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thought ：
A visible point , You can see , The light source can also be seen
An invisible point （ shadow ）, You can see , The light source cannot be seen
classical Shadow Mapping It can only deal with point light （ Hard shadow , The boundary is either black or white ）

1. Starting from the light source , Look at each pixel , Record the depth of each point

2. Starting from the camera , From the point of view Project back to the light source , Record a depth , If this point is visible , Then the two depth effects are the same

3. Starting from the camera , Project back to the light source from the point you see , Record a depth , It is found that the depth is different from the depth from the light source , It's a shadow

If an object has soft shadows , Then its light source must be of size .

The difference and principle between hard shadow and soft shadow , That is, the umbra (Umbra) And penumbra (Penumbra) The difference between .

thus ,shadow maps There are some problems ：

1. Ray tracing ： light path ( Line ) Reversibility of 、 Refraction of light 、 Coloring contribution

1.0 Why Ray Tracing？ Limitations of rasterization

Limitations of rasterization ：

Unable to deal with multiple light ejections , Global illumination
Rasterization is fast , But the quality is relatively low

1.1 Recursive (Whitted-Style) Ray Tracing： Simulate the process of continuous light ejection

Quote teacher Yan's original words in class to make a refinement ：
This is a glass ball , Light from the line of sight , Two things happen ： Refraction and reflection
The coloring has changed ： Originally, we only focused on whether the point can be illuminated by the light source , Calculate only once ;Whitted-Style It is considered that if the light source can illuminate any point where ejection can occur , Just add the value of their coloring to the value of pixels .
Consider the energy loss of light .

The results of each light path that calculates the refraction of light at different times are accumulated into one pixel .

At every ejection point , Will calculate the value of its coloring . Finally, it will be added to the pixel points through which the line of sight passes .

1.2 Intersection of light and surface Ray-Surface Intersection

1.2.1 Ray equation ：r(t) = o + td

1.2.2 The light intersects with the sphere ：

(o+td)^2-R^2=0

1.2.3 Intersection of light and implicit surface ：

1.2.4 Intersection of light and triangular mesh ： The two methods

decompose ： First find the intersection of light and surface , Then judge whether the intersection point is in the triangle
The plane normal and a point in the plane determine the plane equation , Equal to the ray equation
Möller Trumbore Algorithm： The coordinates of the center of gravity , Directly judge the intersection of triangle and light , Clem's law

2. Ray tracing acceleration ：

The most basic ray tracing will intersect all triangular faces in the scene , And calculate the value of each pixel ,Very Slow！

2.1 Bounding Volumes：AABB（ Axis alignment bounding box ）

And x,y,z Three faces with parallel axes , Therefore, it is called axis alignment .

2.2.1 Judge whether the light is consistent with AABB The intersection ： Take the intersection of the maximum entry time and the minimum departure time

When the light Align faces through all axes , Enter AABB
When the light Just leave one side , Just leave AABB
When light enters Box Time is less than leaving Box Time , That is, the light source is Box Inside

Here we consider the following premises ： Light is a ray ！, With several special cases of concentration ：

$t_{exit}<0$ ：Box After the light source , That is, light and Box There are no intersections ！
$t_{exit}\geqslant 0$ && $t_{enter}< 0$ ： The light source is Box Inside , There are intersections ！

All in all , If and only if ： $t_{enter}<t_{exit}$ && $t_{exit}\geq 0$ , Light and Box There are intersections ！

Then why use axial symmetry This nature ？ The reason is simple , It avoids the trouble of calculating the normals that are not parallel to the axis plane ！

2.2 Space division

2.2.1 Oct-Tree、KD-Tree、BSP-Tree

Oct-Tree： Similar to cutting tofu , Cut three knives for three-dimensional objects , Divided into eight pieces ; Cut two knives for two-dimensional objects , Divided into four pieces ; And so on .

How to cut ？ For example, cut two knives in two-dimensional space （ More 4 block ）, I found that three pieces were empty after this part was cut , Then there is no need to continue cutting . Of course, this is just an example of intuitive understanding , In fact, when to stop depends on the specific constraints , For example, how many objects in this part can stop , Or until this grid is empty .

But as dimensions increase , Tangent space is called exponential rise .

KD-Tree： Each time along a horizontal or vertical axis (xyz) Divide , In two

2.2.2 KD-Tree Light intersection judgment process ：

but

How to judge triangle and AABB It is difficult to have intersections , therefore KD-Tree Gradually reduce the use .
An object may appear in multiple leaf nodes , That is, in multiple grids .

2.3 Object division ：Object Partitions & Bounding Volume Hierarchy(BVH)

Divide the objects into two piles , Recalculate the bounding box

Fast selection algorithm ： Fast selection algorithm - Salty_Fish - Blog Garden (cnblogs.com)

3. Radiometry (Radiometry)

3.0 Motivation

The intensity of light Intensity？etc...

