On Signed Distance Functions

Aug 11, 2020

Reading time: 12 minutes and 41 seconds.

Introduction

Formal Definition

In more rigorous terms, signed distance functions (aka SDFs) are defined as functions that satisfy a particular form of the eikonal equation¹. The general eikonal equation is of the form:

$$||\nabla u(x)|| = \frac{1}{f(x)},\quad x \in \Omega,$$

where $\Omega$ is an open set in $\mathbb{R}^n$, and $f(x) : \mathbb{R}^n \to \mathbb{R}$ is a function with positive values. In physical terms, the solution $u(x)$ to this nonlinear partial differential equation can be interpreted as the shortest time needed to travel from the boundary $\partial\Omega$ to $x \in \Omega$, with $f(x)$ being the speed at $x$.

Now, for a signed distance function, the speed will be unity throughout the domain, and $x$ will not be constrained to lie in $\Omega$. With $f(x) \equiv 1$, the time to travel from the boundary $\partial\Omega$ to any $x \in \mathbb{R}^n$ will simply be equal to the shortest distance from $x$ to $\partial\Omega$. We may then obtain a signed distance function $\textsf{sdf}(x)$, provided that the following conditions are satisfied:

Eikonal equation: $$\tag{EE} ||\nabla\textsf{sdf}(x)|| = 1,\quad x \in \mathbb{R}^3,$$
Zero distance at the boundary: $$\tag{BZ} \textsf{sdf}(x) = 0,\quad \forall \ x \in \partial\Omega,$$
Inward-facing normal at the boundary: $$\tag{IN} \nabla\textsf{sdf}(x) = N(x),\quad \forall \ x \in \partial\Omega.$$

If these conditions are satisfied, a function representing the shortest Euclidean distance from $x \in \mathbb{R}^n$ to a surface $\partial\Omega \subset \Omega \subset \mathbb{R}^n$ is obtained. This distance is signed, meaning that when $x \in \Omega$, $\textsf{sdf}(x)$ is nonnegative ($\geq 0$), while for $x \in \mathbb{R}^n \setminus \Omega$, $\textsf{sdf}(x)$ it is negative ($< 0$).

To prove that these conditions produce a distance field (disregarding the sign for now), consider points $x, y$ on a path of steepest-descent, that is a path along gradient $\nabla \textsf{sdf}$. As established previously $||\nabla \textsf{sdf}(x)|| = 1$ (condition (EE)). The Euclidean distance between these two points will be at most equal to the length of the steepest-descent path, denoted here by $\gamma$. We find that:

$$ \tag{G} ||x - y|| \leq |\gamma| = \left| \int_\gamma \nabla \textsf{sdf}(\xi) \cdot \mathrm{d}s \right| = |\textsf{sdf}(x) - \textsf{sdf}(y)|. $$

Then, considering a straight line path from $x$ to $y$, i.e., the shortest path between the two points, denoted here by $\zeta$, we find the following bound:

$$ \tag{Z} \begin{aligned} ||x - y|| &= |\zeta| = \int_\zeta ||\nabla \textsf{sdf}(\xi)|| \ ||\mathrm{d}s|| \cr &\geq \left| \int_\zeta \nabla \textsf{sdf}(\xi) \cdot \mathrm{d}s \right| = |\textsf{sdf}(x) - \textsf{sdf}(y)|. \end{aligned} $$

The latter inequality is proven straightforwardly by application of the Cauchy–Schwartz inequality.

Proof ↕

Consider the Cauchy–Schwartz inequality as applied to two vectors $u, v \in \mathbb{R}^n$:

$$ \tag{C–S} |u \cdot v| \leq ||u|| \ ||v||. $$

We found $|\zeta|$ to be equal to $$ \int_\zeta ||\nabla \textsf{sdf}(\xi)|| \ ||\mathrm{d}s||. $$

Recalling the definition of the signed distance function, we have that $|\nabla \textsf{sdf}(\xi)| = 1, \forall \ \xi \in \mathbb{R}^n $ (condition (EE)). The following can then be shown by invoking (C–S):

$$ \begin{aligned} \int_\zeta ||\nabla \textsf{sdf}(\xi)|| \ ||\mathrm{d}s|| &= \int_\zeta ||1|| \ ||\mathrm{d}s|| \cr &\geq \left|\int_\zeta 1 \cdot \mathrm{d}s\right| = \left|\int_\zeta \nabla \textsf{sdf}(\xi) \cdot \mathrm{d}s\right| \cr &= |\textsf{sdf}(x) - \textsf{sdf}(y)|. \end{aligned} $$

□

Combining inequalities (G) and (Z), we find that the following must hold:

$$ ||x - y|| = |\textsf{sdf}(x) - \textsf{sdf}(y)|. $$

Finally, from the boundary conditions ((BZ) and (IN)), it is ensured that the function will in fact be a signed distance function for a given object $\Omega$.

