Alice in Wonderland

[Mr. P-1]

Originally Posted by ZFR

But in this case there is no reason for all of them not to go on the first bell.

On the seat of the pants of the four people who have been told their own colour.

Unless they take different amount of time to get the clues from the seat of their pants...

Nobody can see the back of their own pants, or the pants of anyone before the other person has got up to go to the tree.

One possibility is that a message has been sewn to the seat of the pants of the four saying "everyone who is still seated can see at least two spots the same colour as their own". This leads to the same solution I gave earlier. Another is that the message reads "There are exactly three people with red spots. Everyone else can see at least one spot the same colour as their own." This leads to a different solution. There is no way for Alice to distinguish between these two possibilities (assuming she is not in a position to read the pants), or an infinity of others.

So what if the problem has the following:
[...]
The bell doesn't sound every minute. You have all the time in the world to think about it and look around. Only when each of you says whether he knows the colour or not will the bell sound. Restrictions on communicating and reflective surfaces still apply.
[...]
"It is possible to solve this task because I put the following restriction on the colours..." Unfortunately Alice could not hear the rest of his sentence[...][

Now the restriction that the professor put, could be "they are the same as the colour of your socks" or something similar, but in such case they would all go at the first bell.

They all go at the first bell anyway, because there is only one bell. You really do need the bell ringing at intervals, though. This is because the logicians make interences about the failure of other logicians to make inferences, and without a bell, there's no indication that a logician has failed to make a particular inference.

You could try to fix it this way: Every minute the bell rings, at which time each logician who know's their own colour raises their hand, but otherwise does not communicate. Unfortunately your "fix" still leaves the problem cooked, because the professor could have said "It is possible to solve this task because I put the following restriction on the colours: I have used all seven colours of the rainbow, plus black and white, and no others."

I honestly don't think this approach will work.

[ZFR]

Ah I see, I missed the significance of it being under the pants of the people who leave (and not under their own pants). I see what you meant now

Originally Posted by Mr. P-1

They all go at the first bell anyway, because there is only one bell. You really do need the bell ringing at intervals, though. This is because the logicians make interences about the failure of other logicians to make inferences, and without a bell, there's no indication that a logician has failed to make a particular inference.

Absolutely, I realize that. I didn't mean that there is only one bell. What I meant to say is:
The bell doesn't sound every minute. You have all the time in the world to think about it and look around. Only when each of you says whether he knows the colour or not will the next bell sound. Restrictions on communicating and reflective surfaces still apply.

So what I meant is that the next bell will ring only after each person has made one of the following statements:
_I know the colour of my dot.
_I don't know the colour of my dot and there is no way for me to know until I see who leaves at the next bell.
(the statements made privately to the professor who rings the bell).


My point was to remove a specific time interval (like 1 minute) to prevent a following type of solution:
_They need to do a physical task to find out their colour (like unsweing the answer from their pants).
_They each might take different amounts of time to perform the task.


But like you said, this approach will not work. In fact, the professor could say:
"It is possible to solve this task because I put the following restriction on the colours: they must agree with the individual instruction sheets I gave you earlier."
And each person having instruction of the type "your colour is x" or "if n people leave at the mth bell, then your colour is y".

[rogue]

The problem is only solvable if each color is used at least twice. The logisticians were told that the problem is solvable so they know that.

At the first bell, four people leave. That tells us that two of them had one color and two of them had another color as those four only saw only one other person with a specific color.

That leaves 27 logisticians after the first bell.

The remaining determine that each color must be on at least three foreheads, but we don't know how many of them have left the table. Let's assume that that there are three with the color red. They leave because they see only two other reds and they know that red is on at least three foreheads.

That leaves 24 logisticians after the second bell.

Since only three left the table, then the remaining know that each color must be on at least four foreheads.

Nobody leaves at the third bell.

Some leave at the fourth bell. If I were to infer anything, I would infer that each color must be on at least five foreheads otherwise some would have got up and left. That means that left the table.

That leaves 19 logisticians after the fourth bell.

We know that at least two colors remain and that each color is used at least 6 times. Since Johnny and his sister have different colors and since the last bell was run AFTER they left the table, then one can assume that the color on Johnny's head was on six foreheads and that the one on his sister's head was also on six foreheads. That means that 12 would leave the table at the fifth bell.

That leaves 7 logisticians after the fifth bell.

They figure out that they must all have the same color and leave the table at the sixth bell.

BTW, if more than three have red, then I would assume that nobody would leave at the second bell.

[wblipp]

Originally Posted by Mr. P-1

To take this to it's logical conclusion: The professor draws a dot on both heads, and asks them to determine whether the two dots are the same colour or a different colour, and says that this problem is solvable. Are we and the logicians to deduce that the colours must be the same?

Yes.

The logicians reason that there are at most two colours. If there are exactly two colours, then the problem is unsolvable for the reasons you and I have both stated. But the problem is known to solvable, so they cannot be in one of the situations where there are exactly two colours. That leaves the situations where there is exactly one colour, and they know what that one colour is.

I'll make my final stand on this analysis for the simplified problem of two people. Show me the fault and I'll concede. Where is the fault in this logic chain?

A. There are at most two colours.
B. If there are exactly two colours, the problem is unsolvable.
C. The problem is solvable.
D. There must be exactly one colour.
E. The one colour must be the colour I see on the other logicians head.
F. My dot is the colour of the other logician's.

