[Games 101] Lecture 13-16 Ray Tracing

Ray Tracing

Why Ray Tracing

• 光栅化不能得到很好的全局光照效果
  • 软阴影
  • 光线弹射超过一次（间接光照）
• 光栅化是一个快速的近似，但是质量较低

• 光线追踪是准确的，但是较慢
  • Rasterization: real-time, ray tracing: offline
  • 生成一帧需要一万CPU小时

Basic Ray-Tracing Algorithm

• 图形学上的光线定义
  • 光线是沿直线传播的
  • 两个或多个光线不会发生碰撞
  • 光线一定会从光源出发到达我们的眼镜
    • light reciprocity 光路的可逆性

Pinhole Camera Model

1. 对于每一个像素，从眼睛投射出一个光线（eye ray）
2. 和场景相交，求最近的交点
3. 和光源连线判断是否对光源可见
4. 计算着色，写回像素的值

以上算法还是只考虑了光线只弹射一次的情况

Recursive (Whitted-Style) Ray Tracing

• 在任意一个点可以继续传播光线（反射和折射）
• 所有点的着色都会被加到像素中（需要计算能量损失）

• 对于从眼睛出发的光线定义为 Primary Ray
• 由 Primary Ray 折射和反射的光线可以称作 Secondary Ray
• 判断可见性的光路被称作 Shadow Ray

Ray-Surface Intersection

Ray Equation

• 光线被定义为有一个起点和方向的向量
• Equation: $\mathbf{r}(t)=\mathbf{o}+t\mathbf{d},0\le t<\infty$

Ray Intersection With Sphere

• Ray: $\mathbf{r}(t)=\mathbf{o}+t\mathbf{d},0\le t<\infty$
• Sphere: $\mathbf{p}:(\mathbf{p}-\mathbf{c}{)}^2-R^2=0$

由于点需要满足两个方程，所以
$$(\mathbf{o}+t\mathbf{d}-\mathbf{c}{)}^2-R^2=0$$
展开后也就有，
$$\begin{array}{rl}& a{t}^2+bt+c=0,\text{ where }\\ & a=\mathbf{d}\cdot \mathbf{d}\\ & b=2(\mathbf{o}-\mathbf{c})\cdot \mathbf{d}\\ & c=(\mathbf{o}-\mathbf{c})\cdot (\mathbf{o}-\mathbf{c})-R^2\\ & t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\end{array}$$
注意，对于解出后的结果 $t$，需要满足 $t\ge 0$

Ray Intersection With Implicit Surface

• Ray: $\mathbf{r}(t)=\mathbf{o}+t\mathbf{d},0\le t<\infty$

对于任意的隐式表面，有

• General implicit surface: $\mathbf{p}:f(\mathbf{p})=0$
• Substitute ray equation: $f(\mathbf{o}+t\mathbf{d})=0$

解出 Substitute ray equation 即可

注意，解出的结果需要是实数并且为非负数

Ray Intersection With Triangle Mesh

Convert It to Ray Intersection With Plane

• Applications

  • Rendering: visibility, shadows, lighting
  • Geometry: inside/outside test （计算是否一个点是否在形状内部或者是外部，封闭物体）
    • 在形状上取一个点找一个光线求交点，如果是奇数个交点，那么点就一定在物体内，否则在物体外

• Compute

  • Simple idea: just intersect ray with each triangle [Simple, but slow (acceleration?) ]
  • Acceleration: 将光线和三角形求交转化为光线和平面求交，然后再判定交点是否在三角形内

Plane Equation

• Plane is defined by normal vector and a point on plane

Ray Intersection With Plane

• Ray equation: $\mathbf{r}(t)=\mathbf{o}+t\mathbf{d},0\le t<\infty$

• Plane equation: $\mathbf{p}:\left(\mathbf{p}-{\mathbf{p}}^{\prime }\right)\cdot \mathbf{N}=0$

• Solve for intersection

$$\begin{array}{rl}& \text{set}\mathbf{p}=\mathbf{r}(t)\text{and solve for}t\\ & \left(\mathbf{p}-{\mathbf{p}}^{\prime }\right)\cdot \mathbf{N}=\left(\mathbf{o}+t\mathbf{d}-{\mathbf{p}}^{\prime }\right)\cdot \mathbf{N}=0\\ & t=\frac{\left({\mathbf{p}}^{\prime }-\mathbf{o}\right)\cdot \mathbf{N}}{\mathbf{d}\cdot \mathbf{N}}\end{array}$$

