Conference rusure::math

    first, since all women know that there are 46 unfaithful husbands, and
    each knows who they are, by process of elimination, each woman knows
    if her husband is faithful or not ! 
    example: woman counts all unfaithful men (except her husband if she finds 
    only 45, but she knows there must be 46, then her husband (one she did not
    count) is the remaining one , she shoots him, and the number of
    unfaithfuls men goes down to 45, and one couple is removed from village.
    
    and since each woman does the same, by morning, there will be 46 less 
    couples in town. and all unfaitful men dead.
    
    so, the question is wrong. it cant be that each knows who is the
    faithful, and know how many there are of them and yet not know if it
    her husband or not..
    

    Re .1:
    
    The people involved don't know, initially at least, how many unfaithful
    spouses there are.
    
    
    				-- edp

.0>    What effect, if any, does this announcement have?
    
    The mayor's husband shoots her in the morning.

If, in addition to what we know from .0, all the men in town are perfect 
logicians and don't travel, we can determine what will happen by induction.

Suppose only one wife were Unfaithful (U). Her husband hears for the first 
time that someone is U, and since he knows all other wives are non-U, he 
shoots his wife.

Suppose there are two U wives. On the first day no one can conclude 
anything, but on the second day each husband of a U can argue, "Why didn't
the other guy figure out that his wife is U? Only because he knew of some
other U wife. But I only know of one U wife, so my own must be U." So 
each such husband shoots his wife.

Suppose there are N U wives. On the N-1st day no one can conclude 
anything, but on the Nth day each husband of a U can argue, "Why didn't
those N-1 guys figure out that their wives are U? Only because they knew of
N-1 other U wives. But I only know of N-1 U wives, so my own must be U."
So each such husband shoots his wife. 

Plug N=46 into the above sentence and you have solved the problem.

As a corollary problem, what would have happened if, after making that 
announcement, the mayor had shot his own wife on the K[<46]th day?

    Lynn's got the solution.  Or the part I heard.  But I'm still puzzled.
    
    The mayor's announcement seems completely superfluous and extraneous
    -- each and every husband ALREADY KNOWS that some wife is unfaithful
    (and can name at least 45 of them).  So why is it necessary?
    
    Without the announcement, the husbands effectively behave like this:
    if a husband knows of N unfaithful wives (with N >=2, to ensure that
    everyone knows there is some unfaithfulness), he waits N days;
    if no wives are shot in that time, he then shoots his own wife.
    
    But something seems strange about that.  Consider one husband who
    knows of two unfaithful wives, and his own happens to be faithful.
    Within two days, one or both of those two wives must be shot, lest
    a faithful one die.

>    Without the announcement, the husbands effectively behave like this:
>    if a husband knows of N unfaithful wives (with N >=2, to ensure that
>    everyone knows there is some unfaithfulness), he waits N days;
>    if no wives are shot in that time, he then shoots his own wife.

Two comments:
1) I think there may be some need to synchronize the starting point for 
everybody to be able to draw any conclusion - i.e. what happens if someone 
new moves into town?

2) If some husband of a U-wife can't find his gun, it has terrible 
consequences: the next day all other husbands conclude that the reason he 
didn't shoot his wife is because their wives are U, so every other wife in 
town gets shot.

>>    Without the announcement, the husbands effectively behave like this:
>>    if a husband knows of N unfaithful wives (with N >=2, to ensure that
>>    everyone knows there is some unfaithfulness), he waits N days;
>>    if no wives are shot in that time, he then shoots his own wife.

Actually, the husband must wait until he knows of N >= 3 unfaithful wives,
since with only two, everyone doesn't know that everyone knows that some
wife(s) is unfaithful.  Suppose a husband knows of N=2 (and his own wife is
faithful).  Then the husband of one of those two wives sees only one
unfaithful wife -- so he can't assume everyone knows that some wife is
unfaithful: that other wife's husband might (he hopes :^) know of *no*
unfaithful wives.

>1) I think there may be some need to synchronize the starting point for 
>everybody to be able to draw any conclusion - i.e. what happens if someone 
>new moves into town?

I think the problem has problems even without this curve ball!

I would assume that if these husbands are such good logicians, they'd also
take into account the possibility that someone can't find their gun, rather
than hasty rash decisions.


This whole problem is similar to Eye Island, discussed in rec.games.puzzles
two years ago.

Eye Island has a very strange set of rules.  No one may discuss eye color,
except for the one green-eyed guru.

The island contains 200 blue-eyed people, 300 brown-eyed people, and the
green-eyed guru.

Everyone's an excellent logician, and if anyone ever can logically deduce
their own eye color, they must commit suicide the night of the day on which
they deduce their own eye color.

Everyone can see everyone else's eyes, but there are no mirrors.

One day, the green-eyed guru stands up and announces:

	There is a blue-eyed person here !


What effect does this seemingly non-news statement have ?

/Eric

    Seems like the same solution Lynn outlined applies.  This is a better
    (i.e., less sexist) phrasing of the same problem, and in fact looks
    to be related to the "red hat"-"green hat" problem.
    
    It could be made a bit more colorful by having the people commit
    suicide by jumping off the mile-high cliff at the Northern tip of
    Eye Island.
    
    ON THE OTHER HAND
    
    Since everybody is such an excellent logician, and presumably prefers
    to kill rather than commit suicide, it would seem MORE logical to
    assume that the inhabitants en masse kill the green-eyed guru before
    he has a chance to announce anything.  Thus preventing the inevitable
    and fateful 200th Night.
    
      John

>Everyone's an excellent logician, and if anyone ever can logically deduce
>their own eye color, they must commit suicide the night of the day on which
>they deduce their own eye color.

This rule doesn't apply to the green-eyed guru.  Right?


>What effect does this seemingly non-news statement have ?

Non-news shouldn't have an effect.  That why there seems to be something
wrong with these problems.

Peter, are you asking a serious question about the problem ?  The truth is,
there *is* more information being revealed by the guru.  If you can't
figure it out, I'll give you a hint.

In answer to someone else's question, yes, I suppose it should be said
explicitly that the green-eyed guru is allowed to discuss eye color (although
perhaps he isn't as good a logician as the rest, in which case he probably
would have kept his mouth shut for the good of the tribe)