1, because we don't know whether the abnormal ball is light or heavy, no matter how we weigh it, there will be three different results, that is, the weight on the left is heavier, lighter or equal to the weight on the right. In order to find out the unqualified table tennis balls after three times weighing, the balls must be divided into three groups (four balls in each group). Now, for the convenience of solving the problem, we will number these three groups of table tennis as group A, group B and group C respectively.
First, choose any two groups of balls and weigh them on the balance. For example, we call group A and group B on the balance. There will be two situations:
In the first case, both sides of the balance are balanced. Then, the unqualified bad ball must be in group C.
Secondly, two balls (such as C 1 and C2) are randomly selected from Group C and placed on the left and right disks respectively, which is called the second time. At this time, two situations may occur:
1 Balance on both sides. In this way, the bad ball must be in C3 and C4. This is because only one of 12 table tennis players is an unqualified bad ball. Only when one of C 1 and C2 is a bad ball will the two sides of the balance be unbalanced. Since both sides of the balance are balanced, it can be seen that C 1 and C2 are qualified blows.
When weighing for the third time, you can take any ball from C3 and C4 (for example, C3), and put another qualified ball (for example, C 1) on both sides of the balance, and you can deduce the result. At this time, there may be two results: if the two sides of the balance are balanced, then the bad ball must be C4; If the balance is unbalanced, then the bad ball must be C3.
2. The balance between the two sides is unbalanced. In this way, the bad ball must be in C 1 and C2. This is because only when one of C 1 and C2 is a bad ball can the two sides of the balance be balanced. This is the second time.
When weighing for the third time, you can take any ball from C 1 and C2 (for example, C 1) and put another qualified good ball (for example, C3) on both sides of the balance, and the result can be deduced. For the same reason.
The above is the analysis of the first situation after the first weighing.
In the second case, after the first weighing, both sides of the balance are unbalanced. This shows that Group C must be a qualified strike, and the unqualified bad ball must be in Group A or Group B. ..
We assume that Group A (four balls A 1, A2, A3 and A4) is heavy, while Group B (four balls B 1, B2, B3 and B4) is light. At this time, it is necessary to take A 1 out of the heavy plate and put it aside, take A2 and A3 out of the light plate and leave A4 in the heavy plate. At the same time, take B 1 and B4 out of the optical disc and put them aside, take B2 out and put it in the heavy disc, leave B3 in the optical disc, and then take a standard ball C 1 and put it in the heavy disc. After this exchange, each group has three balls: A4, B2 and C 1 in the original reorganization, A2, A3 and B3 in the original light group.
At this point, it can be called the second time. There are three possible situations after weighing:
1 Balance on both sides. This shows that A4B2C1= AAA3B3, that is to say, these six balls are just good balls, so the bad balls must be outside the A 1 or B 1 or B4 set. It is known that disk A is heavier than disk B, so A 1 is either a good shot or more important than a good shot; While B 1 and B4 are either good shots or lighter than good shots.
At this time, you can put B 1 and B4 at one end of the balance, which is called the third time. At this time, there may be three situations: (1) If both sides of the balance are balanced, it can be inferred that A 1 is an unqualified bad ball, because 12 ball has only one bad ball. Because B 1 and B4 have the same weight, it can be seen that these two balls are good balls, while A 1 is bad. (2) B 1 is lighter than B4, so B 1 is a bad ball; (3) B4 is lighter than B 1, so B4 is a bad ball. This is because B 1 and B4 are either good balls or lighter than good balls, so the third weighing is actually lighter than one of the two light balls, and the light ball must be a bad ball.
2. The plates of 2.A4, B2, C 1 (originally in group A) are heavier than those of A2, A3 and B3 (originally in group B). In this case, the bad ball must be in non-exchange A4 or B3. This is because the exchanged B2, A2 and A3 balls did not affect the weight, so it can be seen that these three balls are all good.
The above shows that both A4 and B3 are bad balls. At this time, just compare A4 or B3 with standard ball C 1. For example, A4 is placed at one end of the balance, and C 1 is placed at the other end of the balance. This is called the third time. If both sides of the balance are balanced, then B3 is a bad ball; If the balance is uneven, then A4 is a bad ball (A4 is heavier than C 1 at this time).
3. The plates of 3.A4, B2, C 1 (original group A) are lighter than those of A2, A3 and B3 (original group B). In this case, the bad ball must be in the balls A2, A3 and B23 just exchanged. This is because if A2, A3 and B2 are all good balls, then the bad ball must be A4 or B3. If A4 or B3 is a bad ball, then the plates of A4, B2 and C 1 must be heavier than those of A2, A3 and B3. Now the situation is just the opposite, so not all A2, A3 and B2 are good shots.
The above shows that one of A2, A3 and B2 is a bad ball. At this time, we only need to compare A2 and A3, and call for the third time, which is the bad ball. When A2 and A3 are placed at one end of the balance for the third time, there may be three situations: (1) both sides of the balance are balanced, so it can be inferred that B2 is a bad ball; (2) A2 is more important than A3, so it can be inferred that A2 is a bad ball; (3) A3 is more important than A2, so it can be inferred that A3 is a bad goal.
According to the different weights of group A and group B after the first weighing, we just assume that group A is heavier than group B, and do the above analysis to explain how to infer which ball is a bad ball in this case. If we now assume that group A is lighter than group B, the reasoning process is the same as above.
2, two sides of the corresponding three weighing method:
Left 5, 7, 9, 1 1: right 6, 8, 10,12;
Left 2,9, 10, 12: right 3,4,8,11;
Left 1 4, 1 1, 12: right 3, 6, 7, 9.
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Ball 1, and heavy flat, flat left ball 1, and light flat, flat right.
No.2 ball, Heping Ping, Zuoping No.2 ball, Heping Ping, Right Ping.
The third ball, and the balance, right, right three balls, and the light flat, left, left.
No.4 ball, Heping, left and right No.4 ball, Heping, left and right.
The No.5 ball is flat on the left and flat on the right.
Ball 6, heavy right and flat right, light left and flat left.
No.7 ball is heavy left and flat right, light right and flat left.
No.8 ball, heavy right and flat right, light left and flat left.
The ninth ball is heavy left, left and right, light right, right and left.
The ball 10 is heavy right and left flat, and light left and right flat.
Ball 1 1, and heavy left, right and left balls 1 1, and light right, left and flat.
Ball 12, and heavy-right, left and left balls 12, and light-left, right and right.