Is there a way to simplify $\Big|\;|x+y|+|x-y|+z\;\Big|+\Big|\;|x+y|+|x-y|-z\;\Big|=r$?

Question

Is there a way to simplify this $$\Big|\;|x+y|+|x-y|+z\;\Big|+\Big|\;|x+y|+|x-y|-z\;\Big|=r$$

so that it turns out something like: $$|ax+by+cz|+|dx+ey+fz|+\cdots=ur$$
where $a$, $b$, $c$, ..., $u$ is a constant?

Or is there a way to at least simplify it so that there is only 1 layer deep of abs() function?

Side notes: This is for a fast collision detection of a AABB and a ray. If you graph the formula above (with r related to the size of the box), it will show a 3d box. I have used the same method to make a fast 2d box collision detection but have no idea how to make it 3d as I was stuck on this step...

Edit: Yes, I know I can use max(a,b,c) =ur to discribe this. And I'm turning it around so that I can find all the "break" points (should be 8 of them??) of this function:

$$y=\Big|\;|P_x+P_y|+|P_x-P_y|+P_z\;\Big|+\Big|\;|P_x+P_y|+|P_x-P_y|-P_z\;\Big|$$ $$where$$ $$P=(V_1+(V_2-V_1)x)$$ (V1 and V2 is a given 3D vector.)

so that I can calculate quickly what x has to be so that the function above return the smallest value possible.

(Also, everything is real number. Don't want to deal with complex number today...)

Actually... should I just ask what the break point is...Errrr... Sorry. New user here.

Answer 1

I would suggest working through different cases -

i) For $x \ge y \ge 0, x \ge \frac{|z|}{2},\Big|\;|x+y|+|x-y|+z\;\Big|+\Big|\;|x+y|+|x-y|-z\;\Big|=r$ becomes (where $r \ge 0$)

$2x + z + 2x - z = r, x = \frac{r}{4}$

ii) For $x \ge y \ge 0, x \lt \frac{|z|}{2}$, $2x + z - (2x - z) = r, z = \pm \frac{r}{2}$

iii) For $y \ge x \ge 0, y \ge \frac{|z|}{2}$, $4y = r, y = \frac{r}{4}$

iv) For $y \ge x \ge 0, y \lt \frac{|z|}{2}$, $2z = r, z = \pm \frac{r}{2}$

Checking all other cases, the final set of equations become -

i) For $|x| \ge |y|, |x| \ge \frac{|z|}{2}, x = \pm \frac{r}{4}$

ii) For $|x| \ge |y|, |x| \lt \frac{|z|}{2}$, $z = \pm \frac{r}{2}$

iii) For $|y| \ge |x|, y \ge \frac{|z|}{2}$, $y = \pm \frac{r}{4}$

iv) For $|y| \ge |x|, |y| \lt \frac{|z|}{2}$, $z = \pm \frac{r}{2}$

So you get $6$ planes, $x = \pm \frac{r}{4}, y = \pm \frac{r}{4}, z = \pm \frac{r}{2}$. It is obvious what the shape of the bound region would be.

Answer 2

There is a way to remove both layers of absolute values, but I wouldn't consider it a simplification since they just get replaced by two layers of the $\max(x,y)$ function.

The maximum of $x$ and $y$ is given by the formula $\max(x,y)=\frac{x+y+|x-y|}{2}$, so we have $$2\max(x,y)+2\max(x,-y)=x+y+|x-y|+x-y+|x+y|=2x+|x-y|+|x+y|\Rightarrow\\ |x+y|+|x-y|=2(\max(x,y)+\max(x,-y)-x),$$ so now the term $||x+y|-|x-y|\pm z|$ can be replaced with $$|\max(x,y)+max(x,-y)-x\pm\frac 12z|$$ and the right-hand side has to be divided by $2$.

To remove the second layer of the absolute value, let $a=\max(x,y)+max(x,-y)-x$. Now we have $$|a+\frac 12z|+|a-\frac12 z|,$$ so the same as above can be applied with $\max(a,\frac12 z)$ and $\max(a,-\frac 12 z)$.

Answer 3

I claim that for all $a, b \in \mathbb{R}$, we have

$$\frac{|a+b| + |a-b|}{2} = \max(|a|,|b|)$$

To see that, let's denote $f(a,b) = \frac{|a+b| + |a-b|}{2}$, and separate 3 cases : $a < -|b|$, $-|b| \le a \le |b|$ and $a > |b|$ :

1. If $a < - |b|$, then $a+|b| < 0$ and $a - |b|< -2|b| < 0$ so $f(a,b) = f(a,|b|) = \frac{-(a+|b|) - a + |b|}{2} = -a = \max(|a|,|b|)$.

2. If $-|b| \le a \le |b|$, then $a + |b| \ge 0$ and $a - |b|\le 0$ so $f(a,b) = f(a,|b|) = \frac{(a+|b|) - a + |b|}{2} = |b| = \max(|a|,|b|)$.

3. If $|b| \le a$, then $a + |b| \ge 2|b|\ge 0$ and $a - |b| \ge 0$ so $f(a,b) = \frac{(a+|b|) + a-|b|}{2} = a = \max(|a|,|b|)$

Knowing that, your equation becomes :

$$2 \max(2\max(|x|,|y|), |z|) = r$$

Or in other words

$$\max(\, 2|x|, \ 2|y|,\ |z| \,) = \frac{r}{2}$$

So I think the equation describes a cuboid of lengths $\frac{r}{2}$, $\frac{r}{2}$ and $r$ (along the $x$, $y$ and $z$ axis respectively) centered at the origin.