• Check: $0\le t<\infty$，判断这个点是否在三角形内

Möller Trumbore Algorithm

A faster approach, giving barycentric coordinate directly （以另外一个形式描述平面）
$$\mathbf{O}+t\mathbf{D}=\left(1-b_1-b_2\right){\mathbf{P}}_0+b_1{\mathbf{P}}_1+b_2{\mathbf{P}}_2$$

• 联立光线和重心坐标
• 使用克莱姆法则解这个线性方程组，解出 $b_1,b_2,t$
• 如果 $1-b_1-b_2,b_1,b_2$ 都是非负的，那么点在三角形内

Accelerating Ray-Surface Intersection

• Simple ray-scene intersection
  • Exhaustively test ray-intersection with every triangle
  • Find the closest hit (i.e. minimum t)
• Problem
  • Naive algorithm = #pixels ⨉ # triangles (⨉ #bounces)
  • Slow!!!

Bounding Volumes

• Quick way to avoid intersections: bound complex object with a simple volume
  • Object is fully contained in the volume
  • If it doesn’t hit the volume, it doesn’t hit the object
  • So test BVol first, then test object if it hits (如果一个光线连包围盒都碰不到，就也不可能碰到里面的物体)
• Understanding: box is the intersection of 3 pairs of slabs (认为长方体是三组对面形成的交集)
• Specifically: We often use an Axis-Aligned Bounding Box (AABB) [轴对齐包围盒]

Ray Intersection with Axis-Aligned Box

• 考虑二维的情况（两组对面）: Compute intersections with slabs and take intersection of $t_{min}/t_{max}$ intervals
  • 对得到的两组线段求交集即可

• Recall: a box (3D) = three pairs of infinitely large slabs
• Key ideas
  • The ray enters the box only when it enters all pairs of slabs（如果三个对面都有光线进入，才能说光线进入了盒子）
  • The ray exits the box as long as it exits any pair of slabs （只要光线离开任意一个对面，光线就算离开了这个盒子）
• For each pair, calculate the $t_{min}$ and $t_{max}$ (negative is fine)
• For the 3D box, $t_{enter}=\max \{t_{min}\},t_{exit}=\min \{t_{max}\}$
• If $t_{enter}<t_{exit}$, we know the ray stays a while in the box (so they must intersect!)
• physical correctness
  • $t_{exit}<0$, The box is “behind” the ray — no intersection
  • $t_{exit}\ge 0,t_{enter}<0$, The ray’s origin is inside the box — certainly have intersection

In summary, Ray and AABB intersect iff
$$t_{enter}<t_{exit}\text{and}{t}_{exit}\ge 0$$

Why Axis-Aligned

• 在求交时计算方便

Using AABBs to accelerate ray tracing

Uniform Spatial Partitions (Grids)

• Precomputation

1. Find bounding box
2. Create grid
3. Store each object in overlapping cells
   • 只考虑物体的表面

• Ray Tracing

1. Step through grid in ray traversal order
2. For each grid cell, Test intersection with all objects stored at that cell
   • 让光线经过各个包围盒中的区块，让光线和盒子求交（假设光线和盒子求交的运行速度远大于和物体求交的运行速度）
   • 如果发现和光线相交的盒子内有物体，再让物体和盒子求交，通过这样找到所有的交点

• Grid Resolution?

  • Heuristic
    • #cells = C * #objs
    • C = 27 in 3D

• Pros & Cons

Spatial Partitions

• 基于 Grid 方法的优化，物体稀疏的地方不需要用很多格子

• Examples

  • Oct-Tree: 以下的八叉树是二维的情况，在二维空间中就是四叉的，把空间切成了不均匀的结构
  • KD-Tree: KD-Tree 永远是找到一个轴把空间分成两个部分，和维数无关，把空间划分成了类似二叉树的结构，水平划分和数值划分是交替的以保证空间的划分是均匀的，计算简单
  • BSP-Tree: 对空间的二分方法，每一次选择一个方向把节点的分开，和 KD-Tree 的区别是不是横平竖直的划分，会随着维度的升高计算变得复杂

  

