Twelve Coins - Three Measures

F

Fluidity

Registered Senior Member
This is one of the hardest word puzzles I have ever heard.
It is an ancient puzzle, apparently from a real life dilemma.

You only have a balance for a scale. You are taking money at the bazaar, and the only way you can verify the silver coins are genuine, is by weight.

You do not have a standard for the coinage. So, if there is a bad
coin in the lot, you cannot tell if it is light or heavy in a single measure.

You must find the bad coin by process of elimination.

You are being tested by your Father.
You are given twelve coins.

By law, you only have three measures, or you will be accused of disrespecting the merchants.

You know there is one bad coin in this lot of twelve.

Given three chances to find the bad coin, which you cannot tell is light or heavy, you must find the bad coin, or take the chance of telling your Father the test is not fair, in which case he must prove you wrong.

Hint: Is there a difference in your answer if the coin is light?
 
Tough problem... I think this works... but checking it over again right now...

See attachement... writing it out doesn't work
 
Yup, it works... here is the solution with half-proof
E means it's equal, ? means unknown.
 
Persol

If I read your graph correctly, you still have two coins on the scale, and you don't know which one is good or bad. There is a complete solution. You can find 'the' bad coin of the 12 in just 3 measures. And only in two cases is it not known to be heavy or light. BTW, this question was posed by my Calc II teacher, and he never gave us the answer. Over a weekend, no one in the class of 37 students got it right. Close, but no cigar. Need another hint? The one I gave was supposed to suggest that you can determine which is heavy or light.

Four out of six times you know whether the coin is light or heavy.
 
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I have worked this out only for one variant (false coin is lighter)

if 1..4 = 5.. 8 then if 7,8 = 9,10 then if 8=11 (12 false)
if 1..4 = 5.. 8 then if 7,8 = 9,10 then if 8<>11 (11 false)
if 1..4 = 5.. 8 then if 7,8 <> 9,10 then if 8=9 (10 false)
if 1..4 = 5.. 8 then if 7,8 <> 9,10 then if 8<>9 (9 false)
if 1..4(heavier) <> 5.. 8 then if 123 =456 then if 9=7 (8 false)
if 1..4(heavier) <> 5.. 8 then if 123 =456 then if 9<>7(lighter) (7 false)
if 1..4(heavier) <> 5.. 8 then if 123<>456 then if 5=6 (4 false)
if 1..4(heavier) <> 5.. 8 then if 123(heavier)<>456 then if 5<>6(lighter) (6 false)
if 1..4(heavier) <> 5.. 8 then if 123(heavier)<>456 then if 1=2 (3 false)
if 1..4(heavier) <> 5.. 8 then if 123(heavier)<>456 then if 1(heavier)<>2 (2 false)
 
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With 9 vs 10 if it doesn't ballance the light coin is bad if 9-11 was light. If 9-11 was heavy then the heavy coin is bad.

The 4 vs 7 should change to 4 vs 6. If it balances 7 is bad.... otherwise 4 is bad. You can tell heavy or light by the original 1-4 vs 5-8 test.

Same with the 2vs5. It should be 2vs 6. If it balances, 5 is bad. If it doesn't balance 2 is bad. You can tell heavy light by the original 1-4 vs 5-8 test.
 
Here's my solution:

I can't find anyting better than this. Let me know if you think it's a rip-off or not. Four of six times, given any scenario, you find the bad coin, and you know if it is light or heavy. This 'proof' seems to cover all probabilities, all scenarios.

<I>If you start with four on each side, and you get an imbalance, you only have four known good coins, and no idea whether the bad coin is light or heavy.</I>

Don't look unless you think you have a perfect solution. I said 'complete,' not perfect. ;)

http://www.austinmassagetherapy.com/balance.jpg
 
Mine'll give you the answer each time with the changes listed above.
 
Since you have a balance scale,
1) I would put 6 on one side and 6 on the other, whichever side weighed less the lighter coin is in that 6
2) Split the 6 into 3 and 3, whichever side weighed less the lighter coin is in that 3
3) Take 1 coin out of your decided 3 coins from step 2, put it aside, then weigh the last 1 on one side and 1 on the other

The lighter coin will be obvious if it is on the scale. If they weigh the same, then the light coin is the one you put aside before the final weighing.

That's 3 measurements and I've found the coin with no graphs.
What am I missing here?
 
If

you have lifted the 'lighther' coin, you still don't know if the bad coin is heavy or not. I specifically stated you don't know the case. You have to find the coin, light or heavy.
 
I think I have heard it before where you were asked to find the light coin in the bunch. I assumed it was the same question so I half ass read it, or read it the way I wanted it to read so it woud be easy. :cool:
 
Does this work or has it already been suggested?

If you divide into two groups of six, then further divide into two groups of 3 each pile ( a total of 4 groups). Measure groups one and two against each other. If they are not equal pick the lighter group to measure against one of the other groups of 3 which are now known to be good. If the two are equal then the bad coin is in the first group of three which was set aside. Otherwise, the one group should still be lighter and contains the bad coin. Once you have determined which group has the bad coin, you will also know if it is lighter or heavier. Pick two coins from the group to measure. If they are equal, then the bad coin is the one left over, otherwise it will be the coin which matches either lighter or heavier.
 
