Author |
Topic: Logical Signs II (Read 2144 times) |
|
potter
Newbie
Posts: 1
|
 |
Logical Signs II
« on: Jul 29th, 2002, 3:03am » |
|
This one seems simpler than logical sign I. There are 6 possible cases of True and False for the three chests. 4 of them are contradictory as in each one of them point to more than one chest as containing the treasure. The rest 2 cases both point towards the BRONZE one. so the treasure lies in the BRONZE chest.
These 2 cases are: Silver - false false
Gold - true true or false
Bronze- true or false true
|
| « Last Edit: Jul 29th, 2002, 3:04am by potter » |
 IP Logged |
|
|
|
jc
Guest
|
|
Re: Logical Signs II
« Reply #1 on: Aug 3rd, 2002, 11:23pm » |
|
Well, I got the same answer, only I found only one of the six possible answers pointing to bronze.
The six were:
1: S - True, G - False, B - False
Which wouldn't work because G and B both being false mean that the treasure is in G, and yet S is also true. Since the treasure is only in one, this possibility becomes impossible.
2: S - True, G - True, B - False
Which wouldn't work because G and B contradict each other.
3: S - True, G - False, B - True
Which wouldn't work because G and B contradict each other.
4: S - False, G - True, B - False
Which wouldn't work because G and B contradict each other.
5: S - False, G - False, B - True
Which wouldn't work because G and B contradict each other.
6: S - False, G - True, B - True
Which is the only solution that works out, and it points to B, since S being false means the treasure isn't in S and G and B being true means that it's not in G.
So, 4 contradictions, one case of the treasure being in 2 chests, and one solution that works. 
I admit this was an awfully methodical way to do it, and you could shorten it by quite a bit by assuming that S must be false in the "true" solution because you can infer that all three possibilities with S being true must be ruled out without actually listing them. Still, it worked.
|
|
IP Logged |
|
|
|
Melkor
Guest
|
|
Re: Logical Signs II
« Reply #2 on: Aug 5th, 2002, 9:06pm » |
|
There is an easier approach: begin by assuming that the silver chest holds the treasure. Then check the resultant veracity of the chests' labels against the known information - i.e., that at least one chest's label is true and at least on is false. Thus the silver chest cannot contain the treasure since then all three statements would be true. Continue with the other chests until you find an assumption that fits the given information.
|
|
IP Logged |
|
|
|
Icarus
wu::riddles Moderator
Uberpuzzler
Boldly going where even angels fear to tread.
Gender: 
Posts: 4863
|
|
Re: Logical Signs II
« Reply #3 on: Oct 8th, 2002, 7:50pm » |
|
Usually, the best way to solve a logic problem is by examining the cases and using them to evaluate the statements. Not only is it often easier, but you can more easily avoid the mistake of assuming a statement has to be true or false. It can also be a contradiction (true if it is false, and false if it is true) or indeterminate (true only if it is true, and false only if it is false). This problem does not have that situation, but it is very big in Logical Signs I.
|
|
IP Logged |
"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Notify of replies
Send Topic
Print