CG-Assignment-2

Question-1

Give a sequence of 4x4 matrices that transforms the parallelogram in the left figure to the unit square in the right. Write down the name of the transformation for each matrix. Also write down the transformation matrix of each step.

Firstly，Translate the parallelogram to the origin. The translation matrix is:

$$M_{Translation}=\left[\begin{array}{cccc}1& 0& 0& -0.6\\ 0& 1& 0& 1\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]$$

Then sheared the parallelogram to a rectangle. The shear matrix is:

$$M_{Sheer}=\left[\begin{array}{cccc}1& 0.2& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]$$

Finally, scale the rectangle to a unit square. The scale matrix is:

$$M_{Scale}=\left[\begin{array}{cccc}2.5& 0& 0& 0\\ 0& 0.5& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]$$

The final transformation matrix is the product of the three matrices:

$M_{Composite}=M_{Scale}\cdot M_{Sheer}\cdot M_{Translation}$

Question-2

The following converts an oblique frustum into an orthogonal one where the z-axis coincides with the frustum’s central line. Note that $x_a=\frac{x_r-x_l}{2}$ and $y_a=\frac{y_t-y_b}{2}$. Give the matrix of the transformation.

To transform the oblique frustum into an orthogonal frustum, we use a shear transformation matrix that aligns the centerline with the $z$-axis.

• Original frustum coordinates:
  • Right-top: $(x_r,y_t,z_d)$,
  • Left-bottom: $(x_l,y_b,z_d)$.
• New orthogonal frustum:
  • Right-top: $(x_a,y_a,z_d)$,
  • Left-bottom: $(-x_a,-y_a,z_d)$, where $x_a=\frac{x_r-x_l}{2}$ and $y_a=\frac{y_t-y_b}{2}$.

Shear Factors: To remove the offsets caused by the oblique frustum:

$$s_x=-\frac{x_r+x_l}{2z_d},s_y=-\frac{y_t+y_b}{2z_d}.$$

Shear Transformation Matrix:

$${\mathbf{M}}_{\text{shear}}=\left[\begin{array}{cccc}1& 0& -\frac{x_r+x_l}{2z_d}& 0\\ 0& 1& -\frac{y_t+y_b}{2z_d}& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right].$$

Question-3

Consider the figure below. i is the direction of the incident light, n is the normal of the surface at S and r is the direction of the reflection. All of them are unit vectors. Derive a formula for r in terms of i and n.

Decompose the incident vector $\mathbf{i}$:

$$\mathbf{i}={\mathbf{i}}_{\parallel }+{\mathbf{i}}_{\perp },$$

where:

• ${\mathbf{i}}_{\parallel }=(\mathbf{i}\cdot \mathbf{n})\mathbf{n}$ (component parallel to the normal),
• ${\mathbf{i}}_{\perp }=\mathbf{i}-{\mathbf{i}}_{\parallel }$ (component perpendicular to the normal).

When reflected:

• The parallel component ${\mathbf{i}}_{\parallel }$ is reversed: $-{\mathbf{i}}_{\parallel }$,
• The perpendicular component ${\mathbf{i}}_{\perp }$ stays the same.

So the reflected vector becomes:

$$\mathbf{r}={\mathbf{i}}_{\perp }-{\mathbf{i}}_{\parallel }.$$

Substitute ${\mathbf{i}}_{\parallel }$:

$$\mathbf{r}=(\mathbf{i}-(\mathbf{i}\cdot \mathbf{n})\mathbf{n})-(\mathbf{i}\cdot \mathbf{n})\mathbf{n}.$$

Simplify:

$$\mathbf{r}=\mathbf{i}-2(\mathbf{i}\cdot \mathbf{n})\mathbf{n}.$$

Question-4

(a)

Derive the $4\times 4$ homogenous matrix representation of a rotation transformation in 3D space of 30 degree about the directed line through the origin with direction vector $(0,1,1)$.

To derive the homogeneous $4\times 4$ rotation matrix for a rotation of ${30}^{\circ }$ about the direction vector $(0,1,1)$, using Rodrigues' formula:

$$R(\mathbf{n},\alpha )=\cos (\alpha )\mathbf{I}+(1-\cos (\alpha ))\mathbf{n}{\mathbf{n}}^T+\sin (\alpha )\mathbf{N}.$$

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Normalize the Direction Vector

$$\mathbf{n}=(0,1,1)\Rightarrow \mathbf{n}=(0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}).$$

