2
$\begingroup$

At first glance, it seems obvious that the score is limited, but trial and error has led me to fairly high scores that may suggest otherwise. Thank you for any suggestions.

Draw a continuous curve on a flat surface according to the following

Rules:

Draw a curve that self-intersects to create a loop. Call the loop so far segment $S_0$. Continue drawing the curve outside the loop. Call this new segment of the curve $S_1$. Have $S_1$ intersect $S_0$ twice. At the latest intersection, mark the end of $S_1$ and the start of $S_2$. Have $S_2$ intersect $S_0$ twice, then $S_1$ twice. At the latest intersection, mark the end of $S_2$ and the start of $S_3$.

In general, have segment $S_n$ intersect $S_0$ twice, then $S_1$ twice, then $S_2$ twice, … , then $S_{n-2}$ twice, then $S_{n-1}$ twice. At the latest intersection, mark the end of $S_n$ and the start of $S_{n+1}$. Repeat this rule until the game ends.

A segment $S_n$ may only intersect other segments according to the previous rule. The game ends when the most recent segment can no longer intersect another segment.

At the end of the game, your score is the number of intersections on the curve.

Question:

What is the highest possible score?

$\endgroup$
  • 2
    $\begingroup$ Can you please show us an example of drawing of such a game? $\endgroup$ – Crostul May 11 '17 at 9:41
  • $\begingroup$ Does the order of the intersections matter? $\endgroup$ – user2520938 May 11 '17 at 9:42
  • $\begingroup$ @user2520938 Indeed. Intersect in increasing order of $n$ where $n$ is the index of a segment. $\endgroup$ – Ola May 11 '17 at 9:44
  • $\begingroup$ Why would the score be limited? $\endgroup$ – Ivan Neretin May 11 '17 at 9:57
7
$\begingroup$

If I understood the question correctly, there is no maximum score. Here is a picture that shows a method that can be extended indefinitely.

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.