response 62 of 132:

Jul 15 05:15 UTC 1997

L(I) => !(L(I) = L(II))    therefore  L(I) => T(II)
T(I) =>  (T(I) = T(II))    therefore  T(I) => T(II)

So we can conclude T(II) without even looking at sign II.

T(II) => L(I) so pick room I.

response 64 of 132:

Jul 15 15:12 UTC 1997

Very nice, gentlemen - on to

The Eighth Trial

   "There are no signs above the doors!" exclaimed the prisoner.
   "Quite true," said the king.  "The signs were just made, and I haven't
had time to put them up yet."
   "Then how do you expect me to choose?" demanded the prisoner.
   "Well, here are the signs," replied the king.

                     THIS ROOM
                     CONTAINS
                      A TIGER


                     BOTH ROOMS
                      CONTAIN
                       TIGERS

   "That's all well and good," said the prisoner anxiously, "but which
sign goes on which door?"
   The king thought awhile.  "I needn't tell you," he said.  "You can
solve this problem without that information."
   "Only remember, of course," he added, "that a lady in the lefthand room
means the sign which should be on that door is true and a tiger in it
means the sign should be false, and that the reverse is true for the
righthand room."
   What is the solution?

response 66 of 132:

Jul 16 19:20 UTC 1997

I'm afraid that's not right.  (But Valerie probably didn't want to marry the
lady anyway.  However, since she's allergic to cats, the tiger wouldn't be
any better, I don't think.)  Statements 2 and 5 can be taken farther, and
statement 8 is incorrect.

response 68 of 132:

Jul 17 15:58 UTC 1997

Here we go again.

if sign(I)  then L(I)  else T(I).
if sign(II) then T(II) else L(II).

"This room contains a tiger" cannot go on door I, because this would
establish

(sign(I) AND T(I)) OR ( (NOT sign(I)) AND L(I))

which evaluates as FALSE. So this sign goes on the other door.

"Both rooms..." then goes on door I, and it can't be true, because
it would put a kitty in room I as well as II, violating the
"if sign(I) then L(I)" clause. So NOT sign(I), and the prisoner
doesn't have to end up as cat food.

However, since NOT sign(I), T(I). But fortunately, sign(II) isn't
true either which leads to L(II).

response 70 of 132:

Jul 17 18:56 UTC 1997

Signs II and III are mutually exclusive.  That is, if one is true, the other
must be false.  Therefore, one of them must be true, no matter what.  Since
at most one sign is true, then sign I must be false, and the Lady is in room
I.

response 72 of 132:

Jul 18 00:04 UTC 1997

Yes, but I think dang has only mis-articulated his analysis.  II XOR III is
true, so I must be false (to avoid having 2 true signs).

response 74 of 132:

Jul 18 22:23 UTC 1997

It is in fact the case that if one of the two signs (II and III) is false, 
then the other must be true.  *That* is what you need to conclude that one of 
them must be true.  It's not enough to say that "if one is true, then the 
other is false", because that doesn't eliminate the possibility that they are
both false.  (And it's *not* the case that one sign must be true.  All we know
is that *at most one* of the signs is true.)

With that correction, though, dang's (and i's) solution is a very nice one.
I'll enter the next trial when i get home.

response 76 of 132:

Jul 19 12:41 UTC 1997

If the lady were in Room (II), then sign (II) would be false,
contradicting the fact that the sign on the room containing the
lady is true. Therefore, the lady is not in room (II).

If the lady were in room (III), then tigers would be in the
other two rooms, implying that signs (I) and (II) are both true,
contradicting the fact that at least one of the signs on the
tiger rooms is false. Therefore, the lady is not in room (III).

Hence the lady is in room (I). (Note that this implies that
sign (I) is true and sign (III) is false, consistent with the
king's preconditions.)

response 78 of 132:

Jul 19 17:22 UTC 1997

The lady can't be in room III, because this would make sign(III) false.

If she were in room II, this would put the cat in room I with sign(I) being
true.

So she has to go in room I, making sign(I) true and room III empty, leaving
the cat in room II.

response 80 of 132:

Jul 20 15:56 UTC 1997

Ugh.  I got as far as figuring that if room VIII is not empty then the lady
is in room one or seven, but then I decided to get a life.

response 82 of 132:

Jul 20 18:24 UTC 1997

In order for sign(VIII) to yield any useful information, it must be
ascertained to have some boolian value. Thus room VIII can't be empty.
It can't have the lady either, since sign(VIII) would have to be true, and
it would not be. So there's a kitty in room VIII. This is permitted by the
AND function in sign(VIII) so long as there is something in room IX.

Room IX also has to have a kitty, for similar reasons that room VIII does.
This makes sign(VI) *right* (true).

This of course makes sign(III) false.

Sign(III) is an OR function (it was stated that either-or was *inclusive*-or
in an earlier problem) which requires *both* clauses to be false.

so, sign(VII), meaning that the lady is not in room I.

Also, NOT sign(V), which is another OR function, both of whose switches must
be turned off.

Thus, NOT sign(IV); sign(I) is right, and the lady must be in either room I,
III, V, VI, and IX.

We've already eliminated rooms III, V, and IX, due to their signs being false,
which leaves rooms I and VII; and we had already eliminated room I due to
sign(VII) being true. This leaves the lady in room VII.

response 84 of 132:

Aug 29 17:47 UTC 1997

Ok, so I've started reading Jack Chalker's Well World books.  The Well
world is supposed to be a planet which is divided into 1560 environments,
each of which is shaped like a hexagon.  "Each hex is identical in size -
each one of the six sides is just a shade under three hundred fifty-five
kilometers, and they're a shade under six hundred fifteen kilometers
across." 

So the planet is tiled with hexagons.  It's not clear whether 1560 is the
total number of hexagons, though, because Chalker implies later that some
hexes are cut in half by the equator, and I'm not sure whether those count
in the 1560 number (I think not).

So the question is: did Chalker think this out before he wrote it?  Is it
possible to tile a sphere with hexagons in a way that fits the
description?

response 86 of 132:

Aug 30 03:48 UTC 1997

Hi, Math always the worst experience in college . I get E 
on both Calculus I & II. The only way to study it is
to do 10,000 exercises. (take a long time). The 2nd year,
I got A and B. It is a nightmare! Remembering all those
Integral formula etc. They just prove that Newton is 
wrong in Gravitation theory. There is a WEAK force
that cause in-accurate result. Fortunately physic is
an interesting subject (nuclear physic) Regards AW