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How to find the fake coin from a set of 10 coins?

 
 
Reply Fri 28 Aug, 2015 04:57 pm
You have 10 gold coins. All 10 coins look exactly the same but 1 coin is a fake which is either lighter or heavier than the other 9 coins. You have a scale - balance type with 2 trays - but can only load it thrice. How do you find the fake gold coin?
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Type: Question • Score: 1 • Views: 1,300 • Replies: 8
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fresco
 
  1  
Reply Sat 29 Aug, 2015 07:46 am
@raam2209,
Too easy. !
The original puzzle has 12 coins.
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Kolyo
 
  2  
Reply Sat 29 Aug, 2015 08:13 pm
@raam2209,
raam2209 wrote:

You have 10 gold coins....1 coin is a fake which is either lighter or heavier than the other 9 coins.


Weigh 3 against another 3. If the scale balances out you're dealing with Case 1. Otherwise, you're dealing with Case 2.

Case 1: You have 4 iffy coins left, one of which is a fake, either lighter or heavier than the others. You also have 6 perfectly good coins. Weigh 3 iffy coins against 3 of the perfectly good coins. If the iffy coins are too light, you're dealing with Case 1a. If the iffy coins are too heavy, you're dealing with Case 1b. Otherwise, the fourth iffy coin is the dud, so just weigh it against a known good coin to determine whether it is too heavy or too light.

Case 1a: You have three coins that are collectively too light. Weigh two of them against each other. If they don't balance, the lighter one is the dud. If they balance, the one not weighed is the dud.

Case 1b: Similar to case Case 1a -- just replace references to "light" with references to "heavy".

Case 2: You have 3 coins (call this "Set X") which are collectively lighter than 3 other coins (call this "Set Y"). You also have 4 perfectly good coins (select 3 of these at random and call them "Set Z"). Weigh X against Z. If X is lighter than Z, X is too light, and you're dealing with Case 2a. Otherwise, Y is too heavy, and you're dealing with Case 2b.

Case 2a: Set X contains the dud, which is too light. Weigh two coins from X against each other, and if one is lighter it's the dud. Otherwise, the coin not weighed is the dud.

Case 2b: Set Y contains the dud, which is too heavy. Weigh two coins from Y against each other, and if one is heavier it's the dud. Otherwise, the coin not weighed is the dud.
fresco
 
  1  
Reply Sat 29 Aug, 2015 10:37 pm
@Kolyo,
Good.
Now try it with 12.
Kolyo
 
  1  
Reply Sun 30 Aug, 2015 10:39 am
@fresco,
I didn't think it could be done at first, but there are only 24 possibilities, which is fewer than 27, the critical number. I think I have a way, but you have to remember a coin's "weighings history" once it develops one. Basically, you can't lose track of which coin is which, although initially there is nothing to distinguish them.
0 Replies
 
Kolyo
 
  1  
Reply Sun 30 Aug, 2015 10:56 am
@Kolyo,
With 12...

First, weigh 4 against 4. If the scale balances you are down to 4 iffy coins, and I've dealt with that scenario already. So let's say one side is lighter. In your mind, label two of the lighter coins "X1", one of the lighter coins "X2", and the last "X3". Also, label two of the heavier coins "Y1", one of the heavier coins "Y2", and the last "Y3". On the left side of the scale, put the two X1 coins and the two Y1 coins together; on the right side put the X2 coin, the Y2 coin, and two known good coins, together. If the scale balances, see "Case A". If the left side is heavier, see "Case B". If the right side is heavier, see "Case C".

Case A: Either X3 is too light or Y3 is too heavy. Weigh X3 against a known good coin to see if it is too light.

Case B: The two X1 and two Y1 coins were heavier than the X2 and Y2 coins and the two good coins. Either one of the Y1 coins was too heavy, or the X2 coin was too light. Take one of the Y1 coins ("Y1a" as opposed to "Y1b") and put it on the left side of the scale with the X2 coin, and weigh them against two good coins. If they (on the left) are too heavy, then Y1a is too heavy. If they are too light, then X2 is too light. Otherwise, "Y1b" is too heavy.

Case C: Similar to "Case B", just change the roles of X and Y, and the roles of "heavy" and "light".
fresco
 
  1  
Reply Sun 30 Aug, 2015 12:36 pm
@Kolyo,
I agree on the 4-4 balance.
For an imbalance you seem to be on the right track but the normal solution requires three against three where two coins swap sides and another coin in one pan is replaced by a 'good' one.
i.e. 3 have been removed, 3 remain in place , 2 have swapped sides.
CASES
1 The pans balance - it is one of the 3 removed
So weigh the two that came out from the same pan against each other and chose the one that gave the 4-4 direction or if they balance choose the third removed one.
2 The pans go the same way as before - it is one of the three originals
so weigh the two originals in the same pan against each other .....follow the logic of step 1 if they balance.
3 The pans go the other way - it is one of the two which swapped sides
Weigh either against a 'good' one.
Kolyo
 
  2  
Reply Sun 30 Aug, 2015 06:01 pm
@fresco,
There's no reason this problem needs to have a unique solution algorithm.

The normal solution is more elegantly phrased than mine was, but my solution still works, as far as I can see. Also, they're doing the same thing I did (that thing which required me to remember the weighing-history of the coins)--that is, they're throwing potentially heavy coins in with potentially light coins, as well as known good ones, and not getting them all mixed up.

Thanks for pointing out that my solution was abnormal. That just makes it more likely I was the first to find my particular solution to the problem (although I seriously doubt I was first, considering how few distinct solutions there are to this problem and how many people have looked at it.)

In each case--both in the normal solution and in mine--observe that we're narrowing down the possibilities first from 24 to 8, then from 8 to (2 or 3), and finally from (2 or 3) to 1. As long as you narrow the possibilities to 3^n or fewer, with n moves left, you may still have a chance of narrowing down the possibilities to 1 by the time you finish. (You're not guaranteed a path to success, though, as the impossibility of solving this problem with 13 coins shows.)

Feel free to show me where I went wrong.

Some people may object to the labels I gave the coins in my answer, but in places where I labelled them arbitrarily, my labelling scheme made no assumptions about which coin was potentially light or potentially heavy. Mathematicians number the elements of sets arbitrarily all the time to make proofs more readable.
fresco
 
  1  
Reply Tue 1 Sep, 2015 01:04 am
@Kolyo,
Subsequent research indicates that you are correct in saying that the solution is not unique.
ref: '12 ball problem'
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