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Grex > Tutoring > #14: Algebra, Geometry, Calculus, Trig, all that good stuff |  |
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| 25 new of 132 responses total. |
aruba
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response 37 of 132:
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Jul 10 15:31 UTC 1997 |
Right:
Both false => (I) both rooms contain tigers and (II) room I does not
contain a tiger, two statements which are clearly in conflict. So they can't
both be false, which means (according to the king) they must both be true.
Both true => (I) either there is a lady in room I or there is a lady in
room II, and (II) there is not a lady in room I. So there must be a lady in
room II.
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aruba
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response 38 of 132:
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Jul 10 15:41 UTC 1997 |
The Third Trial
In this trial, the king explained that, again, the signs were either both true
or both false. Here are the signs:
I II
EITHER A TIGER IS IN A LADY IS IN
THIS ROOM OR A LADY IS THE OTHER ROOM
IN THE OTHER ROOM
Does the first room contain a lady or a tiger? What about the other
room?
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aruba
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response 39 of 132:
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Jul 10 15:47 UTC 1997 |
BTW the symbol "=>" that remmers used in #33 and I used in #37 means
"implies". So "A => B" can be read "A implies B", or in other words,
"If A is true then B must be true as well".
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aruba
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response 40 of 132:
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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"
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drew
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response 41 of 132:
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Jul 10 23:42 UTC 1997 |
I am assuming either...or to mean exclusive-or (XOR).
If room I has a cat, then II is false. In that case, if there were a lady
in room II, then the either/or statement (TRUE XOR TRUE) would evaluate FALSE.
If room I has a lady, then II is true. In order for I to evaluate TRUE, then,
there must also be a lady in room II. (Kitty_in_I XOR Lady_in_II ==
FALSE XOR TRUE == TRUE).
So they're either both chicks or both cats, and it makes no difference which
door you pick.
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nsiddall
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response 42 of 132:
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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.
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aruba
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response 43 of 132:
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Jul 11 06:01 UTC 1997 |
You should assume that "either A or B" is an *inclusive-or* - that is,
"either A or B" means "either A is true or B is true or both are true."
(That's the way mathematicians and logicians use the word "or", as a rule,
and that's definitely the way Smullyan means it here. (I can tell from his
solution.))
Care to have another go, drew?
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aruba
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response 44 of 132:
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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?
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remmers
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response 45 of 132:
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Jul 11 13:24 UTC 1997 |
If room (II) contained a tiger, then by the king's condition,
sign (II) would be true and room (II) would contain a lady, a
contradiction. Hence room (II) cannot contain a tiger and so
must contain a lady. The prisoner should pick room (II).
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aruba
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response 46 of 132:
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Jul 11 18:44 UTC 1997 |
Quite right. (And room I must contain a tiger.) I'll post the next one
tonight.
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aruba
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response 47 of 132:
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Jul 12 05:29 UTC 1997 |
The Fifth Trial
The same rules apply, and here are the signs:
I II
AT LEAST ONE ROOM THE OTHER ROOM
CONTAINS A LADY CONTAINS A LADY
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srw
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response 48 of 132:
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Jul 12 07:26 UTC 1997 |
View hidden response.
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srw
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response 49 of 132:
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Jul 12 07:28 UTC 1997 |
resp:48 is my solution. I hid it so others could have a go at it .
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mcnally
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response 50 of 132:
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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..
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srw
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response 51 of 132:
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Jul 12 19:38 UTC 1997 |
same conclusion arrived at slightly differently.
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aruba
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response 52 of 132:
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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?
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aruba
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response 53 of 132:
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Jul 13 17:47 UTC 1997 |
(Note that the rules in #44 still apply, through the 8th trial.)
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mcnally
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response 54 of 132:
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Jul 13 18:26 UTC 1997 |
This response has been erased.
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mcnally
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response 55 of 132:
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Jul 13 18:29 UTC 1997 |
solution scribbled in deference to Steve's precedence of # 49
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drew
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response 56 of 132:
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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.
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aruba
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response 57 of 132:
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Jul 13 21:58 UTC 1997 |
More good solutions, Mike and Drew.
The Seventh Trial
Here are the signs:
I II
IT DOES MAKE A YOU ARE BETTER
DIFFERENCE WHICH OFF CHOOSING THE
ROOM YOU PICK OTHER ROOM
What should the prisoner do?
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mcnally
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response 58 of 132:
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Jul 14 17:00 UTC 1997 |
View hidden response.
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aruba
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response 59 of 132:
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Jul 14 19:22 UTC 1997 |
How about a little more explanation, Mike? And don't worry, they get harder -
we're just warming up.
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richard
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response 60 of 132:
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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
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suzie
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response 61 of 132:
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Jul 14 21:47 UTC 1997 |
Ooooh, I hate algebra too! I'm so *stoo* pid! Richard you must be my kind
of guy!
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