response 40 of 132:

Jul 10 16:23 UTC 1997

Here's some more notation, just so we're all talking the same language:

A & B   means "A is true and B is true"
A | B   means "either A is true or B is true (or both are true)"
!A      means "A is false"
A <=> B means "A implies B and B implies A", which is the same as saying
              "A is true if and only if B is true"
L(a)    means "there is a lady in room a"
T(a)    means "there is a tiger in room a"

response 42 of 132:

Jul 11 05:53 UTC 1997

Both signs can't be false.  For sign I to be false there must be a
lady in room I and a tiger in in room II.  In that case sign II must be true.
Since they cannot both be false, they must be true.  If sign II is true, the
first part of sign I cannot be true.  Therefore the truth of sign I must be
fulfilled by the second part, meaning there is a lady in room II.  Guaranteed
ladies in both rooms.

response 44 of 132:

Jul 11 06:13 UTC 1997

Nathaniel slipped in, and is quite correct.  Which means we should move on
to...

THE SECOND DAY

   "Yesterday was a fiasco," said the king to his minister.  "All three
prisoners solved their puzzles!  Well, we have five trials coming up
today, and I think I'll make them a little tougher." 
   "Excellent idea!" said the minister.
   Well, in each of the tirals of this day, the king explained that in the
lefthand room (Room I), if a lady is in it, then the sign on the door is
true, but if a tiger is in it, the sign is false.  In the righthand room
(Room II), the situation is the opposite: a lady in the room means the
sign on the door is false, and a tiger in the room means the sign is true.
Again, it is possible that both rooms contain ladies or both rooms contain
tigers, or that one room contains a lady and the other a tiger.

THE FOURTH TRIAL

After the king explained the above rules to the prisoner, he pointed to
the two signs:


      I                      II
  BOTH ROOMS             BOTH ROOMS
CONTAIN LADIES         CONTAIN LADIES

Which room should the prisoner pick?

response 46 of 132:

Jul 11 18:44 UTC 1997

Quite right.  (And room I must contain a tiger.)  I'll post the next one
tonight.

response 50 of 132:

Jul 12 17:39 UTC 1997

  Room I must contain a lady and room II a tiger.  Let's begin with 
  the statement on door I.  In order for it to be true (and for there
  to be at least one lady) there must be a lady behind door I.  If
  the statement is not true then there is a tiger behind both doors
  (for the statement to be false) but this is not possible because
  if a tiger is behind door II then the statement on door II must be
  true, a condition which is not satisfied if there's a tiger behind
  door I..

response 52 of 132:

Jul 13 17:27 UTC 1997

Good solutions, Steve and Mike.

The Sixth Trial

The king was particularly fond of this puzzle, and the next one too.  Here
are the signs:

          I                     II
IT MAKES NO DIFFERENCE    THERE IS A LADY
 WHICH ROOM YOU PICK     IN THE OTHER ROOM

What should the prisoner do?

response 56 of 132:

Jul 13 21:38 UTC 1997

Okay, here goes. From #44:
        if L(I) then sign(I)
        if T(I) then not sign(I)
        if L(II) then not sign(II)
        if T(II) then sign(II)

sign(I) requires L(I) and L(II), and L(II) in turn requires NOT sign(II) and
consequently  (sign(I) AND NOT sign(I)). Contradiction.

Therefore, NOT sign(I). This means it *does* make a difference!
Since sign(I), T(I) leading to NOT sign(II) ==> L(II).

Room I has the cat; room II has the chick.

response 60 of 132:

Jul 14 20:59 UTC 1997

This item is far andaway the frontrunner for
geekiest grex item of the year.  I hated alegrbra...
Geometry was interesting though...I like proving

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.