Give a precise definition of all aspects of the physical meaning of light ！

3.1 Radiant Energy & Flux（Power）

3.1.1 Radiant Energy： energy

3.1.2 Radiant Flux：Lumen( Lumen )、 power

3.2 Radiant Intensity： Intensity of light emitted per solid angle

3.3 Irradiance： How much energy is received per unit surface

3.3.1 Lamber's Cosint Law： The energy received by the surface is only related to the vertical direction

3.4 Radiance： Per unit solid angle 、 Per unit surface Accept 、 How much energy is emitted （ Describe light ）

3.4.1 Incident Radiance： Per unit solid angle reaching the surface Irradiance

3.4.2 Exiting Radiance： Emitted per unit area away from the surface Intensity

Irrandiance And Radiance The connection and difference between ： The relation of integral , Different units

3.5 Bidirectional Reflectance Distribution Function (BRDF)： Describe the energy of light incident in a certain direction and reflected in a certain direction

A light enters from a certain direction , It will reflect in all directions , What we need is the energy carried by the reflected light in a specific direction .

It can also be understood as ： Light carries energy into a certain point , This point will absorb the energy of light , Then reflect and release this part of energy from this point . But we don't know how much energy is emitted in a specific direction , At this time, we need such a function , That is to say BRDF.

BRDF Describes how light interacts with objects , It is this concept , Determines the material of the object ,BRDF Items define different materials .

3.5.1 Reflection Equation ： The contribution of all incident light to a reflected light

3.5.2 Rendering equations The Rendering Equation： Reflection Equation + Self luminous items

3.5.3 Global illumination ： Understanding of reflection equation — Direct light + Indirect care

Reflected light = Self luminous +Sum（（ Point light emitting ）× BRDF( texture of material ) × Angle of incident light cos ）

If it is an area light source , Then it can be obtained by integrating the point light source on the surface ：

A linear representation of the reflection equation , Its purpose is to understand that under ray tracing , Light is a combination of direct and indirect light , Indirect illumination includes one direct illumination and several bounces （ It's a little vague here , I still have to go back and listen to the teacher's explanation ）：

4. Monte Carlo path tracking

4.1 Monte Carlo integral ： Find the definite integral of a complex function — Analytic solution --> Numerical solution

Want to find an integral , But the function form is particularly complex , Cannot provide a mathematical expression , That is, the analytic form , To calculate the definite integral .

How to solve ？ Approximate the definite integral by averaging the randomly sampled values .

Monte Carlo integral formula ：

4.2 Path tracking

reflection ：Whitted Style Must ray tracing be right ？

Constantly eject light , Connect a light anywhere .

Hit objects such as glass , Reflection or refraction will occur ;

Hit the diffuse object , And it stopped .

Are these two things really right ？

not always . Here's the picture , Specular reflection Whitted It may be right , but Glossy It's wrong. , He should not be with specular The same reflection

The following example illustrates a problem more ： The light hits diffuse Objects should never stop

Whitted It's wrong. , But the rendering equation is right ：

Rendering equations need to solve two problems ：

How to integrate it ？（ In the hemisphere ）
How to deal with recursive process ？（ Because every point of ejection comes from the light received by the previous point ）

4.2.1 Monte Carlo method for definite integral （ Simple case ： Direct light ）

Consider a simple case first , This point does not glow , Find the integral result of direct illumination , That is to say, starting from this point, start tracking , See how much light directly reaches the light source ：

In this simple case , We do not consider self illumination at this point , And also know the hemisphere PDF, Then we can use Monte Carlo integral to approximate the result ：

The pseudo code is as follows , Well understood. , From this point out wi Root light , If a ray of light reaches the light source , Then add the result to the value of this part of the energy integral .

4.2.2 Monte Carlo definite integral （ Global illumination ）

So how to consider the global illumination ？

Suppose the traced light reaches an object , Not the light source , At the same time, the hit object is also affected by the light source , Then at this time, let's consider the problem of transformation ：

Convert the global illumination into direct illumination between objects , That is to treat the object as a light source , It becomes a local direct illumination , Recursively find the direct illumination of the next object .

The depth of recursion , That is, the number of times light is ejected from the light source to the pixel .

thus , Convert the total ejection times of global illumination into the number of local objects “ Direct light ”

The pseudocode is as follows ：

But there are still two problems ：

If a point is ejected once , issue 100 Root light , At the next ejection 100 Root light , Only two ejections , Light from just one point , There will be 10000 The root light emits . This exponential level algorithm is unacceptable .
When recursion stops ？ As mentioned above, recursion stops a certain number of times , But in nature, light can be ejected countless times , How to solve this problem ？

At this time, the basic situation of path tracking is introduced ： When N=1 when , That is, each ejection point emits only one light , It's route tracking . Remove from pseudo code for Circulation is enough ：

At this time, each pixel is calculated radiance The pseudo-code is as follows ：

that N=1 The situation is path tracking , Solved the problem of explosion of light quantity index , So how to solve the problem of recursion depth ？

Russian Roulette ！

in other words , We stop and continue to track the path with a certain probability .

Its principle is also relatively simple ： Finally, the expectation of probability is still the result of definite integral ：

It's right in principle , But there are still some problems ： Whether you can reach the light source depends entirely on luck ！

Because it is a probability based event , The amount of light emitted by the point and the size of the light source have a great relationship with probability ！

A lot of light is wasted ！

At this time, other sampling methods need to be considered .

We consider the , No more sampling from all directions , Only consider sampling in the area of the light source .

But there's a problem here , I sample on the light source area , But the solid angle of the hemispherical region of the point is integrated ,

Then we have to write the rendering equation as an integral on the light source . as follows , Need a corresponding relationship ： Unit solid angle and unit area

Finally, consider the last case ： Whether there are objects blocking the direct illumination

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