There are twelve ping pong balls of the same size.

First, divide the 12 balls into three equal portions of four.

Take out two of them and place them on both sides of the scale (for the first time).

Situation 1: The balance is balanced.

Then the eight balls that are weighed are normal, and the special ones are in the four.

Take out three of the remaining four balls and put them aside, and put three normal balls on the other side (the second time) as the balance is balanced, and the special thing is the one left.

If it's not balanced, it's in the three above the scale. And know whether it's heavy or light.

Two of the remaining three are weighed, because you already know the weight, so you can know the special. (for the third time).

Scenario 2: The scales are tilted.

Special balls are inside those eight on the scale.

The four balls on the heavier side are counted as a1a2a3a4 and the light ones are counted as b1b2b3b4.

The remainder is determined to be four normal, denoted as C.

Set a1b2b3b4 aside, b1 and the three normal c-balls aside. Scenario 1: The scales are balanced.

The special ball is in a2a3a4, and you know that the special ball is heavier.

Weigh a2a3 and you'll know which of the three is special. (3rd) Situation 2: The balance is still heavier on the side of A1.

Special balls are between A1 and B1.

Just take one and the normal scale, and you will know which one is special. (3rd) Situation 3: The balance is reversed, and B1 is heavier.

The special ball is in the middle of b2b3b4, and you know that the special ball is lighter.

Weigh b2b3 and you'll know which one is special. The answer from (the third time)!!

What are you talking about, what are you talking about?

I can only guarantee that this process is detailed and understood, but not that it is not complicated.

Divide the table tennis balls into 3 groups of 4 balls each, take the two groups and compare them (the first time), and then there are two cases.

1. If it's the same.

Then the different spheres exist in the third group, set to (a, b, c, d) [compared to the judgment of two, here the letter case is related to the result], the standard ball is t, then the next take a+b+t:c+t+t (the second time).

If a+b+t=c+t+t, then d is a different ball, then d:t (third), - if d t, then d is a heavy ball, - if d t, then d is a light ball.

If a+b+t c+t+t, then there is a heavy ball in (a, b), or c is a light ball, take a:b (the third time), - if a=b, then c is a light ball, - if a≠b, then the heavy ball is a ball.

If a+b+t c+t+t, then there is a light ball in (a, b), or c is a heavy ball, take a:b (third), - if a=b, then c is a double ball, - if a≠b, then the light ball is a different ball.

Second, if it is different.

Then define these two groups as a+b+c+d a+b+c+d [case rules: from here, it can be seen that in the following case, if the ball is an uppercase letter, it must be heavy, and if it is a lowercase letter, it must be light], and the standard ball is t, take a+b+c+a: d+t+t+t (the second time).

If a+b+c+a=d+t+t+t+t, then the heterosphere exists in (b, c, d), take b:c (the third time), - if b=c, then d is a light ball, and if b≠c then the light one is a heterosphere (lowercase).

If a+b+c+a d+t+t+t, then (a, b, c, a) has a heavy ball, or d is a light ball. From the case rules, it can be seen that the heterosphere can only exist in (a, b, c), take a:b (the third time), - if a=b, then c is a heterogeneous ball;

If A≠B, it is a different ball.

If a+b+c+a d+t+t+t+t, then (a, b, c, a) there is a light ball, or d is a heavy ball, from the capitalization rules, the ball can only exist in (a, d), take a:t (the third time), - if a=t, d is a heavy ball, - if a≠t, a is a light ball.

There are 12 table tennis balls, which are nicknamed the same. There was only one weight anomaly, and it was now required to weigh three times on a weightless balance, starting with four on each side of the scale, and four more.

Situation 1: If the two sides are flat, then the bad one must be in the 4 that are left. Number the 4 balls as 1, 2, 3, 4

Take out 1 and 2 first, and weigh it, if it is flat, then it means that the bad ones are in 3 and 4. Then since 1 and 2 are good, then 1 and 3 are called, and if 1 and 3 are flat, then 4 is bad. If 1 and 3 are not equal, then it must be 3.

