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Fatalis
July 28th, 2012, 03:33 AM
You have twelve coins. You know that one is fake. The only thing that distinguishes the fake coin from the real coins is that its weight is imperceptibly different. You have a perfectly balanced scale. The scale only tells you which side weighs more than the other side.

What is the smallest number of times you must use the scale in order to always find the fake coin?

Use only the twelve coins themselves and no others, no other weights, no cutting coins, no pencil marks on the scale. etc.

These are modern coins, so the fake coin is not necessarily lighter.

Presume the worst case scenario, and don't hope that you will pick the right coin on the first attempt.


P.S Lets see how fast you can finish this Glip

creedkingsx
July 28th, 2012, 03:45 AM
4
^.^
You could get lucky and do 3, but 4 will guarantee finding it.

Fatalis
July 28th, 2012, 03:50 AM
4
^.^
You could get lucky and do 3, but 4 will guarantee finding it.

Please include your procedure
Oh and thats wrong :p

Gingerape
July 28th, 2012, 03:50 AM
its 3

Fatalis
July 28th, 2012, 03:57 AM
its 3
Ding ding! We has a winner
No I lied you need a procedure *insert evil math teacher face*

Fatalis
July 28th, 2012, 03:59 AM
Thanks to ginger I will now abuse the secrets of the darkness

Fatalis
July 28th, 2012, 04:05 AM
I see you reading this Glip
Whats the matter? confused? MWAHAHAHAHAHAHA

Fatalis
July 28th, 2012, 04:06 AM
Apparently I got another weird O symbols
Still having no idea where it came from

JSaint
July 28th, 2012, 04:08 AM
If you know if the coin is heavier or lighter, than the answer is 3.

But if you do not... that's a different topic altogether.

Worst case is about 7 steps if you do not know if the coin is heavier or lighter.

Fatalis
July 28th, 2012, 04:09 AM
If you know if the coin is heavier or lighter, than the answer is 3.

But if you do not... that's a different topic altogether.

You don't :)

Fatalis
July 28th, 2012, 04:11 AM
Oh and Gingerape already mentioned the answer
But I'm a math teacher and I want solutions

JSaint
July 28th, 2012, 04:16 AM
It's the divide the problem into

6/6 >>> 3/3 >>> 1/1

If you get what I mean... but that makes the presumption that you KNOW if the coin is heavier/lighter so you can pick the right pile.

6/6 >>> 3/3 (balanced pile)>>> 3/3(Pile Change) >>> 1/1

I see how this would work if you pick the wrong pile now. Brain is not working too well. And I think you are lying about being a math teacher. You probably just want us to do your homework for you :p

Fatalis
July 28th, 2012, 04:19 AM
It's the divide the problem into

6/6 >>> 3/3 >>> 1/1

If you get what I mean... but that makes the presumption that you KNOW if the coin is heavier/lighter so you can pick the right pile.

6/6 >>> 3/3 (balanced pile)>>> 3/3(Pile Change) >>> 1/1

I see how this would work if you pick the wrong pile now. Brain is not working too well. And I think you are lying about being a math teacher. You probably just want us to do your homework for you :p

:O my secrets are revealed

No seriously, I am not a math teacher but I said that cuz everybody knows that math teachers are super annoying and wants the solution
Again, you don't know whether its lighter or heavier :p


P.S If you want to help me with my homework, if tan A is x then what is sin A in terms of x

JSaint
July 28th, 2012, 04:22 AM
You did say least number of times, but as a programmer, I am trained to assume worst case scenarios to cover and cater to a wide variety of cases.

In the best case and the least number of times needed, yes the answer is 3.

But to be able to find the coin all the time, you need to have at least four weighs.

For the purposes of practicality as a dev, I would always reserve 4 weighs because it would always find me the coin. Reserving 3 weighs might save some time, but it almost guarantees me a 50% failure rate.

Again my brain does not work properly, because we have a chance of guessing WRONG on the 3/3 and WRONG on the 1/1 because we do not for sure know if the marble is heavier or lighter.

In other words, Best-case scenarios are rather retarded, whether in maths or programming.

