任意找四人A，B，C，D。按假设，其中必有一人与另外三人通过电话，假定这人

是A。如果问题的结论不成立，则对应于A，B，C，D中的每个人都可以找到另外

一个人没有与他们通过电话。假设对应的人是a，b，c，d。

Thank you, Professor Wan for the solution. I would like to post some comments and

an alternative solution.

With an assumption, the professor suddenly came up with additional 4 people, a, b, c, d, which

brought the problem to life. With 8 people, the professor has more pieces to play with.

The other good point of this solution is the professor made no assumption what happened

among B, C, D - they could called each, and none at all, or things in between.

But what is difficult for me to understand, which is

*not* the professor's fault, is that based on the

his assumption, the professor first has to prove

*positively* that a, b, c, d are the

same person, and then

*negatively* that even this single person does not exist, and

therefore the assumption can not be true. There are some logical jump back and forth.

Here is my alternative solution:

"任意找四人A，B，C，D。按假设，其中必有一人与另外三人通过电话，假定这人是A。"

Now I begin to deviate from the professor. What happened with B, C, and D?

There are two possibilities: (1)

*At least* one pair did not communicate with each

other; (2) They

*all* communicated with each other.

Interestingly, case (2) is hard to handle (I do not know why).

Now let us deal with case (1). For simplicity of explanation and without losing generality,

let us assume C and D did not take to each other. Now let us choose

*any* person a

from the class besides A, B, C, D, and consider a, A, C, D. Since C and D did not take to each other, then

a and A must have talked to each other. Problem solved under the case (1): A talked to anybody

in the class.

Now let us look at case (2). We

*try to find* person b besides A, B, C, D who did not talk to

at least one of A, B, C, D.

The result of this trial has two possibilities: (2A) failed, we can not find such person. This means

any of the A, B, C, D has talked to anyone of the class. The problem is solved under this case.

The other possibility (2B) is we found this person: assume b and D did not talk.

Under the condition (2B), there two possibilities: (2B1) b has talked with

* at least* one of A, B, C;

and (2B2) b did not talk to any of the A, B, C. The possibility (2B2) does not exist.

Now consider (2B1), assume b has talked to B. Consider B, C, b, D. This is the same as case (1) which

has been proved. Therefore under all possible conditions, the problem is proved.

I realized my solution is like writing a computer program which consider all possibilities, and is not elegant,

but I hope at least the proof is correct. What do you think, Professor Wan?

Thanks for everyone who reached this line

.