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?
For n lines, the following are always possible:
0 intersections -- all lines parallel
1 intersection -- all lines go through a single point
n-1 points -- this can be done in at least two ways
a) n-1 parallel lines, and the nth is not parallel to any of them
b) n-1 lines go through a single point, and the nth is parallel to one of the others
n points -- n-1 lines go through a single point, and the nth is not parallel to any of the others and does not go through that single point
n*(n-1)/2 -- no shared intersections, no parallel lines, so every line forms a unique intersection with any other line
So far:
With 4 lines one can have 0,1,3,4 and 6 intersections.
Also, 5 is possible if there are no shared intersections, and two of the lines are parallel.
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With 5 lines one can have 0,1,4,5 and 10 intersections. Not sure how many others are available.
6 is possible, if 3 lines are parallel to each other, and the other two are parallel to each other, so there are 3*2 intersections.
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With 6 lines one can have 0,1, 5, 6 and 15 intersections. Not sure how many others are available.
6 can be partitioned as 2+4 and 3+3 and 2+2+2, so 8 and 9 intersections are also possible as a result of intersecting parallel lines. because 8 = 2*4 = 2*2*2 and 9 = 3*3.
Also, if k lines share a single intersection (where k > 3) and the other (n-k) lines are parallel to each other but to none of the other lines, then there are 1+k(n-k) intersections.
(n,k) = (6,4) is another way of getting to 9
(n,k) = (6,3) gives rise to 10 intersections
(n,n-1) = n, which I have listed above
Also, if k lines share a single intersection (where k > 2) and the other (n-k) lines are parallel to each and to one of the first k lines, then there are 1+(k-1)(n-k) intersections.
(n,k) = (6,3) gives rise to 7 intersections
(n,k) = (6,4) is another way of getting to 7
(n,n-1) = n-1, which I have listed above
So far, for 6 lines, I have 0,1, 5, 6, 7, 8, 9,10 and 15 intersections, but I feel that I have not yet exhausted the possibilities. Gotta go to sleep
Edited on October 9, 2014, 7:37 am
To sleep, perchance to dream
