Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.2 (MU) 9/23/84; site munnari.OZ Path: utzoo!watmath!clyde!bonnie!akgua!whuxlm!harpo!decvax!mulga!munnari!lizs From: lizs@munnari.OZ (Liz Sonenberg) Newsgroups: net.math Subject: Re: Lists of points clarification Message-ID: <614@munnari.OZ> Date: Thu, 20-Dec-84 03:11:33 EST Article-I.D.: munnari.614 Posted: Thu Dec 20 03:11:33 1984 Date-Received: Mon, 24-Dec-84 03:10:30 EST References: <206@cmu-cs-g.ARPA> Organization: Comp Sci, Melbourne Uni, Australia Lines: 27 > >> 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<=icontains 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