A balance, 12 balls of the same size and appearance. The weight of a ball is different from that of other balls, and the ball is identified three times.

An interesting smart question: given 12 balls of the same size, only one ball has a different weight from other balls. How can we use tianyao to name it three times to find the unique ball, in addition, it is determined that the emphasis is light or not.

Number the ball, 1 ~ 12. extract 1 ~ The 4th ball is a group of A, and 5 ~ The ball on the 8 th is a group of B, and the remaining 9 ~ The ball on the 12th is a group of C

Weigh the balls in group A and group B with three results:

1. Balance. The problematic ball may only exist in four balls in group C 9 ~ In 12, the ball 9 is taken from C, and another ball is randomly taken from A or B to form a group of D. The ball 10 and 11 are taken from C to form a group of E, weighing D and E has three results:

1) balance, then the ball No. 12th is the problem ball. You only need to find one of the remaining balls and weigh the 12th ball to determine whether the ball is heavy or light.

2) If D is heavy and E is light, the problem ball may only appear in ball 9 and 10 and 11. Weigh the balls 10 and 11. If the ball 9 is balanced, the ball 9 is the problematic ball and heavy. Otherwise, the ball 10 and 11 is lighter than the problematic ball.

3) D is light and heavy, then the problem ball may only appear in the 9 and 10, 11 ball. Weigh the balls 10 and 11. If the ball 9 is balanced, the ball 9 is the problematic ball and lighter. Otherwise, the ball 10 and 11 is heavier than the problematic ball.

2. If a is light-weight B, the problem ball appears in Group A or Group B. Remove the No. 1 ball from a, take the No. 5 ball from B, and then take any two balls from C to form a group of F; remove the balls 2 and 3 from a, and take the balls 6 and 7 from B to form a group of G. Weighing F and G has three results:

1) balance, the problem ball is the 4 or 8 ball. Take a normal ball and ball 4 for weighing. If the balance is reached, the ball 8 is the problem ball and light; otherwise, the ball 4 is the problem ball and heavy.

2) If the weight of F is light, the weight of the ball No. 1 is light or the weight of the ball No. 7 is light. The weight of the ball No. 6 and the ball No. 7 is heavy. If the balance is reached, the weight of the ball No. 1 is heavy, otherwise, the smaller ball on the 6th and 7th is the problematic ball and lighter.

3) if the weight of F is light, the weight of ball No. 5 is light or that of ball No. 2 and No. 3 is heavy. Weigh the balls 2 and 3. If balanced, the ball 5 is lighter. Otherwise, the ball on the 2 and 3 is more focused on the problematic ball.

3. If a is light and B is heavy, the problem ball appears in Group A or Group B. Remove the No. 1 ball from a, take the No. 5 ball from B, and then take any two balls from C to form a group of H; remove the 2 and 3 balls from a, and take the 6 and 7 balls from B to form a group of I. There are three results for weighing H and I:

1) balance, the problem ball is the 4 or 8 ball. Take a normal ball and a 4 ball for weighing. If the balance is reached, the 8 ball is the problem ball and the emphasis, otherwise the 4 ball is the problem ball and the light

2) If the weight of H is light, the weight of the ball No. 5 or the weight of the ball No. 2 and No. 3 is light, if the balance is set, the ball No. 5 is the problem ball and the emphasis is on it. Otherwise, the light ball in the ball No. 2 is the problem ball and the light ball.

3) if the weight of H is light, the weight of the ball No. 1 is light or the weight of the ball No. 6 and No. 7 is heavy. Weigh the ball No. 6 and No. 7. If the ball is balanced, the ball No. 1 is the problematic ball and light. Otherwise, the ball on the 7 th is the problematic ball and heavy.