Actually,

The method you describe is very close to my solution. This thing is hard to solve. I rested at a degree of success. I kept trying to figure out a way to tell if the bad coin is light or heavy, the second measure, everytime. The only way to do it costs you a measure, as far as I can tell. I'd say starting with four groups is the only way. If there is means better than my solution, I don't know what it is yet.
 
Haynewp,

I think your last one would work. The problem with the first solution was that if the scale doesn't balance, than either one could be the bad coin. The last one you gave would tell if it is lighter or heavier, so it would work.
 
The problem with my last solution would be the case where:

I am naming the groups (each with 3 coins) 1 2 3 and 4

Measurement #1) Group 1 and group 2 are equal

Measurement #2) Group 1 and group 3 are equal

At this point I know the bad coin is in group 4, but I still don't know if it is light or heavy

Measurement #3) Take 2 coins from group 4, weigh them against each other, set the other coin aside. Their weight turns up different. But which one is the bad coin? The light or the heavy one?
 
There

You don't measure two of the coins from group 4 against each other. You measure 2 of the coins in group four against 2 of the coins that are known good. If the scales tip, you will know whether the bad coin is light or heavy. If they do not tip, the only remaining coin is bad, but you don't know if it's light or heavy.
(in this puzzle, knowing whether the bad coin is light or heavy is irrelavent, but it helps you find the bad coin in 3 measures)

So, here's the argument about whether this is a solution or not.
If the scales tip, either way, you know the bad coin is on the scale, and you know if it's light or heavy, but you still have one choice to make of two remaining coins. You pick up one coin from each side of the scale. (You have to remove the coins, anyway.)
If the scales balance, the bad coin is in your hand. If they do not move, the bad coin is the only coin it coud be--the one on the scale. This way, you find the bad coin in the 3rd measure, and before you complete the transaction with your merchant, you can show him/her the bad coin before making a fourth complete measure.

????
 
Re: There

You don't measure two of the coins from group 4 against each other. You measure 2 of the coins in group four against 2 of the coins that are known good. If the scales tip, you will know whether the bad coin is light or heavy. If they do not tip, the only remaining coin is bad, but you don't know if it's light or heavy.
(in this puzzle, knowing whether the bad coin is light or heavy is irrelavent, but it helps you find the bad coin in 3 measures
Click to expand...

OK you just completed your 3rd measurement, correct?


You pick up one coin from each side of the scale. (You have to remove the coins, anyway.)
If the scales balance, the bad coin is in your hand. If they do not move, the bad coin is the only coin it coud be--the one on the scale. This way, you find the bad coin in the 3rd measure, and before you complete the transaction with your merchant, you can show him/her the bad coin before making a fourth complete measure.????
Click to expand...

I know you have to remove the coins anyway, but wouldn't taking only 1 off of each side of the scale, and seeing what the scale does, be considered a measurement? If only the act of placing coins on the scale, then reading the scale, is considered a measurement then yours works.
 
Yup

I consider taking two of the coins off the scale and looking at the results a fourth measurement. But, at the time this puzzle was created, it was a real-life dilemma, and there is a limit to the amount you can learn from leaving the coins on the scale. So, it isn't a 'complete' fourth measure, like moving the coins around or replacing them.

To whit: If we put six coins on each side, and the scales tip, then tip again when we remove 3 from each side, we have learned nothing, except that the bad coin is still one of six. When we finally get down to two coins, we learn something from removing one. (each side)

Also, you do find the bad coin in 3 measures to a probability around 85%, if not higher, because the bad coin has to elude your grasp taking half the coins, then half again, then it has to be 2 of 3, leaving only 2 of 12 coins to 'hide' in. So, in effect, you find the bad coin in 3 measures 'almost' every time.

If there is a better solution, I haven't found it.
 
Identify the coins as 1 through 12

A -- Weign 1,2,3,4 against 5,6,7,8

AA -- If relult of A is equal weigh 1,10 against 11,12

If equal then 9 is the coin -- weigh against one of others to determine if heavy or light

If not equal weigh 11 against 12 (final answer is obvious)

AB -- If result of A is not equal weigh 1,2,5,6 against 3,7,9,10

ABA -- If AB equal it must be either 4 or 8. Weigh either against a know coin and refer to original (A) result (which way the tip went)

ABB -- If AB not equal and left side tips same as result of A it is either 1,2, or 7. Weigh 1 against 2 and compare with result of A (meaning whichever of 1 or 2 tips up or down to correspond with result of A is the coin and indicates heavy or light) and if 1 and 2 are equal then 7 is the coin and the results of A show heavy or light.

ABC -- If AB not equal and the left side tips opposite as result of A then it is either 5,6 or 3. Weigh 5 against 6 and follow the same logic conclusion as in case ABB just preceding.

Note: You cannot solve using any starting group other than four against four.
 
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