圆桌问题

乱弹

This strategy is obvious: everyone takes the average of his number and next number as his new number.    www.ddhw.com 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.  www.ddhw.com 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.       www.ddhw.com

乱弹

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.www.ddhw.com   My strategy is just a quick answer. Maybe there are better ones. www.ddhw.com