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Compute $\mathbf{N}$ (Skew-Symmetric Matrix)

$$\mathbf{N}=\left[\begin{array}{ccc}0& -n_z& n_y\\ n_z& 0& -n_x\\ -n_y& n_x& 0\end{array}\right]=\left[\begin{array}{ccc}0& -\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& 0& 0\\ -\frac{1}{\sqrt{2}}& 0& 0\end{array}\right].$$

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Outer Product $\mathbf{n}{\mathbf{n}}^T$

$$\mathbf{n}{\mathbf{n}}^T=\left[\begin{array}{c}0\\ \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{array}\right]\left[\begin{array}{ccc}0& \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\\ 0& \frac{1}{2}& \frac{1}{2}\\ 0& \frac{1}{2}& \frac{1}{2}\end{array}\right].$$

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Plug into Rodrigues' Formula Substitute into the formula for rotation, with $\alpha ={30}^{\circ }$:

• $\cos ({30}^{\circ })=\frac{\sqrt{3}}{2}$,
• $\sin ({30}^{\circ })=\frac{1}{2}$,
• $1-\cos ({30}^{\circ })=1-\frac{\sqrt{3}}{2}=\frac{2-\sqrt{3}}{2}$.

$$R(\mathbf{n},{30}^{\circ })=\cos (\alpha )\mathbf{I}+(1-\cos (\alpha ))\mathbf{n}{\mathbf{n}}^T+\sin (\alpha )\mathbf{N}.$$

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Compute Each Term

1. $\cos (\alpha )\mathbf{I}$:

$$\cos ({30}^{\circ })\mathbf{I}=\frac{\sqrt{3}}{2}\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right].$$

2. $(1-\cos (\alpha ))\mathbf{n}{\mathbf{n}}^T$:

$$(1-\cos ({30}^{\circ }))\mathbf{n}{\mathbf{n}}^T=\frac{2-\sqrt{3}}{2}\left[\begin{array}{ccc}0& 0& 0\\ 0& \frac{1}{2}& \frac{1}{2}\\ 0& \frac{1}{2}& \frac{1}{2}\end{array}\right].$$

3. $\sin (\alpha )\mathbf{N}$:

$$\sin ({30}^{\circ })\mathbf{N}=\frac{1}{2}\left[\begin{array}{ccc}0& -\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& 0& 0\\ -\frac{1}{\sqrt{2}}& 0& 0\end{array}\right].$$

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Add Terms Together Sum the above three terms to get:

$$R(\mathbf{n},{30}^{\circ })=\left[\begin{array}{ccc}\frac{\sqrt{3}}{2}& -\frac{1}{2\sqrt{2}}& \frac{1}{2\sqrt{2}}\\ \frac{1}{2\sqrt{2}}& \frac{\sqrt{3}}{2}+\frac{2-\sqrt{3}}{4}& \frac{2-\sqrt{3}}{4}\\ -\frac{1}{2\sqrt{2}}& \frac{2-\sqrt{3}}{4}& \frac{\sqrt{3}}{2}+\frac{2-\sqrt{3}}{4}\end{array}\right].$$

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Create the $4\times 4$ Homogeneous Matrix

Embed the $3\times 3$ rotation matrix into a $4\times 4$ matrix by adding a bottom row and column for homogeneity:

$$R(\mathbf{n},{30}^{\circ })=\left[\begin{array}{cccc}\frac{\sqrt{3}}{2}& -\frac{1}{2\sqrt{2}}& \frac{1}{2\sqrt{2}}& 0\\ \frac{1}{2\sqrt{2}}& \frac{\sqrt{3}}{2}+\frac{2-\sqrt{3}}{4}& \frac{2-\sqrt{3}}{4}& 0\\ -\frac{1}{2\sqrt{2}}& \frac{2-\sqrt{3}}{4}& \frac{\sqrt{3}}{2}+\frac{2-\sqrt{3}}{4}& 0\\ 0& 0& 0& 1\end{array}\right].$$

(b)

What is the image of point $(0,\sqrt{3},\sqrt{3})$ under the transformation in $(a)$?

$$v=\left[\begin{array}{c}0\\ \sqrt{3}\\ \sqrt{3}\\ 1\end{array}\right]$$$$v^{\prime }=R(\mathbf{n},{30}^{\circ })\cdot v$$

The image of the point $(0,\sqrt{3},\sqrt{3})$ under the given transformation (including homogeneous coordinates) is:

$$v^{\prime }=\left[\begin{array}{c}0\\ \sqrt{3}\\ \sqrt{3}\\ 1\end{array}\right]$$

Question-5

(a)

