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标题: 野菜花' problem posted at WXC [打印本页]

作者: fzy 时间: 2005-3-30 18:21
标题: 野菜花' problem posted at WXC

8位歌手参加艺术节，为他们安排 M 场演出。每场由其中 4 位登台表演，要求 8 位歌手中任意两位同时演出的场次一样多，设计一种方案，使演出的场数 M 最小。

The problem was posted by, well, by 野菜花. The poster said she is not the real 野菜花, it turns out that the account was closed and the screen name is available. So I dont know if you (the real 野菜花) want to do something about it.

作者: 野菜花 时间: 2005-3-31 21:04
标题: 回复：野菜花' problem posted at WXC

I remember I posted this question somewhere before. I just went to WXC and saw that somebody (I guess is 乱弹 )wanted to try my ID to see whether it worked now, and found it did not work but using the name without password through his computer was ok. I know they have locked my IP, I tried once just for curiosity (Somebody asked me to do so), I could not post anything to WXC on any forum in any name.

I don't know whether all computers in the whole school have the same IP? If this is true, I feel sorry for the Chinese (if any) in our school who want to but can not post on WXC just because of me. It is very funny!

I really appreciate your and others' constant concerns and supports, but don't worry, I really don't care whether they unlock my IP there. Thank you again!

作者: fzy 时间: 2005-3-31 22:45
标题: 回复：回复：野菜花' problem posted at WXC

I was not trying to persuade you to go back to WXC. Although I would love to see it, I accepted you explanation. But you could go there from your home computer to claim the screen name and put a password so that other people would not be able to do it.

Now the problem. Anybody care to solve it? Or this generalization: Suppose there are n performers performing in groups of k (n > k >= 3). What is the minimum performances so that the number of joint performances by any two performers is a constant?

Sorry I generalized your problem again. But this time I do not know the answer. I have a conjecture but cannot prove it.

作者: 野菜花 时间: 2005-4-1 01:34
标题: 回复：回复：回复：野菜花' problem posted at WXC

For your generalized problem, I could only get the following:

Let m be the number of the performances, r be the constant. Then

mC(k,2)=rC(n,2)

m=r*n(n-1)/[k(k-1)], choose r such that m is the minimum interger.

作者: fzy 时间: 2005-4-1 18:40
标题: That is my conjecture

and it is obviously a lower bound. But I cannot prove it is always achievable. (8, 4) is easy. Then I tried (7, 4), which is less easy:

ABCD, ABEF, ACEG, ADFG, BCFG, BDEG, CDEF

At that time I realized that for (15, 3) this is the Kirkman Schoolgirl problem. So I gave up.

作者: 乱弹 时间: 2005-4-5 08:25
标题: The admin told me that

your IP was not blocked. (if your IP is 1*:1*:20:1*) Maybe they have unblocked it.

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