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标题: 天平称球[:B] [打印本页]

作者: ob 时间: 2005-6-21 09:58
标题: 天平称球[:B]

对10个互不相同重量的小球，用一台无砝码天平，最少经过多少次称量，可以排出轻重顺序？

作者: tcpip004 时间: 2005-6-22 18:07
标题: 回复：天平称球[:B]

should be 25

作者: math59 时间: 2005-6-23 06:57
标题: 能不能讲一下方法，我最多只能到36

我的方法是每次都分两组取最大最小

作者: ob 时间: 2005-6-23 23:50
标题: 回复：能不能讲一下方法，我最多只能到36

1. 3 balls take three times to get them in order from Heavy to Light. Say the order is

1, 2, 3

2. Now use the 4th ball to compare with 2, if it is heavier than 2, compare it with 1, otherwise compare it with 3. So it take 5 times to get 4 balls in order, say the order is 1, 2, 3, 4.

3. Now compare 5th ball with 2, and base on the results compare it with either 1 or 3, 4.

It takes 8 times to get 5 balls in order. say the order is 1, 2, 3, 4, 5.

4. Compare 6th with 3 first and then compare it with either 1, 2 or 4, 5. It takes 3 more times to get 6th balls into the order.

... 1, 2, 3, 4, 5, 6, 7

Compare 8th ball with 4 first and then compare 8th ball with 2 or 6. Then compare 8th with the others based on the previous two comparison results. So it take 3 more time to get 8th in order.

You can go on like this and get 25 times to make 10 balls in order.

I think there should be a formula to calculate this. For example:

Number of balls 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

Number of times 0 1 3 5 8 11 14 17 21 25 29 33 37 41 45 49 54

It add one more time once the number of balls is 2^n+1 and stays as that until it reaches 2^(n+1)+1.

作者: ob 时间: 2005-6-23 23:55
标题: The question is to use minimum # times to

gurantee to get the balls in order. 9 times can not make sure the balls are in order.

作者: math59 时间: 2005-6-24 04:32
标题: 简化到了30[:D)]

简化到了30

作者: math59 时间: 2005-6-24 04:40
标题: 答案已经出来了[:((]刚刚有些进步

答案已经出来了

刚刚有些进步

作者: ghgh 时间: 2005-6-24 04:56
标题: ob 和苹果的题大部分在文学城都被解了。

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