Differentiability

Since for certain geometries, the derivative at a point where there exist multiple minima (e.g., along the diagonal of a rectangle), is not defined. The points at which this is true are interesting in their own right; as a matter of fact, this set of points is known as the medial axis, or when this set is closed, the cut locus.

In the box below, an example of this phenomenon is given for a 2D square SDF.

Differentiability Example: Square ↕

Let us take one of the signed distance functions presented below to show that these points do indeed exist. Take for example the rectangle SDF, which can be written on one line as follows: $$ \begin{aligned} \textsf{sdf}(p, b) &= \sqrt{\max(|p_x|-b_x, 0)^2 + \max(|p_y|-b_y, 0)^2} \cr &+ \min(\max(|p_x|-b_x, |p_y|-b_y), 0). \end{aligned} $$

Taking the derivative w.r.t. $x$, where we replaced the absolute value by an equivalent expression ($|x| = \sqrt{x^2}$ for $x \in \mathbb{R}$), we obtain: $$ \begin{aligned} \frac{\partial \textsf{sdf}(x, y, w, h)}{\partial x} &= \left(\begin{cases} \frac{x}{\sqrt{x^2}} & \text{if } \sqrt{x^2} < w \text{ and } h + \sqrt{x^2} \geq w + \sqrt{y^2} \cr 0 & \text{else} \end{cases}\right) \cr &+ \frac{ \max(0, -w + \sqrt{x^2}) \left(\begin{cases} \frac{x}{\sqrt{x^2}} & \text{if } \sqrt{x^2} > w \cr 0 & \text{else} \end{cases}\right) }{ \sqrt{\max(0, -w + \sqrt{x^2})^2 + \max(0, -h + \sqrt{y^2})^2} }, \end{aligned} $$

where $p = [\begin{smallmatrix} x & y \end{smallmatrix}]^\intercal$ and $b = [\begin{smallmatrix} w & h \end{smallmatrix}]^\intercal$. For ease of exposition, we will consider a particular rectangle, namely a square ($w = h = l$). After some simplification we find: $$ \begin{aligned} \frac{\partial \textsf{sdf}(x, y, l)}{\partial x} &= \left(\begin{cases} \frac{x}{\sqrt{x^2}} & \text{if } \sqrt{y^2} \leq \sqrt{x^2} < l \cr 0 & \text{else} \end{cases}\right) \cr &+ \frac{ \max(0, -l + \sqrt{x^2}) \left(\begin{cases} \frac{x}{\sqrt{x^2}} & \text{if } \sqrt{x^2} > l \cr 0 & \text{else} \end{cases}\right) }{ \sqrt{\max(0, -l + \sqrt{x^2})^2 + \max(0, -l + \sqrt{y^2})^2} }. \end{aligned} $$

Now things start to become a little tricky. To able to make sense of all the $\min$s and $\max$es, we need to replace them with equivalent expressions in terms of Heaviside unit step functions ($H(x)$): $$ H(x) \equiv \begin{cases} 0 &\text{for } x < 0 \cr 1 &\text{for } x \geq 0 \end{cases}. $$ It can readily be shown that: $$\begin{aligned} \max(a,b) &= H(a-b)a + [1-H(a-b)] b = H(a-b)a + H(b-a) b, \cr \min(a,b) &= H(b-a)a + [1-H(b-a)] b = H(b-a)a + H(a-b) b. \end{aligned}$$

We shall only consider the second term in the partial derivative, since the first term is known to have a limit to zero, and will therefore be determinate within our region of interest. Through substitution of the previous expressions, we obtain: $$\begin{aligned} D(x, y, l) &= \frac{D_n(x, l)}{D_d(x,y,l)} \cr &= \frac{ x (-l + \sqrt{x^2}) H(-l + \sqrt{x^2})^2 }{ \sqrt{x^2}\sqrt{(l-\sqrt{x^2})^2 H(-l + \sqrt{x^2}) + (l-\sqrt{y^2})^2 H(-l+\sqrt{y^2})} }. \end{aligned}$$

