# Larsen problem

1. Dec 29, 2007

### ehrenfest

1. The problem statement, all variables and given/known data
Can someone just explain to me what this problem is asking? The words that got cut off are "number" and "resulting". For example, why T_1 = 1?

2. Relevant equations

3. The attempt at a solution

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2. Dec 30, 2007

### HallsofIvy

Staff Emeritus
The sum is a sum of pairs of numbers, the first of each pair starts at $T_0[$ up to $T_{n-1}$ while the second number goes the other way. For example that formula asserts that $T_1= T_0T_0= 1$ and $T_2= T_0T_1+ T_1T_0=1+ 1= 2$.

$T_1$ is "the number of ways it is possible to join the vertices of a triangle (1+ 2= 3 vertices) in pairs so that they do not intersect each other". We can connect two vertices by drawing a side of the triangle. If we drew another side, it would have to intersect the first at an endpoint so we can draw only one line. $T_2$ would be the number of ways we can connect the vertices of a square so that they do not connect: $T_2= 2$ as the formula said because we can draw to parallel sides that do not connect.

I will agree that the wording, "the number of ways", is misleading. If we have triangle ABC, I would have thought we could have drawn AB and no other side, or BC and no other side, or AC and no other side, giving 3 "ways" but that is clearly not what is intended.

3. Dec 30, 2007

### ehrenfest

But in the triangle you can join AB, BC, or AC, right? Why are there not three ways?

4. Dec 30, 2007

### HallsofIvy

Staff Emeritus
Yes, that was why I said the phrasing was misleading. Since the three points A, B, and C could be labeled in different ways, so that any side could be labeled "AB", they are not considering those as "different ways" of drawing a line. You can pick any one of the three sides to draw first- then there are no other lines to be drawn that do not intersect the side already drawn. One way of drawing lines that do not intersec.

5. Dec 30, 2007

### ehrenfest

Gosh. Sorry, I just reread your first post and saw how you literally answered that question in the last paragraph.

So, if I understand this right, it is really the maximum number of non-intersecting line segments that you can draw in the circle each of which joins two vertices, right?

Last edited: Dec 30, 2007
6. Dec 30, 2007

### HallsofIvy

Staff Emeritus
I think that is a much better way of stating the problem!

7. Dec 30, 2007

### ehrenfest

Wait. T_3 = 5 means that it is possible to draw five such lines, but that is impossible since there are a maximum of two disjoint pairs of vertices--so there can only be two such lines! I mean that in ABCDE, you can take two pairs of letters and draw lines between them, and then there is nothing else you can do since there is only one letter left!

Last edited: Dec 30, 2007
8. Dec 30, 2007

### Dick

I think the sense of the problem is that you just keep drawing segments until you can't draw any more without crossing another. In the n=3 case (pentagon) you connect the consecutive vertices to get the pentagon and then you can draw two more interior diagonals. However you do this one vertex has four edges attached to it. There are five choices for this vertex, so T_3=5.

9. Dec 30, 2007

### ehrenfest

I think HallsofIvy suggested that the question meant something different in the previous posts. From your explanation, T_2 would equal 3 because you could join any to vertices. I think that the problem does not distinguish between the different vertices.

EDIT: I meant T_1

Last edited: Dec 30, 2007
10. Dec 30, 2007

### Dick

No, I would say T_2 (square)=2. Because there are two different internal diagonals to draw. I don't think that's what Halls meant. The vertices are distinguishable. The order in which you choose them to draw isn't.

11. Dec 30, 2007

### ehrenfest

So, T_1 (triangle) = 1 because there is one way to draw internal vertices i.e. drawing none at all?

Dick, I really think your explanation is different than HallsofIvy because if you reread post #2, some of which I have quoted, it seems like HallsofIvy is assuming that the polygon is not already drawn--only the vertices around the circle are. Also, HallsofIvy talks about drawing parallel sides for the T_2 case while you talk about drawing diagonals.

Last edited: Dec 30, 2007
12. Dec 30, 2007

### Dick

Mmm. Maybe we are saying two different things. I was taking it to mean draw without INTERNAL intersections. In which case you can always draw the usual polygon lines for free and then just worry about number of ways to draw the internal lines. I checked that T_4=14 and thought I had counted all of the diagrams correctly. Try it for T_3 and see what you conclude.