The 12 Coins Puzzle

[Tiberius]

Part A:

What if 11 or 12 are lighter coins? All you checked for were the anomalies. That's only one part of the question.

Part B:

In i) 7,8 can't possibly be the anomaly, since they would have caused an imbalance in the initial weighing. If you've already determined that the anomaly is in 9,10,11,12 then weighing 1,9,10 against 2,11,12 will never result in a balance.

In ii) 9,10 are the anomaly *only if* the anomaly is a heavy coin rather than a lighter coin. You haven't determined this. 11 or 12 could easily be the lighter coin.

This was your undoing in the previous step. You made assumptions about whether the coin was heavy or light, and you cannot do that, because I will always be able to find a scenario where the coin you choose isn't the anomaly.

[theVOID]

I used an example of one scenario, i was trying to work out how to narrow it down to 1 in 3 steps, which i did didn't i?. the anomaly being heavier in this example was not important, i was just trying to work out how to do it, rather than write a rule for it.

[Tiberius]

The anomaly being heavier in this example is important, since you have made it an assumption rather than a result of solving the puzzle.

The puzzle isn't about assuming the anomaly is lighter or heavier, and then working out which coin it is in 3 steps. It's about making no assumptions and finding out which coin is the anomaly and whether it is heavier or lighter.

It is easy to write out scenarios involving 3 weighings that depend upon knowing whether the coin is heavier or not, but the question asks for more.

Perhaps I'm misunderstanding your methods because you keep writing down weird notations. For instance:

a) If za = zb, discard za and zc.

Why do you discard zc? Surely you meant zb?

and

If zb = zd. z1 is heavier and the anomaly.

Where on earth did z1 come from???

I must seem like I'm nitpicking, but given the complexity of the problem, it is very confusing when working through a potential solution to come across variables that appear for a second and don't explain themselves.

As I stated before, if you could, please write step by step work in a similar form to the following:

(1) Separate 12 coins into 3 sets (X,Y,Z) of 4 coins each. GOTO (2)
(2) Weigh X against Y. If X = Y, GOTO (3). If X > Y, GOTO (4)
(3) Discard X and Y. Z contains the anomaly. Separate Z into 4 sets (Z1,Z2,Z3,Z4) of 1 coin each. GOTO (3a)
(3a) Weigh Z1 against Z2...
(4) ...

There are numerous answers to the question, but if you want me to put time into marking them I'd appreciate some format that can be read easily!

[leo-rcc]

http://www.mapsofconsciousness.com/12coins/

I solved that one when I was 14 or so. Took me a while to figure out, but once you get it it is not that hard.

[Ephrium]

I have mentioned a number of times my high IQ and now I prove it. My solution can not only solve 12 coins but thirteen.

1.Weight 4 against each other, abcd vs efgh. If they are not balanced, go to 2.
2.Replace f and g by the coins you know now are safe, and swap cd with gh.

If the scales are now balanced, you know the odd one is in the one removed. If the scale remains the same, you know the odd one lies in abe. If they switch from heavy to light, you know it is in cdh.

Take cdh. Now you know cd is lighter, and h heavier, not not know which is the odd one. weigh c and d. if c is lighter, it means c is the light one. if they are balanced, h is the odd one out.

[Tiberius]

How can you swap cd with gh in step (2) if you've just replaced f and g? Which coin do you swap with g?

[Ephrium]

Okay, we have the remaining coins ijkl. use i and j to swapwith f and g. Now swap the positions of cd with j and h, and scale abjh on one side with eicd on the other. This is weight number 2.

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Actually, we have remaining coins ijklm

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After the first weighting, if the odd coin is in abe, the scale will remain the same as the process affects merely the rest of the same coins. If the odd coin is in f ang g which are taken out, it will be balanced on second weighting. If the odd coin is in cdh, the scale will go opposite,. This covers all the possiboilities of after the first unbalanced weighting

[Tiberius]

Should I post the answer to this? I still don't get Ephrium's method, and it doesn't have any method for solving if the scales are balanced at the start.

[Ephrium]

Mine is the answer. I did not post the other easy options.

I do not know why you cannot get it but, I post a simplified version. I took out replacing the coins since it did not matter. You have coins ABCDEFGHIJKL

Algorithm:
1.Weight ABCD vs EFGH. If unbalanced, go to 2. If balanced, go to 5.
2.Weight ABH vs ECD. If it is now balanced, the odd coin must be in FG which you have taken out. If the scales switch, it must be in CDH. If it remains, it must be in ABE.
3. Whether it is in ABE or CDH, Use Two coins in the same side and same group against each other. For instance, If ABH AND ABCD Had been heavier from the start, (As concluded, the Odd coin must be in ABE) use B against A. The heavier coin is the odd one out. If balanced, it is E which is the odd one

5. If ABCD vs EFGH were balanced, weight ABC vs IJK. Say IJK is heavier , go to 6.
6. Weigh I vs J. The heavier coin is the one. If not , the odd coin is K.

As I have said, my method can do it for 13 Coins. High IQ is it not.

Some simple cases I have left out, such as if two coins were left, weigh any coins which you are sure of against one of them.

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Question any weighting combination and out come you are not sure, Etc. ABCD heavier in step 1, Then ABH was lighter than ECD, Then The last weighting of C vs D was equal. It shows H was the lighter coin.

[Tiberius]

Thanks the simplified version is better to read and work through. It works as far as I can see. Not the usual method for solving the problem, but a good one nonetheless.