csci5607/exam-2/exam2.md

---
geometry: margin=2cm
output: pdf_document
title: Exam 2
subtitle: CSCI 5607
date: \today

author: |
  | Michael Zhang
  | zhan4854@umn.edu $\cdot$ ID: 5289259
---

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## Reflection and Refraction

1. \c{Consider a sphere $S$ made of solid glass ($\eta$ = 1.5) that has radius $r =
   3$ and is centered at the location $s = (2, 2, 10)$ in a vaccum ($\eta =
   1.0$). If a ray emanating from the point $e = (0, 0, 0)$ intersects $S$ at a
   point $p = (1, 4, 8)$:}

   a. \c{(2 points) What is the angle of incidence $\theta_i$?}

   The incoming ray is in the direction $I = p - e = (1, 4, 8)$, and the normal at
   that point is $N = p - s = (1, 4, 8) - (2, 2, 10) = (1, -2, 2)$. The angle can
   be found by taking the opposite of the incoming ray $-I$ and using the
   formula $\cos \theta_i = \frac{-I \cdot N}{|I| |N|} = \frac{(-1, -4, -8)
   \cdot (1, -2, 2)}{9 \cdot 3} = \frac{-1 + 8 - 16}{27} = -\frac{1}{3}$. So the
   angle $\boxed{\theta_i = \cos^{-1}(-\frac{1}{3})}$.

   b. \c{(1 points) What is the angle of reflection $\theta_r$?}

   The angle of reflection always equals the angle of incidence, $\theta_r =
   \theta_i = \boxed{cos^{-1}(-\frac{1}{3})}$.

   c. \c{(3 points) What is the direction of the reflected ray?}

   The reflected ray can be found by first projecting the incident ray $-I$ onto
   the normalized normal $N$, which is $v = N \times |-I|\cos(\theta_i) =
   (\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}) \times 9 \times \frac{1}{3} = (-1,
   2, -2)$. Then, we know the point on N where this happened is $p' = p + v =
   (1, 4, 8) + (-1, 2, -2) = (0, 6, 6)$.

   Now, we can subtract this point from where the ray originated to know the
   direction to add in the other direction, which is still $(0, 6, 6)$ in this
   case since the ray starts at the origin. Adding this to the point $p'$ gets
   us $(0, 12, 12)$, which means a point from the origin will get reflected to
   $(0, 12, 12)$.

   Finally, subtract the point to get the final answer $(0, 12, 12) - (1, 4, 8)
   = \boxed{(-1, 8, 4)}$.

   d. \c{(3 points) What is the angle of transmission $\theta_t$?}

   Using Snell's law, we know that $\eta_i \sin \theta_i = \eta_t \sin \theta_t
   = 1.0 \times \sin(\cos^{-1}(-\frac{1}{3})) = 1.5 \times \sin(\theta_t)$. To
   find the angle $\theta_t$ we can just solve: $\theta_t =
   \sin^{-1}(\frac{2}{3} \times \sin(\cos^{-1}(-\frac{1}{3}))) \approx
   \boxed{0.6796}$ (in radians).

   e. \c{(4 points) What is the direction of the transmitted ray?}

## Geometric Transformations

2. \c{(8 points) Consider the airplane model below, defined in object
   coordinates with its center at $(0, 0, 0)$, its wings aligned with the $\pm
   x$ axis, its tail pointing upwards in the $+y$ direction and its nose facing
   in the $+z$ direction. Derive a sequence of model transformation matrices
   that can be applied to the vertices of the airplane to position it in space
   at the location $p = (4, 4, 7)$, with a direction of flight $w = (2, 1, -2)$
   and the wings aligned with the direction $d = (-2, 2, -1)$.}

   The translation matrix is

   $$
   \begin{bmatrix}
     1 & 0 & 0 & x \\
     0 & 1 & 0 & y \\
     0 & 0 & 1 & z \\
     0 & 0 & 0 & 1 \\
   \end{bmatrix}
   =
   \begin{bmatrix}
     1 & 0 & 0 & 4 \\
     0 & 1 & 0 & 4 \\
     0 & 0 & 1 & 7 \\
     0 & 0 & 0 & 1 \\
   \end{bmatrix}
   $$

   Since the direction of flight was originally $(0, 0, 1)$, we have to
   transform it to $(2, 1, -2)$.

3.