• KD-Tree Pre-Processing

  • 此处只考虑每个格子划分了一次，实际中别的格子可能也需要划分
  • 只在叶子节点存储三角形

• Data Structure for KD-Trees

  • Internal nodes store
    • split axis: x-, y-, or z-axis
    • split position: coordinate of split plane along axis
    • children: pointers to child nodes
    • No objects are stored in internal nodes
  • Leaf nodes store
    • list of objects

• Cons of KD-Tree

  • 很难判定一个三角形是否和一个立方体有交集，算法很难实现
  • 如果一个物体和多个盒子都有交集（出现在多个叶子节点中），KD-Tree 在这点上并不直观

Object Partitions & Bounding Volume Hierarchy (BVH)

• 划分物体而不是划分空间
• 对于下图，把三角形划分成蓝色和黄色三角形，并重新计算蓝色和黄色三角形的包围盒
• 终止条件是一堆里最多有 $N$ 个三角形

• Pros of BVH

  • 节省了三角形和包围盒求交问题
  • 保证了一个三角形只在一个盒子里

• Cons of BVH

  • 不同的 Bounding Box 是有可能相交的
  • 不能解决动态的物体，BVH 需要重新计算

• Summary

  • Find bounding box
  • Recursively split set of objects in two subsets
  • Recompute the bounding box of the subsets
  • Stop when necessary
  • Store objects in each leaf node

• Methods to subdivide a node

  • Choose a dimension to split
  • Heuristic #1: Always choose the longest axis in node
    • 让最长的轴变短，让最终的划分是比较均匀的
  • Heuristic #2: Split node at location of median object
    • 取中间的物体指的是，如果有 $n$ 个三角形，找的是 $\frac{n}{2}$ 处的三角形，保证切分后的三角形数量差不多，以保证生成的树是平衡的，减少最终的搜索时间
    • 给任意一列无序的数，找到它第 $i$ 大的数，可以使用快速选择算法，在 $O(n)$ 内解决

• Termination criteria

  • Heuristic: stop when node contains few elements (e.g. 5)

• Data Structure for BVHs

  • Internal nodes store
    • Bounding box
    • Children: pointers to child nodes
  • Leaf nodes store
    • Bounding box
    • List of objects
  • Nodes represent subset of primitives in scene
    • All objects in subtree

BVH Traversal

fun Intersect(Ray ray, BVH node) {
     if (ray misses node.bbox) return;
    
     if (node is a leaf node)
         test intersection with all objs;
         return closest intersection;
    
     hit1 = Intersect(ray, node.child1);
     hit2 = Intersect(ray, node.child2);
    
     return the closer of hit1, hit2;
}

Basic Radiometry

• Chinese Name: 辐射度量学

Why Radiometry

• 在计算光强的时候，我们没有准确的物理定义

• 辐射度量学：定义了一系列方法和单位去描述光照

• Accurately measure the spatial properties of light

  • New terms: Radiant flux, intensity, irradiance, radiance

• Perform lighting calculations in a physically correct manner

• Light Measurements

Radiant Energy and Flux (Power)

• Radiant Energy: 光源辐射出来的能量 (Barely used in CG)

$$Q[\mathrm{J}=\text{ Joule }]$$

• Flux (Power): 光源辐射出的单位时间的能量，目的是分析和比较两个和多个能量（去除时间的影响）
  • lm = lumen (Flux 的单位，翻译是流明)
  • Flux 也可以理解为单位时间通过一个感光平面的光子数目

$$\Phi \equiv \frac{\mathrm{d}Q}{\mathrm{d}t}[\mathrm{W}=\mathrm{W}\mathrm{a}\mathrm{t}\mathrm{t}][\mathrm{lm}=\text{ lumen }{]}^{\ast }$$

Radiant Intensity

• Radiant Intensity: The radiant (luminous) intensity is the power per unit solid angle (立体角) emitted by a point light source.
  • 可以理解为能量除以立体角
  • 定义了光源在任何一个方向上的亮度

$$\begin{array}{c}I(\omega )\equiv \frac{\mathrm{d}\Phi }{\mathrm{d}\omega }\\ \left[\frac{\mathrm{W}}{\mathrm{s}\mathrm{r}}\right]\left[\frac{\mathrm{lm}}{\mathrm{s}\mathrm{r}}=\mathrm{c}\mathrm{d}=\text{ candela }\right]\end{array}$$

