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bonanova
Daedalian Member

 Posted: Tue Nov 27, 2012 5:01 pm    Post subject: 1 What is the average length of a random chord constructed on a unit circle? I'm looking for creative definitions of random that produce the greatest and the smallest answers._________________ Vidi, vici, veni.
MNOWAX
0.999... of a Troll

 Posted: Tue Nov 27, 2012 6:12 pm    Post subject: 2 A unit. why? It's halfway between 0 and 2 units. Go ahead. disprove that. _________________The Man The Myth The Legend MNOWAX
The Potter
Feat of Clay

 Posted: Tue Nov 27, 2012 8:35 pm    Post subject: 3 Funny... I was thinking sqrt(2) because it is halfway._________________Artwork | Fractals | Don't ignore your dreams; don't work too much; say what you think; cultivate friendships; be happy.
Zag
Unintentionally offensive old coot

 Posted: Tue Nov 27, 2012 8:56 pm    Post subject: 4 Creative definition of random: 1. Choose any point on the circle. 2. Choose any point on the intersection of the circle and the line that passes through the first point and the center of the circle, and isn't coincident with the first point. My average chord is pi! edit: I meant 2, of course. pi would be the arc. Do I win for the random function that produces the maximum?Last edited by Zag on Tue Nov 27, 2012 10:47 pm; edited 2 times in total
The Potter
Feat of Clay

 Posted: Tue Nov 27, 2012 9:09 pm    Post subject: 5 chord length as a function of theta is 2sin(theta/2). If we integrate over the range of 0 to 2π -4cos(theta/2) |(2π and 0) = 8 For an average of 8/2π =1.2732_________________Artwork | Fractals | Don't ignore your dreams; don't work too much; say what you think; cultivate friendships; be happy.
Zag
Unintentionally offensive old coot

 Posted: Tue Nov 27, 2012 10:57 pm    Post subject: 6 You've made an assumption about what "a random chord" means: that you choose two points on the circle where every point on the circle has an equal chance of being selected. I realize that my definition of "a random chord" above was silly, but I do think it would be reasonable to define it as: Choose two points within a circle such that every point in the circle has equal probability of being selected, and draw the line which connects them. Where that line intersects the circle defines the chord. I don't know what the average length will be, but I can trivially show that it is longer than the one you've chosen. The random function you chose is the same as saying, "Choose two points within a circle such that every point in the circle has equal probability of being selected, draw the two radii which pass through these two points. The two points where the radii intersect the circle define the chord." Note that this chord is always smaller than the one that my random function makes, given the same two points within the circle selected.
The Potter
Feat of Clay

 Posted: Wed Nov 28, 2012 12:36 am    Post subject: 7 As far as I know he was seeking a cord on the unit circle. Lets pretend there are just 10 points equally spaced around the circle. The first point will be at 0 (we will define it that way). The second point is equally likely to fall on any of 0 to 9. There is no reason for me believe any number is more likely then the next. Then the average cord length can be found by summing the lengths and dividing by the number of points. With 10 discrete points the average cord length will be 1.2628. 0 0.618033988749895 1.17557050458495 1.61803398874990 1.90211303259031 2 1.90211303259031 1.61803398874990 1.17557050458495 0.618033988749895_________________Artwork | Fractals | Don't ignore your dreams; don't work too much; say what you think; cultivate friendships; be happy.
Thok
Oh, foe, the cursed teeth!

 Posted: Wed Nov 28, 2012 12:46 am    Post subject: 8 The minimum is 0. This uses the xkcd random function: randomly choose a point, and then whenever somebody asks for a chord, pick the chord connecting that point to itself.
Zag
Unintentionally offensive old coot

 Posted: Wed Nov 28, 2012 2:04 am    Post subject: 9 The Potter, you've missed my point entirely. I agree that your integral solution was correct, for the problem that you were solving. However, I don't agree that you're solving the right problem. The definition of 'a random chord on a unit circle' is not as clear-cut as you seem to think. I could say, "choose a random integer from 0 to 10" and you might assume that I meant to choose such that each of the 11 integers has the same chance of being selected. But you satisfy the requirements by rolling two dice and subtracting 2 from the result, making a result of 5 far more likely than 0 or 10. Similarly, there are many ways to choose your random chord, and the different selection processes will yield different results. Some of these selection processes (like my first one and like Thok's) are ridiculous if you want to maintain the term "random" but some are perfectly reasonable.
The Potter
Feat of Clay

 Posted: Wed Nov 28, 2012 2:51 am    Post subject: 10 The chord is constructed on the unit circle... So both the points we care about have to fall on the unit circle. How far apart they are is only a function of the angle they form with the center point._________________Artwork | Fractals | Don't ignore your dreams; don't work too much; say what you think; cultivate friendships; be happy.
The Potter
Feat of Clay

 Posted: Wed Nov 28, 2012 3:01 am    Post subject: 11 Ah, I see what you are saying now. But I don't feel that is a very good way of (I guess that is part of the point I missed) cord construction for random numbers. A picture will clear things up _________________Artwork | Fractals | Don't ignore your dreams; don't work too much; say what you think; cultivate friendships; be happy.
bonanova
Daedalian Member

Posted: Wed Nov 28, 2012 5:13 am    Post subject: 12

 Thok wrote: The minimum is 0. This uses the xkcd random function: randomly choose a point, and then whenever somebody asks for a chord, pick the chord connecting that point to itself.

