Warrior General Games

Suppose you have 12 coins and 1 is counterfeit (the other 11 are exactly the same). They all look the same so you originally do not know which one is counterfeit but you do know that it has a different mass. But, you do not know if it s heavier or lighter.

Suppose all you have are those 12 coins and a balance scale (so a way to tell if things are heavier or lighter). Then, what is the minimum number of weighings needed to guarantee you know which coin is counterfeit and whether it is heavier or lighter. Explain

The answer is not simple and requires an explanation for your steps.

don't worry if this takes several hours or days. It can be done quickly, but it's not likely.

14

Divide the coins into 3 stacks of 4.

Weigh Stack A and Stack B. (Weigh #1)
Weigh Stack A and Stack C. (Weigh #2)

Stack A will either be of unequal weight to both stacks (meaning the fake is in A) or it will be equal to either B or C meaning that the other has the fake. Let's say stack A had the fake.

Split stack A into 2 stacks, D and E. Take 2 coins from one of the other stacks, making Stack F. Stack F is a control stack and known to contain 2 real coins.

Weigh Stack F and Stack E (Weigh #3)
Weigh Stack F and Stack D (Weigh #4)

Whichever stack is the imbalance, would show either D or E to contain the Fake. Let's say it is D.

Split stack D. You know have Coin #1 and Coin #2. Take one coin from stack F for Coin #3, your control coin.

Weigh Coin 3 and 1 (Weigh #5)
Weigh Coin 3 and 2 (Weigh #6)

The imbalance would show which coin was fake, either 1 or 2.

So at most, this can be determined in 6 weighings. Since you have a control item in the second and third set of weighings, however, it could also be determined in as few as 4.

Originally posted by pabloex
Divide the coins into 3 stacks of 4.

Weigh Stack A and Stack B. (Weigh #1)
Weigh Stack A and Stack C. (Weigh #2)

Stack A will either be of unequal weight to both stacks (meaning the fake is in A) or it will be equal to either B or C meaning that the other has the fake. Let's say stack A had the fake.

Split stack A into 2 stacks, D and E. Take 2 coins from one of the other stacks, making Stack F. Stack F is a control stack and known to contain 2 real coins.

Weigh Stack F and Stack E (Weigh #3)
Weigh Stack F and Stack D (Weigh #4)

Whichever stack is the imbalance, would show either D or E to contain the Fake. Let's say it is D.

Split stack D. You know have Coin #1 and Coin #2. Take one coin from stack F for Coin #3, your control coin.

Weigh Coin 3 and 1 (Weigh #5)
Weigh Coin 3 and 2 (Weigh #6)

The imbalance would show which coin was fake, either 1 or 2.

So at most, this can be determined in 6 weighings. Since you have a control item in the second and third set of weighings, however, it could also be determined in as few as 4.


How do you know that the fake weighs more or less in weighings 1 and 2?

10990

Originally posted by judnnc
14

HAHAHAHAHAHA

Your question is irrelevant Awolf02, just matters that its different.

Originally posted by pabloex
Divide the coins into 3 stacks of 4.

Weigh Stack A and Stack B. (Weigh #1)
Weigh Stack A and Stack C. (Weigh #2)

Stack A will either be of unequal weight to both stacks (meaning the fake is in A) or it will be equal to either B or C meaning that the other has the fake. Let's say stack A had the fake.

Split stack A into 2 stacks, D and E. Take 2 coins from one of the other stacks, making Stack F. Stack F is a control stack and known to contain 2 real coins.

Weigh Stack F and Stack E (Weigh #3)
Weigh Stack F and Stack D (Weigh #4)

Whichever stack is the imbalance, would show either D or E to contain the Fake. Let's say it is D.

Split stack D. You know have Coin #1 and Coin #2. Take one coin from stack F for Coin #3, your control coin.

Weigh Coin 3 and 1 (Weigh #5)
Weigh Coin 3 and 2 (Weigh #6)

The imbalance would show which coin was fake, either 1 or 2.

So at most, this can be determined in 6 weighings. Since you have a control item in the second and third set of weighings, however, it could also be determined in as few as 4.


good attempt (you left out the explanation on what to do when A=C or A=B), but no. The problem can be done with a max that is much smaller than 6.

to answer AWolf02, in his case you know whether it is heavier or lighter by whether the balance for stack A raised or lowered compared to both B and C

14

Originally posted by Bilal
Your question is irrelevant Awolf02, just matters that its different.

no, the problem also asks to determine whether the counterfeit is more or less than the regular coins

Originally posted by judnnc
14

that was hilarious

Yeah it was that's why i'm on his bandwagon!

2

Originally posted by Butternuts
2

heh, no

3 weighings minimum, and 4 max:

1st weighing: Weigh 6 of 12, 3 on each side
If equal: These are all real; set aside and grab other 6 coins
If imbalanced: The counterfeit is in these 6; use in next weighing

2nd weighing: Weigh 4 of 6, 2 on each side
If equal: These 4 are real; one of the other two is counterfeit (see option 1)
If unequal: One of these 4 is counterfeit; the other two are real (lets call these two pairs A and B; see option 2)


Option 1:
3rd weighing: Take one guaranteed real coin (coin Z) in with the pair that has the counterfeit. Weigh coin Z with coin A of the counterfeit pair; leave coin B out
If equal: Coin B is counterfeit
If unequal: Coin A is counterfeit


Option 2:
3rd weighing: Weigh pair A, 1 on each side
If equal: These are both real; the counterfeit is in pair B
If imbalanced: The counterfeit is in pair A

4th weighing: Take one guaranteed real coin (coin Z) in with the pair that has the counterfeit. Weigh coin Z with coin A of the counterfeit pair; leave coin B out
If equal: Coin B is counterfeit
If unequal: Coin A is counterfeit


This was more difficult than anticipated. I've heard of something like this before (which definitely helped), but it was 10 balls, one of them is heavier, and you try to figure out which one.

So the answer (which I hope is right, and I'm staying late at work to figure it out dammit), is a minimum of 3 weighings, and a max of 4.