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United States47024 Posts
On April 24 2009 11:57 p3numbra wrote:
Actually, doesn't Slithe's say that it needs 4 weighings if the coin is lighter? Mine needs exactly 3 weighings regardless of the weight of the coin.
Nah. Slithe's weighings are:
EFGH vs IJKL
BFGK vs CDHL
CEHK vs ADGL
The 3 weighings can uniquely determine which coin is fake, and what weighings you make are entirely independent of the results of previous weighings, which makes it much cleaner.
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United States2822 Posts
+ Show Spoiler +EFGH vs IJKL
BFGK vs CDHL
CEHK vs ADGL
Case 1: EFGH = IJKL
ABCD contains the miscreant coin, weight undetermined.
Case 1.1: BFGK = CDHL
A is the miscreant coin, BCD are all determined.
Case 1.1.1: CEHK = ADGL
Impossible.
Case 1.1.2: CEHK > ADGL
A is light.
Case 1.1.3: CEHK < ADGL
A is heavy.
Case 1.2: BFGK > CDHL
B is heavy or CD is light.
Case 1.2.1: CEHK = ADGL
B is heavy.
Case 1.2.2: CEHK > ADGL
D is light.
Case 1.2.3: CEHK < ADGL
C is light.
Case 1.3: BFGK < CDHL
B is light or CD is heavy.
Case 1.3.1: CEHK = ADGL
B is light.
Case 1.3.2: CEHK > ADGL
C is heavy.
Case 1.3.3: CEHK < ADGL
D is heavy.
Case 2: EFGH < IJKL
EFGH is light or IJKL is heavy.
Case 2.1: BFGK = CDHL
E is light or IJ is heavy.
Case 2.1.1: CEHK = ADGL
IJ is heavy. Cannot determine which.
Case 2.1.2: CEHK > ADGL
Impossible.
Case 2.1.3: CEHK < ADGL
E is light.
Case 2.2: BFGK > CDHL
H is light or K is heavy.
Case 2.2.1 CEHK = ADGL
Impossible.
Case 2.2.2: CEHK > ADGL
K is heavy.
Case 2.2.3: CEHK < ADGL
H is light.
Case 2.3: BFGK < CDHL
FG is light or L is heavy.
Case 2.3.1 CDHK = ADGL
F is light.
Case 2.3.2 CDHK > ADGL
G is light.
Case 2.3.3 CDHK < ADGL
L is heavy.
Case 3: EFGH > IJKL
EFGH is heavy or IJKL is light.
Case 3.1: BFGK = CDHL
E is heavy or IJ is light.
Case 3.1.1: CEHK = ADGL
IJ is light. Cannot determine which.
Case 3.2.2: CEHK > ADGL
E is heavy.
Case 3.3.3: CEHK < ADGL
Impossible.
Case 3.2: BFGK > CDHL
FG is heavy or L is light.
Case 3.2.1: CEHK = ADGL
F is heavy.
Case 3.2.2: CEHK > ADGL
L is light.
Case 3.2.3: CEHK < ADGL
G is heavy.
Case 3.3: BFGK < CDHL
H is heavy or K is light.
Case 3.3.1: CEHK = ADGL
Impossible.
Case 3.3.2: CEHK > ADGL
H is heavy.
Case 3.3.3: CEHK < ADGL
K is light.
Found two flaws through rigor.
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United States47024 Posts
On April 24 2009 12:28 p3numbra wrote:
Found two flaws through rigor.
Gah, I carried through his typo, the correct one was:
EFGH vs IJKL
BFGK vs CDHL
CEHJ vs ADGL
Also, rigor is a REALLY INEFFICIENT way of testing whether it works. I did an analysis in spoilers previously.
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United States2822 Posts
Aah, sick. That method is way more efficient than mine. Good job!
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Weigh two coins against one coin and hope you get lucky on the first try.
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United States24745 Posts
On April 24 2009 13:01 Chef wrote:
Weigh two coins against one coin and hope you get lucky on the first try.
This answer is specifically incorrect in case you weren't aware since you can factor in the odds and payout of each case and you'll realize it is inferior to some of the other solutions already presented.
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What if you only have time to weigh once, or you DIE?
Then it doesn't matter how efficient the other methods are, because anymore than one try kills you.
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United States24745 Posts
On April 24 2009 13:05 Chef wrote:
What if you only have time to weigh once, or you DIE?
Then it doesn't matter how efficient the other methods are, because anymore than one try kills you.
I suppose if you randomly add in convenient circumstances unrelated to the original problem, then you could make a case for just about any possible solution.
What a waste of time this conversation has been :p
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The task is to use the scale the least number of times to determine which coin is fake.
I didn't add anything. He asked what scenario will give you the least number of tries.
Suppose there were a contest with 1000 people. Whoever determines which coin is fake in the least number of tries wins. You're not going to win by being slow and steady, because enough other people are going to be using the quick and dirty method that at least one will get lucky.
Hah ha!
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United States24745 Posts
On April 24 2009 13:09 Chef wrote:Show nested quote +The task is to use the scale the least number of times to determine which coin is fake. I didn't add anything. He asked what scenario will give you the least number of tries. Suppose there were a contest with 1000 people. Whoever determines which coin is fake in the least number of tries wins. You're not going to win by being slow and steady, because enough other people are going to be using the quick and dirty method that at least one will get lucky. Hah ha!
Then why don't you just set the condition that 'whoever is first wins' and then just pick a random coin and you might be right?
This is so ridiculous...
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Because you have to determine it's fake. You won't know if it's fake or not if you just pick up a coin.
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chef, how would weighing 2 coins against 1 coin even work if a. the fake is lighter b. the fake is less heavy than 2 normal coins
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I'm assuming we live in 2009 and have scales that tell you the actual weight of the coin. IE: If the one coin's weight multiplied by two is not equal to the weight of the other two coins, it's the fake.
Unless of course you get fucked and the fake is one of the two coins you're measuring together.
Oops 
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On April 24 2009 13:20 Chef wrote:I'm assuming we live in 2009 and have scales that tell you the actual weight of the coin. IE: If the one coin's weight multiplied by two is not equal to the weight of the other two coins, it's the fake. Unless of course you get fucked and the fake is one of the two coins you're measuring together. Oops
ummmmm stop making shit up
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United States47024 Posts
Chef, do you really have nothing better to do than to troll blogs right now?
Hell, this blog isn't worth trolling anymore, because the OP's question has been answered in multiple ways already.
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On April 24 2009 13:25 SpiritoftheTunA wrote:Show nested quote +On April 24 2009 13:20 Chef wrote:I'm assuming we live in 2009 and have scales that tell you the actual weight of the coin. IE: If the one coin's weight multiplied by two is not equal to the weight of the other two coins, it's the fake. Unless of course you get fucked and the fake is one of the two coins you're measuring together. Oops ummmmm stop making shit up
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This can be solved with 3 weighs I'm quite sure since I double-checked with the help of my math teacher, took me a week to figure out at some random point in highschool.
Trying to explain it all here seems like a titanic task, since there are many possibilities. But the answer is 3.
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When I first heard this problem I did it p3numbra's way. I didn't write anything down but it was clear in my head. I like Slithe's solution a lot more though, it's more obvious how the problem generalises too I think.
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on second thought, I'm heavily trying to do it again, but there are a few evolutions where it's impossible to make to without the fourth weigh.
I'm a little confused, will probably try again later
My solution was basically Slithe's solution
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EPT
