天平称球[:B]

ob

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？

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ob

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.www.ddhw.com 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.www.ddhw.com 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.     www.ddhw.com

ob

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