## The Camera/Viewing Transformation

4. \c{Consider the viewing transformation matrix $V$ that enables all of the
   vertices in a scene to be expressed in terms of a coordinate system in which
   the eye is located at $(0, 0, 0)$, the viewing direction ($-n$) is aligned
   with the $-z$ axis $(0, 0, -1)$, and the camera's 'up' direction (which
   controls the roll of the view) is aligned with the $y$ axis (0, 1, 0).}

   a. \c{(4 points) When the eye is located at $e = (2, 3, 5)$, the camera is
   pointing in the direction $(1, -1, -1)$, and the camera's 'up' direction is
   $(0, 1, 0)$, what are the entries in $V$?}

   First we can calculate $n$ and $u$:

   - Viewing direction is $(1, -1, -1)$.
   - Normalized $n = (\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}})$.
   - $u = up \times n = (-\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}})$.
   - $v = n \times u = (\frac{\sqrt{6}}{6}, \frac{\sqrt{6}}{3}, -\frac{\sqrt{6}}{6})$

   $$
   \begin{bmatrix}
     -\frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} & d_x \\
     1 & -1 & -1 & d_y \\
   \end{bmatrix}
   $$

   \todo

   b. \c{(2 points) How will this matrix change if the eye moves forward in the
   direction of view? [which elements in V will stay the same? which elements
   will change and in what way?]}

   If the eye moves forward, the eye _position_ and everything that depends on it
   will change, while everything else doesn't.

   | $n$  | $u$  | $v$  | $d$       |
   | ---- | ---- | ---- | --------- |
   | same | same | same | different |

   The $n$ is the same because the viewing direction does not change.

   c. \c{(2 points) How will this matrix change if the viewing direction spins
   in the clockwise direction around the camera's 'up' direction? [which
   elements in V will stay the same? which elements will change and in what
   way?]}

   In this case, the eye _position_ stays the same, and everything else changes.

   | $n$       | $u$       | $v$       | $d$  |
   | --------- | --------- | --------- | ---- |
   | different | different | different | same |

   d. \c{(2 points) How will this matrix change if the viewing direction rotates
   directly upward, within the plane defined by the viewing and 'up' directions?
   [which elements in V will stay the same? which elements will change and in
   what way?]}

   In this case, the eye _position_ stays the same, and everything else changes.

   | $n$       | $u$       | $v$       | $d$  |
   | --------- | --------- | --------- | ---- |
   | different | different | different | same |

5. \c{Suppose a viewer located at the point $(0, 0, 0)$ is looking in the $-z$
   direction, with no roll ['up' = $(0, 1 ,0)$], towards a cube of width 2,
   centered at the point $(0, 0, -5)$, whose sides are colored: red at the plane
   $x = 1$, cyan at the plane $x = -1$, green at the plane $y = 1$, magenta at
   the plane $y = -1$, blue at the plane $z = -4$, and yellow at the plane $z =
   -6$.}

   a. \c{(1 point) What is the color of the cube face that the user sees?}

   \boxed{\textrm{Blue}}

   b. \c{(3 points) Because the eye is at the origin, looking down the $-z$ axis
   with 'up' = $(0,1,0)$, the viewing transformation matrix $V$ in this case is
   the identity $I$. What is the model matrix $M$ that you could use to rotate
   the cube so that when the image is rendered, it shows the red side of the
   cube?}

   You would have to do a combination of (1) translate to the origin, (2) rotate
   around the origin, and then (3) untranslate back. This way, the eye position
   doesn't change.

   $$
   M =
   \begin{bmatrix}
      1 & 0 & 0 & 0 \\
      0 & 1 & 0 & 0 \\
      0 & 0 & 1 & -5 \\
      0 & 0 & 0 & 1 \\
   \end{bmatrix} \cdot
   \begin{bmatrix}
      0 & 0 & -1 & 0 \\
      0 & 1 & 0 & 0 \\
      1 & 0 & 0 & 0 \\
      0 & 0 & 0 & 1 \\
   \end{bmatrix} \cdot
   \begin{bmatrix}
      1 & 0 & 0 & 0 \\
      0 & 1 & 0 & 0 \\
      0 & 0 & 1 & 5 \\
      0 & 0 & 0 & 1 \\
   \end{bmatrix}
   =
   \boxed{\begin{bmatrix}
      0 & 0 & -1 & -5 \\
      0 & 1 & 0 & 0 \\
      1 & 0 & 0 & -5 \\
      0 & 0 & 0 & 1 \\
   \end{bmatrix}}
   $$