For a square, if $x=y$ and $|x| < l$, it is known that there are to edges that lie equally close to $(x,y)$. To prove this, we will use L’Hôpital’s rule to show that now derivative exists for $x < l$. Roughly speaking, L’Hôpital’s rule states that for a function of the form $f(x)/g(x)$, its limit for $x \rightarrow c$ exists only if $\lim_{x \rightarrow c} f^\prime(x)/g^\prime(x)$ exists. In fact, in such a case we will find: $$ \lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \lim_{x \rightarrow c} \frac{f^\prime(x)}{g^\prime(x)}. $$

Let us take $f(x) \equiv D_n(x, l)$ and $g(x) \equiv D_d(x, x, l)$. Computing the derivatives, we obtain: $$\begin{aligned} f^\prime(x) &= 2 \sqrt{x^2} \left(\sqrt{x^2}-l\right) \delta \left(\sqrt{x^2}-l\right) H \left(\sqrt{x^2}-l\right) \cr &+\frac{\left(2 x^2-l \sqrt{x^2}\right) H \left(\sqrt{x^2}-l\right)^2}{\sqrt{x^2}}, \end{aligned}$$

$$\begin{aligned} g^\prime(x) &= \frac{\sqrt{x^2} \left(\frac{x \left(l-\sqrt{x^2}\right)^2 \delta \left(\sqrt{x^2}-l\right)}{\sqrt{x^2}}+\frac{2 x \left(\sqrt{x^2}-l\right) H \left(\sqrt{x^2}-l\right)}{\sqrt{x^2}}\right)}{2 \sqrt{2} \sqrt{\left(l-\sqrt{x^2}\right)^2 H \left(\sqrt{x^2}-l\right)}} \cr &+\frac{\sqrt{2} x \sqrt{\left(l-\sqrt{x^2}\right)^2 H \left(\sqrt{x^2}-l\right)}}{\sqrt{x^2}}. \end{aligned}$$

While $f^\prime(x)$ appears to hold no singularities at first glance (other than the Dirac delta, which is not triggered since $x < l$), $g^\prime(x)$ does have a clear singularity. For its second term, it is clear that the Heaviside function in the denominator will always be zero on the interior, giving cause to a singularity of the form $1/\sqrt{0}$. With this, we have shown that on the interior diagonals, the derivative SDF of a square shape will indeed not exist. (Since the shape is symmetric in both axes, a derivation in $y$ will yield the exact same result.)

Boolean Operations (Combinations)

In order to create more complex shapes, it is reasonable to want to run combination operatons on primitives. To get a better handle on what these operations might be, consider the following basic examples:

Union
Difference
Intersection

Below, we consider these, and other, operations on two shapes (sets) $P$ and $Q$, resulting in a set $R$. The SDFs belonging to these sets will be subscripted by the name of the set to which they belong. For more information on this topic, see ²^,³.

Union

$$ R = P \cup Q \Leftrightarrow \textsf{sdf}_R (p) = \min[\textsf{sdf}_P (p), \textsf{sdf}_Q (p)]. $$

Intersection

$$ R = P \cap Q \Leftrightarrow \textsf{sdf}_R (p) = \max[\textsf{sdf}_P (p), \textsf{sdf}_Q (p)]. $$

Note: This does not produce a proper SDF on the interior.

Difference

$$ R = P \setminus Q \Leftrightarrow \textsf{sdf}_R (p) = \max[\textsf{sdf}_P (p), -\textsf{sdf}_Q (p)]. $$

Note: This does not produce a proper SDF on the interior.

Symmetric Difference

Also known as the disjunctive union. Expressed in words, this is the union of the difference of the two sets.

$$ R = P \oplus Q = (P \setminus Q) \cup (Q \setminus P) $$ $$\Leftrightarrow$$ $$ \textsf{sdf}_R (p) = \min\{\max[\textsf{sdf}_P (p), -\textsf{sdf}_Q (p)], \max[\textsf{sdf}_Q (p), -\textsf{sdf}_P (p)]\}. $$

Note: This does not produce a proper SDF on the interior.

