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Consider $n$ line segments in the Cartesian plane. For $1\leq k\leq n$, the $k$-th line segment is drawn from $(k,0)$ to $(x_k,1)$, where $\{x_1,x_2,...,x_k\}$ is a permutation of $\{1,2,...,n\}$. Define $a(n)$ to be the maximal number of distinct points at which two or more line segments intersect.

For example, take $n=3$. A maximal construction for $a(n)$ would be a segment from $(1,0)$ to $(3,1)$, $(2,0)$ to $(1,1)$, and $(3,0)$ to $(2,1)$, giving $2$ distinct intersection points.

I have simulated the maximal values up to $n=13$ (and my friend has posted it on OEIS already here: https://oeis.org/A332774), but I am limited by the extreme time complexity of my program (around $n*n!*\log n$). The values from $n=1$ to $16$ are ($n=14$ to $16$ provided by Misha Lavrov):

$0,1,2,5,8,13,17,23,30,39,47,57,67,79,90,103$

I was wondering if a formula for $a(n)$ could exist, either recursive or explicit. Could it have something to do with the number of intersections in certain types of bipartite graphs? If no formula can be found, can there be any good estimates for upper / lower bounds?

Edit: Misha Lavrov has commented that there could be a repeating pattern of 8 in the second differences, though we do not have significant data supporting this claim.

Tracy Morgan
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    I was able to get a bit further and compute $a(14)=79$, $a(15)=90$, $a(16)=103$. (My approach is a branch-and-bound algorithm with the following upper bound on partial permutations: if, say, $n=10$ and $1,2,3,4,5$ are matched to $2,4,6,8,10$ in any order, then in addition the intersections found so far, the unused values $1,3,5,7,9$ will intersect $5,4,3,2,1$ of the existing segments, respectively, and will intersect each other at most $\binom 52$ times.) – Misha Lavrov Dec 25 '21 at 20:41
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    If we take $a(2), \dots, a(16)$, then the second differences are $2, 0, 2, -1, 2, 1, 2, -1, 2, 0, 2, -1, 2$, which looks like it's started repeating every $8$ steps, but there's not enough data to be sure. – Misha Lavrov Dec 25 '21 at 20:46
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    Here are some easy bounds. On one hand, the maximum number of distinct intersections is at most the number of pairs of lines, so $a(n)\le\binom{n}{2}\le n^2/2$. On the other hand, the permutation $\pi(i)=(\lfloor n/2\rfloor+i)\bmod n$ yields $\lfloor n/2\rfloor\lceil n/2\rceil\ge\lfloor n^2/4\rfloor$ intersections; thus $a(n)\ge\lfloor n^2/4\rfloor$. – ho boon suan Aug 29 '22 at 02:23
  • Thanks for the insight, that's a pretty cool construction. Could the zone theorem possibly be used to tell us something about maximal constructions? I recall that $9.5n-3$ is a strong upper bound on the complexity of a zone with $n+1$ lines, though I'm not sure if bounded area is too applicable here. – Tracy Morgan Aug 29 '22 at 06:50

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