Microsoft has 12 balls and a balance. I know that only one ball is different from the other. How can I find the ball three times? What about 13?

Divided into three groups: A (4) B (4) C (4)
1. If a (4) = B (4), the bad one is in C (4. C1 (3) is equivalent to a (3), and C2 (1) is bad. Otherwise, if C1 (3)> A (3) is used, take two one terms in C1. We can see that the one (heavy) is bad.
2. Suppose A (4)> B (4), group A1 (1) + B1 (2) and A2 (1) + B2 (2) // The second step requires that two balls be bad !!!
If A1 (1) + B1 (2) = a2 (1) + B2 (2) is bad, it must be in A3 (2. Again, we can see the bad ball.
If A1 (1) + B1 (2)> A2 (1) + B2 (2), A1 (1) is bad and heavy or B2 (2) it is bad and light. B2 (2) is equal to one, and A1 (1) is bad. Otherwise, it is light bad.
If A1 (1) + B1 (2) <A2 (1) + B2 (2), then A2 (1) is bad and heavy or B1 (2) it is bad and light. B1 (2) is equivalent to two, A2 (1) is bad, otherwise it is light.

What about 13 balls? It is actually the same as 12 balls.
Divided into three groups: A (4) B (4) C (5)
1. If a (4) = B (4) that is bad in C (5), C1 (3) is equal to A1 (3, so C2 (2) is bad. If we call one of them a better one, we will know which one is bad.
2. If a (4 )! = B (4) is exactly the same as above.