1
$\begingroup$

Imagine a n_regular polygon that vertex is named by 1 to n. We know can draw (n)(n+3)/2 diagonals in n_regular polygon,Also know if we want to draw Maximum diagonals that not intersecting each other inside the n_regular polygon is n−3 diagonals. For example for Flat hexagon we can draw Maximum 3 diagonals that we can do it with 14 cases.(you can find it easy) Now imagine we want to draw diagonals that have 2 condition like this: 1- any vertex is not connected to 2 before and 2 after vertex. 2- any 3 vertex is not exist that intersecting two by two inside the n_regular polygon. It's easy to understand that the Maximum of diagonals can draw is 2(n−5). But the question is how many cases can draw diagonals that Applicable 2 above condition? For n=6 we can draw 3 cases that necessitate this 2 condition . If we labeled regular when n=7 we have 2(7−5)=4 diagonals and we can not use this vertexes {1,4},{2,5},{4,7},{3,6} because we ignore the second condition.

$\endgroup$
1
$\begingroup$

I don't understand what you mean by your 2 conditions (the english). Let me translate up until that point.

Suppose we have a regular $n$-gon with vertices labeled 1 to $n$. There are $n(n-1)/2$ possible diagonals. The maximum number of non-intersecting diagonals is $n-3$, and if we draw them we get a triangulation. The number of ways to triangulate an $n$-gon is the Catalan number $C_{n-2}$. For example, the number of ways to triangulate a hexagon is $C_4 = \tfrac{1}{5} {8 \choose 4} = 14$.

Now instead of non-intersecting, suppose we impose the following two conditions: 1? 2?

$\endgroup$

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.