Skip to main content
12 events
when toggle format what by license comment
Aug 21 at 13:00 comment added bob @rus9384 Right in the double intersection case, for any two sets of lines every line in one set crosses every line in the other in the worst case which is O(n^2) from graph theory. Nice.
Aug 21 at 1:41 history became hot network question
Aug 20 at 21:38 comment added D.W. Is there any sense in which this requires answers from a computer science perspective? It seems like a matter of pure mathematics.
Aug 20 at 21:27 comment added Bader Abu Radi @rus9384 Right, for the upper bound there is no need to have the parallel assumption.
Aug 20 at 21:19 comment added rus9384 @BaderAbuRadi Though even if the lines are not parallel, there are $O(n^2)$ (double) intersections (also easily provable: at step $n$ there are $n-1$ lines on the plane, so a new line can only intersect at most $n-1$ lines) and since triple intersections are only a subset of intersections, the same upper bound follows for them.
Aug 20 at 20:53 history edited user150715 CC BY-SA 4.0
edited title
Aug 20 at 20:47 answer added Yves Daoust timeline score: 4
Aug 20 at 20:27 vote accept CommunityBot
Aug 20 at 18:28 comment added user150715 @BaderAbuRadi Yes, you consider each set consists of parallel lines for sake of simplicity. Note that almost two vertical lines sets creates quadrilateral which is rhombus, but horizontal lines through vertices of rhombus creates two triangles and four triple intersections.
Aug 20 at 18:18 comment added Bader Abu Radi Can you add other assumptions? All we have is a picture. We cannot bound the number of triples from below by a quadratic if each set of lines follows an edge of a triangle, for example. If each set consists of parallel lines, then it is easy to show that there are at most quadratic number of triples.
Aug 20 at 18:09 answer added Bader Abu Radi timeline score: 3
Aug 20 at 17:40 history asked user150715 CC BY-SA 4.0