Find the coin

You can find the unbalanced coin in 3 weighings as follows:

1. Remove the known balanced coin as it isn't very useful. Divide the remaining 10 coins into 4 sets $A, B, C, D$ of size 3, 3, 3, 1 respectively.
2. Weighing 1: Compare $A$ and $B$ .
3. Weighing 2: Compare $A$ and $C$ .
4. If $A = B = C$ , then we know that the unbalanced coin is in $D$ . To determine whether it is heavier or lighter we compare it with the known coin and we are done in 3 weighings.
5. If $A \ne B$ and $A \ne C$ , we know that the unbalanced coin is in $A$ and we know whether the coin is heavier or lighter depending if $A > B$ or $A < B$ . We choose $A$ and proceed to step 7.
6. If $A \ne B$ and $A = C$ , we know the unbalanced coin is in $B$ . We also know if it is heavier or lighter depending if $A < B$ or $A > B$ . We choose $B$ and proceed to step 7. The case when $A = B$ and $A \ne C$ is analogous as the coin is in $C$ .
7. Weighing 3: Take the 3 coins in the chosen set and put two of them on the scale. If they are equal the the remaining coin is the unbalanced coin. If they are unequal, since we already knew in the previous step whether the unbalanced coin is heavier or lighter, choose the corresponding coin from the scale.

If N is the number of coins, calculate 3 * round( $log_3 (N)$ ) = L, that is the max large set that you can obtain. Now distinguish two case:

$a)$ 3 * L ≤ N

$b)$ 3 * L > N

If you are in the $a)$ case, split N in three set of equal number, that we can call A1, A2, A3. If N mod 3 = 1, 2, the other coin are separated: we call, if this other set exist, A4 that contain one or two coins.

Test on the scale A1 and A2 (that contain the balanced coin). You can have three situation:

$1)$ A1 go UP and A2 go DOWN.

$2)$ A1 go DOWN and A2 go UP.

$3)$ A1 and A2 are balanced.

In the first case, it means that A1 can contain the unbalanced coin and it weights less or A2 contain the unbalanced coin and it weights more. But most important, it means that the set A3 and, if exist A4, are balanced. In the second case, it means that A1 can contain the unbalanced coin and it weights more or A2 contain the unbalanced coin and it weights less. Also in this case, it means that the set A3 and, if exist A4, are balanced. In the third case, it means that the set A1 and A2 are balanced and in the set A3 or A4 there is the unbalanced coin.

In the $1)$ case, now you weight one of the two set, for example A1, with A3 (that now we know is composed from only balanced coin). The possible result are only two:

$1)$ A1 go UP and A3 go DOWN (according to the first case, it means that A1 contains the unbalanced coin and it weights less)

$2)$ A1 and A3 are balanced (according to the first case, it means that A2 contains the unbalanced coin and it weights more)

So now you know if the coin weights more or less, than choose the appropriate set and split it in other three part.

You can have three possible situation if M is the length of the set:

$I)$ M mod 3 = 0 so you have three set of M element.

$II)$ M mod 3 = 1 so you have two set of M element and 1 coin remain out

$III)$ M mod 3 = 2 so you have two set of M element and 2 coin that remain out

If you are in the case $I)$ you can test the first of three subset with the second, and choose the subset that go UP (it weights less), than repeat the splitting algorithm until you have subset with only three element, with which you can find simple the unbalanced coin.

If you are in the case $II)$ you can use the same algorithm but if the two first that you compare are equal you can conclude that the unbalanced coin is the remained coin.

If you are in the case $III)$ , if the two subset are equal you can compare the two coin that remain out and you can conclude the research.

In the $2)$ case, now you weight one of the two set, for example A1, with A3 (that now we know is composed from only balanced coin). The possible result are only two:

$-)$ A1 go DOWN and A3 go UP (according to the second case, it means that A1 contains the unbalanced coin and it weights more)

$--)$ A1 and A3 are balanced (according to the second case, it means that A2 contains the unbalanced coin and it weights less)

The algoritm is the same of the case 1) but with the result that are inverted.

In the third case you can test the remained set A3 with A1 or A2.

There are three possible value:

$.)$ A1 go UP and A3 go DOWN (it means that A1 contain the coin and it weights more)

$..)$ A1 go DOWN and A3 go UP (it means that A1 contain the coin and it weights less)

$...)$ A1 and A3 are balanced (it happens only if A4 is not empty)

In the case $.)$ and $..)$ you can apply the splitting algorithm to find the coin, while in the $...)$ case you must distinguish two case:

$i)$ A4 contain one coin: so you can weight directly with one of the other balanced coin and you can determine the nature of the coin

$ii)$ A4 contain two coins: so you can weight A4 with two of balanced coin and reveal if A4 weights more or less then the two balanced, then you can weight the two coins and estabilish which is unbalanced.

In the case $b)$ when N > 3 * L, you obtain three set and the other money costitute A4. If A1 = A2 = A3 (to prove this you need only two weights), you can weight the set A4 with other balanced money and you can determine if A4 weight more or less, and then you can apply the splitting algorithm.

So, the total complexity of the algorithm is ceil ( $\log_3 (N-3^{round(\log_3 (N) -2)}))$ +1 weight, with N number of coins.

Finally, is interesting to see that this algorithm work fine with N > 5. For N ≤ 5, the best algorithm need two only weights, try it :D

When N=9, knowledge about balanced coin is not required to identify unbalanced coin. Group to three sets of 3 coins each. Compare twice and pick the odd set. Split the odd set into three coins & repeat comparisons. So in the fourth step odd coin and it's weight relativity can be determined.

Generalizing the scenario to real and odd values of N... The answer is

2 * ceil(log3 (N))

Recursive process with total of two comparisons between three sets of coins in each level as below.

Algorithm:

Step#1 If N is multiple of 3 (N mod 3 = 0), Split it into three sets.

Step#2 Else if N mod 3 = 2, Step#2.1 If balanced coin is in the set, remove balanced coin and one extra coin and split the rest to three sets. Step#2.2 If not, either add balanced coin and split to three sets of (N+1)/2 each; or remove two extra-coins and split the rest to three sets of (N-2)/3 each. Both have same effect.

Step#3 Else If N mod 3 = 1, Remove one extra-coin and split rest into three sets of (N-1)/3 coins each.

Step#4 Perform two comparisons between three sets to pick the unbalanced set. Repeat from step#1.( If all three sets weigh the same, the extra-coin we removed is the unbalanced one)