@ Mr_rb: You're clearly on the right track. The next steps get trickier. Remember that the reds all leave at once after the first bell but before the second. That means that something about the groups of two being gone allows all of the red people to figure out who they are. It also means that something about all of the other color groups makes it impossible for them to do so (hence, why nobody is able to figure it out in round three).

As a final hint, keep in mind what we have said about no single person having a unique color and the possible significance of someone "reacting" in the fourth round. Hopefully, this will be all you need to figure out the rest. I will check back later to see if somebody figured it out.

And no, the answer is not 6.

I may be a million miles off here as I thought it was to do with a synonym and a well know sequence which would give the answer as 7. Am I on the right lines at all?

@moskowitz
How did you interpret the forth round clue. I took it to mean that at least one person left that round. Was I supposed to interpret that as something else? I'm also assuming that no one made any logical mistakes.

I have a consistent explanation that gives the answer as 6 and I understand the prerequisite background regarding the blue-eyed islanders puzzle.

What answer did you arrive at?

Yea, that problem it written very poorly, unfortunately. As I said Mr_rb, step 1 turns out to be pretty simple, and you feel pretty silly about it figure that part out.

I think I'll give this a shot later to solve the rest of it and see what I come up with.

once you figure that part out*

@yebellz: You gotta take the problem real slow, and consider what you absolutely know at each round, so that you slowly whittle away the number of people at the table.

Consider the red group who leave in round two and how they might all know to leave at the same time. One round before, the people with red dots (though they didn't know it) looked around the table and noticed two pairs. Lets just call them the green pair and the blue pair, to make things easier. Before the green pair and the blue pair leave, the red dot people cannot be sure that they themselves do not have a green dot or a blue dot.

Once the pairs are gone, however, it becomes clear to the red dot people that they could not possibly have a green or blue dot. They know that the only way the green and blue pairs identified themselves was by looking around the table and seeing only a single person with a particular color (I mentioned above how everyone knows at the outset that there must be at least two people bearing each color).

Since you understand the blue-eyed islander problem, you can probably figure out the rest from here. But I'll tell you a bit more anyway, if you care to keep reading.

The red dot people now know that they do not have the same color as the people who have already left the table. They also know that anyone who is a member of a pair will have already figured out their color and stood up. That means that the smallest color group at the table must at least have 3 members (since the pairs are gone, and no one can have their own color).

You can see where this is going: the red group is a group of three. Looking around the table anyone with a red dot would see only two people sitting at the table with red dots. They would know that if those were the only two red dots they would have figured out that they were a pair and left in the last round with the blues and the greens. "There must be a third red dot," thinks the logician, "and since no one else at the table has a red dot other than those two, the third red dot must be me!"

Thus, the three red dots get up and leave together at the second round, free to do so only because all the pairs left at the first bell. Now there are only 24 people sitting at the table. An interesting number, don't you think?

Incidentally, this is how we know six rounds cannot be the right answer. If at each bell a group that was one person larger was able to figure it out and leave it would still take at least 7 rounds (you can do the math to see that I'm right), and we know that in round 3 nobody is able to leave.

"6" is an attractive answer because of the way the problem is written. We are told 4 rounds go by and we're told of two more people, with different colors, who leave "shortly after." It's natural, after being told that the reds all figured it out at once, to assume that the Novice and his sister, being of different colors, left in the two rounds following the fourth.

Wow, sorry about the length of that post. I kinda got carried away...

FK:

The hermit's place is very unlikely to actually be situation on the North Pole. If you stand there for just a few hours, because the pack ice drifts, you are no longer on it and the compass would point one way again. Idk I guess its still possible lol but.. just a thought I had.

@Thucy: maybe the hermit is Santa Claus, but I wouldn't call him a hermit.

i suppose santa's house is on wheels to stay on geographic north

i wouldnt put it past the old fucker.

@moskowitz
I understand how to arrive at the explanation that there are three players with red dots. There's no need to be so pedantic (and frankly, seemingly condescending).

Did you arrive at an answer yet? I'm interested to see how you got something specifically other than 6? We may be making some different fundamental assumptions in interpreting the statements in the problem.

Note: after round 4, although the novice and his sister have different colored dots, it is still possible that they both leave in the same round (specifically round 5).

Here is how I got my answer of 6:


*** SPOILER BELOW ***

Assumption (as moskowitz already detailed above): the professors statement implies that no one has a unique color. Thus, everyone belongs to a color group of at least two.

Core Lemma:
If there are n players with the same colored dot, they will all leave on the bell ring (n-1).
(The explanation of this lemma is essentially given by the blue-eyed islander puzzle. However, it is essentially the crux of this problem and perhaps the hardest part to understand. I'll be happy to go into more detail, but I think the links provided above explain it far better than I probably could.)