Fatalis
July 28th, 2012, 04:25 AM
No other weights
If you want to know the answer I could PM you ^ ^

Glip
July 28th, 2012, 04:25 AM
I see you reading this Glip
Whats the matter? confused? MWAHAHAHAHAHAHA

1. Place 3 coins on each side of the scale. We're assuming worst case scenario, so the balance will be the same, all these coins are real. Scrap 'em.
2. Place 2 coins from the 6 not used yet on each side of the scale.
3. Worst case scenario the balance is now different. Swap the top two coins of each stack.
4. a. If the balance changed, replace one of the coins you just swapped with one of the coins not used yet. If the balances changes again, the one swapped out was the fake. if no change, then the other of the top two coins is the fake.
4. b. If the balance didn't change, then remove the top two coins, and replace one of the remaining coins on each side with an unused coin. If the balance changes, then the coin just replaced is the fake one. If the balance doesn't change, then the coin remaining on the other side is the fake one.


Currently working on trying to figure out a quicker solution, since you guys have apparently said there is one.

JSaint
July 28th, 2012, 04:28 AM
The quickest solution... assumes that you pick the right pile all the time XD.

It does not guarantee that it works all the time, Glip

Glip
July 28th, 2012, 04:29 AM
The quickest solution... assumes that you pick the right pile all the time XD.

It does not guarantee that it works all the time, Glip

My solution doesn't have anything to do with picking piles, so I don't know what you're talking about.

Fatalis
July 28th, 2012, 04:29 AM
1. Place 3 coins on each side of the scale. We're assuming worst case scenario, so the balance will be the same, all these coins are real. Scrap 'em.
2. Place 2 coins from the 6 not used yet on each side of the scale.
3. Worst case scenario the balance is now different. Swap the top two coins of each stack.
4. a. If the balance changed, replace one of the coins you just swapped with one of the coins not used yet. If the balances changes again, the one swapped out was the fake. No change, then the other of the top two coins is the fake.
4. b. If the balance didn't change, then remove the top two coins, and replace one of the remaining coins on each side with an unused coin. If the balance changes, then the coin just replaced is the fake one. If the balance doesn't change, then the coin remaining on the other side is the fake one.


Currently working on trying to figure out a quicker solution, since you guys have apparently said there is one.

*hint*hint*
There are 12 coins, use more of them

It took me about 2 hours and asking friends to finish this
I simply don't want to look at the answer key

JSaint
July 28th, 2012, 04:31 AM
Basically it's the 6/6 >>> 3/3 >>> 1/1 solution, which guarantees 100% finding the coin assuming you know if the fake coin is heavier or lighter.

Fatalis
July 28th, 2012, 04:32 AM
Basically it's the 6/6 >>> 3/3 >>> 1/1 solution, which guarantees 100% finding the coin assuming you know if the fake coin is heavier or lighter.

Again, you don't

JSaint
July 28th, 2012, 04:34 AM
*goes to watch porn instead*

*would rather have physical than mental masturbation*

Fatalis
July 28th, 2012, 04:42 AM
*goes to watch porn instead*

*would rather have physical than mental masturbation*

This made my day

Glip
July 28th, 2012, 05:01 AM
Meh. I just googled the solution, and I'll just say it's probably something that I never would have come up with. Not without hours of dedication to the problem. But I don't even do that when I have to for schoolwork, so why would I do that on my free time for no reason, when I could be doing other things that don't hurt my brain? XD

JSaint
July 28th, 2012, 05:08 AM
So cher, do I get detention and alone time with you?

I have some good videos in my collection. We can have some practical sessions after that too ^_~.

Anyway, I've found how to determine if the coin is heavier or lighter. It's just how to still get it done in 3 steps if you chose wrong ^^.

Fatalis
July 28th, 2012, 05:19 AM
Meh. I just googled the solution, and I'll just say it's probably something that I never would have come up with. Not without hours of dedication to the problem. But I don't even do that when I have to for schoolwork, so why would I do that on my free time for no reason, when I could be doing other things that don't hurt my brain? XD

Google knows everything :p

Phyr
July 28th, 2012, 05:21 AM
Three piles of 4. Format: left side of the balance/right side of the balance/not on balance
4/4/4
1. unbalance: 2+1/1+2/2
1.1 no switch in unbalance: 1/1/1
1.2 switch in unbalance: 1/1/1
1.3 equal weight: 1/1/1
2. balance: 1/1/2
2.1 unbalance: 1/1/1
2.2 balance: 1/1/1
All those weightings get you a definite result on what the fake coin is.
How to get the fake coin through the result of the weighting should be obvious.

Phyr
July 28th, 2012, 05:26 AM
The quickest solution... assumes that you pick the right pile all the time XD.

