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Mikey 14778
Poke It With A Stick
1631 Posts |
 Posted - 25 Feb 2010 : 9:25:52 PM
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No, you can get it in 3 and be sure whether it is heavier or lighter. But you'd need a mind like a steel trap to work out the method from scratch IMHO.
My method (which took 5 moves) went like this:
1) Put 6 coins on one side and the other 6 on the other. One side goes down.
2) Take 3 coins off each side. If the remaining coins balance then they're all good so put them somewhere else and replace them with the 3 each side you took off. If they don't balance then hang on to them.
3) Take 1 coin off each side. If the remaining coins balance then they're all good so put them somewhere else and replace them with the 1 each side you took off, then skip to step 5. If they don't balance then hang on to them and proceed to step 4.
4) Take 1 coin off each side. If the remaining coins balance then they're both good so put them somewhere else and replace them with the 1 each side you took off
You are now down to 2 coins, call them A and B (1 each side) and they don't balance.
5) Compare the heavier of these (say A) with a known good coin. If it balances then B was the odd one and it was lighter. If it doesn't balance then it's A and it's heavier.
I'd be interested to hear how the other (non-Googled) methods worked.
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Andrew Blee
Wetter than a very wet thing
519 Posts |
Posted - 26 Feb 2010 : 07:32:45 AM
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split the coins into 4 piles of: A, B, C, D
1 Balance A v B If they balance then A & B are good, else C & D.
2 Take one pile from the known good ones (say A). Balance A v C. If they balance then D contains the fake else C contains the fake.
Now you know the 3 coins with the fake 1, 2 and 3.
3 Balance 1 v 2. If they balance then 3 is fake.
4 If they don't balance then balance 1 v 3. If they balance then 2 is fake else 1
So is that 3 maybe 4 steps? |
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Mikey 14778
Poke It With A Stick
1631 Posts |
Posted - 26 Feb 2010 : 07:47:47 AM
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Looks good to me Andrew!
I just noticed that my in method, step 1 is totally pointless - we know that one side will go down, so why do it ?
So skip that and it becomes a 3 or 4 move method. |
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Richard Le Mare
Soaked
688 Posts |
Posted - 26 Feb 2010 : 09:50:51 AM
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Andrew agree 4 pile of 3, 2 moves then the differnt pile needs 2 moves, you have to do 2 as you dont know until you have copmared all 3 as you don't know which one is lighter or heavier
 426  Erica 
all replies are only my opinion not that of the committee |
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robbersdog
Drenched
1427 Posts |
Posted - 26 Feb 2010 : 11:01:25 AM
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Mikey's method will tell you which the fake coin is in three weighings every time, but it won't tell you if it's heavy or light in each case. The trick is to weigh the coins three times, with four coins on each side each time. You just have to set up the scales so that each coin appears in a unique pattern through the weighings so the results of the weighings tell you which coin is counterfeit and whether it's heavy or light. Each coin, if counterfeit, will give a unique set of results of the three weighings.
There are several combinations which work. The one I found was this one:
Number the coins 1,2,3,4,5,6,7,8,9,A,B,C
Weigh three times in these combinations:
1234 - 5678
279B - 468A
358C - 67AB
This combination of weighings gives a unique set of results for each different situation. So, if the wieghings are as follows, scale tilts to the left, then balances, then balances again we can see that coin 1 is heavy. If the scales tilt right, then left, then right we can see that coin 7 is heavy, and so on.
Yes I did work this out for myself, but again with the proviso that I'm used to looking at puzzles and you learn how do approach them. And now that you've seen how this works you all know of another way of approaching the problem so next time you see a weighing puzzle like this, you'll all be closer to solving it before you start.
Here's a simpler puzzle:
A computer scientist creates a machine that always answers truthfully when asked a yes/no type question. The machine looks like a simple box with a red light and a green light on top of it. It answers by flashing either red or green. Two batches of these machines were made, one lot in China, the other in Japan. Unfortunately they weren't wired the same, so one batch answers yes by flashing green and no by flashing red, the other batch answers yes by flashing red and no by flashing green. To add to the confusion it is not known whther it's the Chinese or Japanese machine that answers yes by flashing green.
Someone has given you one of these machines and you want to know if it's Chinese or Japanese. Which single yes/no question could you ask it to find out?
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Chris Smith
Flying Fifteen 797
Sailing Photos
http://www.sushidesign.net/
http://www.gs-illustration.co.uk/ |
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Mikey 14778
Poke It With A Stick
1631 Posts |
Posted - 26 Feb 2010 : 11:18:55 AM
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| Hmmm, this a variant of the old "Which way would your brother tell me to go ?" question.... |
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plumberpete
Damp
101 Posts |
Posted - 26 Feb 2010 : 7:26:52 PM
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talk about complicating things
6 coins on each side - 1 will be lighter
then 3 coins on each side - 1 will be lighter
then 1 coin on each side - 1 will be lighter therfore fake
if not the one not weighed will be fake
you can tell none of you are engineers -
must managers or computer geeks
cheers
pete
Pete Gough
Phantom 1273
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rodgerwebb
Completely dry
29 Posts |
Posted - 26 Feb 2010 : 7:59:06 PM
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Good point Pete! - reminds me of the old engineer in a balloon joke_
A man in a hot air balloon realised he was lost. He reduced altitude and spotted a man below. He descended a bit more and shouted, "Excuse me, can you help me? I promised a friend I would meet him half an hour ago, but I don't know where I am."