Lemma: (Symmetry)
Everyone with the same color will leave the table at the same time.

31 people
2 blue, 2 green, 3 red, 5 brown, 6 yellow, 6 orange, 7 purple
The novice is wearing yellow, his sister is wearing orange.
All of the colors are arbitrary except for the reds in round 3 since that is explicitly mentioned.
Here is what happens on each ring:
1) 2 blues and 2 greens leave
2) 3 reds leave
3) no one leaves
4) 5 browns leave
5) 6 yellows and 6 oranges leave
6) 7 purples leave

Rounds 1 and 2 have already been thoroughly explained so no need to repeat. We know explicitly that no one leaves on round 3.
In round 4, we know that at least one person left, which implies that an entire group of 5 (or multiple groups of 5) must have left.
Since, the novice and his sister belong in two different groups, and since neither of them left in the last round, there must be a third group remaining.
Thus, in order for there to even be enough people to make up the 3 remaining groups (which each must be of size 6 or larger), only 5 (and not 10, 15, ...) could have left in the fourth round.
Thus, we are left with 19 people after round 4, which can only be split up into groups of 6, 6, and 7 (since each group must be larger than 6).
The novice and his sister must each belong to a different group of 6 since they do not leave in the last round.

Does that sound ok? I can provide some more explanation if anything is unclear.

@Thucy
Well, we all know that Santa is magic, so he must be working some wheels or levitation or ice drift magic in order to keep that gig up.

What I really don't buy is the hermit part. I mean he definitely spends one day a year flying all about visiting millions of people. Then, there are all those reports of sightings at local malls, typically occurring in the later months of the calendar year. And how about those naughty and nice lists? Surely he must be doing some groundwork in gathering that level of intel.

:-P

@yebellz - It's really quite simple. The sled is a time machine so Santa has a couple hundred days after Christmas to be at all the malls and do the deliveries Christmas Eve night. He probably gets 'em all done. has the elves do recon for him, and still has time to vacation in the Bahamas.

Alice puzzle link:
http://web.mst.edu/~wjcharat/Alice.pdf

It was already posted above, but I just wanted wanted to make it more visible for those viewing this part of the thread.

Wow, I never thought I would say this, but...

Draugnar must be right

(regarding Santa Claus)

:P

Right, that's also what I got Yebellz. The red dots are fairly straightforward and after that you just have to find a combination of 5, 6, and 7 that matches 24. If you interpret 'shortly' in this sentence "Novice, who was mentioned before and his sister, with dots of different colors, have left shortly after" as 'in one of the following rounds', as is intended I suppose, only 1 solution is possible.

Alternative solution

The grand master wearing a scarlet tunic and a scarlet dot on his forehead left before a bell tolled. The deputy grandmasters wearing purple tunics and purple dots on their foreheads left next, the vice-deputies with red tunics and red dots on their foreheads left on bell 3, the sub-vice-deputy was absent with illness so no-one left on the fouth bell.

The novices left next, followed by the sisters, followed by the plebs

As we don't know wether or not there were any intruders or how many ranks there were we can't tell Alice how many times the bell rang.

*** Another Spoiler ***

In my earlier explanation, the "Core Lemma" that I gave was a bit inaccurate.

In this case, it turns out that each group of size N leaves on the (N-1)th bell ring. However, that's not always the case, as in general scenarios, it may be possible for a group to determine their color before that round is reached.

The Core Lemma should read:
A group of size N will all leave ON OR BEFORE the (N-1)th bell ring.

I believe that the only case that would result in the "before" case of the above lemma would be if you had remaining only one group of size of N after round a that is N-3 or less.

For example, consider if there were 32 people around the table to begin. I think the only working explanation with adding an additional person, but not changing any of the rest of the setup, would be if that person was in the purple group making it of size 8. Now, this would *not* change the solution, as this group of 8 would still leave on the 6th bell ring.
Why? Since after all of the other groups have left in round 5, each of them would only see 7 remaining browns and have to conclude that they must also be brown (they know that they weren't part of any other group since otherwise they would have left with them, and they know that they don't have a unique color).
Note: you can't add the additional person to either the yellow or orange group since then that would result in that group having 7, which would cause it to leave in the last round violating the given statements.

