### Author Topic: 通电话  (Read 7112 times)

#### PuraVidaQQSH

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• Posts: 7
##### 通电话
« on: 六月 28, 2012, 11:53:12 am »

Comment: I saw this problem somewhere. Judging from
its audience, it is not a difficult problem. But I do not think
I could think of is convoluted. Request a simple and strict

#### 万精油

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• Posts: 1831
##### Re: 通电话
« Reply #1 on: 六月 28, 2012, 01:03:05 pm »

#### 万精油

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• Posts: 1831
##### Re: 通电话
« Reply #2 on: 七月 03, 2012, 09:43:33 am »

b 这四人。他们中不存在任何一个人与其余三人都通过电话，与假设矛盾。唯一的

« Last Edit: 七月 03, 2012, 08:28:54 pm by 万精油 »

#### PuraVidaQQSH

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• Posts: 7
##### Re: 通电话
« Reply #3 on: 七月 03, 2012, 11:31:12 am »
Lao Wan,

I believe I have found a bug in your logic. To prove that your conclusion
"A has called everyone" is incorrect, I only need to find an exception which
is as follows:

For the same 4 persons in your answer A, B, C, D, A has called B, C, D.
You did not make any assumptions among B, C, and D, but let us assume
B, C, D have called each other. I can find another person a outside this
group of A, B, C, D, person a has called B, C, D but not A.

Could you explain this exception?

#### 万精油

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##### Re: 通电话
« Reply #4 on: 七月 03, 2012, 12:29:13 pm »
Quote
For the same 4 persons in your answer A, B, C, D, A has called B, C, D.
You did not make any assumptions among B, C, and D, but let us assume
B, C, D have called each other. I can find another person a outside this
group of A, B, C, D, person a has called B, C, D but not A.

#### PuraVidaQQSH

• Newbie
• Posts: 7
##### Re: 通电话
« Reply #5 on: 七月 03, 2012, 02:47:21 pm »
... A与所有人都通过电话。

Prof Wan,

Why do you say that "A与所有人都通过电话"? You could not logically come to the
conclusion (see my exception case), and you do not need to.

#### 万精油

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• Posts: 1831
##### Re: 通电话
« Reply #6 on: 七月 03, 2012, 08:33:57 pm »
Quote
Why do you say that "A与所有人都通过电话"? You could not logically come to the
conclusion (see my exception case), and you do not need to.

#### PuraVidaQQSH

• Newbie
• Posts: 7
##### Re: 通电话
« Reply #7 on: 七月 08, 2012, 11:52:18 pm »

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 .

#### 万精油

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• Posts: 1831
##### Re: 通电话
« Reply #8 on: 七月 09, 2012, 11:27:23 pm »