Angles and Solid Angles

• Angle: ratio of subtended arc length on circle to radius

  • $\theta =\frac{l}{r}$
  • Circle has $2\pi$ radians

• Solid angle: ratio of subtended area on sphere to radius squared

  • 弧度制在三维空间中的延申，在三维空间中找一个球，从球出发形成某个大小的锥，锥会打到球面上，形成一个面积 $A$，立体角是它除以半径的平方
  • 把任何一个物体投影到单位球上，在单位球上框出的范围就是立体角
  • 描述一个空间中角有多大
  • $\Omega =\frac{A}{r^2}$
  • Sphere has $4\pi$ steradians(立体角的单位)

Differential Solid Angles

• Differential Solid Angles: 微分立体角
• 通常使用 $\omega$ 来定义单位方向

Isotropic Point Source

• 对于一个均匀点光源，它的 Radiant Intensity 的相关计算是

  • 对于所有方向上的 Intensity 积分起来可以得到 Flux
    $$\begin{array}{rl}\Phi & ={\int }_{S^2}I\mathrm{d}\omega \\ & =4\pi I\end{array}$$

  • 对于任何一个方向上的 Intensity 有

  $$I=\frac{\Phi }{4\pi }$$

Irradiance

• The irradiance is the power per unit area incident on a surface point.
  • Intensity: power per soild angle
  • 注意，面需要和入射光线垂直

$$\begin{array}{c}E(\mathbf{x})\equiv \frac{\mathrm{d}\Phi (\mathbf{x})}{\mathrm{d}A}\\ \left[\frac{\mathrm{W}}{{\mathrm{m}}^2}\right]\left[\frac{\mathrm{lm}}{{\mathrm{m}}^2}=\mathrm{lux}\right]\end{array}$$

Radiance

Radiance is the fundamental field quantity that describes the distribution of light in an environment

• Radiance: The radiance (luminance) is the power emitted, reflected, transmitted or received by a surface, per unit solid angle, per projected unit area.
  • 考虑某一个确定的微小面和一个方向

$$L(\mathrm{p},\omega )\equiv \frac{{\mathrm{d}}^2\Phi (\mathrm{p},\omega )}{\mathrm{d}\omega \mathrm{d}A\cos \theta }$$

• Recall

  • Irradiance: power per projected unit area
  • Intensity: power per solid angle

• So

  • Radiance: Irradiance per solid angle
  • Radiance: Intensity per projected unit area

Irradiance vs. Radiance

• Irradiance: total power received by area dA
  • 某一个小区域内接收到的所以能量
• Radiance: power received by area dA from “direction” dω
  • 某一个小区域的某一个方向上接受到的能量

$$\begin{array}{rl}dE(\mathrm{p},\omega )& =L_i(\mathrm{p},\omega )\cos \theta \mathrm{d}\omega \\ E(\mathrm{p})& ={\int }_{H^2}{L}_i(\mathrm{p},\omega )\cos \theta \mathrm{d}\omega \end{array}$$

Bidirectional Reflectance Distribution Function (BRDF)

• 双向反射分布函数
• 告诉你不同反射方向上分布的能量情况

Reflection at a Point

Radiance from direction ωi turns into the power E that dA receives Then power E will become the radiance to any other direction ωo

• Differential irradiance incoming: $dE\left({\omega }_i\right)=L\left({\omega }_i\right)\cos {\theta }_{i}d{\omega }_i$
• Differential radiance exiting (due to $dE\left({\omega }_i\right)$ ): $d{L}_r\left({\omega }_r\right)$

BRDF

The Bidirectional Reflectance Distribution Function (BRDF) represents how much light is reflected into each outgoing direction ${\omega }_r$ from each incoming direction

BRDF 描述了光线和物体是如何作用的

$$f_r\left({\omega }_i\to {\omega }_r\right)=\frac{\mathrm{d}{L}_r\left({\omega }_r\right)}{\mathrm{d}{E}_i\left({\omega }_i\right)}=\frac{\mathrm{d}{L}_r\left({\omega }_r\right)}{L_i\left({\omega }_i\right)\cos {\theta }_i\mathrm{d}{\omega }_i}\left[\frac{1}{\mathrm{s}\mathrm{r}}\right]$$