Given a triangle $\mathrm{△}ABC$ in 2D plane, show that any point $P$ in the plane can be represented uniquely as show that

$$P=\alpha A+\beta B+\gamma C,\text{ with }\alpha +\beta +\lambda =1.$$

1. Affine Combination: Any point $P$ in the plane of $A,B,C$ can be written as:$$P=\alpha A+\beta B+\gamma C,$$because $A,B,C$ define the plane (not collinear). To ensure $P$ lies in the affine subspace of the triangle:$$\alpha +\beta +\gamma =1.$$

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2. Coordinate Representation: Let $A=(x_A,y_A)$, $B=(x_B,y_B)$, $C=(x_C,y_C)$, and $P=(x_P,y_P)$. Write:$$x_P=\alpha x_A+\beta x_B+\gamma x_C,$$$$y_P=\alpha y_A+\beta y_B+\gamma y_C,$$with $\alpha +\beta +\gamma =1$. This forms a system of linear equations to uniquely determine $\alpha ,\beta ,\gamma$.

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3. Uniqueness: Assume $P$ has two representations:$$P={\alpha }_{1}A+{\beta }_{1}B+{\gamma }_{1}C,{\alpha }_1+{\beta }_1+{\gamma }_1=1,$$$$P={\alpha }_{2}A+{\beta }_{2}B+{\gamma }_{2}C,{\alpha }_2+{\beta }_2+{\gamma }_2=1.$$Subtract:$$0=({\alpha }_1-{\alpha }_2)A+({\beta }_1-{\beta }_2)B+({\gamma }_1-{\gamma }_2)C.$$Since $A,B,C$ are not collinear (linearly independent), the only solution is:$${\alpha }_1={\alpha }_2,{\beta }_1={\beta }_2,{\gamma }_1={\gamma }_2.$$

Therefore, any point $P$ in the plane can be written uniquely as:

$$P=\alpha A+\beta B+\gamma C,\text{with }\alpha +\beta +\gamma =1.$$

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(b)

$P$ is inside $\mathrm{△}ABC$ or on one of its sides if and only if $\alpha$, $\beta$, $\lambda$ are all nonnegative.

To show that $P$ is inside $\mathrm{△}ABC$ or on its sides if and only if $\alpha ,\beta ,\gamma \ge 0$:

1. Sufficient Condition ($\alpha ,\beta ,\gamma \ge 0$): If $\alpha ,\beta ,\gamma \ge 0$ and $\alpha +\beta +\gamma =1$, $P=\alpha A+\beta B+\gamma C$ is a convex combination.

   • If $\alpha ,\beta ,\gamma >0$, $P$ is inside the triangle.
   • If one of $\alpha ,\beta ,\gamma =0$, $P$ is on an edge of $\mathrm{△}ABC$.
   • If two of $\alpha ,\beta ,\gamma =0$, $P$ is at a vertex.

2. Necessary Condition: If $P$ is inside $\mathrm{△}ABC$ or on its sides, $P$ can be written as $P=\alpha A+\beta B+\gamma C$ where $\alpha +\beta +\gamma =1$. In these cases:

   • $P$ inside implies $\alpha ,\beta ,\gamma >0$.
   • $P$ on edges or vertices implies $\alpha ,\beta ,\gamma \ge 0$.

Thus, $P$ is inside or on $\mathrm{△}ABC$ if and only if $\alpha ,\beta ,\gamma \ge 0$ and $\alpha +\beta +\gamma =1$.

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(c)

Given a triangle $\mathrm{△}ABC$ in 2D plane with

$$A=\left(\begin{array}{c}1\\ 1\end{array}\right),B=\left(\begin{array}{c}4\\ 3\end{array}\right),\text{and}C=\left(\begin{array}{c}7\\ 6\end{array}\right),$$

determine if the point

$$P=\left(\begin{array}{c}5\\ 4\end{array}\right)$$

is inside $\mathrm{△}ABC$ by computing the barycentric coordinates of $P$ with respect to $\mathrm{△}ABC$.

To determine if $P=\left(\begin{array}{c}5\\ 4\end{array}\right)$ is inside $\mathrm{△}ABC$, compute its barycentric coordinates. The vertices of the triangle are:

$$A=\left(\begin{array}{c}1\\ 1\end{array}\right),B=\left(\begin{array}{c}4\\ 3\end{array}\right),C=\left(\begin{array}{c}7\\ 6\end{array}\right).$$

Barycentric coordinate equations

Point $P$ satisfies:

$$P=\alpha A+\beta B+\gamma C,\text{where}\alpha +\beta +\gamma =1.$$

This expands into:

$$5=\alpha \cdot 1+\beta \cdot 4+\gamma \cdot 7,4=\alpha \cdot 1+\beta \cdot 3+\gamma \cdot 6.$$

