about caring for yourself even if you expect to lose the memory of (e.g.) the current hour. AIXI only cares about the tapes that are successors of the current one. Maybe expand the tape from write-only to also have some erasing operations?
I think there are probably some other toy problems that illustrate this issue a lot better than Sleeping Beauty, where AIXI equating memory loss with death might not actually change its decisions much in the bet.
I still don’t see how you’re getting those probabilities. Say it takes 1 bit to describe the outcome of the coin toss, and assume it’s easy to find all the copies of yourself(ie your memories) in different worlds. Then you need:
1 bit to specify if the coin landed heads or tails
If the coin landed tails, you need 1 more bit to specify if it’s Monday or Tuesday.
So AIXI would give these scenarios P(HM)=0.50, P(TM)=0.25, P(TT)=0.25.
I’m not thinking of it like specifying parts of the toy problem. I’m thinking of it as if for each of HM, TM, and TT, the observer is about to recieve 2 bits that describe which situation they’re in, and the only object that matters for the probability of each is the shortest program that reproduces all past observations plus the next 2 bits.
If we assume Sleeping Beauty has lots of information, we might expect that the shortest matching program will look like a simulation of physical law, plus a “bridging law” that, given this simulation, tells you what symbols get written to the tape. It is in this context that is seems like HM and TM are equally complex—you’re simulating the same chunk of universe and have what seems like (naively) a similar bridging law. It’s only for tuesday that you obviously need a different program to reproduce the data.
If we assume Sleeping Beauty has lots of information, we might expect that the shortest matching program will look like a simulation of physical law plus a “bridging law” that, given this simulation, tells you what symbols get written to the tape
I agree. I still think that the probabilities would be closer to 1⁄2, 1⁄4, 1⁄4. The bridging law could look like this: search over the universe for compact encodings of my memories so far, then see what is written next onto this encoding. In this case, it would take no more bits to specify waking up on Tuesday, because the memories are identical, in the same format, and just slightly later temporally.
In a naturalized setting, it seems like the tricky part would be getting the AIXI on Monday to care what happens after it goes to sleep. It ‘knows’ that it’s going to lose consciousness(it can see that its current memory encoding is going to be overwritten) so its next prediction is undetermined by its world-model. There is one program that will give it the reward of its successor then terminates, as I described above, but it’s not clear why the AIXI would favour that hypothesis. Maybe if it has been in situations involving memory-wiping before, or has observed other RO-AIXI’s in such situations.
Oops, that should have been 0.02 :)
Good point about caring for yourself even if you expect to lose the memory of (e.g.) the current hour. AIXI only cares about the tapes that are successors of the current one. Maybe expand the tape from write-only to also have some erasing operations?
I think there are probably some other toy problems that illustrate this issue a lot better than Sleeping Beauty, where AIXI equating memory loss with death might not actually change its decisions much in the bet.
I still don’t see how you’re getting those probabilities. Say it takes 1 bit to describe the outcome of the coin toss, and assume it’s easy to find all the copies of yourself(ie your memories) in different worlds. Then you need:
1 bit to specify if the coin landed heads or tails
If the coin landed tails, you need 1 more bit to specify if it’s Monday or Tuesday.
So AIXI would give these scenarios P(HM)=0.50, P(TM)=0.25, P(TT)=0.25.
I’m not thinking of it like specifying parts of the toy problem. I’m thinking of it as if for each of HM, TM, and TT, the observer is about to recieve 2 bits that describe which situation they’re in, and the only object that matters for the probability of each is the shortest program that reproduces all past observations plus the next 2 bits.
If we assume Sleeping Beauty has lots of information, we might expect that the shortest matching program will look like a simulation of physical law, plus a “bridging law” that, given this simulation, tells you what symbols get written to the tape. It is in this context that is seems like HM and TM are equally complex—you’re simulating the same chunk of universe and have what seems like (naively) a similar bridging law. It’s only for tuesday that you obviously need a different program to reproduce the data.
I agree. I still think that the probabilities would be closer to 1⁄2, 1⁄4, 1⁄4. The bridging law could look like this: search over the universe for compact encodings of my memories so far, then see what is written next onto this encoding. In this case, it would take no more bits to specify waking up on Tuesday, because the memories are identical, in the same format, and just slightly later temporally.
In a naturalized setting, it seems like the tricky part would be getting the AIXI on Monday to care what happens after it goes to sleep. It ‘knows’ that it’s going to lose consciousness(it can see that its current memory encoding is going to be overwritten) so its next prediction is undetermined by its world-model. There is one program that will give it the reward of its successor then terminates, as I described above, but it’s not clear why the AIXI would favour that hypothesis. Maybe if it has been in situations involving memory-wiping before, or has observed other RO-AIXI’s in such situations.