   To verify this, testing with an example point $(1, 1, -4)$ yields:

   $$
   \begin{bmatrix}
      0 & 0 & -1 & -5 \\
      0 & 1 & 0 & 0 \\
      1 & 0 & 0 & -5 \\
      0 & 0 & 0 & 1 \\
   \end{bmatrix}
   \cdot
   \begin{bmatrix}
      1 \\ 1 \\ -4 \\ 1
   \end{bmatrix}
   =
   \begin{bmatrix}
      -1 \\ 1 \\ -4 \\ 1
   \end{bmatrix}
   $$

   c. \c{(4 points) Suppose now that you want to leave the model matrix $M$ as
   the identity. What is the viewing matrix $V$ that you would need to use to
   render an image of the scene from a re-defined camera configuration so that
   when the scene is rendered, it shows the red side of the cube? Where is the
   eye in this case and in what direction is the camera looking?}

   For this, a different eye position will have to be used. Instead of looking
   from the origin, you could view it from the red side, and then change the
   direction so it's still pointing at the cube.

   - eye is located at $(5, 0, -5)$
   - viewing direction is $(-1, 0, 0)$
   - $n = (1, 0, 0)$
   - $u = up \times n = (0, 0, -1)$
   - $v = n \times u = (0, 1, 0)$
   - $d = (-5, 0, -5)$

   The final viewing matrix is $\boxed{\begin{bmatrix}
     0 & 0 & -1 & -5 \\
     0 & 1 & 0 & 0 \\
     1 & 0 & 0 & -5 \\
     0 & 0 & 0 & 1 \\
   \end{bmatrix}}$. Turns out it's the same matrix! Wow!

## The Projection Transformation

6.

7.

8. \c{Consider the perspective projection-normalization matrix $P$ which maps
   the contents of the viewing frustum into a cube that extends from -1 to 1 in
   $x, y, z$ (called normalized device coordinates).}

   \c{Suppose you want to define a square, symmetric viewing frustum with a near
   clipping plane located 0.5 units in front of the camera, a far clipping plane
   located 20 units from the front of the camera, a $60^\circ$ vertical field of
   view, and a $60^\circ$ horizontal field of view.}

   a. \c{(2 points) What are the entries in $P$?}

   The left / right values are found by using the tangent of the field-of-view
   triangle: $\tan(60^\circ) = \frac{\textrm{right}}{0.5}$, so $\textrm{right} =
   \tan(60^\circ) \times 0.5 = \boxed{\frac{\sqrt{3}}{2}}$. The same goes for the
   vertical, which also yields $\frac{\sqrt{3}}{2}$.

   $$
   \begin{bmatrix}
     \frac{2\times near}{right - left} & 0 & \frac{right + left}{right - left} & 0 \\
     0 & \frac{2\times near}{top - bottom} & \frac{top + bottom}{top - bottom} & 0 \\
     0 & 0 & -\frac{far + near}{far - near} & -\frac{2\times far\times near}{far - near} \\
     0 & 0 & -1 & 0
   \end{bmatrix}
   $$

   $$
   = \begin{bmatrix}
     \frac{2\times 0.5}{\frac{\sqrt{3}}{2} - (-\frac{\sqrt{3}}{2})} & 0 & \frac{\frac{\sqrt{3}}{2} + (-\frac{\sqrt{3}}{2})}{\frac{\sqrt{3}}{2} - (-\frac{\sqrt{3}}{2})} & 0 \\
     0 & \frac{2\times 0.5}{\frac{\sqrt{3}}{2} - (-\frac{\sqrt{3}}{2})} & \frac{\frac{\sqrt{3}}{2} + (-\frac{\sqrt{3}}{2})}{\frac{\sqrt{3}}{2} - (-\frac{\sqrt{3}}{2})} & 0 \\
     0 & 0 & -\frac{20 + 0.5}{20 - 0.5} & -\frac{2\times 20\times 0.5}{20 - 0.5} \\
     0 & 0 & -1 & 0
   \end{bmatrix}
   $$

   $$
   = \boxed{\begin{bmatrix}
     \frac{1}{\sqrt{3}} & 0 & 0 & 0 \\
     0 & \frac{1}{\sqrt{3}} & 0 & 0 \\
     0 & 0 & -\frac{41}{39} & -\frac{40}{39} \\
     0 & 0 & -1 & 0
   \end{bmatrix}}
   $$

   b. \c{(3 points) How should be matrix $P$ be re-defined if the viewing window
   is re-sized to be twice as tall as it is wide?}

   c. \c{(3 points) What are the new horizontal and vertical fields of view
   after this change has been made?}