From $\alpha +\beta +\gamma =1$, we have $\alpha =1-\beta -\gamma$. Substitute into the other equations:

$$5=(1-\beta -\gamma )+4\beta +7\gamma ,4=(1-\beta -\gamma )+3\beta +6\gamma .$$

Simplify both equations:

$$5=1+3\beta +6\gamma ,4=1+2\beta +5\gamma .$$

Reduce:

$$4=3\beta +6\gamma ,3=2\beta +5\gamma .$$

Solve the system by elimination: Subtract $2\beta +5\gamma =3$ from $3\beta +6\gamma =4$:

$$\beta +\gamma =1.$$

Substitute $\gamma =1-\beta$ into $2\beta +5\gamma =3$:

$$2\beta +5(1-\beta )=3,2\beta +5-5\beta =3,-3\beta =-2,\beta =\frac{2}{3}.$$

Thus:

$$\gamma =1-\beta =\frac{1}{3},\alpha =1-\beta -\gamma =0.$$

The barycentric coordinates of $P$ are:

$$\alpha =0,\beta =\frac{2}{3},\gamma =\frac{1}{3}.$$

Since $\alpha ,\beta ,\gamma \ge 0$ and $\alpha +\beta +\gamma =1$, point $P$ lies on the edge of $\mathrm{△}ABC$. Specifically, $P$ is on the line segment $BC$.

Question-6

The following code is written to draw the outer surface of a cylinder as a quad-strip but nothing is seen on the screen. Assuming that lighting parameters and reflectance are properly set, explain why and suggest a remedy.

The issue is caused by incorrect vertex winding: the vertices are specified in clockwise order, making the quads back-facing by default. Since glCullFace(GL_BACK) is enabled, these polygons are culled, leaving nothing visible.

Remedy: Change the vertex order to counterclockwise for proper front-facing polygons:

glPolygonMode(GL_FRONT, GL_FILL);
glCullFace(GL_BACK);
glEnable(GL_CULL_FACE);

int nslice = 20;
glBegin(GL_QUAD_STRIP);
double t = 0., dt = 2. * 3.1416 / nslice;
for (int j = 0; j <= nslice; ++j) {
    glNormal3f(cos(t), 0., -sin(t));
    glVertex3f(cos(t), 2., -sin(t)); // Top vertex first
    glVertex3f(cos(t), 0., -sin(t)); // Bottom vertex second
    t += dt;
}
glEnd();

Question-7

Consider the function sphere() that draws the outer surface of a unit sphere. Assume that a cylindrical map of the earth has been set up as the current texture and lighting is enabled. Suggest modifications on the function for supporting lighting calculation and the mapping of the texture on the sphere.

To modify the sphere() function for lighting and texture mapping:

1. Add Normals for Lighting: Use the vertex position as normals:

   glNormal3f(r1 * ct, y1, r1 * st); // For top vertex
   glNormal3f(r2 * ct, y2, r2 * st); // For bottom vertex

2. Implement Texture Mapping: Calculate texture coordinates using spherical coordinates:

   glTexCoord2f(theta / (2.0 * M_PI), alpha1 / M_PI); // For top vertex
   glTexCoord2f(theta / (2.0 * M_PI), alpha2 / M_PI); // For bottom vertex

Updated sphere() Function:

void sphere() {
    double alpha1, alpha2, da, theta, dtheta, y1, y2, r1, r2, ct, st;
    int nslice = 50, nstack = 20;

    da = M_PI / nstack;
    dtheta = 2.0 * M_PI / nslice;

    alpha1 = 0.0; y1 = cos(alpha1); r1 = sin(alpha1);

    for (int i = 1; i <= nstack; ++i) {
        alpha2 = alpha1 + da;
        y2 = cos(alpha2); r2 = sin(alpha2);

        glBegin(GL_QUAD_STRIP);
        theta = 0.0;

        for (int j = 0; j <= nslice; ++j) {
            ct = cos(theta); st = sin(theta);

            glTexCoord2f(theta / (2.0 * M_PI), alpha1 / M_PI);
            glNormal3f(r1 * ct, y1, r1 * st); // Normal for top
            glVertex3f(r1 * ct, y1, r1 * st);

            glTexCoord2f(theta / (2.0 * M_PI), alpha2 / M_PI);
            glNormal3f(r2 * ct, y2, r2 * st); // Normal for bottom
            glVertex3f(r2 * ct, y2, r2 * st);

            theta += dtheta;
        }
        glEnd();

        alpha1 = alpha2; y1 = y2; r1 = r2;
    }
}