Math Quiz 9

[Ether]

Originally Posted by Hitchhiker 42

What is a Monte Carlo simulation?

Aren beat me to it.

One-line AWK script:

BEGIN{for(i=1;i<100000;i++){sum+=sqrt((rand()-rand())^2+(rand()-rand())^2);}print sum/i;}

[Hitchhiker 42]

I ran the Monte Carlo thing in a python script and the answers, each run with 1,000,000 trials, were all around 0.333. I'm gonna keep thinking, though, how to get an exact answer.

Originally Posted by Ether

Your script has an error.

Compare it to the one-line AWK script I posted.

EDIT: I found the error - the value comes out to 0.521

[FiMFanatic]

I am thinking the answer is a number that is very small (close to 0), as you can draw a ton more 0.000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 01 lines than you can 1 inch (or larger on a diagonal) lines.

[Hitchhiker 42]

Originally Posted by FiMFanatic

I am thinking the answer is a number that is very small (close to 0), as you can draw a ton more 0.000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 01 lines than you can 1 inch (or larger on a diagonal) lines.

Yet if you look at the Monte Carlo simulation, it looks closer to 0.521"

[z_beeblebrox]

My simple Python Monte Carlo script gave me an average of 0.521408 (or 0.52141 rounded to 5 digits) in ~5e9 iterations (10 miles of hiking worth).

Code is below:

import numpy as np

iterations = 10000000

avg = 0

for i in range(100000):
    for i in range(iterations):
        pos = np.random.rand(4)
        length = np.sqrt((pos[0]-pos[1])**2+(pos[2]-pos[3])**2)
        avg = (avg*i + length)/(i+1)
    with open("test.txt", "a") as f:
        f.write(str(avg)+'\n')

import numpy as np

f = np.loadtxt('test.txt')
print(np.average(f))
print(len(f))

I split it into two scripts just so I could stop the computation whenever without altering the result.

Edit: Reps to whoever finds the bug in my code and explains what it does.

[Ether]

Originally Posted by z_beeblebrox

0.52141 rounded to 5 digits

That is the correct answer. Reps to you

[Ether]

Originally Posted by Ether

That is the correct answer. Reps to you

Now, how would you get the correct answer accurate to, let's say, 8 decimal places?

[FiMFanatic]

Makes no logical sense unless the question is improperly worded or bounded. You can fit an infinite number of lines in the box, and fact is, more lines closer to 0 in length fit in the box than lines averaging 0.52xxxx

[z_beeblebrox]

Originally Posted by FiMFanatic

Makes no logical sense unless the question is improperly worded or bounded. You can fit an infinite number of lines in the box, and fact is, more lines closer to 0 in length fit in the box than lines averaging 0.52xxxx

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?"

[FiMFanatic]

Originally Posted by z_beeblebrox

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?

[z_beeblebrox]

Originally Posted by FiMFanatic

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?

If the problem is interpreted as you describe (What is the average length of all the line segments that could be packed into a 1" square at once without overlapping?), there would be no meaningful answer, since infinite line segments of any length l < sqrt(2)" could be fit within the square, as line segments have no thickness thus can be packed infinitely densely.

[FiMFanatic]

Exactly - hence why people struggled to answer I think......

[Ether]

Originally Posted by FiMFanatic

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

Originally Posted by FiMFanatic

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

Originally Posted by FiMFanatic

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?

[GeeTwo]

Originally Posted by Ether

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

Whew! I've been torn among several ideas:

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

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

[Caleb Sykes]

Originally Posted by FiMFanatic

Makes no logical sense unless the question is improperly worded or bounded. You can fit an infinite number of lines in the box, and fact is, more lines closer to 0 in length fit in the box than lines averaging 0.52xxxx

There are uncountably infinite lines of length 0.01, just as there are an uncountably infinite number of lines of length 0.52.

It makes no logical sense to say that there are "more" or "less" of one uncountably infinite thing than another uncountably infinite thing.