The h Bar

Anonymous

10:15 AM

@Secret I was reading through the transcript. Tbh I find the concept of soulmate very naive and illogical.

Sure, all real life relationships need some degree of understanding and balance between both parties, to accept the weakeness of the personality of the other while to adjust oneself to achieve mutaul understanding

This is why maintain a relationship takes a lot of effort, but it is worth the trouble

Anonymous

@Secret Right, I will agree to that. It is impossible that you'll ever find a perfect partner in the absolute sense of the word.

Secret

and of course, there are a lot of things outsiders like me don't really understood outside a relationship status, these will likely become clearer to me later on...

That's true, but anyway, I am probably way too lazy to start a relationship with anyone else, so (insert phrase to complete the sentence)

Anonymous

Same here :D

ACuriousMind

10:32 AM

Sometimes I get the feeling Randall is not actually a fan of physicists :P

Anonymous

@Kaumudi.H Oh, don't worry. They are young adults blinded by their oxytocin levels. :P If I were in your place I'd just go and enjoy the movie without giving a damn about the people beside me. :D

Dawood ibn Kareem

@ACuriousMind Haha! "I said brains. All they've got are string theorists". I love xkcd.

ACuriousMind

Luckily for you, this string theorist also finds that funny ;P

Dawood ibn Kareem

Why is that lucky for me?

ACuriousMind

Well, I guess it might not matter to you if you don't care about insulting me...

John Rennie

10:45 AM

@ACuriousMind he's just quoting from the XKCD comic you linked ...

Yashas

How do I check if a point (x,y, z) is inside a cube (8 corners are given)?

John Rennie

In my days working for Unilever Research I knew a chap who spent his whole career working on the physics of granular flows.

ACuriousMind

@JohnRennie I know! But that doesn't make the statement less potentially insulting (and I actually know at least one guy who doesn't like xkcd because of that strip, which seems petty to me, but...whatever)

John Rennie

@Yashas a randomly oriented cube or a cube with the edges aligned with the axes?

Yashas

@JohnRennie random

actually it isn't a cube, it is a cuboid

John Rennie

10:47 AM

I'd probably rotate the axes to line up with the cuboid. Once you've done that the problem is trivial.

Yashas

no easy way? :(

I was looking for something to do with distances

distance b/w the point and every corner is less than the length body diagonal?

John Rennie

Well I suppose you can measure the normal distance to each of the six faces ...

Yashas

I need to write code for it

distance looks simple :p

and usually other people won't understand if I use too much of math

Secret

Anonymous

@Yashas That should be simple if the coordinates of vertices are given

Secret

10:50 AM

Distance is insufficient. For this example, all absolute value of distances from the corners are less than the body diagonal

Yashas

ah

ACuriousMind

I think the two ways John has already proposed are the "simplest".

@Secret You need to use signed distance.

I.e. set up all the normal vectors either pointing into or out of the cube, then you can tell by the signs of the distances whether the point is inside or outside.

Anonymous

I'd first of all rotate the axes so that they align with the sides of the cube and put one vertex to (0,0,0)

Anonymous

I guess JR suggested that

Yashas

@Blue the code is going to look dirty and trig functions are expensive operations

Anonymous

10:53 AM

@Yashas Could you tell me what your code is going to input?

Anonymous

Will it input the vertices?

Yashas

@Blue 8 corners

even though 8 aren't needed to define a cuboid, u get all 8 of them

Secret

Acuriousmind's suggestions of normal vectors might help you to avoid the trig stuff

ACuriousMind

@Yashas that's why I'd go with the signed distance - you pre-compute the plane equations for the cube's faces once, and then you just have 6 linear algebra operations, which should be fairly inexpensive.

Anonymous

Another way could be that if the equations of two opposite planes are $ax+by+cz=d_1$ and $ax+by+cz=d_2$ then the point should satisfy one of the planes whose form is $ax+by+cz=d_3$ where $d_1<d_3<d_2$

Anonymous

10:56 AM

Similarly for two other sides

John Rennie

It does feel as if there ought to be some cunning method involving the distances to the vertices ...