Because 1 is intact, 1 and 2 are the same weight).If 1 and 2 are not even, then 3 and 4 must be intact, weigh 1 and 3 again, if 1 and 3 are tied, then it is 2, if 1 and 3 are not even, then it is 1

Scenario 2: If the two sides are uneven, then group the two sides. The heavier ones are divided into 1, 2, 3, and 4, and the lighter ones are divided into a, b, c, and dThen they exchanged the scales, weighing 1,2,a and 3,4,b.

If 1,2,a and 3,4,b are drawn, then in other words, 1,2,3,4 and a,b are equal weights, which means that there are no bad balls in 1,2,3,4, that is, the bad balls are light. (Because the bad ball appears in the light ball group!) So that is to say, the light one in c and d is bad, and then it is called c, d can get the bad ball, and the light one is.

If 1,2,A and 3,4,b are not even, then it depends on which side is heavier. Let's say it's 1,2,a weight. (This is interchangeable with 3, 4, and b.) , then weigh 1 and 2.

If 1 and 2 are drawn, then it means that B is bad, because 1 and 2 are equal weight, that is, there is no bad ball in 1,2 (which is also a heavy ball), and A is from the light ball group, and A cannot be heavier than the other balls. So why is it 1,2,A heavy, the reason is obvious, 3,4,b has bad balls in it, and bad balls are light! But 3 and 4 come from the heavy ball group, that is, there can't be a light ball in 3 and 4, (otherwise 1, 2, 3, 4 will be light at the beginning!)

So B is a bad ball, and it's also a light ball.

If 1 and 2 are not drawn, then one of the 1,2 must be a bad ball, and since 1,2 is from the heavy ball group, the heavy one is bad. In the same way, if 3, 4, and b are the heavier side, then the reasoning process is the same as above.

Because knowing whether it's heavier or lighter.

Situation 1: heavier.

Divide the twelve balls into three groups of four, with group numbers 1 2 3.

1.Weigh groups 1 and 2, and if they weigh the same, the anomalous ball is in group 3. (If one of the groups is heavier when weighing groups 1 and 2, the anomalous orbs are in that group.) ）

2.Then divide the 3 into two piles, and then weigh the abnormal balls in the heavy pile.

3.Place two more balls at the ends of the scale, and the heavier balls are anomalous balls.

(If one of the groups is heavier when weighing groups 1 and 2, the anomalous orbs are in that group.) The same goes for lighter.

A group of 4, divided into 3 groups, can be weighed once to determine which group the ball is in.

Take 4 balls in this group and weigh 2--2.

Last 1--1 weighing.

Use the specific gravity of 6 and 6 to know which of the 6 balls has a different weight, and also know whether the weight of the ball is heavier or lighter than the others. In 3 to 3, find out which of the three have the weight difference. Finally, use 1 to 1 to see if there is a difference in the weight of these two balls, if there is no difference, it is not compared to that.

There is no answer to this question, and it is impossible to know it in just three times.

Not knowing the severity requires a certain amount of logical reasoning.

Step 1: Divide into three groups, 444, take two of the groups, and there will be two situations here:

a is the balance balance;

b is an imbalance in the scale.

They are discussed separately as follows:

For case A:

Step 2: One of the remaining 4 is non-standard, and three of them and three of the standard are extracted to be weighed.

If it is not balanced, you can judge whether the ball is light or heavy, in this case it is a1;

If the balance is balanced, the remaining ball is not standard, but the weight is not known, and this case is a2.

Step 3: For A1, you only need to take any two of the three unbalanced balls to weigh, if the balance of the remaining balls is naturally non-standard, and the weight is also known;

For A2, you only need to take a standard ball and weigh it against this non-standard to know whether it is light or heavy.

End of case a.

For case B:

First of all, we label the three groups in the first step as x, y, and z groups, and the balls are represented by x1, x2, x3, x4, and so on.

From 1, it can be seen that the non-standard balls are in groups X and Y, and the Z group is full of standard balls.

Step 2: Take out three balls from the X and Y groups, put the balls of the Y group into the tray where the X group is, take three from the Z group and put them in the tray where the Y group is, then the balance X group is Y1, Y2, Y3, X4; Group Y is Z1, Z2, Z3, Y4.