I came up with a zero average chord length by another method. Inscribe a regular polygon of n sides (n>1) and take one of the sides as the chord. So that n=2 gives the diameter, e.g. If n is chosen at random, the mean value is zero. I think.
_________________
Vidi, vici, veni.
LordKinbote
Daedalian Member

Posted: Wed Nov 28, 2012 5:45 am    Post subject: 13

bonanova wrote:
 Thok wrote: The minimum is 0. This uses the xkcd random function: randomly choose a point, and then whenever somebody asks for a chord, pick the chord connecting that point to itself.

I came up with a zero average chord length by another method. Inscribe a regular polygon of n sides (n>1) and take one of the sides as the chord. So that n=2 gives the diameter, e.g. If n is chosen at random, the mean value is zero. I think.

Okay, then. Inscribe any right triangle. Take the hypotenuse as your chord.
L'lanmal
Daedalian Member

 Posted: Wed Nov 28, 2012 12:19 pm    Post subject: 14 Consider any method of choosing a random chord on a circle, where every chord has a non-zero probability of being selected. Observe that this has a (not necessarily well-behaved) density function over the space of chords in a circle. Now consider some method of choosing a random chord on a circle (every chord has a non-zero probability of being selected) whose pdf is equal to the original scaled by the chord length of each outcome and renormalized. Alternatively, scale by (2 - chord length). Iterate. It seems like you can generate methods (where every chord has a non-zero probability of being selected) which have an average chord length arbitrarily close to 2. Alternatively, 0. I'm guessing you'd want one that is easily described.
L'lanmal
Daedalian Member

 Posted: Wed Nov 28, 2012 12:27 pm    Post subject: 15 Obviously individual chords can have a 0 probability of being selected. I meant that distribution function is non-zero on that chord.
bonanova
Daedalian Member

Posted: Wed Nov 28, 2012 1:59 pm    Post subject: 16

 L'lanmal wrote: Obviously individual chords can have a 0 probability of being selected. I meant that distribution function is non-zero on that chord.

My zero-mean method is hokey in that regard: it samples only countably infinite chords (if one node is fixed.) Thok's space with one point fixed is even smaller(!) LK's maximal length method is much superior. Even when the right-angle node is fixed, it samples an uncountably infinite set of diameters. Kudos.
_________________
Vidi, vici, veni.
Nsof
Daedalian Member

 Posted: Fri Nov 30, 2012 2:34 pm    Post subject: 17 removed a non creative idea_________________Will sell this place for beer
Jack_Ian
Big Endian

 Posted: Fri Nov 30, 2012 2:59 pm    Post subject: 18 Choose two chords A & B. Toss a coin to select the appropriate chord. Select initial chords depending upon whether you want a large or small final average.
Zag
Unintentionally offensive old coot

 Posted: Fri Nov 30, 2012 5:42 pm    Post subject: 19 Ah ha! I've just realized that I can win this game easily, at least for longest average chord length. Step 1. Choose any process for randomizing which chord you choose such that there is a non-zero chance of choosing a diameter. Step 2. Draw your circle around a black hole. Your average chord length is infinity!
Nsof
Daedalian Member

 Posted: Sat Dec 01, 2012 9:59 pm    Post subject: 20 choose a point outside the circle. there are two tangent lines from that point to the circle. the "random" chord is the one connecting the two points where the tangent lines intersect the circle. depending on how you "randomly" choose that point, you can get different average results. e.g.: if unit circle is centered on origin and point is defined as (n,0) where n, an integer, is uniformaly taken from [1,1000] the average result would be close to two._________________Will sell this place for beer
groza528
No Place Like Home

Posted: Sun Dec 02, 2012 5:04 am    Post subject: 21

 Nsof wrote: depending on how you "randomly" choose that point, you can get different average results. e.g.: if unit circle is centered on origin and point is defined as (n,0) where n, an integer, is uniformaly taken from [1,1000] the average result would be close to two.

I think the "most random" way to do it would be to not limit the choice at all; i.e. "Choose any exterior point sharing the plane of the circle."
The farther you get from the center, the closer you get to 2, and the higher that distance gets the more likely you will choose a point at that distance.
Amb
Amb the Hitched.

 Posted: Sun Dec 02, 2012 8:44 am    Post subject: 22 What happens if you draw a circle on a sphere? Then either side of the line, is "inside" the circle depending on your perspective. Could you think shrink the circle to small, and increase the sphere to infinite? Then you could get interesting large and small chords?

Posted: Sun Dec 02, 2012 9:39 am    Post subject: 23

 Amb wrote: What happens if you draw a circle on a sphere? Then either side of the line, is "inside" the circle depending on your perspective. Could you think shrink the circle to small, and increase the sphere to infinite? Then you could get interesting large and small chords?
How many lines can be drawn from a point outside the circle, tangent to the circle, if you are on a sphere? I don't think it's 2.
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