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I kinda like that viral riddle from a few years ago with cheryl's birthday

Today i tried to craft a similar one which one would be solved with the same technique. I especially like how the last line of it communicates information only to the reader and not between the characters.

Anyway I failed to make one which did not fall apart in the last steps. My restrictions where 3 people which know their own variable. I tried with several dialogue chains, the following seemed the most promising:

A: I don't know but you guys don't know either

B: I did not not know, but now i know!

C: Then i know as well!

A: Then i know as well!

I'd post my 'possibility-set' as well, but they get so bloated quickly.

For the following e.g. i realized that it is impossible for the second line to add anything no matter how you set up the posibilities

A: I don't know but you guys don't know either

B: I still don't know, but neither of you does either!

C: I did not not know, but now i know!

B: Then i know as well!

A: Then i know as well!

The goal was to not have the last dimension/person feel superfluous, and just another layer to the existing riddle and maybe change the theme. I also tried to avoid having some person exclaiming that they know that someone else knows something.

Is there some logical constraint which i don't see which makes this not doable? Maybe i'm too focused on having the last two 'I don't know's' to convey information to only the reader

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    $\begingroup$ In the original problem, the first person knows the day and the second one knows the month. In your version, what does the third person know? $\endgroup$ – Mhmd Jan 16 '18 at 14:00
  • $\begingroup$ @Mhmd could be anything, at first i went with car-model, color and pet, but as it was not as simple as i thought i abstractified it to just variables $\endgroup$ – Adam Jan 20 '18 at 15:06
  • $\begingroup$ I think this type of puzzle was already created (and answered). puzzling.stackexchange.com/questions/58641/… $\endgroup$ – Mhmd Jan 20 '18 at 17:09
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You can make such a puzzle.

Let's assume Alice, Bob and Charlie are given a set of 3-tuples, one 3-tuple is chosen from the set and they are told the first, the second, and the third elements respectively. They need to figure out what the 3-tuple is.

If

The set is $\{(0,0,0), (0,0,1), (0,4,0), (1,0,1), (1,1,2), (1,1,4), (1,1,5), (1,3,1), (2,1,2), (2,2,1), (2,4,1), (3,1,0), (3,1,3), (4,0,0), (5,3,4), (5,3,5)\}$

Then the puzzle will work, and have a unique solution. The completed puzzle is,

Alice, Bob and Charlie are provided a list of 3-tuples. They are told that a 3-tuple is chosen from the list and they are told the first, the second, and the third elements respectively. They need to figure out what the 3-tuple is with logical deduction.

The dialog then goes like:
Alice: I don't know what the tuple is, and I am sure you guys don't know either.
Bob: I don't know what the tuple is either, and I am sure Charlie doesn't know even after hearing what Alice has said.
Charlie: I didn't know what the tuple is, but now I do.
Alice & Bob: So do I.

The logic is,

Since Alice doesn't immediately know, Bob and Charlie can safely cross out the tuple $(4,0,0)$. If Alice were told the number $2$, her statement would fail if the actual 3-tuple is $(2,2,1)$ in which case Bob will immediately know it. So Alice cannot be told the number $2$. Using the same logic for Charlie, Alice cannot be told the number $3$. So after the first round they can all safely cross out the tuples $(2,1,2), (2,2,1), (2,4,1), (3,1,0), (3,1,3), (4,0,0)$ from the 3-tuple list.

Bob cannot deduce the 3-tuple after hearing what Alice says. If Bob were told the number $4$ he would know the tuple is $(0,4,0)$ once Alice's statement has finished and the $(2,4,1)$ is crossed out. Since he doesn't know, they can all safely cross out $(0,4,0)$. He also claims Charlie doesn't know, and this claim will be false if he were told number $1$, in which case Charlie will immediately get the answer $(1,1,2)$ if told the number $2$(since $(2,1,2)$ is already crossed out). Hence $(1,1,2),(1,1,4),(1,1,5)$ can be safely crossed out from the list as well.

Now the tuples left are $(0,0,0), (0,0,1), (1,0,1), (1,3,1), (5,3,4), (5,3,5)$. Since Charlie knows the 3-tuple at this stage, The final tuple must be one of $\{(0,0,0), (5,3,4), (5,3,5)\}$, and since both Alice and Bob burst that they know the tuple right after Charlie says he knows, it can only be $(0,0,0)$.

This is definitely not the minimal 3-tuple set that makes the puzzle work, but it's certainly a working one.

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  • $\begingroup$ Hmm what enables you to cross out 3,1,0 after the first statement? $\endgroup$ – Adam Jan 20 '18 at 15:24
  • $\begingroup$ @Adam Alice cannot be told the number $3$, otherwise Charlie will immediately know it is $(3,1,3)$ if he were told the number $3$, and Alice can not make the claim that Charlie does not know the tuple. $\endgroup$ – Weijun Zhou Jan 20 '18 at 15:27
  • $\begingroup$ Ofcourse, got confused there. Ran it through succesfully now! BUT, it doesn't build quite as nice as one would hope. In the original riddle the last line is crucial, in this version you can omit both alice's and bob's 'i know' exclamations and still work out the riddle. The hard part is to get both those lines to be meaningful? $\endgroup$ – Adam Jan 20 '18 at 15:33
  • $\begingroup$ @Adam I see what you mean. I will try to work out a better version if I have time, and I really hope this one can inspire someone else to give a version that is tailored to your needs. $\endgroup$ – Weijun Zhou Jan 20 '18 at 15:39
  • $\begingroup$ @Adam Updated the puzzle. $\endgroup$ – Weijun Zhou Jan 20 '18 at 16:44

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