The Reflection Equation

• Reflection Equation 描述了任何一个着色点在各种不同光照环境下对出射光线的贡献

$$L_r\left(\mathrm{p},{\omega }_r\right)={\int }_{H^2}{f}_r\left(\mathrm{p},{\omega }_i\to {\omega }_r\right)L_i\left(\mathrm{p},{\omega }_i\right)\cos {\theta }_i\mathrm{d}{\omega }_i$$

Challenge: Recursive Equation

• 能够到达着色点的光线不止有光源，还有其他物体的反射出去的 Radiance

The Rendering Equation

Re-write the reflection equation:
$$L_r\left(\mathrm{p},{\omega }_r\right)={\int }_{H^2}{f}_r\left(\mathrm{p},{\omega }_i\to {\omega }_r\right)L_i\left(\mathrm{p},{\omega }_i\right)\cos {\theta }_i\mathrm{d}{\omega }_i$$
by adding an Emission term to make it general（考虑物体会发光的情况），这样就得到了 Rendering Equation
$$L_o\left(p,{\omega }_o\right)=L_e\left(p,{\omega }_o\right)+{\int }_{{\Omega }^{+}}{L}_i\left(p,{\omega }_i\right)f_r\left(p,{\omega }_i,{\omega }_o\right)\left(n\cdot {\omega }_i\right)\mathrm{d}{\omega }_i$$
Note: now, we assume that all directions are pointing outwards!

Understanding the rendering equation

Reflection Equation

• 当有一个点光源时

• 当有很多点光源时

• 当有面光源时（看成点光源的集合）

• 当有其他物体反射的 Radiance 时

Rendering Equation as Integral Equation

• 对渲染方程进行简写后的结果

Linear Operator Equation

• 简化写成算子形式，目的是对渲染方程进行求解，中间省略了很多步骤

Ray Tracing and extensions

• General class numerical Monte Carlo methods
• Approximate set of all paths of light in scene

Ray Tracing

• 将最后看到的图写成直接看到光源、弹射一次、弹射二次…、弹射 $n$ 次的和，结果是全局光照

• 光栅化只解决了 $E+KE$ 部分

Monte Carlo Path Tracing

Monte Carlo Integration

Why&What Monte Carlo Integration

• 对于给定函数的定积分，直接通过解析计算求解难度太大
• Monte Carlo Integration 是一种数值方法，求出的只是一个数
• 在积分域内不断采样，假设积分区域是长方形，对于所有采样得到的结果求平均

Define Monte Carlo Integration

• Definite integral

$${\int }_a^{b}f(x)dx$$

• Random variable

$$X_i\sim p(x)$$

• Monte Carlo estimator

$$F_N=\int f(x)\mathrm{d}x=\frac{1}{N}\sum _{i=1}^N\frac{f\left(X_i\right)}{p\left(X_i\right)}$$

• Note that
  • The more samples, the less variance.
  • Sample on $x$, integrate on $x$.

• 如果随机变量均匀采样（均匀分布），Monte Carlo Integration 有以下形式

  • Definite integral
    $${\int }_a^{b}f(x)dx$$

  • Uniform random variable

  $$X_i\sim p(x)=\frac{1}{b-a}$$

  • Basic Monte Carlo estimator

  $$F_N=\frac{b-a}{N}\sum _{i=1}^{N}f\left(X_i\right)$$

Path Tracing

Motivation: Problems of Whitted-Style Ray Tracing

For Whitted-style ray tracing, it has some cons/ simplifications

• Always perform specular reflections / refractions

• Stop bouncing at diffuse surfaces

Whitted-Style Ray Tracing: Problem 1

• Whitted-Style Ray Tracing 对于 glossy materials 还认为光线是镜面反射的话是不对的

Whitted-Style Ray Tracing: Problem 2

• 对于 Whitted-Style Ray Tracing, No reflections between diffuse materials
  • 体现在接触不到光的漫反射面反射出了接触到光的红色面的光 (color bleeding)

Problem Summary: Whitted-Style Ray Tracing is Wrong

But the rendering equation is correct
$$L_o\left(p,{\omega }_o\right)=L_e\left(p,{\omega }_o\right)+{\int }_{{\Omega }^{+}}{L}_i\left(p,{\omega }_i\right)f_r\left(p,{\omega }_i,{\omega }_o\right)\left(n\cdot {\omega }_i\right)\mathrm{d}{\omega }_i$$
But it involves

