Each of 10 friends knows some item of gossip not known to the others.   They communicate by telephone, and in each call the two friends on the line   share everything they have heard thus far. What is the smallest number of   calls that can be made, such that by the end, everyone knows everything?

I came across this interesting problem today !

This is what i understood.To make the minimum number of calls.Consider   it as a graph or an n-agon. Now,you'll move through the graph for n calls.

1-2-3-4-5-6-7-8-9-10-1

10 calls,now 1 and 10 knows all the information.All other nodes will   communicate with any of the two nodes and accept the data.Which means   n+n-2=2*n-2(18 calls).

I thought this was minimum value.But found out that it is 2*n-4.

And the logic behind that is divide the group as 4 and 6.Now you'll   have

f(4)=4 And f(6)=6.

Now,all the six people have to communicate with all the four people + 2   times so that everyone in the 6 people group will know all the data..which is   16 calls < 18 calls.

What i wanted to know is what is so special about 4 in this problem why   after 4 every problem(n>4) depends on f(4) ? Why splitting 4 seperately ?

题目大致意思是：

10个人中的每一个人都知道一个消息，而且这10个消息都不相同。为了使所有的人都知道一切消息，他们一共至少要打几个电话？
   

最后一段

What i wanted to know is what is so special about 4 in this problem why   after 4 every problem(n>4) depends on f(4) ? Why splitting 4 seperately ?

中文解释：

我想知道的是特别的在这4个问题为什么在4的每一个问题是什么（n＞4）取决于F（4）？为什么分裂4分？

这个中文的大致意思可以理解为：解决这个问题的关键在于4个人之间最少需要打几次电话。因为4个人的时候是最特殊的，只需要4次。所以加以推广的话，任意分组后，每组总有2个人最先知道本组的全部消息，而这样的每组2人只需要互通电话就可以有4个人知晓所有消息。然后的方法就是由他们通知其余所有的人。