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- Path: sparky!uunet!zaphod.mps.ohio-state.edu!saimiri.primate.wisc.edu!ames!agate!biosci!parc!xerox!feuerman
- From: feuerman@parc.xerox.com (Ken Feuerman)
- Newsgroups: sci.math
- Subject: Re: Interesting Discrete Math Question
- Message-ID: <1992Dec17.205217.2719@parc.xerox.com>
- Date: 17 Dec 92 20:52:17 GMT
- References: <3548@carroll1.cc.edu> <BzDIzv.5px@cantua.canterbury.ac.nz>
- Sender: news@parc.xerox.com
- Organization: Xerox PARC
- Lines: 35
-
- [munch...]
- > Here is the proof. (Pigeonhole principle).
- > ~~~~~~~~~~~~~~~~~
- > Partition the integers into 51 subsets:
- >
- > {1,4} {2,5} {3,6}
- > {7,10} {8,11} {9,12}
- > .....
- > {91,94} {92,95} {93,96}
- > {97,100}
- > {98}
- > {99}
- >
- > Now when 52 numbers are drawn, at least 2 must come from one subset, and thus
- > be 3 apart.
- [more munch...]
-
- This is all well and fine, except, how do you know you've found the best partition
- of those integers? That is, how do you know there isn't some other partition Q =
- {q_1, q_2, ... q_n} of the numbers 1-100 such that
-
- (i) for all i, j, and for all a in q_i, b in q_j, |a - b| != 3; and
- (ii) n > 51?
-
- I bring this up since someone else answered 51, having found a way to choose 50
- elements before guaranteeing that two would be 3 apart. At first blush that
- seemed plausible, until others got a little more clever. Is
-
- 1 2 3, 7 8 9, ....., 97 98 99
-
- as clever as we need to be?
-
- For the record, I believe so (QED :-)).
-
- --Ken.
-