However, I believe that attempting to add 2 additional people (bringing it to 33 overall) would create an ambiguous riddle (insufficient info to solve). With 33 people and all of the other clues remaining the same, it seems that the last possible round could have two possible scenarios:
Scenario A: +1 to orange (or equivalently yellow), +1 to purple
5) 6 yellows leave
6) 7 oranges leave
7) 8 purples leave
=> 7 bell rings
Scenario B: +2 to purple
5) 6 yellows and 6 oranges leave
6) 8 purples leave
=> 6 bell rings

Similarly, changing the riddle by adding even more people, while keeping the rest of the clues the same would allow scenarios that makes the riddle lacking sufficient info to solve. However, with either 31 or 32 total people, I believe the solution should be 6 bell rings.

If anyone has any alternative solutions or finds a mistake in my logic, I would be very interested to hear about it.

@Moskowitz, I'm quite intrigued to hear more about why you think that 6 is ruled out as an incorrect solution. I think that, perhaps, you made an erroneous assumption that the novice and his sister could not leave on the same round since they have different colors, but it is in fact possible (and not ruled out by the clues) for two groups to leave in the same round.

I see a problem with your solution Yebellz and it is this.

Before 4 people get up to leave on the first bell how does every logican know that the lowest number leave first? And also if there was an imposter the chief logican would have known that and placed a unique colour dot on his forehead, which would defeat the leap of logic that each colour must appear more than once. Also, why hold the test if it only succeeds when no one is an imposter.

@Maniac
The riddle has not been stated very cleanly and its presentation does have many issues where one could poke holes in its ambiguity and grammar, in order to break the solvability of the riddle. Complex logic puzzles like these do require idealized and often convoluted setups. It is trivial to find problems in the presentation or look technical breaks that make the riddle invalid, however the interesting part of this puzzle is trying to solve it as intended. In my next post, let me try to restate to riddle clarify any ambiguity.

The "Alice riddle" restated:

31 players are gathered around a table to play a game. The game master (not counted as one of the 31 players) arrives and sets up the game. He places a colored dot on each players head, making sure that each player does not see their own dot, but can see everyone else's.

The game master explains the rules of the game:
1) Once the game begins a bell will ring every minute. If at the moment the bell rings you are aware of the color of the dot on your head, you must stand up and leave the table.
2) You must remain at the table until you are compelled to leave by rule 1.
3) No one is allowed is talk or otherwise communicate during the game.

The game master then states that everyone will eventually leave the table during the course of this game. Then the game begins.

During the course of the game:
1) 4 players leave on the first ring
2) all of the players with a red dot (and no one else) leave on the second ring
3) no one leaves on the third ring
4) at least one player leaves on the fourth ring
5) after the fourth ring, there are two players of the remaining group that have different colored dots and neither of them are part of the last set of players to leave the table.

An important detail of this riddle is the assumption that the following facts are true and "common knowledge"
1) All 31 players and the game master are perfect logicians, which means that they will correctly make any logical deduction that could be possibly be made given the facts that they know.
2) None of the players have any prior knowledge to the distribution of the colors.
3) The game master does in fact sets up the game in a manner that makes his final statement true, but does not otherwise introduce bias into the distribution.
4) All of the players follow the rules correctly.
5) No player is able to see the color of their own dot, but is able to see everyone else's.

A fact is "common knowledge" if all 32 people involved know this fact, they all know that they all know, they all know that they all know that they all know, and so on.

The challenge of the puzzle is to determine how many rings it takes for the table to clear, or more generally, to infer how many players left on each ring throughout the course of the game.

I've tried my best to clean up the presentation of the puzzle. Hopefully, that helps resolves some ambiguities and there are no glaring flaws.

Of course, one can always come up with technical spoilers like "what if some one is colorblind?", "what is someone is deaf?", or "what if the universal rules of causality spontaneously fail?". I'm not going to bother attempting to enumerate and exclude all of these silly possibilities, because logical problems of this nature require "idealized conditions".

It is trivial to find loopholes that make the problem unsolvable, but they are not the interesting part of the problem.

This is actually easy if you think about it. The only caveat is there can't be someone with no match at the table.

Ring 1 - 2 colors of 2 each (4 people) leave
Ring 2 - 3 reds leave (only reds and they could see 2 other reds so knew they had to be the third.
Ring 3 - no one, so no color groupings of 4
Ring 4 - At least 5 people leave leaving at most 19 people
- Because we know ring 5 will have groups of 6 leave the table and ring 6 groups of 7, and because we know there are at least 3 color groups left, we deduce that only 5 people left
Ring 5 has 12 people (2 groups of 6) leaving the last 7 that leave on Ring 6.

And yes, I solved it and didn't google or bing or some other search engine an answer.

And everyone must remember that these are perfectly logical people. They *will* leave on the bell for their color group's size.

And yebellz cleaned up posting was the first time I read the riddle. I really wasn't interested in linking off site and hadn't read any of the discussion before on this one.