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Re: [2013-03-26] 2927

Posted: Tue Apr 02, 2013 8:44 pm
by Hentron
Any line on the graph, any line, straight or curved, presuming that one side of the line (either side) is filled in and considered "not empty" would be correct. On the other hand, if the width of the actual line is considered what is actually filled in the only way it could be accurate would be if the line was one dimensional and had no width, and then the graph, to be accurately drawn would be a line so thin it is invisible on the x axis.

Re: [2013-03-26] 2927

Posted: Tue Apr 02, 2013 9:07 pm
by DonRetrasado
augh why are we still having this conversation

Re: [2013-03-26] 2927

Posted: Tue Apr 02, 2013 9:49 pm
by ThePeople
I present to you pig poo balls

.GIF

Image

Re: [2013-03-26] 2927

Posted: Tue Apr 02, 2013 11:40 pm
by Lethal Interjection
I... just... can't... look away.

Re: [2013-03-26] 2927

Posted: Wed Apr 03, 2013 11:10 am
by Apocalyptus
It's almost like these commenters don't even care whether people actually want to hear what they're saying, rather they gain some sort of perverse satisfaction in just achieving a certain level of pedantry and putting it in writing on the internet. I have no other explanation for this to keep happening despite the obvious hostility that is usually the reward of such a comment.

Re: [2013-03-26] 2927

Posted: Wed Apr 03, 2013 4:38 pm
by Sahan
Having basic social and conversational skills is so last decade Apoc, get with the times!

Seriously, I've said this before, but reading these discussion threads is like having people drive past your house and shout things at you. What they say warrants a response, but you know it's pointless because they'll never hear your reply and you know they didn't want to discuss anything anyway, they just want to shout. And then another one drives by and shouts the exact same thing. It's maddening.

Re: [2013-03-26] 2927

Posted: Thu Apr 04, 2013 2:26 am
by graaahh
Isn't it just a negative slope?

Re: [2013-03-26] 2927

Posted: Thu Apr 04, 2013 2:45 pm
by Sahan
We've seriously gotten to the point where you can get a better conversation out of spambots than most of the unregistered posters.

Re: [2013-03-26] 2927

Posted: Fri Apr 05, 2013 7:58 am
by Apocalyptus
We're just going to have to accept the fact that this is the fandom that Zach seems to be cultivating now. That, or post pooballs into infinity.

I know what my choice is.

Re: [2013-03-26] 2927

Posted: Fri Apr 05, 2013 11:47 am
by Sahan
The pooballs posting has really come along really nicely, I must admit, but I can only see it getting better from here. Soon we will have a golden age, and patrons of the arts will come from all over to gaze at the wonder that is a turd resting gently on a pig's gigantic balls.

Re: [2013-03-26] 2927

Posted: Sat Apr 06, 2013 1:46 am
by ThePeople
I'm working on something that should be pretty good. I'll finish it if we truly need it.

Re: [2013-Mar-26] 2927

Posted: Sun Feb 25, 2018 5:11 am
by ribbonmaster
I believe I have the answer. I was surprised to go through four pages of comments and not see a definitive answer, and instead it was full of pig poop.

My interpretation of the problem: draw a line and fill below it with black. If you truncate the graph at any width, the width = %black of the graph that is visible (don't truncate the y value, keep the height the same).

A simple summary: the solution is any function ax^b
a, b must be positive
plot under the interval 0 to xmax, where xmax = exp(ln(1/(1+b)/a)/b).
plot with the height from 0 to ymax, where ymax = 1/(1+b)

Example 1: 1*x., plot from 0 to 0.5 in x and 0 to 0.5 in y.
Taking the entire x range, it makes a triangle so 50% fill. And the x value at the right is 50%. Works.
Trying one intermediate value: width = 0.1. Then it makes a smaller triangle with an area 0.1*0.1*1/2. The total area is just width * height = 0.1*0.5. Dividing the black area by the total area gives 10%, same as x=0.1.

Example 2: 1*x^2, plot from 0 to sqrt(1/3) in x and 0 to 1/3 in y.
I'll just show the right hand value: width=sqrt(1/3).
Finding the area in the black using the integral, got 1/3^(5/2).
Total area = width * height = 1/3^(3/2).
Dividing the black by the total area gives 1/3, same as width = 1/3. Works!

Derivation: start with the rule: f(w) = int_0^w f(x) / w / h. That is just stating width = black/total area.
Rearrange h*w*f(w) = int_0^w f(x)
Realize that this is exactly how polynomials integrate.
Derive rules for the appropriate xmax and height in terms of a and b, pretty simple algebra.

Re: [2013-Mar-26] 2927

Posted: Thu Mar 08, 2018 4:31 pm
by NeutralSamaritan
Image

You're welcome. *disappears coughing in a smoke bomb explosion*

Re: [2013-Mar-26] 2927

Posted: Fri Mar 09, 2018 5:25 am
by Jack
NeutralSamaritan wrote:Image

You're welcome. *disappears coughing in a smoke bomb explosion*
"I just wanted to talk about your poor smoke-bomb techniques"

Re: [2013-Mar-26] 2927

Posted: Wed Mar 14, 2018 6:01 pm
by unnecessary answerer
I think any plot-answer is only as valid as the explanation that goes with it. As many have pointed out, there are multiple ways to define emptiness. This includes self-referencing or not. Also, there are multiple ways to include data in the plot. Do you take a line that is filled below? do you take data points? crosses? or a violin plot?

Personally I would define emptiness as "the amount of white of the plot as the fraction of the total area left of x", with the plotting being a single line (without filling below it). then at x=inf the plot is probably 98% empty (depending on line width), because it will just be a line and an axis surrounded by whitespace. At x = 0 it would probably still be around 70% empty, as the axis description doesn't fill much of the void. It will rather quickly go to ~90% while there is some text below the abscissa. After the text ends it will go to the aforementioned 98%.

As to how the slopes look like: i would go for a-exp(-k*x) type of slopes, to get the nice asymptotes. But since it is plotted, you would not see a difference between them and 1/x^2 for example.