A solitaire game is played with the following rules:
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On a line segment (of arbitrary length, set it as long as you wish, but for convenience/reference sake, let's say it extends from 0 to 1 on the number line), you place a point anywhere you like on it.

Now place a second point, such that either of the two points is within a different half of the line segment. (The halves are taken to be "open intervals", which means that the end points are not considered "inside" the interval.)

Place a third point so that each of the three is in a different third of the line.

At this point, you may notice that the first two points can't be just anywhere. They cannot, for example, be close together in the middle of the line or close together at one end. They must be carefully placed so that when the third point is added, each will be in a different third of the line.

You proceed in this way, placing every nth point so that the first n points always occupy different 1/nth parts of the line.
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If you choose locations carefully, how many points can you put on the line?

(In reply to Solution (Infinite Points) by Paul)

A valiant effort, and on first look satisfying, but as exoticorn points out, it fails at 5 points. I suspect it fails similarly at every level k where k=(2^n)+1 points.

I have not yet tested it out yet, but I'd like to suggest that point A be just ε. Point B would be Paul's Point A (ε+(1/2)). Point C would be Paul's Point B (ε+(1/4), etc. This would relieve the "strain" on the first segment at levels k=(2^n)+1 points. It may just move that strain to a different point, (It does -- at level 3 with points ε, ε+.25 and ε+.5, ε must lie between .16 and .33, a very high value for a "vanishingly small" number) and the points may need to be added in each level range 2^(n-1)<k≤2^n in an order other than simply with numerators 1,3,5... to relieve that strain.

Another possibility might be to start with A=ε, B= 1-ε, and then start adding Paul's sequence with C=ε+.5, D=ε+.25, etc.

Posted by TomM on 2004-01-20 06:17:55