fix error

Author: C-none

.github/workflows/jekyll.yml

@@ -34,7 +34,7 @@ jobs:
       - name: Checkout
         uses: actions/checkout@v4
       - name: Setup Ruby
-        uses: ruby/setup-ruby@v1 # v1.146.0
+        uses: ruby/setup-ruby@v1 # latest version
         with:
           ruby-version: "3.1" # Not needed with a .ruby-version file
           bundler-cache: true # runs 'bundle install' and caches installed gems automatically

_posts/2021-12-31-easy-software-rasterizer.md

@@ -22,7 +22,7 @@ show_sidebar: false
 <div class="more"><a href="https://github.com/C-none/easy-software-rasterizer">code</a></div>
 
 ## Introduction
-This simple rasterizer is based on modern CPP, mainly before c++20. It is a good example for learning how to write a software rasterizer, including the basic rasterizing pipeline--vertex processing, primitive assembly, clipping, rasterization, visibility test, and fragment processing, . 
+This simple rasterizer is based on modern CPP, mainly before c++20. It is a good example for learning how to write a software rasterizer, including the basic rasterizing pipeline--vertex processing, primitive assembly, clipping, rasterization, visibility test, and fragment processing. 
 Additionally, it supports `Blinn-Phong shading model`, `texture mapping`, and `SSAA`.
 
 <center>

_posts/2024-1-11-screen-probes.md

@@ -44,23 +44,23 @@ The implementation of Spherical Harmonics refers to [this](https://github.com/Ep
 To project the irradiance to the SH basis,
 
 $$
-\Large f_l^m=\int_{\Omega}I(\omega)Y_l^m(\omega)d\omega
+\huge f_l^m=\int_{\Omega}I(\omega)Y_l^m(\omega)d\omega
 $$
 
 where $l$ indicates the band, $m$ indicates the order, $\omega$ is the direction of the incident light, $I(\omega)$ is the irradiance, $Y_l^m(\omega)$ is the SH basis, $\Omega$ is the solid angle, and $f_l^m$ is the corresponding SH coefficient.
 
 To estimate coefficients,
 
 $$
-\Large f_l^m=\frac{1}{N}\sum_{i=1}^NI(\omega_i)Y_l^m(\omega_i)/pdf(\omega_i)
+\huge f_l^m=\frac{1}{N}\sum_{i=1}^NI(\omega_i)Y_l^m(\omega_i)/pdf(\omega_i)
 $$
 
 where $N$ is the number of samples, and $pdf(\omega_i)$ is the probability density function of the direction $\omega_i$.
 
 To reconstruct the irradiance from the SH basis,
 
 $$
-\Large I(\omega)=\sum_{l=0}^L\sum_{m=-l}^lf_l^mY_l^m(\omega)
+\huge I(\omega)=\sum_{l=0}^L\sum_{m=-l}^lf_l^mY_l^m(\omega)
 $$
 
 ### Uniform sampling on a hemisphere
@@ -72,19 +72,19 @@ Imagine that we divide the hemisphere into infinite small rings, the probability
 To normalize the pdf, we need to divide the pdf by the integral of the pdf:
 
 $$
-\Large pdf(\theta)=\frac{\sin\theta}{\int_0^{\frac{\pi}{2}}\sin\theta d\theta}=\sin\theta
+\huge pdf(\theta)=\frac{\sin\theta}{\int_0^{\frac{\pi}{2}}\sin\theta d\theta}=\sin\theta
 $$
 
 Thus, the cumulative distribution function of $\theta$ is:
 
 $$
-\Large cdf(\theta)=\int_0^{\theta}pdf(\theta)d\theta=1-\cos\theta
+\huge cdf(\theta)=\int_0^{\theta}pdf(\theta)d\theta=1-\cos\theta
 $$
 
 The inverse function of $cdf(\theta)$ is:
 
 $$
-\Large \theta=acos(1-u)=acos(u)
+\huge \theta=acos(1-u)=acos(u)
 $$
 
 where $u$ is a random number in $[0,1)$.
@@ -96,25 +96,25 @@ The pdf of direction is $\frac{1}{2\pi}$, because it is uniform sampling on a he
 Considering the rendering equation:
 
 $$
-\Large L_o(p,\omega_o)=L_e(p,\omega_o)+\int_{\Omega}f(p,\omega_i,\omega_o)L_i(p,\omega_i)(n\cdot\omega_i)d\omega_i
+\huge L_o(p,\omega_o)=L_e(p,\omega_o)+\int_{\Omega}f(p,\omega_i,\omega_o)L_i(p,\omega_i)(n\cdot\omega_i)d\omega_i
 $$
 
 In coding, I set the pdf of $\theta$ to be proportional to $(n\cdot\omega_i)=cos\theta$, which is the cosine term in the rendering equation. Thus, the pdf of $\theta$ is:
 
 $$
-\Large pdf(\theta)=\frac{\cos\theta\sin\theta}{\int_0^{\frac{\pi}{2}}\cos\theta\sin\theta d\theta}=2\cos\theta\sin\theta=sin(2\theta)
+\huge pdf(\theta)=\frac{\cos\theta\sin\theta}{\int_0^{\frac{\pi}{2}}\cos\theta\sin\theta d\theta}=2\cos\theta\sin\theta=sin(2\theta)
 $$
 
 The cumulative distribution function of $\theta$ is:
 
 $$
-\Large cdf(\theta)=\int_0^{\theta}pdf(\theta)d\theta=\frac{1}{2}(1-\cos(2\theta))=sin^2(\theta)
+\huge cdf(\theta)=\int_0^{\theta}pdf(\theta)d\theta=\frac{1}{2}(1-\cos(2\theta))=sin^2(\theta)
 $$
 
 The inverse function of $cdf(\theta)$ is:
 
 $$
-\Large \theta=asin(\sqrt{u})=acos(\sqrt{1-u})=acos(\sqrt{u}) \tag{*}
+\huge \theta=asin(\sqrt{u})=acos(\sqrt{1-u})=acos(\sqrt{u}) \tag{*}
 $$
 
 where $u$ is a random number in $[0,1)$.