There are three ways in which the balance changes at this step:

The first is that the direction of the balance imbalance does not change, in this case B1;

The second is that the balance has become balanced, which is B2;

The third is when the direction of the balance imbalance changes, in this case B3.

Step 3: For B1, it means that the ball moved above has no effect on the balance of the balance, that is to say, only X4, Y4 two balls that have not changed have the existence of non-standard balls, only need to take one of them out and the standard ball (take Z4 well) to weigh the third time, if the balance of the remaining ball is not standard, judge the weight by the direction of the previous balance, if it is not balanced, you can judge the weight directly.

For B2, it means that x1, x2, x3 are not standard, and all of the Y group are standard, combined with 1 can get the weight of the non-standard ball, and then only need to take two arbitrary scales from x1, x2, x3, if the balance means that there is one non-standard left, if it is not balanced, you can judge which is not standard according to the weight.

For B3, it means that the moving Y1, Y2, and Y3 have an impact on the balance of the balance, and the X group is all standard, and the combination of 1 can also get the weight of the non-standard ball, and the rest is the same as the situation of B2, only need to take two arbitrary scales from Y1, Y2, and Y3, and if the balance means that there is one non-standard left, if it is not balanced, it can be judged which is not standard according to the weight of the balance.

End of case B.

The scales are not bad, but if they are not bad, they are called so.

Put 6 on one side and see which side sinks.

I divide the six on the sinking side evenly and three on one side.

Take the 3 on the sinking side, take one and put the remaining 2 on each side, and judge that if the last one is flat, the one taken away is the heaviest, if not, the sinking side is the heaviest.

Take any 8 out, 4 on both sides of the scale, if balanced: see analysis a, if unbalanced, see analysis b.

Analysis A: You can make sure that the anomaly ball is in the remaining 4, take 2 out of the 4 and put it on both sides of the scale, if it is balanced, see Analysis A1: if it is unbalanced, look at A2.

Analysis A1: It can be determined that the anomalous ball is in the remaining 2, take out any one on the scale, and then put one of the remaining 2, if it is balanced, it can be determined that the abnormal ball is the last remaining one, if it is unbalanced, it can be determined that the abnormal ball is the one that has just been put up.

Analysis A2: It can be determined that the abnormal ball is in these two, take either one on the scale, and then put any of the remaining two, if the balance is balanced at this time, you can determine that the abnormal ball is the one that has just been removed, if it is unbalanced, it is the one left on the balance after removal.

Analysis B: You can determine the abnormal ball among these 8 balls, and note the balance between the left and right sides of the balance at this time, whether it is heavy on the left or right. Take 1 ball from the left and put it outside, let's say it's ball 1, take 2 balls from the right and put it outside, let's say it's ball 2, 3, take out a ball from the left and put it on the right, let's say it's ball 4, let's say the ball on the left is ball 5, and the two balls on the right are balls 6 and 7.

At this time, take out one of the 4 small balls from the outside and put it on the left to observe the balance of the balance at this time. At this time, it is estimated that there will be the following situations: 1, the balance is balanced, see the analysis B1; 2. The balance is unbalanced, and the balance remains unchanged, that is, the weight of both sides has not changed, see the analysis B2; 3. The balance is unbalanced, and the weight of the two sides has changed, see the analysis B3.

Analysis B1: At this time, it can be determined that the abnormal ball is one of the balls 1, 2 and 3, and the 2 and 3 are placed at both ends of the balance respectively, and the balance is observed, if it is balanced, it is obvious that the abnormal ball is the ball 1, if it is unbalanced, and the severity of the situation has not changed, it can be determined that the one on the right side of the balance is the abnormal ball at this time, and if the severity of the situation changes, it can be determined that the one on the left side of the balance is the abnormal ball at this time.

Analysis B2: The anomalous sphere can be determined to be one of the 5, 6, and 7 spheres, in which case the analysis method is similar to that of analysis B1.

Analysis B3: At this point, it can be determined that the anomalous ball is ball 4.