Timeline for Are there eighteen or twenty bars in my castle?
Current License: CC BY-SA 4.0
21 events
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| Jan 14 at 0:19 | comment | added | tehtmi | I would agree that the accepted solution requires both Alice and Bob to know that an actor that can deduce the answer is certain to give that answer. If this is your objection, I think that's okay, although I don't think it ruins the question/answer. Is there something beyond this that you mean by strategy? In (16, 2), Bob believes either (18, 2) or (16, 2). If he knows Alice chose not to answer, you think he can still believe (18, 2)? How does he reconcile (18, 2) with Alice not answering? | |
| Jan 14 at 0:01 | comment | added | Vento | Also I don't think that Bob can conclude with absolute certainty that Alice hasn't answered first because of the (16,2) case: that again would require Bob and Alice to devise a common strategy by which they stay silent because of some interpretation of some node in these trees. In such a case, given that they can draw their tree from the get go, Bob could theoretically solve the riddle directly on the night of the first day. But once again, they are not allowed to strategize, so I don't se how they would decide to interpret their silence in this specific way | |
| Jan 13 at 23:50 | comment | added | Vento | I think I can see where you are going here, but I am not sure the real Alice and Bob can use the beliefs of the hypothetical Alices and Bobs down the tree they way you are suggesting. I agree that Alice can rule out part of the branches of her tree based on what she sees, but I still don't understand how she can climb the tree in a unique way back to $b_A=8$. | |
| Jan 13 at 21:12 | comment | added | tehtmi | I would suggest looking at the cases where the numbers of bars are (18, 2) and (16, 2). In the first case, I think you will agree because there are no hypotheticals, Alice immediately knows there are 20 bars because Bob has at least one. In the second case, cannot Bob conclude that because Alice has not answered that she doesn't have 18 bars, thus she has 16? If so, is this not information Bob has only gained after learning Alice has already been visited, information that your answers suggests cannot exist? | |
| Jan 13 at 20:55 | comment | added | tehtmi | That hypothetical Bob believes $a$ could be 18. The next step of inference would be to look at the parent node in the tree. A hypothetical Bob with 2 bars, knowing Alice can't have 18 bars, could conclude Alice has 16 bars. If Bob does not answer, then on the next visit, Alice (and all hypothetical Alices) will know Bob doesn't have 2 bars. Your construction seems to admit the presence of these hypothetical actors -- your tree is a nice construction of them. | |
| Jan 13 at 20:55 | comment | added | tehtmi | "Bob can only believe that Alice has 10 or 12 bars", yes but only because Bob has 8 bars; the hypothetical Bob in question has 2 bars, so it logical for him to believe Alice could have 18 (or 16) bars. "[They] know the [...] from the get go" -- this is true. But they don't know from the get go that every hypothetical actor doesn't have enough info. There are hypothetical actors, like Alice with 18 bars, that do have enough info. "[A] hypothetical Bob that believes $a_{BABAB}$ is 18;" this not a belief of a hypothetical Bob, but a belief by the real Bob about a hypothetical Bob. | |
| Jan 13 at 18:20 | comment | added | Vento | If you are talking about a hypothetical Bob that believes $a_{BABAB}$ is 18, sure: Alice can rule out the existence of this hypothetical Bob. But there are 5 other hypothetical Bobs each believin in a different realistic value of $a_{BABAB}$ and branching from both $a_B$ being 8 or 6. How can Alice figure out which of these other 5 Bobs is the real one? This is a genuine question, I am not sure what kind of denial would I have... can you explain to me why Alice is capable of ruling out 4 of the 5 remaining Bobs and correctly climb the tree back to the right guess of $a_{B}$? | |
| Jan 13 at 16:56 | comment | added | Vento | Also, both Alice and Bob already know that the other doesn't have enough information to solve the riddle from the get go just by pure logics. So the visits of the evil logician would not add any more information that they don't already know. Unless of course they are following a strategy where they pretend not to know what they already know, so that they can create artificial information with each visit of the logician. But given that they didn't have the possibility to strategize beforehand, they cannot know for sure which strategy the other is following, if any. | |
| Jan 13 at 16:43 | history | edited | Vento | CC BY-SA 4.0 |
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| Jan 13 at 16:40 | history | edited | Vento | CC BY-SA 4.0 |
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| Jan 13 at 16:11 | history | edited | Vento | CC BY-SA 4.0 |
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| Jan 13 at 15:58 | history | edited | Vento | CC BY-SA 4.0 |
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| Jan 13 at 15:53 | history | edited | Vento | CC BY-SA 4.0 |
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| S Jan 13 at 15:12 | review | First answers | |||
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| S Jan 13 at 15:12 | history | edited | Vento | CC BY-SA 4.0 |
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| Jan 13 at 14:26 | comment | added | Vento | Yeah, I read this rationale in other comments and honestly I disagree with it. This hypothetical Bob that believes Alice could have 18 bars is simply not a perfect logician, because by following perfect logic Bob can only believe that Alice has either 10 or 12 bars. Any other deduction is just not supported by logics and it's a mere consequence of Alice and Bob following a strategy to keep count of each other's bars | |
| Jan 13 at 11:18 | comment | added | tehtmi | The important information that is supposed to be gained by visits of the evil logician is that the other captive did not have enough information deduce the answer. For example: Knowing the premise, but before any visit, you agree $a_{BABAB}$ includes 18 -- there is a hypothetical Bob that believes Alice could have 18 bars. But, after Bob learns Alice has been visited and not guessed the answer, this is no longer consistent because that hypothetical Alice would have known there were 20 bars and guessed as such. I don't see any argument against this besides denial. | |
| Jan 13 at 6:15 | history | edited | Vento | CC BY-SA 4.0 |
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| Jan 13 at 6:11 | review | Late answers | |||
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| S Jan 13 at 5:52 | review | First answers | |||
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| S Jan 13 at 5:52 | history | answered | Vento | CC BY-SA 4.0 |