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Path: utzoo!utcsrgv!dalcs!holmes
From: holmes@dalcs.UUCP (Ray Holmes)
Newsgroups: net.math
Subject: Re: Re: Lists of points clarification
Message-ID: <1400@dalcs.UUCP>
Date: Thu, 27-Dec-84 14:47:35 EST
Article-I.D.: dalcs.1400
Posted: Thu Dec 27 14:47:35 1984
Date-Received: Fri, 28-Dec-84 00:42:45 EST
References: <206@cmu-cs-g.ARPA> <614@munnari.OZ>
Organization: Dalhousie University, Halifax, N.S., Canada
Lines: 32

> > >> 	It seems to me that I must be missing something because 
> > >> this looks easy. If I understand it correctly we are asked to
> > >>
> > >>		Greg Rawlins.
> > 
> > ... solve a rather ill-stated problem.  Here is a more precise formulation:
> > 
> > Suppose we call a list of points ( p_1, p_2, ... , p_n ) in [0,1)
> > *evenly distributed* if each segment of the form [i/n,(i+1)/n) for 0<=i > contains a point in the list.
> > 
> > The problem asks for a list of points ( p_1, p_2, ... , p_n ) such that
> > ( p_1, p_2, ... , p_k ) is an evenly distributed list for *every* k between
> > 1 and n, inclusive.  What is the largest n for which one can find such a list?
> 
>   I still don't see the problem.  After an *evenly distributed* list of k points
> has been chosen, exactly  k  of the subintervals
> [0, 1/(k+1) ),  [1/(k+1), 2/(k+1)), ... , [ k/(k+1), 1)  will contain one of
> the chosen points.  By the pigeon-hole principle, there must be an interval
> [i/(k+1), (i+1)/(k+1)) that doesn't contain one of the points.   Choosing the
> (k+1)th  point in this interval will produce an *evenly distributed* list of
> k+1  points, won't it ?   Since the first point can be chosen arbitrarily to
> get an *evenly distributed* list of  1  point, it seems to me that the list
> can be made arbitrarily large.
> 
> 				David Wilson

No - no - no
At MOST k of the intervals will contain one of the points, probably several
contain more than one (violating the condition) point.

					Ray