|
Grex > Tutoring > #14: Algebra, Geometry, Calculus, Trig, all that good stuff |  |
|
| Author |
Message |
| 25 new of 132 responses total. |
srw
|
|
response 48 of 132:
| 
Jul 12 07:26 UTC 1997 |
View hidden response.
|
srw
|
|
response 49 of 132:
|
Jul 12 07:28 UTC 1997 |
resp:48 is my solution. I hid it so others could have a go at it .
|
mcnally
|
|
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..
|
srw
|
|
response 51 of 132:
|
Jul 12 19:38 UTC 1997 |
same conclusion arrived at slightly differently.
|
aruba
|
|
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?
|
aruba
|
|
response 53 of 132:
|
Jul 13 17:47 UTC 1997 |
(Note that the rules in #44 still apply, through the 8th trial.)
|
mcnally
|
|
response 54 of 132:
|
Jul 13 18:26 UTC 1997 |
This response has been erased.
|
mcnally
|
|
response 55 of 132:
|
Jul 13 18:29 UTC 1997 |
solution scribbled in deference to Steve's precedence of # 49
|
drew
|
|
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.
|
aruba
|
|
response 57 of 132:
|
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?
|
mcnally
|
|
response 58 of 132:
|
Jul 14 17:00 UTC 1997 |
View hidden response.
|
aruba
|
|
response 59 of 132:
|
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.
|
richard
|
|
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
|
suzie
|
|
response 61 of 132:
|
Jul 14 21:47 UTC 1997 |
Ooooh, I hate algebra too! I'm so *stoo* pid! Richard you must be my kind
of guy!
|
srw
|
|
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.
|
mcnally
|
|
response 63 of 132:
|
Jul 15 06:01 UTC 1997 |
My reasoning was essentially the same as Steve's though less
formally expressed. With sign I reading as it does there's
no reason to open door II -- either you get a lady behind door
I or you get a tiger behind both doors but never a lady behind
door II. Luckily, for sign II to satisfy the conditions the only
possible setup is a lady behind door I and a tiger behind door II.
|
aruba
|
|
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?
|
valerie
|
|
response 65 of 132:
|
Jul 15 15:50 UTC 1997 |
This response has been erased.
|
aruba
|
|
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.
|
valerie
|
|
response 67 of 132:
|
Jul 17 08:11 UTC 1997 |
This response has been erased.
|
drew
|
|
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).
|
aruba
|
|
response 69 of 132:
|
Jul 17 17:45 UTC 1997 |
Looks good, Drew, except that you need to invoke "process of elimination"
in the final sentence to draw the final conclusion. As in:
The fact that sign(I) is false implies both
1) There is a tiger in Room I, and
2) One of the rooms does not contain a tiger.
By process of elimination, Room II must contain a lady. (Sign(II) doesn't
help you at all.)
Ok, then, it's on to
THE THIRD DAY
"Confound it!" said the king. "Again all the prisoners won! I think
tomorrow I'll have *three* rooms instead of two; I'll put a lady in one
room and a tiger in each of the other two rooms. Then we'll see how the
prisoners fare!"
"Excellent idea!" replied the minister.
"Your conversation, though flattering, is just a bit on the repetitious
side!" exclaimed the king.
"Excellently put!" replied the minister.
The Ninth Trial
Well, on the third day, the king did as planned. He offered three rooms
to choose from, and he explained to the prisoner that one room contained a
lady and the other two contained tigers. Here are the three signs:
I II III
A TIGER A LADY A TIGER
IS IN IS IN IS IN
THIS ROOM THIS ROOM ROOM II
The king explained that at most one of the three signs was true. Which
room contains the lady?
|
dang
|
|
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.
|
aruba
|
|
response 71 of 132:
|
Jul 17 22:33 UTC 1997 |
Hmmm... The fact that "if one is true, the other must be false" does not
imply "one of them must be true".
|
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).
|