[CRGreathouse]

Originally Posted by wblipp

I'll make my final stand on this analysis. Show me the fault and I'll concede. Where is the fault in this logic chain?

A. There are at most two colours.
B. If there are exactly two colours, the problem is unsolvable.
C. The problem is solvable.
D. There must be exactly one colour.
E. The one colour must be the colour I see on the other logicians head.
F. My dot is the colour of the other logician's.

B means, essentially: if the problem is as stated, but additionally it is common knowledge that there are exactly two colors, it is unsolvable by the participants. But there are other things which could be common knowledge instead which would make the problem solvable -- for example, everyone knows the [two] colors that everyone has. So I reject your analysis, in particular B (which is essentially right, but missing some information which makes it appear that D follows from it and modus ponens).

[rogue]

Originally Posted by wblipp

The logicians reason that there are at most two colours.

I missed something. How could they reason that? Given the puzzle as stated, if there are only two colors then logisticians would have left the table in a different order.

[wblipp]

Originally Posted by rogue

I missed something. How could they reason that? Given the puzzle as stated, if there are only two colors then logisticians would have left the table in a different order.

I'm making my final stand on the simplified problem of only two people. Quite a few posts intervened, so that isn't so clear. I'll edit the post to clarify.

[wblipp]

Originally Posted by wblipp

I'll make my final stand on this analysis for the simplified problem of two people. Show me the fault and I'll concede. Where is the fault in this logic chain?

A. There are at most two colours.
B. If there are exactly two colours, the problem is unsolvable.
C. The problem is solvable.
D. There must be exactly one colour.
E. The one colour must be the colour I see on the other logicians head.
F. My dot is the colour of the other logician's.

Originally Posted by CRGreathouse

B means, essentially: if the problem is as stated, but additionally it is common knowledge that there are exactly two colors, it is unsolvable by the participants.

I don't know what "common knowledge" means. I don't believe you need to know the actual number of colours in order to know that IF there are two colours, THEN the problem is unsolvable.

The total number of colours is known to the person who says the problem is solvable. The person who says the problem is solvable is signalling to the logicians that there are NOT two colours.

[CRGreathouse]

Originally Posted by wblipp

I don't know what "common knowledge" means.

"X is common knowledge among the participants P" means that all P know X, all P know that all P know X, all P know that all P know that all P know X, and so on.

Originally Posted by wblipp

I don't believe you need to know the actual number of colours in order to know that IF there are two colours, THEN the problem is unsolvable.

If you don't know the number of colors, the problem is unsolvable. If you know that the number of colors is two, but nothing else, then the problem is unsolvable. But you could introduce other knowledge other than the number of colors, and you have no way to privilege that knowledge over other pieces of information which would also make the problem solvable.

[wblipp]

Originally Posted by CRGreathouse

If you don't know the number of colors, the problem is unsolvable. If you know that the number of colors is two, but nothing else, then the problem is unsolvable. But you could introduce other knowledge other than the number of colors, and you have no way to privilege that knowledge over other pieces of information which would also make the problem solvable.

I assume you are still explaining to me why "B" is incomplete in a way that invalidates my analysis. I'm still not getting it. My alleged error appears to be somehow related to the issues of common knowledge and privileged knowledge. I'll ramble on a bit about these things - perhaps you can then point out the flaw in a way I can grasp.

A1. It's common knowledge there are either one or two colours.
B1. Before speaking, it's privileged to the wizard to simultaneously know the colours of both logicians. It is common knowledge that the wizard has this privileged knowledge.
B2. Before speaking, it's privileged to the wizard to know the actual number of colours. It is common knowledge that the wizard has this privileged knowledge.
B3. Before the wizard speaks, neither logician knows his own colour - so knowledge of these colours is privileged to the other two participants. The existence of this privileged information is common knowledge.
B4. It is common knowledge that IF there are exactly two colors and they are unknown, the problem is not solvable.
B5. It is common knowledge that IF it is common knowledge that there is exactly one color, and the color is known, then the problem is solvable.

It seems to me that the wizard uses his privileged information about the number of colours and the common knowledge about when the problem is solvable to make his statement.

AHH - is this your point of disagreement? That perhaps the reason the wizard states "it is solvable" is because it is common knowledge that the wizard has only red and blue markers and never repeats? Hence it is invalid for the participants to conclude the wizard knows that there is only one colour?

[CRGreathouse]

That's not my point of disagreement, no.

You've given one world-state in which the problem is solvable, and concluded that this must be the world-state because the problem is solvable. But there are other world-states in which the problem is solvable, even though the number of colors used is not 1.

World-state (unsolvable): The wizard has three colors of paint, and uses them randomly.
World-state (solvable): The wizard has only one color of paint (and this is common knowledge).
World-state (solvable): The wizard has two colors of paint, red and blue, but never repeats a color (and this is common knowledge).
World-state (unsolvable): The wizard has three colors of paint, red, green, and blue, and never repeats a color (and this is common knowledge).
World-state (solvable?): The wizard has two colors of paint, red and blue, and always uses red in the morning and blue otherwise (and this is common knowledge).
World-state (unsolvable): The wizard has only one color of paint, but the participants do not know this.
World-state (solvable): The wizard has an unlimited number of colors of paint, and always matches the eye color of the person painted (and this is common knowledge).
etc.

Since there are multiple world-states which are solvable, the participants cannot conclude which one they are in with certainty.