Don't touch me

The points 1 , 2 , , 9 1,2,\ldots, 9 are labelled clockwise around a circle. How many ways can a subset of the chords be drawn between these points so that no pair of chords intersects?

Details and assumptions

Chords are not allowed to intersect at vertices nor at interior points.

It is implicit in the question that since an odd number of points are given, not all the points will correspond to a chord.

The empty set of chords satisfies the above conditions.


The answer is 835.

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3 solutions

Johan de Ruiter
May 20, 2014

Let a k a_k be the number of valid configurations for k k points. Pick an arbitrary point p 1 p_1 . The point p 1 p_1 can either not be used as the end point of a chord, in a k 1 a_{k-1} ways, or it can be connected to any other point p 2 p_2 , in which case it splits the remaining points into two groups without chords between them. Both smaller groups can be considered to be independent smaller instances of the same problem, so we can multiply their number of configurations.

This can be written as a k = a k 1 + i = 0 k 2 a i × a k 2 i a_k = a_{k-1} + \sum_{i=0}^{k-2} a_i \times a_{k-2-i} , which, starting with a 0 = 1 a_0 = 1 , gives the sequence:

1 , 1 , 2 , 4 , 9 , 21 , 51 , 127 , 323 , 835 , 1, 1, 2, 4, 9, 21, 51, 127, 323, 835, \ldots

Noticing that this problem can be solved via a recurrence relation makes the question much easier to solve. Trying to count configurations of chords and the number of ways each configuration occurs was prone to errors in calculation, missing some configurations and double-counting other cases. This was the only correct solution submitted.

Calvin Lin Staff - 7 years ago
Calvin Lin Staff
May 13, 2014

Let S k S_k be the number of ways to draw chords as described above with points labelled 1 , 2 , k . 1,2, \ldots k. Then S 0 = S 1 = 1 , S_0 = S_1 = 1, and we see that S 2 = 2 S_2 = 2 since we can either draw the chord between the points or not.

We claim that S k = S k 1 + i = 0 k 2 S i S k 2 i . S_k = S_{k-1} + \sum\limits_{i=0}^{k-2} S_{i}S_{k-2-i}. First note that there are S k 1 S_{k-1} ways to draw chords such that no chord uses the point k . k. The number of ways to draw the chords so that there is a chord from k k to j j is S j 1 S k j 1 , S_{j-1}S_{k-j-1}, since this chord divides the remaining points into two sets of size j 1 j-1 and k j 1 k-j-1 respectively. As j j ranges from 1 1 to k 1 k-1 we get the claimed sum.

Using the recursive formula S k = S k 1 + i = 0 k 2 S i S k 2 i , S_k = S_{k-1} + \sum\limits_{i=0}^{k-2} S_{i}S_{k-2-i}, we construct the following table of values for S k . S_{k}.

k 0 1 2 3 4 5 6 7 8 9 S k 1 1 2 4 9 21 51 127 323 835 \begin{array}{|r|cccccccccc|} \hline k & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9\\ \hline S_k & 1 & 1 & 2 & 4 & 9 & 21 & 51 & 127 & 323 & 835\\ \hline \end{array}

Thus, there are 835 835 ways to choose a set of chords such that none of them intersect.

Aditya Parson
May 20, 2014

The number of ways of drawing non-intersecting chords between k k points is given by the Motzkin number M k M_k .

As such the number of non-intersecting chords which we can draw between 9 9 points is the 9 t h 9th Motzkin number which is given by M 9 M_9 , which is equal to 835 \fbox{835} .

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