• Solving an integral over the hemisphere, and
• Recursive execution

A Simple Monte Carlo Solution

因此，对于着色点 $p$ 的 Monte Carlo Integration 为
$$\begin{array}{rl}{L}_o\left(p,{\omega }_o\right)& ={\int }_{{\Omega }^{+}}{L}_i\left(p,{\omega }_i\right)f_r\left(p,{\omega }_i,{\omega }_o\right)\left(n\cdot {\omega }_i\right)\mathrm{d}{\omega }_i\\ & \approx \frac{1}{N}\sum _{i=1}^N\frac{L_i\left(p,{\omega }_i\right)f_r\left(p,{\omega }_i,{\omega }_o\right)\left(n\cdot {\omega }_i\right)}{p\left({\omega }_i\right)}\end{array}$$

shade(p, wo)
	Randomly choose N directions wi~pdf
	Lo = 0.0
	For each wi
		Trace a ray r(p, wi)
		If ray r hit the light
			Lo += (1 / N) * L_i * f_r * cosine / pdf(wi)
	Return Lo

Introducing Global Illumination

Naive Algorithm

• 考虑物体反射过来的光线，如果打到物体，就递归计算

shade(p, wo)
    Randomly choose N directions wi~pdf
    Lo = 0.0
    For each wi
        Trace a ray r(p, wi)
		If ray r hit the light	
			Lo += (1 / N) * L_i * f_r * cosine / pdf(wi)
		Else If ray r hit an object at q
			Lo += (1 / N) * shade(q, -wi) * f_r * cosine / pdf(wi)
	Return Lo

Problem #1: Explosion of #rays as #bounces go up - let N = 1!

• 光线仅仅弹射两次的话数量级就已经不可以接受了
• Key observation: #rays will not explode iff $N=1$
  • 当使用 $N=1$ 进行 Monte Carlo Integration 时 $\to$ 路径追踪 (path tracing)
  • 当使用 $N\ne 1$ 进行 Monte Carlo Integration 时 $\to$ 分布式光线追踪
• 使用 $N=1$ 进行 Monte Carlo Integration 噪声很大，但只要使用多个路径穿过一个像素对它们求平均即可

修改刚刚的算法

shade(p, wo)
	Randomly choose ONE direction wi~pdf(w)
	Trace a ray r(p, wi)
	If ray r hit the light
		Return L_i * f_r * cosine / pdf(wi)
	Else If ray r hit an object at q
		Return shade(q, -wi) * f_r * cosine / pdf(wi)
		
ray_generation(camPos, pixel)
	Uniformly choose N sample positions within the pixel
	pixel_radiance = 0.0
	For each sample in the pixel
		Shoot a ray r(camPos, cam_to_sample)
		If ray r hit the scene at p
			pixel_radiance += 1 / N * shade(p, sample_to_cam)
	Return pixel_radiance

Problem #2: The recursive algorithm will never stop - Russian Roulette!

• 但是在真实世界中，光线是不会停止弹射的，限制弹射次数并不合理，因为对于弹射次数的削减相当于直接削减了能量，这违背了能量守恒定律
• Solution: Russian Roulette (RR) （俄罗斯轮盘赌）
  • 在一定的概率下停止追踪，方法如下
    1. 假设得到的理想追踪结果为 $L_o$
    2. 人工设定一个概率 $P(0<P<1)$
    3. 以概率 $P$ shoot a ray，返回 the shading result divided by $P:L_o/P$
    4. 以概率 $1-P$ shoot a ray，返回 the shading result is $0$
  • 在这种情况下的期望 $E=P\times (L_o/P)+(1-P)\times 0=L_o$

代码如下

shade(p, wo)
    Manually specify a probability P_RR
    Randomly select ksi in a uniform dist. in [0, 1]
	If (ksi > P_RR) return 0.0;
	
	Randomly choose ONE direction wi~pdf(w)
	Trace a ray r(p, wi)
	If ray r hit the light
		Return L_i * f_r * cosine / pdf(wi) / P_RR
	Else If ray r hit an object at q
		Return shade(q, -wi) * f_r * cosine / pdf(wi) / P_RR