Complement

The complement of an SDF is simply its negative, since its interior then becomes its exterior.

$$ R = P^\complement \Leftrightarrow \textsf{sdf}_R (p) = -\textsf{sdf}_P (p). $$

This does not produce a proper SDF on the interior? ↕

I bet you are puzzled as to why you won’t get a proper SDF on using some of the combination operators introduced above. Let’s shed some light on this by considering an intersection operator:

$$ \textsf{sdf}_R (p) = \max[\textsf{sdf}_P (p), \textsf{sdf}_Q (p)]. $$

Now, before we get to differentiate, let’s try to get rid of that $\max$ function by introducing Heaviside unit step functions instead:

$$ \textsf{sdf}_R (p) = \textsf{sdf}_P (p) H(\textsf{sdf}_P (p) - \textsf{sdf}_Q (p)) + \textsf{sdf}_Q (p) [1 - H(\textsf{sdf}_P (p) - \textsf{sdf}_Q (p))]. $$

Now, let’s take the gradient of this to see if it satisfies the eikonal condition (cond. (EE)):

$$ \begin{aligned} [\nabla \textsf{sdf}_R (p)]_i &= H_{P-Q} \partial_i \textsf{sdf}_P (p) + \delta_{P-Q} \textsf{sdf}_P (p) \partial_i \textsf{sdf}_{P-Q} (p) \cr &- \delta_{P-Q} \textsf{sdf}_Q (p) \partial_i \textsf{sdf}_{P-Q} (p) + (1 - H_{P-Q}) \partial_i \textsf{sdf}_Q (p), \end{aligned} $$

where we have introduced some compact but truthfully nasty shorthand:

$H_{P-Q} \equiv H(\textsf{sdf}_P (p) - \textsf{sdf}_Q (p))$,
$\delta_{P-Q} \equiv \delta(\textsf{sdf}_P (p) - \textsf{sdf}_Q (p))$,
$\textsf{sdf}_{P-Q}(p) \equiv \textsf{sdf}_P (p) - \textsf{sdf}_Q (p)$,
$\partial_i \textsf{sdf}(p) \equiv \frac{\partial \textsf{sdf}(p)}{\partial p_i}$.

Let’s consider some point $p$ where $\textsf{sdf}_P (p) \neq \textsf{sdf}_Q (p)$ (so as not to deal with the Dirac delta). We can can identify two cases: (1) $\textsf{sdf}_{P-Q} (p) > 0$, and (2) $\textsf{sdf}_{P-Q} (p) < 0$. Let’s look at the first case:

$$ [\nabla \textsf{sdf}_R (p)]_{i,1} = \partial_i \textsf{sdf}_P (p). $$

Hey, now that’s cool: the gradient is exactly that of $\textsf{sdf}_P$, therefore satisfying the eikonal condition. It does not take much reasoning to find out that the same holds for case (2), except this time the gradient will correspond to that of $\textsf{sdf}_Q$. All things seem to point to a proper SDF … or is there a catch?

The catch

There is indeed a catch; we have not considered the case where $\textsf{sdf}_P (p) = \textsf{sdf}_Q (p)$. What we get is:

$$ \begin{aligned} [\nabla \textsf{sdf}_R (p)]_i &= \partial_i \textsf{sdf}_P (p) \cr &+ \delta_{P-Q} \partial_i \textsf{sdf}_{P-Q} (p) \left( \textsf{sdf}_P (p) - \textsf{sdf}_Q (p) \right), \end{aligned} $$

where $\delta_{P-Q} = \delta(0) = \infty$. Therefore, the norm of the gradient of this function is indeterminate, let alone equal to 1. For this very reason, we cannot say that the resulting function (after intersection) is a ‘real’ SDF.

What about the interior?

All technicalities aside, it seems at first glance that there are no problems with the distance itself. We haven’t taken a look at the interior yet though. While on the interior, the eikonal condition is satisfied, something arguably more important is violated: the shortest distance is not given.

Let’s do a little thought experiment: imagine two circles, with a spacing between the centers. Now take the difference between the two shapes: since we are taking the $\max$, the distance on the boundary will be given from the farthest circle, instead of the closest circle. Quite obviously, this is not the shortest distance, even though the sign matches. Now this will pose problems when raymarching on the interior, but a similar problem will not be encountered on the exterior, since the distance will monotonically decrease towards the center.

Transformations

Translation

Consider a translation by distance $t \in \mathbb{R}^3$. The translated SDF will be of the form:

$$ \textsf{sdf}_R (p) = \textsf{sdf}_P (p - t). $$

Scaling

We distinguish uniform and nonuniform scaling. In the first case, the scaling factor is the same across all axes, while in the latter case different scaling factors may be used on different axes.

