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- Newsgroups: sci.math
- Path: sparky!uunet!snorkelwacker.mit.edu!mintaka.lcs.mit.edu!hal.gnu.ai.mit.edu!jamie
- From: jamie@hal.gnu.ai.mit.edu (Andrew John Radcliffe)
- Subject: Re: Combinatorics Problem (probably easy)
- Message-ID: <1992Oct16.185841.26130@mintaka.lcs.mit.edu>
- Summary: It's Sperner's lemma
- Sender: jamie@gnu.ai.mit.edu
- Organization: /etc/organization
- References: <1992Oct15.232815.26763@u.washington.edu>
- Date: Fri, 16 Oct 1992 18:58:41 GMT
- Lines: 25
-
- In article <1992Oct15.232815.26763@u.washington.edu> cxu@milton.u.washington.edu (Chongguang Xu) writes:
- >
- >I have a conjecture about a combinatorics question which I have been
- >unable to prove. I'm hoping someone on the net knows the answer or
- >can give me a reference.
- >
- >Given a set S of k elements, let A be the largest family of subsets of S
- >such that for every two elements a and b of A, b is not a substet of a.
- >My conjecture is that the cardinality of A is C(k, [k/2]), where C(p,q)
- >is the binomial coefficient, and [c] is the floor of c. I can prove that
- >the cardinality of A is at least that big, but I can't prove it can't
- >be bigger.
- >
-
- Your conjecture is true, and the result is known as Sperner's lemma.
- (Somewhat unfairly Sperner has >two< lemmas to his name.) There
- are several lovely proofs, none really short enough to give here.
- A good book that talks about Sperner's lemma, among other things, is
- B. Bollobas, Combinatorics, Cambridge University Press. I'll email
- you several proofs if you want.
-
- Jamie Radcliffe
-
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-
-