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(mathematics) Twelve steel balls, one of which is broken (I don't know whether it is heavy or light), can only be weighed three times with a balance. How can we find the bad ball?
Put four on each side of the balance for the first time, such as 1234 | 5678. There are three possibilities:

1. Both ends are balanced. It shows that the target ball is normal in ABCD 12345678.

Second time: 123 | ABC. There are also three possibilities:

(1) Both ends are balanced. This means that the goal is d.

(2) the left is heavy and the right is light. It shows that the target ball is in ABC, which is lighter than the normal ball. Just call it A | B for the third time.

(3) the left is light and the right is heavy. It shows that the target ball is in ABC, which is heavier than the normal ball. Just call it A | B for the third time.

2. The left side is heavy and the right side is light. It means ABCD is normal.

The second call: 34567 | ABCD8. There are also three possibilities:

(1) Both ends are balanced. Explain that the target ball is in 12, and it is enough to weigh 1 | D for the third time.

(2) the left is heavy and the right is light. Remember that the result of the first weighing is1234,5678 light. The weight of 34,567 this time means that 567 must be normal ("567 weight" contradicts the first statement, and "567 light" contradicts the second statement). The target ball must be at 348. The third time it is called 3 | 4, and the heavier one is the target ball (if it is balanced, 8 is the target ball, which is lighter than the normal ball).

(3) the left is light and the right is heavy. Remember that the result of the first weighing is1234,5678 light. This time 34567 is light, which means that 34 must be normal ("34 is light" and the first contradiction, "34 is heavy" and the second contradiction), and 8 must be normal ("8 is heavy" and the first contradiction, and "8 is light" and the second contradiction). The target ball must be at 567, which is lighter than the normal ball. The third time it weighed 5 | 6.

3. light left and heavy right. It means ABCD is normal.

The second call: 34567 | ABCD8. There are also three possibilities:

(1) Both ends are balanced. Explain that the target ball is in 12, and it is enough to weigh 1 | D for the third time.

(2) the left is heavy and the right is light. Remember that the result of the first weighing is that 1234 is 5678 lighter. This time, the weight of 34567 means that 34 must be normal ("34 weight" and the first contradiction, "34 light" and the second contradiction) and 8 must be normal ("8 light" and the first contradiction, "8 heavy" and the second contradiction). The target ball must be at 567, which is heavier than the ordinary ball. The third time it weighed 5 | 6.

(3) the left is light and the right is heavy. Remember that the result of the first weighing is that 1234 is 5678 lighter. This time, 34567 is light, which means that 567 must be normal ("567 is light" contradicts the first statement and "567 is heavy" contradicts the second statement). The target ball must be at 348. The third time it is called 3 | 4, and the lighter one is the target ball (if it is balanced, 8 is the target ball, which is heavier than the normal ball.