Conference rusure::math

        Ooooh looks like fun!  Thank you for posting this.
        
        Dan

This is probably too succinct, since I spent a lot of effort compressing 
cases and eliminating fruitless paths.

For the moment, call the actual colors by their frequency, i.e. 4444 333 22. 

(1) Adam must be able to see 4444 to be able to draw an inference about two
of his own colors. (If he sees RRRBBG, for example, both GGG and RBG are
possible on his own head, so he can only infer one color.) 
Now Adam sees distributed on Alex and Andy, either
(1A)	4444xy, and knows he has x,y,?
(1B)	4444xx, and knows he has y,y,?

(2) 
Case 1A: Alex sees on Andy/Adam
	443 / 332 - he holds 442
	44x / xyy - he holds 44?	(valid case 1)
	432 / xy? - he holds 444
	44x / xxy - he holds 44y
Case 1B: Alex sees on Andy/Adam
	444 / xxy - he holds 4y? 	(valid case 2)
	44x / xyy - he holds 44?	(valid case 3)
	4xx / xyy - he holds 444

In all the valid cases, if Andy can't distinguish Adam's colors (i.e. GR)
he can never resolve whether his own odd color is G or R. (This may take a 
while to assimilate.) So either Andy has no odd color (he holds 444) or at
least one of Adam's is B. 

Checking cases:
	valid case Adam	Alex Andy
	1	   BG?	BRR  RR?
Can never be resolved. The ?'s are BG in either order. 
 	3a	   BBR  BGR  GGG
 	3b	   BBG  BGR  RRR
3a and 3b are mutually unresolvable.
	2a	   BRR	BG?  GGG	?=B/R
 - also unresolvable. 
Case	2b	   BGG	BR?  RRR	?=B/G
is resolvable, since Andy can distinguish ?=G from ?=B. 

So BGG/BGR/RRR appears to be the solution.

        re .2,
        
>For the moment, call the actual colors by their frequency, i.e. 4444 333 22. 
>
>(1) Adam must be able to see 4444 to be able to draw an inference about two
>of his own colors. (If he sees RRRBBG, for example, both GGG and RBG are
>possible on his own head, so he can only infer one color.) 
        
        Adam can also see three each of two of the colors.  With
        colors A, B, and C, if Adam sees AAABBB then he knows he
        has CC?, but he doesn't know which of A or B the ? is.
        
        Dan

Am sees 6 stickers.   Let A,B,C denote the three colours.

what Am sees:	what Am knows he has:	how many Am knows:	Verdict:
------------------------------------------------------------------------
AABBCC			?			0		Not possible
AAABBC			C			1		Not possible
AAAABC			BC (& B or C)		2		Possible
AAABBB			CC (& A or B)		2		Possible
AAAABB			CC (& B or C)		2		Possible

Now look at things from Ax's perspective.   Let D,E,F denote the three colours.
From the above reasoning, Am must have (1) DDD or (2) DDE.

Case (1) Am has DDD
If Ax sees Am to have DDD, then he knows from the above, that Ax&Ay share
EEEEFF or EEFFFF.

what Ax sees on Ay:   what Ax knows he has:   how many Ax knows:    Verdict:
----------------------------------------------------------------------------
EEE			EFF				3	   Not possible
EEF			EEF or FFF			1	   Not possible
EFF			EEE or EFF			1	   Not possible
FFF			EEF				3	   Not possible

Hence we know that case (2) holds:
Case (2) Am has DDE
Ax knows from the above that Ax&Ay share DEFFFF or EEEFFF or EEFFFF.   If Ay 
has D, then Ax will know immediately which of the 3 options in the previous line
is correct, and hence by subtraction works out his own 3 stickers.   So Ay has
no D:

what Ax sees on Ay:   what Ax knows he has:   how many Ax knows:    Verdict:
----------------------------------------------------------------------------
EEE			FFF				3	   Not possible
EEF			EFF or FFF			2	   Possible
EFF			DFF or EEF or EFF		1	   Not possible
FFF			DEF or EEE or EEF		1	   Not possible

Thus in summary, so far from Ax's point of view:

Am has DDE
Ax has FF (and E or F)
Ay has EEF

Now we get to the complex part of Ay.   

what DEF correspond to:   what Ay sees on Am:	what Ay sees on Ax:	 unique:
--------------------------------------------------------------------------------
     brg		    bb(rg)		gg (and g or (rb))	    Y
     bgr		    bb(rg)		(rb)(rb) (and g or (rb))    Y
     rgb		    (rg)(rg)(rg)	(rb)(rb) (and g or (rb))    N/Y
     grb		    (rg)(rg)(rg)	(rb)(rb)(rb)		    N
     rbg		    (rg)(rg)b		gg (and g or (rb))	    Y
     gbr		    (rg)(rg)b		(rb)(rb)(rb)		    Y

Ay has EEF.   So to know his own colours is equivalent to knowing what D,E,F
each correspond to.   The only way that Ay could avoid knowing the colour
correspondence is if he can see:
			    (rg)(rg)(rg)	(rb)(rb)(rb)
which comes from either the third or fourth row.

