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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.
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