There are $$12$$ gold coins with exactly the same appearance, including $$11$$ real coins and $$1$$ fake coin. The weight of the fake coin is different from that of the real coin, and whether the fake coin is lighter or heavier than the real coin is unknown. How many times at least do you need to weigh the coins using a balance to ensure that you can find the fake coin? 
Overseas Competition->Knowledge Point->Combinatorics->Strategies and Operations->Operational Problem
In the first weighing with the balance, put coins $$1$$, $$2$$, $$3$$, and $$4$$ on one end of the balance and coins $$5$$, $$6$$, $$7$$, and $$8$$ on the other end of the balance.  The balance has two situations: balanced or not.  Analyze the situation of the balance: if balanced, the fake coin is among the remaining $$4$$ coins.  In the second weighing with the balance, randomly take $$3$$ coins from coin $$1$$ to coin $$8$$ and put them on the left end of the balance and randomly take $$3$$ coins from coin $$9$$ to coin $$12$$ and put them on the right end of the balance (such as $$9$$, $$10$$, $$11$$). The balance also has two situations: balanced or not.  If balanced, coin $$12$$ is the coin of different weight. In the third weighing with the balance, comparing No. $$12$$ coin with any other coin, whether the coin is lighter or heavier can be known.  If not, it can be known that the coin of different weight is among the three coins $$9$$, $$10$$, and $$11$$, and that whether it is lighter or heavier than other coins can be known. In the third weighing with the balance, randomly take two of the coins (such as $$9$$ and $$10$$) and put them on the both ends of the balance. If balanced, the remaining coin (coin $$11$$) is the one we are looking for; if not, based on the previous judgement that whether the coin is lighter or heavier, it can be determined that which one of the coins on the balance is what we are looking for.  Analyze the first imbalanced situation as follows:  There are two situations: the right end weighs more or the left end weighs more. Assume the left end weighs more (which is the same for the situation that the right end weighs more.)  In the second weighing with the balance, take off $$3$$ coins randomly from the left end (such as $$1$$, $$2$$, and $$3$$) and move $$3$$ coins from the right end to the left end (such as $$5$$, $$6$$, and $$7$$), then take $$3$$ coins randomly from the $$4$$ coins left in the first weighing (such as $$9$$, $$10$$, and $$11$$) to the right end, and there sees $$3$$ situations for the balance: ① the left end weighs more, ② the two ends strike a balance, ③ the right end weighs more. Analyze the situations one by one as follows:  ① If the left end weighs more, the coin we are looking for must be coin $$4$$ or coin $$8$$. In the third time weighing with the balance, take one of the coins (such as coin $$4$$) and put it on the left end of the balance. Randomly take one of the remaining $$10$$ coins and put it on the right end. There are also $$3$$ situations.  $$a$$: If balanced, coin $$8$$ is the one we are looking for. Based on the result of using the balance for the second time, it is known that the coin weighs less than other coins.  $$b$$: If the left end weighs more, coin $$4$$ is the one we are looking for and it weighs more than other coins.  $$c$$: If the right end weighs more, coin $$4$$ is the one we are looking for and it weighs less than other coins.  ② If the two ends strike a balance, the coin we are looking for is among the three coins ($$1$$, $$2$$, and $$3$$) taken from the left end. Since the left end weighs more in the first weighing, it is known that the coin weighs more than other coins. The following analysis is the same as previous one and will not be repeated.  ③ If the right end weighs more, the coin we are looking for is among the three coins ($$5$$, $$6$$, and $$7$$) moved from the right end to the left end. Based on the results of weighing in the first two times (the left end weighs more in the first weighing and the right end weighs more in the second weighing), it is known that the coin weighs less than other coins. The following analysis is the same. 