# Cheryl's birthday with one more dimension/person

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

• 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? Jan 16, 2018 at 14:00
• @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
Jan 20, 2018 at 15:06
• I think this type of puzzle was already created (and answered). puzzling.stackexchange.com/questions/58641/… Jan 20, 2018 at 17:09

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.

• Hmm what enables you to cross out 3,1,0 after the first statement?
• @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. Jan 20, 2018 at 15:27