Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.2 9/5/84; site dalcs.UUCP 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