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- Newsgroups: sci.math
- Path: sparky!uunet!gatech!usenet.ins.cwru.edu!agate!stanford.edu!nntp.Stanford.EDU!ilan
- From: ilan@leland.Stanford.EDU (ilan vardi)
- Subject: Re: condoms
- Message-ID: <1993Jan6.122524.341@leland.Stanford.EDU>
- Sender: news@leland.Stanford.EDU (Mr News)
- Organization: DSG, Stanford University, CA 94305, USA
- Date: Wed, 6 Jan 93 12:25:24 GMT
- Lines: 30
-
- I have solved the general problem of the least number of condoms it
- takes for m men and n women to have all mxn heterosexual encounters.
- The answer is
-
- m = n = 2 => 2 condoms
-
- m = 2k+1, n = 1 => k+1 condoms
-
- otherwise, it's the smallest integer greater or equal to m/2 + 2n/3.
- I assume that m >= n, otherwise interchange m and n.
-
- The same method solves the problem for m homosexual men all having
- sex, and m bisexual men and n heterosexual women (posed by A. Orlitzky
- and L. Shepp).
-
- The problem was almost solved by Hajnal and Lovasz [``An algorithm to
- prevent the propagation of certain diseases at minimum cost,'' in
- ``Interfaces between computer science and operations research,''
- edited by J.K. Lenstra, A.H.G. Rinnoy Kan, and P. van Emde Boas,
- Matematisch Centrum, Amsterdam 1978] who showed that the number was
- within 2 of m/2 + 2n/3. In any case, showing that 6 men and 6 women
- require only 7 condoms pretty characterizes the algorithm (Hajnal and
- Lovasz required 8 condoms).
-
- My solution appeard in my book ``Computational Recreations in
- Mathematica'', Addison--Wesley, 1991. The reason it appeared there is
- that I suspected that interest in this problem might last longer than
- interest Mathematica.
-
- -ilan
-