Anonymous

That's one alternative to the distance method which ACM sugested

Yashas

@JohnRennie yea exactly :D

ACuriousMind

@JohnRennie Nope, the distances to the vertices just constrain the particle to spheres around the verticles, and intersections of spheres don't give a cube.

Secret

I have no idea how we can make use of inequalities without computing the equations of the planes

Yashas

10:58 AM

@Blue finding equations and playing with them isn't easy coding-wise

Secret

This problem is one of the examples where we humans are good at finding the solution (we can literally tell by inspection whether the point will be in the cuboid), but computers need to take a few steps. Somehow computers cannot "see the full picture"...?

Anonymous

@Yashas I can think of only two methods as of now. I'll let you know if I get something easier later on. BTW the method I suggested has only 3 condition checks.

ACuriousMind

@Secret I think that commits the fundamental error of implicitly assuming that "tell by inspection" is somehow less computationally intensive than a few linear algebra operations.

Anonymous

This ^ :)

Secret

That's true, but is sometimes confuse me, how and what type of calculations are our brains doing to do all these inspection type solutions

ACuriousMind

11:02 AM

@Secret Another fundamental error: Assuming our brain does "calculations" :P

Secret

Everytime in an unfamilar field when I saw people solve problems by inspection, I always thought "magic happens" and I then spent a lot of time to no avail to understand why and how that magic happens

but when I finally get used to the field, I stop questioning about it, but deep down, why the magic happens is never resolved, its just ignored, almost subconsciously

Anonymous

@Yashas Okay, I found an easier method using dot products.

Anonymous

2

A: Check if a point is inside a rectangular shaped area (3D)?

The three important directions are $u=P_1-P_2$, $v=P_1-P_4$ and $w=P_1-P_5$. They are three perpendicular edges of the rectangular box. A point $x$ lies within the box when the three following constraints are respected: The dot product $u.x$ is between $u.P_1$ and $u.P_2$ The dot product $v...

Secret

and it will take events like these (somebody else ask about a related question) to force the brain to think about the question it silently ignored

I think I really have a low tolerance for unexplained things

Anonymous

0

A: Check if a point is inside a rectangular shaped area (3D)?

Given $p_1,p_2,p_4,p_5$ vertices of your cuboid, and $p_v$ the point to test for intersection with the cuboid, compute: $$\begin{matrix} i=p_2-p_1\\ j=p_4-p_1\\ k=p_5-p_1\\ v=p_v-p_1\\ \end{matrix}$$ then, if $$\begin{matrix} 0<v\cdot i<i\cdot i\\ 0<v\cdot j<j\cdot j\\ 0<v\cdot k<k\cdot k \end...

ACuriousMind

11:08 AM

@Blue Basically the same as the plane approach, since computing these three vectors is the same as computing the normal vectors to the faces.

Anonymous

@ACuriousMind Yup, but this will be a bit shorter if you are going to write a code. Finding equation of planes and dealing with them will be more difficult.

Anonymous

But, yes. They are essentially the same

Dawood ibn Kareem

We can "tell by inspection" only if the point isn't close to one of the faces.

ACuriousMind

@Blue Well, if we're going to try to make efficient code we'd have to know what language he's programming in, whether it has good native support of vectors or not, etc... Short code is not always best code.

Dawood ibn Kareem

It does feel like there ought to be an easy way though. I think you subtract the co-ordinates of the point from each of the eight vertices, then check whether there's one vertex in every octant.

But I don't know how to prove that it's correct.

Anonymous

11:12 AM

@ACuriousMind Oh. I assumed he is using something like C++ or Java. :) Anyhow, let him decide for himself :D

Secret

4

Q: Can our brain do trigonometry?

I was sitting in college today today doodling around as I was bored, when I drew an arrow onto my paper, Looking at the arrow, I predicted that the arrow would move in the direction it was facing. Then, I thought "How did I know it would move in that direction?". Being a programmer/game developer...