Problems of Path Tracing

Problem #1: Not very efficient

浪费现象产生的原因是我们均匀的在半球上采样，很大一部分采样没有进入光源，如果找到一个好的概率密度函数，可以提高效率 $\to$ 直接在光源上进行采样

Solution #1: Sampling the Light (pure math)

• Assume uniformly sampling on the light: $\text{pdf}=1/A$ (because $\int \text{pdf}\mathrm{d}A=1$)

• But the rendering equation integrates on the solid angle: $L_o=\int L_{i}frcos\mathrm{d}\omega .$

  • Recall Monte Carlo Integration: Sample on $x$ & integrate on $x$

  • we sample on the light, so we must integrate on the light

  • 我们只要知道 $\mathrm{d}\omega$ 和 $\mathrm{d}A$ 的关系即可

    • Recall the alternative def. of solid angle: Projected area on the unit sphere
    • 将 $\mathrm{d}A$ 往单位球上投影即可 ( $\mathrm{d}A\cos {\theta }^{\prime }$ 就是把面转到朝向中心的方向，根据立体角的定义来理解下面的式子)

    $$\mathrm{d}\omega =\frac{\mathrm{d}A\cos {\theta }^{\prime }}{{\Vert x^{\prime }-x\Vert }^2}$$

• 我们重写渲染方程即可（变量替换，改变积分域）
  $$\begin{array}{rl}{L}_o\left(x,{\omega }_o\right)& ={\int }_{{\Omega }^{+}}{L}_i\left(x,{\omega }_i\right)f_r\left(x,{\omega }_i,{\omega }_o\right)\cos \theta \mathrm{d}{\omega }_i\\ & ={\int }_A{L}_i\left(x,{\omega }_i\right)f_r\left(x,{\omega }_i,{\omega }_o\right)\frac{\cos \theta \cos {\theta }^{\prime }}{{\Vert x^{\prime }-x\Vert }^2}\mathrm{d}A\end{array}$$

Previously, we assume the light is “accidentally” shot by uniform hemisphere sampling

Now we consider the radiance coming from two parts:

1. light source (colored blue, direct, no need to have RR)
2. other reflectors (colored orange, indirect, RR)

• 代码如下（拆分直接光照和间接光照）

shade(p, wo)
	# Contribution from the light source.
	Uniformly sample the light at x’ (pdf_light = 1 / A)
	L_dir = L_i * f_r * cos θ * cos θ’ / |x’ - p|^2 / pdf_light
	
	# Contribution from other reflectors.
	L_indir = 0.0
	Test Russian Roulette with probability P_RR
	Uniformly sample the hemisphere toward wi (pdf_hemi = 1 / 2pi)
	Trace a ray r(p, wi)
	If ray r hit a non-emitting object at q
		L_indir = shade(q, -wi) * f_r * cos θ / pdf_hemi / P_RR
	
	Return L_dir + L_indir

Problem #2: Light is not blocked or not

• 在之前的计算中，假设了光线不会被挡到

Solution #2: Give a ray from P

• 考虑点 $p$ 到 $x^{\prime }$ 的连线，从 $p$ 点打一根光线，看看有没有被挡到

# Contribution from the light source.
L_dir = 0.0
Uniformly sample the light at x’ (pdf_light = 1 / A)
Shoot a ray from p to x’
If the ray is not blocked in the middle
	L_dir = …

Some Side Notes

• Previous

  • Ray tracing == Whitted-style ray tracing

• Modern (my own definition)

  • The general solution of light transport, including
  • (Unidirectional & bidirectional) path tracing
  • Photon mapping
  • Metropolis light transport - VCM / UPBP…

• Uniformly sampling the hemisphere

  • How? And in general, how to sample any function? (sampling)

• Monte Carlo integration allows arbitrary pdfs

  • What’s the best choice? (importance sampling)
  • 重要性采样：针对某一种特定形状进行采样

• Do random numbers matter?

  • Yes! (low discrepancy sequences)

• I can sample the hemisphere and the light

  • Can I combine them? Yes! (multiple imp. sampling)
  • 结合从半球和光源的采样方法

• The radiance of a pixel is the average of radiance on all paths passing through it

  • Why? (pixel reconstruction filter)

• Is the radiance of a pixel the color of a pixel?

  • No. (gamma correction, curves, color space)