4
$\begingroup$

The following problem seems a very hard one, is it known? It has a resemblance to the lonely runner conjecture. I am guessing.

In the plane let $v_i$ be $n$ unit vectors no two of them are colinear. Take $P_0$ any point in the plane and construct a successive set of points $P_i$ such that for every $i$, $1\le i\le n$ the segments $P_{i-1}P_{i}=v_j$ for some $j$. Every vector $v_j$ is used once. Can we choose the vectors so that the obtained path from $P_0\to P_n$ is either a simple cycle or has no crossings (planar)?

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ Do you mean you want to characterize the $n$-tuples of pairwise not collinear vectors for which you can make that choice? And if so, for every $n$? I don't really see the link with the lonely runner, by the way. $\endgroup$
    – Alessandro Della Corte
    Commented Jul 30, 2021 at 16:20
  • $\begingroup$ Hey there is no link (just imagining) the $n$ vectors are given. You take a vector for each edge in the path... $\endgroup$
    – Toni Mhax
    Commented Jul 30, 2021 at 17:08
  • 1
    $\begingroup$ So you are given $n$ unit vectors, and $n$ doesn't matter. You wonder whether you can always choose one out of $n$! in principle distinct permutations of your vectors so that the corresponding concatenation doesn't self-intersect. Is this your question? If so the answer is no: take 3 vectors parallel to the sides of an equilateral triangle oriented suitably. $\endgroup$
    – Alessandro Della Corte
    Commented Jul 30, 2021 at 19:12
  • 1
    $\begingroup$ Crossing here is strictly speaking as X rather than V. Thanks $\endgroup$
    – Toni Mhax
    Commented Jul 30, 2021 at 23:23

1 Answer 1

Reset to default
11
$\begingroup$

Let $S=\Sigma v_i$. If $S=0$, sort the vectors according to their angle along the unit circle. Then the corresponding closed path traces the boundary of a convex polygon.

In fact, the vectors $v_i$ can be of arbitrary length.

If $S\neq 0$, then add an auxiliary vector $v_{n+1}=-S$ and proceed as in the first case. Finally remove the segment given by the vector $v_{n+1}$.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Tx but you didn't follow the conditions the points need to be constructed successively in a one way directed path, here when you add a vector you add a point from nowhere.... $\endgroup$
    – Toni Mhax
    Commented Jul 31, 2021 at 2:11
  • 2
    $\begingroup$ I don't understand your comment, Toni. Each new vector $v$ starts where the last one ended, and goes to wherever $v$ goes from there. Not adding any points from nowhere. It's just ordering the vectors in a clever way as $u_1,u_2,\dots$ and then going from the origin (say) to $u_1$ to $u_1+u_2$ and so on. $\endgroup$
    – Gerry Myerson
    Commented Jul 31, 2021 at 6:03
  • 1
    $\begingroup$ Yes i see . Thank you $\endgroup$
    – Toni Mhax
    Commented Jul 31, 2021 at 6:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .