progress

exam-2/exam2.md

@@ -12,6 +12,7 @@ author: |
 
 \renewcommand{\c}[1]{\textcolor{gray}{#1}}
 \newcommand{\now}[1]{\textcolor{blue}{#1}}
+\newcommand{\todo}[0]{\textcolor{red}{\textbf{TODO}}}
 
 ## Reflection and Refraction
 
@@ -99,31 +100,156 @@ author: |
 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
+   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. (4 points) When the eye is located at $e = (2, 3, 5)$, the camera is
+   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 $V$?
+   $(0, 1, 0)$, what are the entries in $V$?}
 
-   $$\begin{bmatrix}
+   First we can calculate $n$ and $u$:
-     0 & 1 & 0 & d_x \\
-   \end{bmatrix}$$
 
-   b. (2 points) How will this matrix change if the eye moves forward in the
+   - 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?]
+   will change and in what way?]}
 
-   c. (2 points) How will this matrix change if the viewing direction spins in
+   If the eye moves forward, the eye _position_ and everything that depends on it
-   the clockwise direction around the camera's 'up' direction? [which elements
+   will change, while everything else doesn't.
-   in V will stay the same? which elements will change and in what way?]
 
-   d. (2 points) How will this matrix change if the viewing direction rotates
+   | $n$  | $u$  | $v$  | $d$       |
+   | ---- | ---- | ---- | --------- |
+   | 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?]
+   what way?]}
 
-5.
+   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
 
@@ -147,32 +273,36 @@ author: |
    \tan(60^\circ) \times 0.5 = \boxed{\frac{\sqrt{3}}{2}}$. The same goes for the
    vertical, which also yields $\frac{\sqrt{3}}{2}$.
 
-   $$\begin{bmatrix}
+   $$
+   \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}$$
+   \end{bmatrix}
+   $$
 
-   $$= \begin{bmatrix}
+   $$
+   = \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}$$
+   \end{bmatrix}
+   $$
 
-   $$= \boxed{\begin{bmatrix}
+   $$
+   = \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}}$$
+   \end{bmatrix}}
+   $$
 
    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?}
 

exam-2/exam2.py

@@ -37,6 +37,62 @@ def problem_1():
 
 def problem_4():
     print("part 4a.")
+    up = np.array([0, 1, 0])
+    viewing_dir = np.array([1, -1, -1])
+    n = unit(viewing_dir)
+    print(f"{n = }")
+
+    u = unit(np.cross(up, n))
+    print(f"{u = }")
+
+    v = np.cross(n, u)
+    print(f"{v = }")
+
+    print(math.sqrt(1 / 6.0))
+    print(math.sqrt(2 / 3.0))
+
+def build_translation_matrix(vec):
+    return np.array([
+        [1, 0, 0, vec[0]],
+        [0, 1, 0, vec[1]],
+        [0, 0, 1, vec[2]],
+        [0, 0, 0, 1],
+    ])
+
+def problem_5():
+    b1 = build_translation_matrix(np.array([0, 0, 5]))
+    theta = math.radians(-90)
+    sin_theta = round( math.sin(theta), 5)
+    cos_theta = round(math.cos(theta), 5)
+    b2 = np.array([
+        [cos_theta, 0, sin_theta, 0],
+        [0, 1, 0, 0],
+        [-sin_theta, 0, cos_theta, 0],
+        [0, 0, 0, 1],
+    ])
+    b3 = build_translation_matrix(np.array([0, 0, -5]))
+
+    print("b1", b1)
+    print("b2", b2)
+    print("b3", b3)
+    M = b3 @ b2 @ b1
+    print("M", M)
+
+    ex1 = np.array([1, 1, -4, 1])
+    print("ex1", ex1, M @ ex1)
+
+    up = np.array([0, 1, 0])
+    n = np.array([1, 0, 0])
+    u = unit(np.cross(up, n))
+    v = np.cross(n, u)
+
+    print(f"{up = }, {n = }, {u = }, {v = }")
+
+    eye = np.array([5, 0, -5])
+    dx = -(np.dot(eye, u))
+    dy = -(np.dot(eye, v))
+    dz = -(np.dot(eye, n))
+    print(f"{dx = }, {dy = }, {dz = }")
 
 def problem_8():
     def P(left, right, bottom, top, near, far):
@@ -67,3 +123,4 @@ def problem_8():
 print("\nPROBLEM 4 -------------------------"); problem_4()
 print("\nPROBLEM 8 -------------------------"); problem_8()
 print("\nPROBLEM 1 -------------------------"); problem_1()
+print("\nPROBLEM 5 -------------------------"); problem_5()