It does not guarantee that it works all the time, Glip
Quickest solution: pick the fake coin and other one and weight them. Then you weight one of those coins with a third one. Done. Two steps - best case scenario.

Btw great riddle. Gonna add it to my collection.

creedkingsx
July 28th, 2012, 05:29 AM
Go to a bank.
Try to deposit your three dollars.
Look in the evidence bag when you are getting arrested for counterfeit money.
Found it :D

Fatalis
July 28th, 2012, 05:32 AM
Go to a bank.
Try to deposit your three dollars.
Look in the evidence bag when you are getting arrested for counterfeit money.
Found it :D

Phyr's solution was great
But this was better

Nick
July 28th, 2012, 06:27 AM
Assuming the fake A is heavier, just invert the sign if lighter to give the same results.

Step1: 1 time
ABCDEF > GHIJKL
If weigh more it is pile ABCDEF but if weigh it is less pile GHIJKL
But we do not know, thus leading to Step 2.

Step2: 2 times
ABC > DEF or GHI = JKL
Two possibilities. Might weigh the wrong pile.

Step3: 1 time
A > B
A > C
B = C

Answer is 4.

Nick
July 28th, 2012, 06:29 AM
Erm... I demand that the host post the solution.

Fatalis
July 28th, 2012, 06:33 AM
Somebody called?

Fatalis
July 28th, 2012, 06:34 AM
1. Weigh coins 1,2,3,4 against coins 5,6,7,8.

1.1. If they balance, then weigh coins 9 and 10 against coins 11 and 8 (we know from the first weighing that 8 is a good coin).

1.1.1. If the second weighing also balances, we know coin 12 (the only one not yet weighed) is the counterfeit. The third weighing indicates whether it is heavy or light.

1.1.2. If (at the second weighing) coins 11 and 8 are heavier than coins 9 and 10, either 11 is heavy or 9 is light or 10 is light. Weigh 9 against 10. If they balance, 11 is heavy. If they don't balance, you know that either 9 or 10 is light, so the top coin is the fake.

1.1.3 If (at the second weighing) coins 11 and 8 are lighter than coins 9 and 10, either 11 is light or 9 is heavy or 10 is heavy. Weigh 9 against 10. If they balance, 11 is light. If they don't balance, you know that either 9 or 10 is heavy, so the bottom coin is the fake.

1.2. Now if (at first weighing) the side with coins 5,6,7,8 are heavier than the side with coins 1,2,3,4. This means that either 1,2,3,4 is light or 5,6,7,8 is heavy. Weigh 1,2, and 5 against 3,6, and 9.

1.2.1. If (when we weigh 1,2, and 5 against 3,6 and 9) they balance, it means that either 7 or 8 is heavy or 4 is light. By weighing 7 and 8 we obtain the answer, because if they balance, then 4 has to be light. If 7 and 8 do not balance, then the heavier coin is the counterfeit.

1.2.2. If (when we weigh 1,2, and 5 against 3,6 and 9) the right side is heavier, then either 6 is heavy or 1 is light or 2 is light. By weighing 1 against 2 the solution is obtained.

1.2.3. If (when we weigh 1,2, and 5 against 3, 6 and 9) the right side is lighter, then either 3 is light or 5 is heavy. By weighing 3 against a good coin the solution is easily arrived at.

1.3 If (at the first weighing) coins 1,2,3,4 are heavier than coins 5,6,7,8 then repeat the previous steps 1.2 through 1.2.3 but switch the numbers of coins 1,2,3,4 with 5,6,7,8.

Fatalis
July 28th, 2012, 06:34 AM
^ ^

Pokemon Trainer Red
July 28th, 2012, 06:35 AM
Mind blown.

Nick
July 28th, 2012, 06:48 AM
*froth*

JSaint
July 28th, 2012, 05:49 PM
I DEMAND that the host give me personal detention and one on one time for my bad Math!

Cryptonic
July 28th, 2012, 05:52 PM
P.S If you want to help me with my homework, if tan A is x then what is sin A in terms of x

TanA=x= SinA/CosA
SinA = xCosA

JSaint
July 28th, 2012, 05:56 PM
And I also demand that Crytonoic not be the relief teacher XD

Fatalis
July 28th, 2012, 09:59 PM
TanA=x= SinA/CosA
SinA = xCosA

:o
YAY!
I'll be back to haunt you if this is wrong

JSaint
July 28th, 2012, 10:57 PM
I THRUST! that you found the answer extremely satisfying.