	Both Alex and Adam know the mapping from {D,E,F} to {b,r,g} (since each
can see someone with a "GGH" configuration.)   Based on Ay's claim, then Am &
Ax think the following (independently).
 
what DEF correspond to:     what Am knows:     what Ax knows:	   Verdict
--------------------------------------------------------------------------------
     brg		     Ay truthful	Ay truthful	   Not possible
     bgr		     Ay truthful	Ay truthful	   Not possible
     rgb		     Ay lying		Doesn't know	   Possible
     grb		     Ay lying		Ay lying	   Not possible
     rbg		     Ay truthful	Ay truthful	   Not possible
     gbr		     Ay truthful	Ay truthful	   Not possible

	Thus the only way that Am can know that Ay was lying, whilst Ax doesn't
know is if row 3 is the case.   So D,E,F correspond to r,g,b respectively.
Ax has bb and either g or b.   We know that Ay *saw* Ax as (rb)(rb)(rb).   So
Ax has bbb.

	The complete solution is:
Am has rrg, Ax has bbb, Ay has ggb.

	The final information, that Ay *can* now identify all the stickers, is
redundant.   We could deduce that he now had that power.   However, it rounds
of the story dramatically.


yours non-Daltonianly,
Ay.

>Case (1) Am has DDD
>If Ax sees Am to have DDD, then he knows from the above, that Ax&Ay share
>EEEEFF or EEFFFF.

Why not DEEEFF or DEEFFF?

>>Case (1) Am has DDD
>>If Ax sees Am to have DDD, then he knows from the above, that Ax&Ay share
>>EEEEFF or EEFFFF.
>
>Why not DEEEFF or DEEFFF?

	Because then Am could know none of his stickers, contrary to his
assertion.

Andrew.

    Re .4, .6:
    
    Yes, you're answer does agree with the one I got in the Brain Bogglers
    conference, so I'll assume your reasoning is correct.  It was fun to do
    that problem the first time, but I don't relish trying to figure it out
    again!
    
    
    				-- edp

    Three men are lined up facing a wall, perpindicular to that wall.
    There is a bin with 3 tan hats and 2 black hats.  The men are blindfolded 
    and 3 hats are taken out of the bin and one is placed on each of their 
    heads.  The blindfolds are removed, but each man can only see in front 
    of himself.  The first man is asked what color hat he has on.  He sees 
    the two men in front of him and says, "I do not know what color hat I'm 
    wearing."  The second man is asked the same question and he sees the first 
    man in front of him and replies, "I do not know what color hat I'm 
    wearing."  The third man is asked and he says, "I *know* what color hat 
    I'm wearing!"
    
    1.  List all possible hat color combinations for the three men as an
        ordered triplet (#1,#2,#3).
    
    2. What are the possible combinations after the first man's statement?
    
    3. What are the possible combinations after the second man's statement?
    
    4. What color hat is the third man wearing and why?
    
    If you give up, let me know.
    
    Mike

In order to make this solvable, you have to include these 2 statements:

	The 3 men *know* that the starting assortment is 3 tan and 2 black.

	The 3 men can hear what each other answers.


The men are arranged like this:


	O ->	o ->	o ->
	1	2	3


1 says he doesn't know.  If he were looking at 2 black hats, he'd know his was tan.
Hence he's looking at at least 1 tan hat.

After 1 speaks, 2 and 3 both know they're not both wearing black hats.

So, if 2 is looking at a black hat, he'd know his was tan.  But says he doesn't
know what his is.  Hence 2 is looking at a tan hat.

Now 3 knows he has a tan hat.


(I solved this one years ago, but I don't remember who gave it.  Maybe Martin
Gardner in Scientific American magazine?)

/Eric

    That's correct.  There are 7 total possible combinations.  After #1's
    statement, there are 6.  After #2's there are 4, with #3 all having a
    tan hat.
    
    Mike