Skip to main content
21 events
when toggle format what by license comment
Mar 11, 2021 at 19:25 comment added Penelope Benenati Thanks to you @PietroMajer, it has been an opportunity to clarify with a picture.
Mar 11, 2021 at 18:44 comment added Pietro Majer Thank you, I see I misunderstood the last condition -- c and c' need not to meet $\partial A$
Mar 11, 2021 at 16:00 history edited Penelope Benenati CC BY-SA 4.0
Moved down the picture
Mar 11, 2021 at 16:00 comment added Penelope Benenati Thank you @Glorfindel, I did not know it.
Mar 11, 2021 at 15:58 comment added Glorfindel FYI, the site doesn't support MathJax in image descriptions: cf. meta.stackexchange.com/q/327908/295232
Mar 11, 2021 at 15:57 history edited Glorfindel CC BY-SA 4.0
added 4 characters in body
Mar 11, 2021 at 15:54 comment added Penelope Benenati @PietroMajer I wrote that for $n=3$ we get $m=13$. I just added a pictorial explanation. I am basically interested in proving that $m$ is quadratic in $n$, which now seems obvious to me.
Mar 11, 2021 at 15:53 history edited Penelope Benenati CC BY-SA 4.0
Added pictorial example for n=3
Mar 11, 2021 at 15:29 comment added Pietro Majer @PenelopeBenenati How do you realize the number $m(3)=15$? I can see $12=3^2+3$ pieces of type R, and in general, $n^2+n$ .
Mar 11, 2021 at 15:22 comment added Pietro Majer (Yet I do not see clearly how this can be exactly related with the OP's problem, because it is not clear what is the number of new pieces of type $R$ made by a new pair of parallel lines)
Mar 11, 2021 at 15:10 comment added Pietro Majer The problem with hyperbolas in A058331 is solved, say, looking the configuration from a long distance, i.e., replacing each hyperbole with a pair of incident lines, and recalling they are hyperboles, so adding the n-th hyperbole makes 4n-2 new pieces, one less than a pair of incident lines. This gives the upper bound of $2n^2+1$, which is realized by any generic choice.
Mar 11, 2021 at 15:07 comment added Pietro Majer @Matt F. But any pair of arbitrary disjoint simple open curves is also topologically the same, yes the number of pieces obtained out of two such pairs may be even infinite.
Mar 11, 2021 at 14:32 comment added user44143 Well, topologically the hyperbolas are the same as pairs of parallel lines, and the configuration of the plane most cut up by two hyperbolas is topologically the same as the configuration most cut up by two pairs of parallel lines.
Mar 11, 2021 at 14:15 comment added Joseph O'Rourke @MattF.: Nice conjecture! Could you explain the intuition behind: hyperbolas = parallel lines?
Mar 11, 2021 at 14:05 comment added Joseph O'Rourke The maximum number of cells in an arrangement of $n$ lines is $n(n+1)/2+1$. So your $2n$ lines could make at most $2n^2+n+1$ cells, if all were in $D$. So indeed there is a quadratic upper bound. MattF was seeking an exact formula.
Mar 11, 2021 at 13:56 comment added Penelope Benenati Very interesting! Thank you @MattF. Maybe there is an easy way to prove that it is just quadratic in $n$ (it would be sufficient for my purpose I guess, even if I am working on hyperspheres in high dimensions in the original problem I trying to solve).
Mar 11, 2021 at 13:35 comment added user44143 This may be just $A058331-2n$ (oeis.org/A058331) or $2n^2-2n+1$, if cutting regions by parallel lines is the same as cutting regions by hyperbolas, and there are always $2n$ regions to remove as being outside all the lines.
Mar 11, 2021 at 12:50 history edited Penelope Benenati CC BY-SA 4.0
Question rephrased
Mar 11, 2021 at 12:38 history edited Penelope Benenati CC BY-SA 4.0
Question rephrased
Mar 11, 2021 at 12:15 history edited Penelope Benenati CC BY-SA 4.0
added 47 characters in body
Mar 11, 2021 at 12:09 history asked Penelope Benenati CC BY-SA 4.0