Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.1 6/24/83; site watdaisy.UUCP Path: utzoo!watmath!watdaisy!gjerawlins From: gjerawlins@watdaisy.UUCP (Greg Rawlins) Newsgroups: net.math Subject: (Some solutions to) A problem about lists of points Message-ID: <6823@watdaisy.UUCP> Date: Fri, 14-Dec-84 02:50:37 EST Article-I.D.: watdaisy.6823 Posted: Fri Dec 14 02:50:37 1984 Date-Received: Fri, 14-Dec-84 06:29:42 EST Organization: U of Waterloo, Ontario Lines: 111 >>A friend gave me this problem about a year ago, and I now know the answer, >>but not to my satisfaction. >> >>Consider a (bounded) line-segment. Choose point 1 anywhere on the segment. >>Then choose point 2 so that the first two points lie in different halves of >>the segment; choose point 3 so that the first three points all lie in >>different thirds of the segment; etc. What is the maximum number of points >>you can choose (before further choice becomes impossible)? >> >>It is convenient to consider a half-open segment, say [0,1[, and to regard >>1/2 as belonging to the second half, 2/3 to the third third, etc. >> >>If you get an answer, and a compelling *reason* why the answer is true, I'd >>very much like to hear about it. >> >>Peter Monta >>ARPA: monta@cmu-cs-g >>UUCP: ...!rochester!cmu-cs-pt!cmu-cs-g!monta --------------------------------- It seems to me that I must be missing something because this looks easy. If I understand it correctly we are asked to place n points on the half-open line segment [0,1) such that each segment of the form [i/n,(i+1)/n) for 0<=i<=n contains exactly one of the points. For example with n=3 we must place 3 points such that each of the intervals [0,1/3),[1/3,2/3),[2/3,1) is occupied. I will discuss the three versions of the problem that occurred to me, hopefully the "correct" problem is covered in here somewhere.... (1)If we define the problem to be one of just finding valid points for any given n then the problem is trivial, all we need do is subdivide the line segment into n pieces and place each point at the boundaries. Since we can do this for any n then the answer to the question is n=|R| where R is the set of all reals. (It is well known that the number of reals in any line segment is the same as the number of reals in total). (2)If, on the other hand, we specify that we must find a valid place for the n+1th point given some *prespecified* arrangement of n points then the best we can do is n=2,for, when we are asked to place the third point the "placement algorithm" breaks down if the first 2 points happened to lie in the intervals [1/3,1/2),[1/2,2/3). (3)The final version I could think of was that we were given a prespecified valid arrangement of n points *which we cannot alter* and asked for conditions on our ability to place k more points so as to then have a valid arrangement of n+k points. Since this last version presents some points of interest I'll look at it in more detail. Clearly if k=n (i.e. we wish to double the number of points) then it is always possible. To see why this is so we should take a careful look at the case n=1,k=1. | [----------------)[----------------) 0 1/2 1 Suppose the initial point was placed in the second half since we have no "preconditions" imposed on us we may place the second point anywhere in [0,1/2). Hence solvable. Suppose now that we have some arrangement of n points, viz.: | | | | | [-----)[-----)[-----)...................[-----)[-----) 0 1/n 2/n 3/n (n-2) (n-1) 1 ----- ----- n n (the bars "|" indicate where the initial n points are located). If we wish to double these points then we can do so by considering each segment separately as an "n=1,k=1" problem (of course there probably are other ways to find a valid arrangement but the question only asks *whether one exists*. Similarly we can triple or quadruple ... up to n-tuple the number of points (because we started off with the assumption that we could find a valid arrangement for 1 up to n points). Hence if we can solve the problem for n points we can solve it for 2*n,3*n,...,n*n. Notice that if we were allowed to move the points around then this version essentially reduces to version 1 since we could then do k=n+1,n+2,... etc. hence for any fixed n we could construct all its multiples (in particular n=1!). What happens now if k=1 and n>=2 ? then the answer is **dependent on the initial arrangement's distribution of points**. It is easy to see why since (for example) if the n points were arranged such that there exists some two sets of two consecutive points each of whose separation is *greater* that 1/(n+1) then we *cannot* place a new point because for the n+1 points to form a valid arrangement each segment must contain a point, here at least one of the segments must be empty. Alternatively if there exists two consecutive points which fall on the "rightmost" and "leftmost" of two adjacent boundaries then it is impossible to add a new point to make a valid arrangement on n+1 points because these two points would now be in the *same segment* (since n and n+1 are relatively prime no boundary stays fixed from n to n+1 and since these two points were "infinitely close" in the first place they must now be in the same segment.) Similar considerations apply for any arbitrary k for n>=2 if we cannot move the initial points. Greg Rawlins. -- /-----------------------------------------------------\ |Mail :Greg Rawlins :Department of Computer Science | | allegra\ U.of Waterloo,Waterloo,Ont.N2L3G1| | clyde \ \ | |UUCP :decvax ---- watmath --- watdaisy --- gjerawlins| | ihnp4 / / | | linus / | |CSNET:gjerawlins%watdaisy@waterloo.csnet | \-----------------------------------------------------/