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How many points can you place? (Posted on 2004-01-16) Difficulty: 5 of 5
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?

See The Solution Submitted by SilverKnight    
Rating: 4.2000 (10 votes)

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A possibility | Comment 2 of 25 |
Let's assume that all points on the interval are zero-dimensional. Then, one approach is to work backwards. For example, say you intend to place five points on the interval. Subdivide the interval into five equal intervals:
.01-.19, .2-.39, .4-.59, .6-.79, .8-.99

Round one, place the first point: .01
Round two, place the second point: .8
Round three, place the third point: .6
Round four, place the fourth point: .4
Round five, place the fifth point: .2

Note the strategy used for placing the points. I began with placing the first point at the smallest possible value. I place the second point at the lowest value of the highest interval. In the scenario of five intervals, .8 is the lowest value in the interval with the highest overall values.

After these first two points, I used the intervals as my guides by taking the lowest possible value in the highest group that covers the appropriate interval. For example, in round 3, I had three intervals from which to choose: .2-.39, .4-.59, and .6-.79. I choose .6 as my point since it is the lowest value in the interval with the highest values that satisfied the thirds requirement.

The only limit my theory has is your patience. Subdivide the interval as much as you like, and this theory should(?) hold. Since I have yet to test it at higher orders, I hesitate to make a definitive statement.

I realize that this isn't a rigorous mathematical proof, so please feel free to amend or correct this.
  Posted by draistal on 2004-01-16 16:46:46
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