Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.1 6/24/83 (MC840302); site mcvax.UUCP Path: utzoo!watmath!clyde!burl!ulysses!unc!mcnc!philabs!cmcl2!seismo!mcvax!lambert From: lambert@mcvax.UUCP (Lambert Meertens) Newsgroups: net.math,net.puzzle Subject: Re: A problem about lists of points Message-ID: <6259@mcvax.UUCP> Date: Tue, 18-Dec-84 04:17:46 EST Article-I.D.: mcvax.6259 Posted: Tue Dec 18 04:17:46 1984 Date-Received: Thu, 20-Dec-84 05:35:37 EST References: <203@cmu-cs-g.ARPA> <1558@sdcrdcf.UUCP> Reply-To: lambert@mcvax.UUCP (Lambert Meertens) Organization: CWI, Amsterdam Lines: 43 Xref: watmath net.math:1663 net.puzzle:499 Summary: >> Consider a (bounded) line-segment. Choose point 1 anywhere on the segment. [etc.] > I believe you can go as far as you want. Use the fractional parts of > multiples of the golden ratio (SQRT(5)-1)/2, or about 0.618. The sequence > 0.618, 0.236, 0.854, 0.472, 0.090, 0.708, 0.326, ... seems to satisfy the > requirements. Not so. Take n = 7, and consider p[i] = (i*phi) mod 1 --> floor(p[i]*n): p1 = 0.618 --> 4 p2 = 0.236 --> 1 p3 = 0.854 --> 5 p4 = 0.472 --> 3 p5 = 0.090 --> 0 p6 = 0.708 --> 4 p7 = 0.326 --> 2 So both p1 and p6 fall in segment 4, and the last segment, 6, is not represented. I have not (yet) tried to prove this, but it appears extremely unlikely to me that an infinite sequence could exist for which every initial segment is evenly distributed. Here is as far as I came using a backtracking method: 0 <= p1 < 1/10 1/2 <= p2 < 5/9 3/4 <= p3 < 7/9 1/4 <= p4 < 2/7 7/8 <= p5 < 8/9 3/8 <= p6 < 2/5 5/8 <= p7 < 2/3 1/8 <= p8 < 1/5 9/10 <= p9 < 1 2/5 <= p10 < 1/2 I do not claim that this is the only solution up to 10, nor that it cannot be extended. -- Lambert Meertens ...!{seismo,philabs,decvax}!lambert@mcvax.UUCP CWI (Centre for Mathematics and Computer Science), Amsterdam