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 Counting Crossings (Posted on 2014-10-08)
3 lines in a plane can be easily be drawn such that there are 0, 1, 2 or 3 points where at least 2 of them cross.

What are the possible numbers of crossing points for 4, 5, or 6 lines?

Can any of these results be generalized?

 No Solution Yet Submitted by Jer Rating: 4.0000 (1 votes)

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 Other possibilities for 6 lines, and a theorem (unproven) | Comment 6 of 7 |
6 lines can have 11 intersections.  One way is if three lines share one intersection, the other 3 share another, and none of the lines are parallel.  This gives rise to 1 + 1 + 3*3 = 11

So far, I have identified (for 6 lines) 0,1, 5, 6, 7, 8, 9,10,11 and 15 intersections, but I feel that I have not yet exhausted the possibilities.

For instance, it seems that n(n-1)/2 - 1 is always possible, if all lines cross other lines without shared intersections, but two of the lines are parallel.  That reduces the intersections by 1, giving me 14.

It also seems to me that n(n-1)/2 - 2 is always possible, if no lines are  parallel, but three of the lines share an intersection.  That reduces the intersections by 2, giving me 13.

Hmm.  Calculation by subtraction has some merit.

Here is a bold prediction, without proof.
For 4 lines, all intersections up to 6 are achievable, except for 2.
For 5 lines, all intersections up to 10 are achievable, except for 2 and 3.
For 6 lines, all intersections up to 15 are achievable, except for 2 and 3 and 4.
For n lines, all intersections up to n*(n-1)/2 are achievable, except for 2, 3 ... (n-2)

I could easily be wrong.

 Posted by Steve Herman on 2014-10-09 07:55:09

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