Chief Delphi - Math Quiz 9

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Quote:

Originally Posted by Ether (Post 1602306)

The difference is in how you choose samples.

If you want the average distance between all unordered pairs of points, then you choose coordinates of pairs of points from a uniform random distribution, and compute the corresponding distance.

If you want the average length of all segments, you randomly choose a segment length, a segment orientation, and the coordinates of the center of the segment. Then you discard any chosen segments which are not inside the square. If you run a Monte Carlo sim of that, you'll get 0.3363

I believe this is fallacious, in particular that for a given center point and two different segment lengths, there are the same number of segments. For the center point at (0.5, 0.5) and length 0.2, the candidate segments are the diameters of a circle of radius 0.1. For the center point at (0.5, 0.5) and length 1.0, the candidate segments are the diameters of a circle of radius 0.5. As there are five times as many points on the large circle to serve as either end of a diameter, it logically has five times as many diameters.

When you take this into account, the integral in the denominator becomes:

∫₀^π/4∫₀^1/cosθ (1- Lcosθ) (1- Lsinθ) L dL dθ =
∫₀^π/4∫₀^1/cosθ L - L²cosθ - L²sinθ + L³cosθsinθ dL dθ =
∫₀^π/4[ L²/2 - L³cosθ/3 - L³sinθ/3 + L⁴cosθsinθ/4 ]₀^1/cosθ dθ =
∫₀^π/4sec²θ/2 - sec²θ/3 - sec²θtanθ/3 + sec²θtanθ/4 dθ =
∫₀^π/4sec²θ/6 - sec²θtanθ/12 dθ =
[tanθ/6 - tan²θ/24 ]₀^π/4 =
[1/6 - 0 - 1/24 + 0 ] = 1/8

If you then integrate the adjusted calculation using 1/8 for the deominator, just substitute r for L and α for θ, and you will have the third equation of my most recent post in this thread.

Addition: Also, the idea of the average segment in a square of size 1 being within 1% of the average segment on any of its sides is .. counter-intuitive, to say the least.

For the center point at (0.1, 0.1) and length 0.2, the candidate segments are the diameters of a circle of radius 0.1. For the center point at (0.1, 0.1) and length 1.0, the number of candidate segments is zero.

Quote:

Originally Posted by Ether (Post 1602327)

Though it is not relevant, there are plenty of candidates, though none are selected.

Here's another case which more clearly shows the fallacy.

What is the average distance from the origin to a point in the unit square?
What is the average length of the radials from the origin to points in the first quadrant of a circle of radius √2?

Solving the first question in Cartesian coordinates, we find the answer to be (√2 + ln(1+√2))/3 = 0.765...

Solving the second question using the logic used by Ether to count radials, the answer comes out to √2/2 = 0.707..

As the quadrant of the large circle includes all the same points as the unit square plus others which area all farther away than either average, there is a fallacy here somewhere.

By weighting the density by r, the second question is properly answered 2√2//3 = .942...

Addition:
To more fully explore this point, I propose the following questions:

What is the average length of a segment in a rectangle 1 unit wide and 0.5 units high?
What is the average length of a segment in a rectangle 1 unit wide and 0.1 units high?
What is the average length of a segment in a rectangle 1 unit wide and infinitesimally high? (that is, the limit as height goes to zero from above)
How does that compare to the average length of a line segment within the line segment from 0 to 1?

Question for Gus:

What is the average length of all the vertical line segments in a unit circle?

Quote:

Originally Posted by Ether (Post 1602385)

Question for Gus:

What is the average length of all the vertical line segments in a unit circle?

If I may,

Spoiler for Thoughts:

Work attached below.

EDIT: Gosh darn I only did those with both points touching the the edges. Lemme work on that.

Quote:

Originally Posted by Hitchhiker 42 (Post 1602388)

EDIT: Gosh darn I only did those with both points touching the the edges. Lemme work on that.

Also, don't forget to weight by the number of segments - the x coordinates near the center contain more segments than those near the left and right edges.

In any case, to answer Ether's question, the short answer is "not the same as the average length of segments within a small angle (e.g. 1 second of arc) of vertical within the circle." We've got another build session this evening getting ready for Red Stick Rumble, so I don't know how much I'll get done on this tonight.

Addition:
Hitchhiker, thanks for answering a different question. While walking across the street to get lunch, I think I put some nails in this coffin. Consider these two questions:

What is the average length of a vertical chord of the unit circle?
What is the average length of a chord of the unit circle which passes through (-1,0)?

You answered the first - the area of the unit circle divided by its width: pi/2.
I answered the second in my paper a score or so posts back (first problem), in a form that is not incompatible with Ether's method, provided angle is checked before length: 4/pi.
If you answer the second using Ether's method but checking length first, then picking one of two possible angles, you will get an average value of 1.

Completing reducto ad absurdum is left to the reader - back to work!

Quote:

Originally Posted by Ether (Post 1602392)

I just can't trip you up, can I :)

Well, not that easily.

Quote:

Originally Posted by GeeTwo (Post 1602391)

In any case, to answer Ether's question, the short answer is "not the same as the average length of segments within a small angle (e.g. 1 second of arc) of vertical within the circle."

I just can't trip you up, can I :)

Quote:

Originally Posted by FiMFanatic (Post 1597038)

...you can draw a ton more 0.000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 01 lines than you can 1 inch (or larger on a diagonal) lines.

Quote:

Originally Posted by FiMFanatic (Post 1597049)

...fact is, more lines closer to 0 in length fit in the box than lines averaging 0.52xxxx

Quote:

Originally Posted by FiMFanatic (Post 1597051)

Quote:

Originally Posted by z_beeblebrox (Post 1597050)

I think the problem is more "Choose two random points inside by a 1" square and measure the distance between them. Repeat infinitely. What's the average of those measurements?"

I agree then - your answer would make sense if that were the question. However, it was:

What's the average length of all the line segments which can be drawn inside a 1 inch square?

@FimFanatic: Are you still reading this thread? I've done the best "devil's advocate" job I can arguing your point. Gus has answered every challenge. Has this discussion been helpful for you?

Quote:

Originally Posted by Ether (Post 1602408)

I've done the best "devil's advocate" job I can arguing your point.

Whew! I've been torn among several ideas:

Ether needs coffee.
Ether has started trolling.
Ether's CD account has been hijacked by a technically competent troll.
Ether is up to a deep teaching moment.

I'm so glad to read that it was #4!

I'm not sure why CD wouldn't let me attach a file to a post on this thread, but I've posted my solutions to the cube at this "white paper": https://www.chiefdelphi.com/media/papers/3289.

For any who find this white paper a bit too much, let me simply note here that you don't want to see all of my blind alleys and solutions which worked but were much uglier than what I've shown here.

Edit: And OBTW, if anyone feels that my substitutions showed a bit too much prescience, that's ok, because I'm posting this on my birthday, and we should always expect prescience on our birthdays, no?