派系数量下界
对于具有 n 个顶点和 m 个边的图,可能的最小团数(即最大团大小)是多少?我正在考虑使用图兰定理,但这只是告诉我们给定团数的边数的上限。我已经被困在这个问题上好几天了,需要一些帮助。
What is the smallest clique number (that is maximum clique size) possible for a graph with n vertices and m edges? I was thinking of using Turan's theorem, but that just tells us an upper bound on the number of edges given the clique number. I've been stuck on this for days, and could use some help.
如果你对这篇内容有疑问,欢迎到本站社区发帖提问 参与讨论,获取更多帮助,或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱,如果其他用户或者作者回复了您的评论,将不能在第一时间通知您!
发布评论
评论(1)
我不相信这个问题有一个简单的答案(拉姆齐式问题经常出现这种情况),但这里有一个让你开始的方法(我假设这是家庭作业,所以我不会尝试工作一切都出来了)。
假设该图是连通的(如果没有这个假设,问题只会变得更困难)。令
k
为可能的最小团数。极端情况是:如果
n = m-1
(相当于,该图是一棵树),则k=2
。如果
m = (n select 2)
(相当于图表完整),则k=n
。由此可以合理地推断,随着
m
相对于n
的增加,k
也应该增加。遵循这个想法,看看它会把你带到哪里。我计算出了小
n,m
的数字,但结果不在 OEIS 中可能我犯了一个计算错误(如果您发现错误,请告诉我)。以下是数字:再次,我假设已连接(至少
n-1
边)并且没有循环(最多(n select 2)
边)。I don't believe this question has a simple answer (as is often the case with Ramsey-esque problems), but here is an approach to get you started (I'm assuming this is homework and so I won't try to work it all the way out).
Assume the graph is connected (without this assumption, the problem only gets harder). Let
k
be the smallest clique number possible. The extreme cases are:If
n = m-1
(equivalently, the graph is a tree), thenk=2
.If
m = (n choose 2)
(equivalently, the graph is complete), thenk=n
.From this one might reasonably infer that as
m
increases relative ton
,k
should also increase. Go with this idea and see where it takes you.I worked out the numbers for small
n,m
and the results are not in OEIS, though possibly I made a computational error (please let me know if you find one). Here are the numbers:Again, I'm assuming connected (at least
n-1
edges) and no loops (at most(n choose 2)
edges).