Every Friday, six spies exchange all the information they have gathered during the week. A spy can never be seen with more than one other spy at the same time. So, they have to have several rounds of meetings where they meet up in pairs and share all the information they have at that point.
每周五，六名间谍都会交换他们一周内所收集到的所有情报。一名间谍绝不能在同一时间与超过一名间谍碰面。因此，他们得分多轮进行会面，每次两人一组，分享当时各自所掌握的所有情报。
The group of 6 spies needs only three rounds to distribute all secrets:
这 6 名间谍组成的小组仅需三轮就能分发完所有机密：
Before the meetings each spy holds a single piece of information. (spy 1 knows 'a', spy 2 knows 'b, etc.). In the first round spies 1 and 2 meet and exchange information so now both know 'ab'. The diagrams show which spies meet in each round with a line. It also shows which pieces of information they all have. After three rounds all information has been distributed.
在会议开始前，每个间谍都掌握着一条单独的信息。（间谍 1 知道“a”，间谍 2 知道“b”，以此类推）。在第一轮中，间谍 1 和间谍 2 相遇并交换信息，这样他们俩现在都知道“ab”。图表显示了每一轮中哪些间谍会面，用一条线表示。它还展示了他们所有人所拥有的信息。经过三轮，所有信息都已分发完毕。
Question: 问题:
After an international incident one spy has stopped attending the meetings. What is the minimum number of rounds needed for the five remaining spies to exchange all information?
在一次国际事件之后，一名间谍不再参加会议。那么剩下的五名间谍要交换所有信息，最少需要几轮？
Explanation 解释
Correct answer is 4 rounds.
正确答案是 4 轮。
This is unexpected: the obvious answer is three (or less?) since we have one spy less. This is even stranger if we consider that four spies would quite obviously exchange the information in two rounds.
这出乎意料：显而易见的答案是三轮（或者更少？），因为我们少了一个间谍。如果考虑到四个间谍显然只需两轮就能交换完信息，那就更奇怪了。
However, unsuccessful attempts at solving the task soon show us the root of the problem: since the number of spies is odd, one of them is “inactive” in every round. Say that spy number 5 does not participate in the first round, but he participates in the second. Thus in the second round, only two spies will know his piece information (e). In the third round, these two spies will meet two other spies, so after three rounds (only) four spies will know e. The fourth round is needed to spread this information to the fifth.
然而，几次尝试解决该任务均未成功，很快我们就发现了问题的根源：由于间谍的数量是奇数，所以在每一轮中都会有一个间谍处于“不活跃”状态。比如，间谍 5 在第一轮不参与，但在第二轮参与。因此，在第二轮中，只有两个间谍会知晓他的情报（e）。到了第三轮，这两个间谍会分别与另外两个间谍碰面，所以经过三轮（仅）之后，只有四个间谍知晓 e。需要第四轮才能将此信息传递给第五个间谍。
Therefore, we proved that at least four rounds are needed. To show that they also suffice, we construct a schema with four rounds.
因此，我们证明了至少需要四轮。为了表明四轮也足够，我们构建了一个四轮的模式。
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