在屏幕上画出一个实心三角形.
相较于hw1,多了一些内容,需要好好看下框架.

判断$(x,y)$对应的pixel否在三角形内部.

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static bool insideTriangle(int x, int y, const Vector3f* _v)

{
    // TODO : Implement this function to check if the point (x, y) is inside the triangle represented by _v[0], _v[1], _v[2]
    Vector2f vertex2p[3], vertex2vertex[3];
    int i;
    //储存顶点到测试点的向量
    for (i = 0;i < 3;i++) {
        vertex2p[i] << x - _v[i].x(), y - _v[i].y();
    }
    //储存顶点到顶点之间的向量
    for (i = 0;i < 3;i++) {
        vertex2vertex[i] << _v[(i + 1) % 3].x() - _v[i].x(), _v[(i + 1) % 3].y() - _v[i].y();
    }
    //计算叉乘法.
    int result[3];
    for (i = 0;i < 3;i++) {
        result[i] = vertex2vertex[i].x()*vertex2p[i].y()
                            -vertex2vertex[i].y()*vertex2p[i].x();
    }
    if(result[0]>0 && result[1]>0 && result[2]>0){
        return true;
    }
    else if(result[0]<0 && result[1]<0 && result[2]<0){
        return true;
    }
    else{
        return false;
    }

}

先限定下founding box范围,提高效率

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void rst::rasterizer::rasterize_triangle(const Triangle& t) {
    auto v = t.toVector4();

    // TODO : Find out the bounding box of current triangle.
    // iterate through the pixel and find if the current pixel is inside the triangle
    float xmin,ymin,xmax,ymax;
    xmin=std::min(std::min(v[0].x(),v[1].x()),v[2].x());
    ymin=std::min(std::min(v[0].y(),v[1].y()),v[2].y());
    xmax=std::max(std::max(v[0].x(),v[1].x()),v[2].x());
    ymax=std::max(std::max(v[0].y(),v[1].y()),v[2].y());
    xmin=std::floor(xmin);
    xmax=std::ceil(xmax);
    ymin=std::floor(ymin);
    ymax=std::ceil(ymax);
    
    for(int x=xmin;x<=xmax;x++){
        for(int y=ymin;y<=ymax;y++){
            if(insideTriangle(x+0.5,y+0.5,t.v)){
                // If so, use the following code to get the interpolated z value.
                auto[alpha, beta, gamma] = computeBarycentric2D(x, y, t.v);
                float w_reciprocal = 1.0/(alpha / v[0].w() + beta / v[1].w() + gamma / v[2].w());
                float z_interpolated = alpha * v[0].z() / v[0].w() + beta * v[1].z() / v[1].w() + gamma * v[2].z() / v[2].w();
                z_interpolated *= w_reciprocal;
                    // TODO : set the current pixel (use the set_pixel function) to the color of the triangle (use getColor function) if it should be painted.
                if (z_interpolated<depth_buf[get_index(x,y)]){
                    set_pixel(Vector3f(x,y,z_interpolated),t.getColor());
                    depth_buf[get_index(x,y)]=z_interpolated;
                }
            }
        }
    }

总的来说,hw1是比较简单的.
本次作业的任务是填写一个旋转矩阵和一个透视投影矩阵,只要记得上课时候给的公式,写进去就行.

• 旋转矩阵
  注意,cos,sin等,需要的参数为弧度制(rad),而不是角度制

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  Eigen::Matrix4f get_model_matrix(float rotation_angle) { Eigen::Matrix4f model = Eigen::Matrix4f::Identity(); // TODO: Implement this function // Create the model matrix for rotating the triangle around the Z axis. // Then return it. rotation_angle = rotation_angle / 180 * MY_PI; Eigen::Matrix4f rotate; rotate << cos(rotation_angle), -sin(rotation_angle), 0, 0, sin(rotation_angle), cos(rotation_angle), 0, 0, 0, 0, 1, 0, 0, 0, 0, 1; model = rotate * model; return model; }

先进行挤压,再进行正交投影得到结果

• persp2ortho

  • 透视投影到正交投影的变化矩阵

• ortho

  • 正交矩阵
    projection = ortho * persp2ortho * projection;

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    Eigen::Matrix4f get_projection_matrix(float eye_fov, float aspect_ratio, float zNear, float zFar) { // Students will implement this function Eigen::Matrix4f projection = Eigen::Matrix4f::Identity(); // TODO: Implement this function // Create the projection matrix for the given parameters. // Then return it. Eigen::Matrix4f persp2ortho = Eigen::Matrix4f::Identity(); Eigen::Matrix4f ortho = Eigen::Matrix4f::Identity(); float top = abs(zNear) * tan(eye_fov / 2 / 180 * MY_PI); float bottom = -top; float right = top * aspect_ratio; float left = -right; persp2ortho << zNear, 0, 0, 0, 0, zNear, 0, 0, 0, 0, zNear + zFar, -zNear * zFar, 0, 0, 1, 0; ortho << 2 / (right - left), 0, 0, -(right + left) / (right - left), 0, 2 / (top - bottom), 0, -(top + bottom) / (top - bottom), 0, 0, 2 / (zNear - zFar), -(zNear + zFar) / (zNear - zFar), 0, 0, 0, 1; projection = ortho * persp2ortho * projection; return projection; }

最后得到的结果与官方答案不同,图形倒转了.这是zNear值导致的.
根据课上的内容,看向的是Z轴负半轴,Z<0; 而代码中,看向Z正半轴,Z>0;

小问题懒得修改了.

顺便学习了一下cmake的相关知识
[[07archive/tech/cmake|cmake]]

环境配置

难绷,论坛上给的虚拟机是ubuntu18 并且是virtual box的版本.
想用vmware虚拟机跑.将vdi格式转化为vmdk,并创建对应的虚拟机.
问题在于VMware tools安装失败.每次安装完成之后,再reboot,就会变成没有VMware tools…搞不懂.
还是在windows上配置环境算了…

看了下windows配置eigen和opencv有点麻烦,选择换成虚拟机ubuntu20.04LTS自己配置

更新
breaking news!
官方给的虚拟机,转换为vmware之后,vmwaretools安装成功了…
直接用就行了.配置环境结束.

ubuntu具体设置

参考

遇上vscode编译时候报错

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/usr/bin/gcc -fdiagnostics-color=always -g /home/l4rk/games101/pa0.c -o /home/l4rk/games101/pa0 /usr/bin/ld: /tmp/ccjPHxdQ.o: in function `main': /home/l4rk/games101/pa0.c:6: undefined reference to `sin' collect2: error: ld returned 1 exit status

Lec1 overview of Computer Graphic

判断游戏画面水平如何
可以去看画面的明暗.亮的,一般都是采用了全局光照技术.从技术层面来看更加优秀

特效是最简单的图形学技术应用
特效是特殊的效果.在平常生活中见得很少,即使特效出错,观众不一定看得出来.而对日常东西的渲染更为困难

Visualization
Virtual Reality
Digital Illustration数字图像处理
Simulation
Graphic User Interface
Typography矢量字体?

• course topic
  • Rasterazation
  • Curves and Meshes
  • Ray Tracing
  • Animation / Simulation

CG与CV
一切需要理解,猜测的都是计算机视觉的内容.
no clear boundaries

Lec2 Review of Linear Algebra

A Swift and Brutal Introduction to Linear Algebra

Vectors

$\vec {a}$ 或者$\boldsymbol {a}$