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Topic: Random Triangle Problem
Replies: 57   Last Post: Aug 17, 1997 10:51 PM

 Messages: [ Previous | Next ]
 Bill Taylor Posts: 1,909 Registered: 12/8/04
Re: Random Triangle Problem
Posted: Jul 30, 1997 1:53 AM

eclrh@sun.leeds.ac.uk (Robert Hill) writes of some fascinating simulation data!

|> I generated 1 million pseudo-random triangles

|> (1) vertices independent uniform inside a square:
|> obtuse at first vertex at second at third all acute
|> 241781 241998 241802 274419
|>
|> (2) vertices independent uniform inside a circle:
|> 248534 248198 248619 254649
|>
|> (3) vertices independent uniform inside an equilateral triangle:
|> 249377 249031 249279 252313

These are all *hugely* away from 3/4 to 1/4. The best has a chi-square

|> I can't understand why the answer for a circle comes between
|> that for a square and that for a triangle! Can anyone?

I certainly can't. It seems most counter-intuitive. Did you write them
down in the wrong order maybe?

|> Or is it just a bug in my program?

I'd hesitate to say so. For one thing there's no significant differences
between the three corners; (in spite of that other twit who thinks there
*should* be! :) ) If it's not too difficult, could you repeat the
simulation with a regular pentagon, hexagon, octagon... see if the results
tend toward the circular case, as they ought.

|> (4) x- and y- coordinates of each vertex independent normal N(0,1)
|> (if my memory is correct this distribution is isotropic,
|> i.e. for any fixed a it makes x cos a + y sin a also N(0,1)):
|> 249791 249989 249974 250246

|> answer for this case may be exactly 3/4. So I re-ran it with
|> 20 million triangles and got
|>
|> 4999137 5002204 4999415 4999244

Now this is absolutely amazing! These figures are spot on the 1:3 ratio.
The chi-sq(1) value is 0.15 . Almost too *good*! (Hello Gregor Mendel!)

The normal distribution has a good record of being the most "truly random"
in various contexts; but even so, I'm amazed to see it do the job here!
What I'm wondering is how much is due to mere circular symmetry and
infinite support. So a good one to compare it with would be Cauchy-like.

Robert:- can you run off a few more million simulations using a 2D Cauchy;
====== and report back to us please? You can do it with...

@ is uniform on [0,pi]; U is uniform on [0,1] (and indedendent of course).

Then R = tan(pi(U-0.5)) and x = R cos@ y = R sin@ .

Hope you can. Thanks in advance.

|> Namely, if the vertices of a triangle
|> are chosen independently and unifomly ON a given circle, then the 3/4

|> So the 3/4 answer for the whole plane is essentially equivalent
|> to the statement that the vertices of a random triangle are
|> uniformly and independently distributed on its circumcircle.

|> Or that the points of contact of the sides with the incircle are
|> independently and uniformly distributed on the incircle.

Yes, an amusing coda. :) A disguised form of my three slopes answer.
(Similar identifications hold in the Bertrand problem.)

-------------------------------------------------------------------------------
Bill Taylor W.Taylor@math.canterbury.ac.nz
-------------------------------------------------------------------------------
I do not claim my argument is logical, but simply that I'm right.
-------------------------------------------------------------------------------

Date Subject Author
7/16/97 Mike Housky
7/21/97 Bill Taylor
7/22/97 tony richards
7/24/97 Brian M. Scott
7/23/97 tony richards
7/23/97 T. Sheridan
7/24/97 Bill Taylor
7/24/97 Bill Taylor
7/25/97 Ilias Kastanas
7/23/97 Robert Hill
7/23/97 tony richards
7/27/97 Bill Taylor
7/24/97 Robert Hill
7/28/97 tony richards
7/30/97 Bill Taylor
7/30/97 tony richards
8/1/97 Bill Taylor
7/24/97 Robert Hill
7/24/97 Robert Hill
7/24/97 Robert Hill
7/25/97 Robert Hill
7/30/97 Bill Taylor
8/1/97 Charles H. Giffen
8/1/97 John Rickard
8/1/97 Chris Thompson
8/1/97 John Rickard
8/4/97 Bill Taylor
8/5/97 John Rickard
7/25/97 Charles H. Giffen
7/25/97 Charles H. Giffen
7/28/97 Hauke Reddmann
7/28/97 Robert Hill
7/28/97 Robert Hill
7/28/97 Robert Hill
7/29/97 tony richards
7/30/97 Keith Ramsay
7/30/97 tony richards
8/2/97 Keith Ramsay
7/29/97 tony richards
8/4/97 Bill Taylor
8/5/97 Charles H. Giffen
8/6/97 Terry Moore
8/7/97 Terry Moore
8/16/97 Kevin Brown
8/17/97 Kevin Brown
7/30/97 Robert Hill
7/31/97 tony richards
8/6/97 Terry Moore
7/31/97 John Rickard
7/30/97 Robert Hill
7/31/97 Robert Hill
7/31/97 Robert Hill
8/1/97 R J Morris
8/4/97 Robert Hill
8/4/97 Robert Hill
8/5/97 Charles H. Giffen
8/6/97 Robert Hill