Uniform Scaling

For uniform scaling, consider a scaling factor $s \in \mathbb{R}_+$, i.e, the positive orthant of $\mathbb{R}$. It can be shown that (see ‘Derivations’ below):

$$ \textsf{sdf}_R (p) = \textsf{sdf}_P (p/s) s . $$

Example: Uniformly Scaled Sphere ↕

Consider the SDF of a sphere with radius $r$, centered at the origin:

$$ \textsf{sdf}(p) = ||p|| - r = \sqrt{p_x^2 + p_y^2 + p_z^2} - r. $$

Uniform scaling by a factor $s \in \mathbb{R}_+$ would yield, as expected:

$$ \begin{aligned} \textsf{sdf}_s (p) &= (||p/s|| - r) s \cr &= (\sqrt{p_x^2/s^2 + p_y^2/s^2 + p_z^2/s^2} - r) s \cr &= (\sqrt{p_x^2 + p_y^2 + p_z^2}/s - r) s \cr &= \sqrt{p_x^2 + p_y^2 + p_z^2} - r s = ||p|| - rs. \end{aligned} $$

Here, we simply obtained the SDF of a sphere of radius $r^\prime = r s$, which is equivalent to scaling the original SDF by a factor of $s$. Clearly, the SDF is proper and the distance is not compressed/dilated. But what about a nonuniform scaling factor $s \in \mathbb{R}^3$?

Taking the inverse of the transformation matrix $S$, we find that when naively transforming $p$ in the SDF we obtain:

$$ \begin{aligned} \textsf{sdf}_S (p) &= ||p \oslash s|| - r \cr &= \sqrt{p_x^2/s_x^2 + p_y^2/s_y^2 + p_z^2/s_z^2} - r, \end{aligned} $$

where ‘$\oslash$’ denotes componentwise division.

At first glance, there don’t seem to be any problems with this. Note, however, that the distances obtained in this way do not correspond to the true distances. To see if this indeed is the case, let’s verify the eikonal equation condition (cond. (EE)):

$$ \begin{aligned} ||\nabla \textsf{sdf}_s (p)|| &= ||\nabla (||p \oslash s|| - r)|| \cr &= ||\nabla (\sqrt{p_x^2/s_x^2 + p_y^2/s_y^2 + p_z^2/s_z^2} - r)|| \cr &= \left|\left| \begin{bmatrix} \frac{p_x}{s_x^2} \cr \frac{p_y}{s_y^2} \cr \frac{p_z}{s_z^2} \end{bmatrix}/\sqrt{p_x^2/s_x^2 + p_y^2/s_y^2 + p_z^2/s_z^2}\right|\right| \cr &= \frac{\sqrt{p_x^2/s_x^4 + p_y^2/s_y^4 + p_z^2/s_z^4}}{\sqrt{p_x^2/s_x^2 + p_y^2/s_y^2 + p_z^2/s_z^2}} \cr &\stackrel{?}{\equiv} 1. \end{aligned} $$

It is straightforward to show, in this case, that this final assertion will only be true when $s_x = s_y = s_z = 1$. This trivially corresponds to uniform scaling with a scaling factor of 1, which constitutes no change from the original SDF. Let us now apply a correction factor $c$ to the SDF, such that we obtain $\textsf{sdf}_{s,c} \equiv \textsf{sdf} (p \oslash s) c$. Leaving out much of the derivation, which is much the same as what was shown previously, we have:

$$ \begin{aligned} ||\nabla \textsf{sdf}_{s,c} (p)|| &= ||\nabla (||p \oslash s|| - r) c|| \cr &= ||\nabla (\sqrt{p_x^2/s_x^2 + p_y^2/s_y^2 + p_z^2/s_z^2} - r) c|| \cr &= c \frac{\sqrt{p_x^2/s_x^4 + p_y^2/s_y^4 + p_z^2/s_z^4}}{\sqrt{p_x^2/s_x^2 + p_y^2/s_y^2 + p_z^2/s_z^2}} \cr &\stackrel{?}{\equiv} 1. \end{aligned} $$

If we take $s_x = s_y = s_z = s$ and $c \equiv s$, we can show that:

$$ \begin{aligned} ||\nabla \textsf{sdf}_s (p)|| &= s \frac{\sqrt{p_x^2/s^4 + p_y^2/s^4 + p_z^2/s^4}}{\sqrt{p_x^2/s^2 + p_y^2/s^2 + p_z^2/s^2}} \cr &= s \frac{\sqrt{s^2}}{\sqrt{s^4}} = s \frac{s}{s^2} \cr &\equiv 1. \end{aligned} $$

With this, we have shown how proper uniform scaling can be achieved on a sphere.

