This strategy is obvious: everyone takes the average of his number and next number as his new number. To show that this must work, consider the sum of all |a(i)-a(j)|, where a(i) is the number that the i-th person has right now. It is easy to show that this sum is nonincreasing after one round's number changing. Thus there is a limit. Further discussion will conclude that the limit is zero. To find the converge rate, we need to go further. Probably there are some other strategies that are similar but work better. If the players are allowed to have memory and they can change strategies round by round, then there should be better strategies. For example, one cheating strategy: for the first k rounds, the k-th person increases his old number by the first guy's number, the (k-1)-th person changes his number to zero, other guys take the next number. Then after k rounds, the k-th person has the sum of all numbers, and others have zero. The next is to redistribute the sum. This strategy need every one to know the value of k. |
To show that this must work, consider the sum of all |a(i)-a(j)|, where a(i) is the number that the i-th person has right now. It is easy to show that this sum is nonincreasing after one round's number changing. Thus there is a limit. Further discussion will conclude that the limit is zero. To find the converge rate, we need to go further. 能否证明在有限次内解决问题？ If the players are allowed to have memory and they can change strategies round by round, then there should be better strategies. For example, one cheating strategy: for the first k rounds, the k-th person increases his old number by the first guy's number, the (k-1)-th person changes his number to zero, other guys take the next number. Then after k rounds, the k-th person has the sum of all numbers, and others have zero. The next is to redistribute the sum. This strategy need every one to know the value of k. very interesting.... |
能否证明在有限次内解决问题？ In general, my strategy won't stop in finite steps. For example, if K is odd, it never stops. When will it stop? I guess we need to study binomials. My strategy is just a quick answer. Maybe there are better ones. |
"This strategy need every one to know the value of k." Only two persons need to know the value of k. The others always get number of the next person. It takes 2k - 2 rounds to finish. Another way is every body remembers his initial number. And at round i he needs to do some calculation so his number is the average of the i+1 people starting from him. This way after k-1 rounds every body have the average. this is the quickest. |
Every one remembers his origianal number. At round number i, everybody changes his number to (his origianal number)/(i+1) + (the number next to him)*i/(i+1). |
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