The man below replied, "You are in a hot air balloon hovering approximately 30 feet about the ground. You are between 40 and 42 degrees north latitude and between 58 and 60 degrees west longitude."
"You must be an engineer," said the balloonist.
"I am," replied the man, "but how did you know?"
"Well," answered the balloonist, "everything you told me is technically correct, but I have no idea what to make of your information, and the fact is I am still lost."
The man below responded, "You must be a manager."
"I am," replied the balloonist, "how did you know?"
"Well," said the man, "you don't know where you are or where you are going. You made a promise which you have no idea how to keep, and you expect me to solve your problem. The fact is you are exactly in the same position you were in before we met, but now, somehow, it's my fault." |
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chris
Poke It With A Stick
1847 Posts |
Posted - 27 Feb 2010 : 08:22:20 AM
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quote:
Originally posted by plumberpete
talk about complicating things
6 coins on each side - 1 will be lighter
then 3 coins on each side - 1 will be lighter
then 1 coin on each side - 1 will be lighter therfore fake
if not the one not weighed will be fake
you can tell none of you are engineers -
must managers or computer geeks
cheers
pete
Pete Gough
Phantom 1273
You must be an engineer Pete, you've assumed everyone else is wrong and don't seem to have read the initial brief properly....
It says you dont know if the fake coin is heavier or lighter so you cant just keep choosing the lightest set and narrowing them down as it may be the heavier set you need which would add steps to your solution.
Chris
425  720 "ERICSSON"
sailing pictures My pictures |
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skifftim
Quite dry
72 Posts |
Posted - 27 Feb 2010 : 5:42:15 PM
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Uh logic?
"Here's another one. You're given twelve identical looking coins and are told one is a fake. The fake is a different weight to the real coins but you don't know if it's heavier or lighter. All the real coins weight exactly the same amount.
You're also given a set of balance scales (the type that have two little baskets hanging from each end of a balanced bar).
What's the minimum number of times you have to use the scales to find the fake coin and know if it's heavier or lighter than a real coin?"
The question asks what is the minimum number of times you have to use the scales to find the fake coin.
It does not ask to guarantee the answer nor any other combinations so the minimum number must be 2. With one coin on either side one side goes down the next it either does or does not, you have your answer.
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plumberpete
Damp
101 Posts |
Posted - 28 Feb 2010 : 9:38:25 PM
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point taken chris
but it only adds one move
if you weigh the lighter pile of the first six
and they are equal you know that the fake coin is heavier
apologies for assuming that all managers and it geeks are a waste of space
should have read the statement properley in the first instant and not assumed it was lighter 
apologies again - cant accept that managers and it geeks arent a wast of space - haha 
pete
Pete Gough
Phantom 1273
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Mikey 14778
Poke It With A Stick
1631 Posts |
Posted - 28 Feb 2010 : 11:00:55 PM
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quote:
6 coins on each side - 1 will be lighter
then 3 coins on each side - 1 will be lighter
then 1 coin on each side - 1 will be lighter therfore fake
But what if the last line (the '1 coin on each side' weighing) comes up with an equal balance ?
Then you've got to go back to the 4 coins you just took off and weigh 2 of them to identify the pair that are unequal.
So that's another step... |
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robbersdog
Drenched
1427 Posts |
Posted - 01 Mar 2010 : 08:18:23 AM
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If the 1 against 1 balances then the one you haven't weighed out of the three is the odd one out.
If you're told if the coin is lighter or heavier then Pete's method works. However, we don't know that, so his method is a 4 step answer, and not the minimum we're looking for..
Pedant's prize goes to Tim. Well done. :o)
Any solutions to the Japanese/Chinese machine question? I suspect Mikey's got it from his reply earlier.
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Chris Smith
Flying Fifteen 797
Sailing Photos
http://www.sushidesign.net/
http://www.gs-illustration.co.uk/ |
Edited by - robbersdog on 01 Mar 2010 08:34:22 AM |
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Mikey 14778
Poke It With A Stick
1631 Posts |
Posted - 01 Mar 2010 : 09:19:42 AM
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quote:
If the 1 against 1 balances then the one you haven't weighed out of the three is the odd one out.
Errm, no.
We weighed 3 v 3 (ie, 6 coins) in the previous step and they didn't balance. If we now weigh 1 v 1 (ie 2 coins) and they balance then we know that the duff coin is one of the 4 coins we just took off. Another weighing is then required to get to a 1 v 1 position where one of those coins is definitely duff.
But yes, the red light puzzle was just toooo easy. |
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robbersdog
Drenched
1427 Posts |
Posted - 01 Mar 2010 : 10:24:04 AM
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I agree Mikey, the method only works if you know if the coin is light or heavy. If you know this then you'll know which of the sets of three contains the duff coin. It then becomes the puzzle 'if you have three coins and a balance scale, how do you find out which coin is the heavy one?'
On to the flashing machines again, here is the extended, full on, get your thinking caps on version:
You go into a shop to buy one of these machines. The shopkeeper says he has three of them left, but unfortunately they know that one of them is faulty and flashes red or green randomly and they don't know if the other two are japanese or chinese machines and they don't know which one is the faulty one. You get to ask one of the machinse a question. What single yes/no question would you ask to ensure that you buy a good one?
If anyone wants the answer to the first flashing machine puzzle let me know and I'll post it.
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Chris Smith
Flying Fifteen 797
Sailing Photos
http://www.sushidesign.net/
http://www.gs-illustration.co.uk/ |
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Brain teaser