Nonuniform Scaling

As discussed in the ‘Derivation’ box below, there is no straightforward solution to the nonuniform scaling problem that does not violate the eikonal equation condition. The next best ‘solution’, would be to obtain some sort of lower distance bound. This provides a straightforward approach to obtaining a lower bound to the scale surface, which will appear correctly when rendered, but does not have a proper SDF. More advantages are listed in the ‘Derivation’ box below. For a similar discussion, see ⁴.

Here, we consider a scaling factor $s \in \mathbb{R}_+^3$. As a lower bound, we obtain:

$$ \textsf{sdf}_{s,\text{min}} (p) = \textsf{sdf} (p \oslash s) s_\text{min}, $$

where $s_\text{min}$ is the smallest element of $s$, and ‘$\oslash$’ denotes componentwise division.

Derivation ↕

To be able to perform nonuniform scaling, a first idea would be to see if there is some function $c(p)$ that would allow $\textsf{sdf}_{s,c} (p) \equiv ||\nabla [\textsf{sdf}(p \oslash s) c(p)]||$ to satisfy the eikonal equation condition.

With this idea in mind, let’s apply it to an arbitrary nonuniformly scaled SDF:

$$ \begin{aligned} & ||\nabla \left[\textsf{sdf}(p \oslash s) c(p) \right]|| = \cr &\left[\left(\textsf{sdf}(p \oslash s) \frac{\partial c(p)}{\partial p_x} + \frac{c(p) \frac{\partial\textsf{sdf}(p \oslash s)}{\partial p_x}}{s_x}\right)^2\right. \cr &\left. + \left(\textsf{sdf}(p \oslash s) \frac{\partial c(p)}{\partial p_y} + \frac{c(p) \frac{\partial \textsf{sdf}(p \oslash s)}{\partial p_y}}{s_y}\right)^2\right. \cr &\left. + \left(\textsf{sdf}(p \oslash s) \frac{\partial c(p)}{\partial p_z} + \frac{c(p) \frac{\partial \textsf{sdf}(p \oslash s)}{\partial p_z}}{s_z}\right)^2\right]^{1/2}. \end{aligned} $$

To satisfy the eikonal equation, our initial goal is to try to achieve the following:

$$ \textsf{sdf}(p \oslash s) \frac{\partial c(p)}{\partial p_j} + \frac{c(p) \frac{\partial\textsf{sdf}(p \oslash s)}{\partial p_j}}{s_j} = \frac{\partial\textsf{sdf}(p \oslash s)}{\partial p_j}. $$

If this is achieved, we know for a fact that $||\nabla \left[\textsf{sdf}(p \oslash s) c(p) \right]|| = ||\nabla \textsf{sdf}(p)|| \equiv 1 $.

Uniform scaling

For uniform scaling, as seen before, where $s_i = s$, the conditions required to obtain this result are:

Uniform scaling: $s_x = s_y = s_z = s$,
A constant $c$: $\frac{\partial c(p)}{\partial p_j} \equiv 0$,
$c \equiv s$ to obtain $(c/s) \frac{\partial\textsf{sdf}(p \oslash s)}{\partial p_j} = \frac{\partial\textsf{sdf}(p \oslash s)}{\partial p_j}$.

Nonuniform scaling

Rewriting the equation given above, we obtain:

$$ \textsf{sdf}(p \oslash s) \frac{\partial c(p)}{\partial p_j} + (c(p)/s_j - 1) \frac{\partial\textsf{sdf}(p \oslash s)}{\partial p_j} = 0. $$

This boils down to solving the following system of equations:

$$ \begin{aligned} \frac{\partial c(p)}{\partial p_j} &= q_j(p) - \frac{q_j(p)}{s_j} c(p), \quad j=1\ldots3 \cr \text{where } q_j(p) &\equiv \textsf{sdf}(p \oslash s) \frac{\partial \textsf{sdf}(p \oslash s)}{\partial p_j}. \end{aligned} $$

This time, we cannot get away with taking a constant $c$, since this would not satisfy the differential equation for all $j$. Even for some sort of discontinuous $c$, this effect can’t be achieved, since there is is no indication as to what condition $j$ is being evaluated based only on $p$.

