the wording of this puzzle is puzzling!!
I don't understand it!

Not only puzzling; it is seriously flawed.

The following is an engine to use algorithms like the one described under "Show Solution", and a demonstration of that algorithm, written in the MUMPS programming language:

n (XPZL12,A,B,N) s N=A f q:N'?1"(".e s N=$e(N,2,$l(N)-1) x XPZL12(1,1),XPZL12(1,2) s I=$s(D0:4,1:3),N=$p(N," ",$p(K,",",I-1)+1,$p(K,",",I)) ;XPZL12(1): apply algorithm A to object list B giving N for different $p(B,",",N)
s J=$tr($p(N," "),"-","+"),D=0 f I=1:1:$l(J,"+") s JI=$p(J,"+",I),D=$e(N,$l($p(J,"+",1,I))-$l(JI))_$p(B,",",JI)+D ;XPZL12(1,1)
s K=1 f I=2:1:$l(N," ") s J=$p(N," ",1,I) s:$l(J,"(")=$l(J,")") K=K_","_I ;XPZL12(1,2)
s A="(1+2+3+4-5-6-7-8 (1+2+5-3-6-9 (1-2 1 6 2) (7-8 8 4 7) (3-9 3 5 error)) (9+10-8-11 (9-10 9 11 10) 12 (9-10 10 11 9)) (5+6+1-7-2-9 (5-6 5 2 6) (3-4 4 8 3) (7-9 7 1 error)))" f V=6,4 f I=1:1:12 s B="5,5,5,5,5,5,5,5,5,5,5,5",$p(B,",",I)=V x XPZL12(1) w I,":",N," " ;XPZL12: "the fake coin" puzzle using general engine

Just type in numbers you will find the answer is 3

that one was waaay easier than the ones before it

3 ftw

split the 12 coins into 3 groups of 4. put one set of four coins on one side of the scale and 4 coins on the other side. if they are equal, then none of the eight are fake. After the first weighing, one will find out which set of four coins has the fake, then it only takes two more weighings to find the fake. :-).

the answer to this should be two. If you pick the fake coin first and weigh it against another coin, they will be different weights. You then have the choice to remove either coin. Suppose you pick off the non-fake coin and then replace it with another one of the real coins. You will then be able to tell that, since there is a discrepenscy in weight, it is the first coin....disCHARRG3EE1!!!UP THE PUNX!!!!

No No No!!! The answer is 1!

If it is a teeter totter type balance where the placement of the coin matters (those on the outside produce more gravity than those on the inside) then my solution of 1 works!!

Only need to put 6 on the inside of each side. SLowly but surely move 1 coin to the outside and record the height difference. The oddball one will be lower or higher than the other 11.

Thus, only 1 is my answer.

Can we use a magnet in distinguishing the fake one? Just asking...

Puzzling, yes--flawed, no. Sliding coins around on the scale or any other such hocus pocus isn't kosher. Or necessary. Look, there's no trick here...there's no thinking outside the box or "lateral thinking" or whatever you want to call it. It's just pure logic, so go to work.

It's a shame that the puzzle is rather clumsily worded here, because the Twelve Coins Problem in which it is unknown whether the one counterfeit is lighter or heavier than the others is actually an elegant, classic logic problem that is worthy of anyone's best effort. And it can be done in three weighings, without fail.

Dude who cares just throw him in jail for counterfeiting. there he will be raped by bubba and SATAN!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
BTW CraigD you have no life.

1st 4 and 4, if equal then trivial to find the odd one
if not the I grouped it as LLL LHH HHN, where L=light in 1st weighing, H=heavy and N = normal (other 4)

then weigh the last two sets, if equal then trivial
if not then use the H and L criteria to resolve...

Three weighings, but with a simpler solution:

Round 1

Split the 12 coins into two groups of 6 - A & B. Take group A and weigh 3 against 3. If the coins are of equal weight, defective coin is in set B. If not, it's in set A. here we will come to know whichever is defective group.


Round 2


now take 3 coins from defective group and 3 from normal and compare. two cases can possible.
1)if both weigh equal the we will come to know that the defective one is in the remaining three from B.
2) and if they weigh unequal then the defective one is is among the three taken from B. also we can get whether the coin is heavier or lighter.

Now we are reduced to 3 coins.

round 3

take any 2 coins from these 3 and compare them with each other. if weigh equal the the defective is third one. and if unequal the defective is heavier or lighter accordingly.

Unfortunately, the 6/3/3 weighing does not necessarily tell you which coin is counterfeit since you don't know if it's lighter or heavier beforehand. (i.e. in the last weighing, if they are unequal, you don't know which of the two coins is counterfeit since you don't know if the fake coin is heavier or lighter).

However, there is another solution that doesn't require conditional action and requires only 3 weighings.
If we think about the fact that there are 3 outcomes to each weighing (no tilt, tilts left, tilts right) and only 24 possibilities (12 coins, each can be heavier or lighter) then we should realize there is some possibility of using 3 weighings (3^3 = 27 > 24) to determine which coin is fake. The original statement of the problem also requires you to determine whether the fake coin is lighter or heavier than the real ones, but this does not add much difficulty to the problem since if you know which is fake, and you've done 3 weighings, you also know if it's heavier or lighter (presuming you've weighed it once).

I will outline an example solution so you see what I mean:
1st weighing: 5,6,7,8 on left; 9,10,11,12 on right
2nd weighing: 2,6,7,12 on left; 3,4,8,11 on right
3rd weighing: 2,3,8,9 on left; 1,7,10,11 on right

There are 27 different outcomes to this weighing, 24 of which provide a unique solution to the coin problem, and the other 3 are not possible outcomes of the weighing, and can thus be ignored. I will describe an outcome by a 3 letter code. The first letter is the outcome of the first weighing, 2nd letter is the second weighing, 3rd letter is 3rd weighing. N means there is no tilt. L means it tilts left. R means it tilts right.

NNN - not possible
NNL - 1 is lighter
NNR - 1 is heavier
NLN - 4 is lighter
NRN - 4 is heavier
NLL - 2 is heavier
NRR - 2 is lighter
NLR - 3 is lighter
NRL - 3 is heavier
LNN - 5 is heavier
RNN - 5 is lighter
LNL - 10 is lighter
RNR - 10 is heavier
LNR - 9 is lighter
RNL - 9 is heavier
LLN - 6 is heavier
RRN - 6 is lighter
LLL - 11 is lighter
RRR - 11 is heavier
LLR - 7 is heavier
RRL - 7 is lighter
LRN - 12 is lighter
RLN - 12 is heavier
LRL - 8 is heavier
RLR - 8 is lighter
LRR - not possible
RLL - not possible

So, if you must determine your steps beforehand (without condition), then this is the best solution.

Group,mark and short wires in bunch of 1; 2,2; 3,3,3; 4,4,4,4;.......15,15........;

Go to other end and check for shortings and mark same as 1; 2,2; 3,3,3; 4....... And then open all shortings and short as 1,15; 2,15; 2,14; 3,15; 3,14;3 ; 4,15; 4,14; 4,13; 4 ;.................14,15; 15;.

now walk to the previous end and open earlier shortings and check for new shortings and mark the unique individual 1,2,3,4...15 wires with no shortings and other unique 1,15; ... pairs.
Try writing as

1
2 2
3 3 3
4 4 4 4
........
15,15,15... 15