## Clipping

9. \c{Consider the triangle whose vertex positions, after the viewport
   transformation, lie in the centers of the pixels: $p_0 = (3, 3), p_1 = (9,
   5), p_2 = (11, 11)$.}

   Starting at $p_0$, the three vectors are:

   - $v_0 = p_1 - p_0 = (9 - 3, 5 - 3) = (6, 2)$
   - $v_1 = p_2 - p_1 = (11 - 9, 11 - 5) = (2, 6)$
   - $v_2 = p_0 - p_2 = (3 - 11, 3 - 11) = (-8, -8)$

   The first edge vector $e$ would be $(6, 2)$, and the edge normal would be
   that rotated by $90^\circ$.

   a. \c{(6 points) Define the edge equations and tests that would be applied,
   during the rasterization process, to each pixel $(x, y)$ within the bounding
   rectangle $3 \le x \le 11, 3 \le y \le 11$ to determine if that pixel is
   inside the triangle or not.}

   b. \c{(3 points) Consider the three pixels $p_4 = (6, 4), p_5 = (7, 7)$, and
   $p_6 = (10, 8)$. Which of these would be considered to lie inside the
   triangle, according to the methods taught in class?}

10. \c{When a model contains many triangles that form a smoothly curving surface
    patch, it can be inefficient to separately represent each triangle in the
    patch independently as a set of three vertices because memory is wasted when
    the same vertex location has to be specified multiple times. A triangle
    strip offers a memory-efficient method for representing connected 'strips'
    of triangles. For example, in the diagram below, the six vertices v0 .. v5
    define four adjacent triangles: (v0, v1, v2), (v2, v1, v3), (v2, v3, v4),
    (v4, v3, v5). [Notice that the vertex order is switched in every other
    triangle to maintain a consistent counter-clockwise orientation.] Ordinarily
    one would need to pass 12 vertex locations to the GPU to represent this
    surface patch (three vertices for each triangle), but when the patch is
    encoded as a triangle strip, only the six vertices need to be sent and the
    geometry they represent will be interpreted using the correspondence pattern
    just described.}

    \c{(5 points) When triangle strips are clipped, however, things
    can get complicated. Consider the short triangle strip shown below in the
    context of a clipping cube.}

    - \c{After the six vertices v0 .. v5 are sent to be clipped, what will the
      vertex list be after clipping process has finished?}

    - \c{How can this new result be expressed as a triangle strip? (Try to be as
      efficient as possible)}

    - \c{How many triangles will be encoded in the clipped triangle strip?}

## Ray Tracing vs Scan Conversion

11. \c{(8 points) List the essential steps in the scan-conversion (raster
    graphics) rendering pipeline, starting with vertex processing and ending
    with the assignment of a color to a pixel in a displayed image. For each
    step briefly describe, in your own words, what is accomplished and how. You
    do not need to include steps that we did not discuss in class, such as
    tessellation (subdividing an input triangle into multiple subtriangles),
    instancing (creating new geometric primitives from existing input vertices),
    but you should not omit any steps that are essential to the process of
    generating an image of a provided list of triangles.}

12. \c{(6 points) Compare and contrast the process of generating an image of a
    scene using ray tracing versus scan conversion. Include a discussion of
    outcomes that can be achieved using a ray tracing approach but not using a
    scan-conversion approach, or vice versa, and explain the reasons why and why
    not.}

    With ray tracing, the process of generating pixels is very hierarchical.
    The basic ray tracer was very simple, but the moment we even added shadows,
    there were recursive rays that needed to be cast, not to mention the
    jittering. None of those could be parallelized with the main one, because in
    order to even figure out where to start, you need to have already performed
    a lot of the calculations. (For my ray tracer implementation, I already
    parallelized as much as I could using the work-stealing library `rayon`)

    But with scan conversion, the majority of the transformations are just done
    with matrix transformations over the geometries, which can be performed
    completely in parallel with minimal branching (only depth testing is not
    exactly) The rasterization process is also massively parallelizable. This
    makes it faster to do on GPUs which are able to do a lot of independent
    operations.