A lower bound for nonuniform scaling

In practical situations, it is useful to at least have a lower bound of the SDF. Since the SDF is identically zero at the surface, and the sign is preserved even when scaling, only the interior and exterior distances will be smaller than they would actually be. This is a useful property in cases such as raymarching, where overshoot is not desirable.

Let’s study if such a lower bound can actually be computed. The new eikonal equation for an pseudo-SDF that performs nonuniform scaling but corrects using the smallest scaling factor, takes the form of:

$$ \begin{aligned} & ||\nabla \left[\textsf{sdf}(p \oslash s) s_\text{min} \right]|| = \cr &\left[\left( \frac{s_\text{min} \frac{\partial \textsf{sdf}(p \oslash s)}{\partial p_x}}{s_x} \right)^2 \right. \cr &\left. + \left( \frac{s_\text{min} \frac{\partial \textsf{sdf}(p \oslash s)}{\partial p_y}}{s_y} \right)^2 \right. \cr &\left. + \left( \frac{s_\text{min} \frac{\partial \textsf{sdf}(p \oslash s)}{\partial p_z}}{s_z} \right)^2 \right]^{1/2}, \end{aligned} $$

where $s_\text{min} \equiv \min(s)$.

Considering the indivual elements, it is clear that:

$$ \left( \frac{s_\text{min} \frac{\partial \textsf{sdf}(p \oslash s)}{\partial p_x}}{s_x} \right)^2 \leq \left( \frac{\partial \textsf{sdf}(p \oslash s)}{\partial p_x} \right)^2, \quad j = 1\ldots 3, $$

which follows from the fact that $s_\text{min}/s_j \leq 1$. Therefore, $||\nabla \left[\textsf{sdf}(p \oslash s) s_\text{min} \right]|| \leq 1$, which means that the distance between points will be lower than the actual distance (see the discussion of the eikonal equation above).

With this, we can use the following equation as a lower-bound SDF-like function for nonuniform scaling:

$$ \textsf{sdf}_{s,\text{min}} (p) = \textsf{sdf} (p \oslash s) s_\text{min}. $$

List of 2D Functions

The following signed distance functions assume that the primitive is centered at the origin, unless stated otherwise. Most of these functions are adapted from Inigo Quilez’s excellent 2D SDF functions article, while others are derived from other sources in the literature.

Point

$$ \textsf{sdf}(p) = ||p|| = \sqrt{p_x^2 + p_y^2}. $$

$$ $$

Implementation

GLSL

float point(vec2 p)
{
    return length(p);
}

Haxe

static public inline function point(p:Vec2):FFloat {
    return p.length;
}

Circle

A circle simply offsets the distance to a point by a radius $r$, as follows:

$$ \textsf{sdf}(p, r) = ||p|| - r = \sqrt{p_x^2 + p_y^2} - r. $$

Implementation

GLSL

float circle(vec2 p, float r)
{
    return length(p) - r;
}

Haxe

static public inline function circle(p:Vec2, r:FFloat):FFloat {
    return p.length - r;
}

Rectangle

For a rectangle, consider a 2D vector $d$ consisting of the width (in $x$-direction), and the height. Operators with the subscript $i$ notation (e.g., $\textrm{abs}_i$) will act on a vector in a component-wise fashion (similar to Einstein’s tensor notation); see ⁵ for more information.

$$\begin{aligned} d(b) &= \textrm{abs}_i(p)-b, \cr \textsf{sdf}(p, d(b)) &= ||\max_i(d, 0)|| + \min(\max(d_x, d_y), 0). \end{aligned}$$

Implementation

GLSL

float rectangle(vec2 p, vec2 b)
{
    vec2 d = abs(p)-b;
    return length(max(d,0.0)) + min(max(d.x,d.y),0.0);
}

Haxe

static public function rectangle(p:Vec2, b:Vec2):FFloat {
    final d = p.abs().sub(b);
    return d.max(0).length + Math.min(Math.max(d.x, d.y), 0);
}

Rounded Rectangle

The rounded rectangle follows a similar construction method to the rectangle, although the individual corners are rounded independently, producing a different result compared to the rectangle with uniform rounding applied.