Simple problem of weighing the ball on the scale, the problem of weighing the ball on the scale

Little Cabbage's sister must be Little Cabbage.

I can't look with my eyes

Categories: Entertainment & Leisure >> brain teasers.

Problem description: There are 12 balls with the same size and shape, and one of them has a different weight (I don't know whether it is light or heavy). Give you a balance and weigh (find) the ball with different weights three times.

Analysis: Divide the 12 balls into 3 groups of 4 each.

1st time: Take 1 group and 2 groups to weigh. There are 2 scenarios:

The scales are balanced. (These 2 sets can be excluded at this time, and the target ball is set in the 4 balls of the 3rd group).

On the second occasion, take 2 balls from Group 3 (referred to as No. 1 and No. 2 below). At this time, there are 2 cases: Balance balance (meaning that the target ball is 3 or 4).

For the third time, take the No. 2 ball and the No. 3 ball. If the balance indicates that the 1, 2, and 3 balls are of equal weight, that is, the 4 ball is the target ball. If it is unbalanced, it means that ball 3 is the target ball.

The balance is unbalanced (indicating that the target ball is 1 or 2).

For the third time, take the No. 2 ball and the No. 3 ball. If the balance indicates that the weight of ball 2 and 3 is equal, that is, ball 1 is the target ball. If it is unbalanced, it means that ball 2 is the target ball.

The balance is unbalanced (this phenomenon indicates that the target ball is in groups of 1,2 in 8 balls).

The second time, because it is balanced, there must be a handful of mountains trembling low on the side and high on the other. Set group 1 high and group 2 low. Then if the target ball is in the Wei Tsai 1 group, the target ball is a heavy ball; If the target ball is in 2 groups, it is a light ball.

Take No. 1 of group 1 and No. 2 and No. 1 of group 2 and place it at one end of the balance; At the other end are placed 1 group of 3,4 and 2 groups of 2.

At this time, there will be 2 situations: balance balance (indicating that the 6 balls on the balance are of equal weight, then the target ball may be the 3 and 4 of the 2 sets), and the 3rd time take the 2 sets of 3 and 4. As mentioned above, if the target ball is in 2 groups, then it must be a light ball. So whoever is light in number 3 and 4 is the target ball.

Unbalanced balance (indicates that the target ball is on the scale.) However, in connection with the previous mention: if the target ball is in group 1, the target ball is a heavy ball; If the target ball loses in 2 sets, it is a light ball.

Then only the two 1 balls on the heavy side and the 1 2 balls on the light side can be target balls)

The third time I take the two one set of balls that may be the target ball. If it is balanced, then the remaining 2 sets of balls are the target balls. If it's not balanced, it's the target ball that counts.

This question is the question of an IMO.

There are two ways to do this:

Type 1: 1) Take out 6 of the balls and place them evenly on both ends of the scale.

2) If the balance is balanced, then compare the remaining two balls.

3) If the balance is unbalanced, choose any two balls on the heavy side for comparison.

4) If it is the same weight, then the remaining one is the heaviest, and if the balance is unbalanced, then the balance is heavier on which side it tilts.

The second type: 8 balls are divided into 3 parts, which are 2 balls, 3 balls, 3 balls (1) 3 balls and 3 balls are weighed, if they are the same weight, then it proves that the heavy ball is in that part of the 2 balls, put one on each side, and the heavy ball can be found.

2) If the 3 balls and the 3 balls are not the same weight, then it proves that the heavy ball is in the 3 balls that are heavier, then, take out 2 of the 3 balls that are heavy, and put one on one side, if they are the same weight, the heavy one is the one that is not weighed, if they are not the same, obviously, you will find the heavy one.

There is a formula for calculating the balance balling problem, according to the formula, the maximum number of true and false can be found k times is n=(3 k-1) 2

So the minimum number of times required for 12345 balls is k > = ln(12345*2+1) ln3 =

So it takes at least 10 times before it can be displayed.

Minimum 3 times. It should be known that the ball is too heavy or too light, but now it is assumed that it is too heavy, and the idea is as follows:

Divided into 3 groups, 13, 13, 14

1 time) weigh two groups of the same number, the balance is biased to the side of the dismantling is which group, if the two sides of the balance are equal, it is not weighed a elimination rubber group.

If it is in 13, it is divided into 445, if it is in 14, it is divided into 4552 times) are called two groups of the same number, and the method is the same as 1

If it is in 4, it is divided into 112, if it is in 5, it is divided into 1223 times) Weigh two groups of the same number next to it, if you are lucky, you can weigh it now, and if it is in 2, you can weigh it again.

4, 5 5 1 first weigh two fives, if balanced it is left.

Putting six on both sides at the beginning is definitely not a good idea.

Because this is the same as not putting Qi Luheng, it must be unbalanced, because the ball does not know whether it is heavier or lighter.

Let's do it: 3 on each side at the beginning:

If balanced, replace the 3 on one side with 3 of the remaining 6:

If balanced, it will take two more of the remaining 3 for a total of 4 times.

If it is not balanced, then it is already possible to know whether the ball is heavier or lighter, and it will take one more time, a total of 3 times.

If it is not balanced, replace the 3 on one side with 3 of the remaining 6:

If it is balanced, it will be in the 3 that are replaced, and at this time I will know whether it is heavier or lighter, and it will take one more time, a total of 3 times.

If you don't balance it, you will know whether it is heavier or lighter among the 3 that you haven't replaced, and you need to do it once, a total of 3 times.

This should be the simplest method to do highly, with a high probability of only 3 times.

In fact, if there are only 11 balls, there is a heavier or lighter one to ensure 3 weigh-ins.

1) Take 9 on each side of the scale, if one side sinks, the heavy ball is in the sinking 9. If both sides weigh the same, the heavy ball is in the remaining 9.

2) Out of those 9 that determine where the heavy ball is. Take 3 of each to each side of the scale, and if one side sinks, the heavy ball is in the 3 that sinks. If both sides are equally heavy, the heavy ball is in the remaining 3.

2) Of the 3 that determine where the heavy ball is. Take 1 on each side of the scale, and if one side sinks, the heavy ball is the one that sinks. If both sides are equally heavy, the heavy ball is the one left.