Progress

exam-2/exam2.md

@@ -11,10 +11,11 @@ author: |
 ---
 
 \renewcommand{\c}[1]{\textcolor{gray}{#1}}
+\renewcommand{\r}[1]{\textcolor{red}{#1}}
 \newcommand{\now}[1]{\textcolor{blue}{#1}}
 \newcommand{\todo}[0]{\textcolor{red}{\textbf{TODO}}}
 
-[ 1 2 3 6 8 9 ]
+[ 2 3 8 9 ]
 
 ## Reflection and Refraction
 
@@ -23,31 +24,44 @@ author: |
    1.0$). If a ray emanating from the point $e = (0, 0, 0)$ intersects $S$ at a
    point $p = (1, 4, 8)$:}
 
+   Legend for the rest of the problem:
+
+   ![](1.jpg){width=50%}
+
    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
+   The incoming ray is in the direction $I = v_0 = p - e = (1, 4, 8)$, and the
-   at that point is $N = p - s = (1, 4, 8) - (2, 2, 10) = (1, -2, 2)$. The angle
+   normal at that point is:
-   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)
+   - $N = p - s = (1, 4, 8) - (2, 2, 10) = (-1, 2, -2)$
-   \cdot (1, -2, 2)}{9 \cdot 3} = \frac{-1 + 8 - 16}{27} = -\frac{1}{3}$. So the
+
-   angle $\boxed{\theta_i = \cos^{-1}(-\frac{1}{3})}$.
+   The angle can be found by taking the opposite of the incoming ray $-I$ and
+   using the formula:
+
+   - $\cos \theta_i = \frac{-I \cdot N}{\|I\|_2 \|N\|_2} 
+     = \frac{(-1, -4, -8) \cdot (-1, 2, -2)}{9 \cdot 3} = \frac{1 - 8 + 16}{27} = \frac{1}{3}$.
+
+   So the angle $\boxed{\theta_i = \cos^{-1}\left(\frac{1}{3}\right)}$.
 
    b. \c{(1 points) What is the angle of reflection $\theta_r$?}
 
    The angle of reflection always equals the angle of incidence, $\theta_r =
-   \theta_i = \boxed{\cos^{-1}(-\frac{1}{3})}$.
+   \theta_i = \boxed{\cos^{-1}\left(\frac{1}{3}\right)}$.
 
    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 reflected ray can be found by first projecting the incident ray $v_0$ onto
-   the normalized normal $N$, which is $v = N \times |-I|\cos(\theta_i) =
+   the normalized normal $n$, which is:
-   (\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 =
+   - $proj = n \times |v_0|\cos(\theta_i) = \left(\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}\right) \times 9 \times \frac{1}{3} = (-1, 2, -2)$
-   (1, 4, 8) + (-1, 2, -2) = (0, 6, 6)$.
+
+   Then, we know the point on N where this happened is:
+
+   - $nx = p + proj = (1, 4, 8) + (-1, 2, -2) = (0, 6, 6)$.
 
    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
+   case since the ray starts at the origin. Adding this to the point $nx$ gets
    us $(0, 12, 12)$, which means a point from the origin will get reflected to
    $(0, 12, 12)$.
 
@@ -56,34 +70,45 @@ author: |
 
    d. \c{(3 points) What is the angle of transmission $\theta_t$?}
 
-   Using Snell's law, we know that $\eta_i \sin \theta_i = \eta_t \sin \theta_t
+   Using Snell's law, we know that
-   = 1.0 \times \sin(\cos^{-1}(-\frac{1}{3})) = 1.5 \times \sin(\theta_t)$. To
+
-   find the angle $\theta_t$ we can just solve: $\theta_t =
+   \begin{align}
-   \sin^{-1}(\frac{2}{3} \times \sin(\cos^{-1}(-\frac{1}{3}))) \approx
+   \eta_i \sin \theta_i &= \eta_t \sin \theta_t \\
-   \boxed{0.6796}$ (in radians).
+   \r{1.0} \times \sin \theta_i &= \r{1.5} \times \sin \theta_t \\
+   1.0 \times \sin(\r{ \cos^{-1}(-\frac{1}{3}) }) &= 1.5 \times \sin \theta_t \\
+   1.0 \times \r{ \frac{2\sqrt{2}}{3} } &= 1.5 \times \sin \theta_t \\
+   \r{ \frac{1.0}{1.5} } \times \frac{2\sqrt{2}}{3} &= \sin \theta_t \\
+   \r{ \frac{4}{9} } \sqrt{2} &= \sin \theta_t \\
+   \theta_t &= \sin^{-1} \left(\frac{4}{9}\sqrt{2}\right) \approx
+   \boxed{0.67967 \ldots}
+   \end{align}
 
    e. \c{(4 points) What is the direction of the transmitted ray?}
 
-   We can just add the angle obtained above to the inverse of the normal $n' =
+   We can just add together $s - p$ and $v_5$ in the diagram above, these two
-   (-1, 2, -2)$ (since that will be transmitted through the material). For the
+   make up the orthogonal basis of the transmitted ray $v_4$. The lengths of
-   tangent direction, we can use the vector from the ray's origin to the
+   these two are not important, but related through the angle $\theta_t$.
-   incident ray projected onto the normal, which we'll call $v = (0, 6, 6)$.
 
-   Now we know that the relationship between the coefficients of $n'$ and $v$ is
+   Also note that in the diagram, the sphere is drawn as a 2D circle. But in
-   $\tan(\theta_t)$. We can add these to obtain a point along the transmitted
+   reality, the number of vectors perpendicular to $s - p$ is infinite. The
-   ray (it doesn't actually matter where this occurs since we're going to
+   $v_5$ we're looking for is parallel with $v_1$, so we can just use
-   normalize this anyway, the proportion stays consistent regardless). So using
+   $\frac{v_1}{\|v_1\|_2}$ as our unit vector for $v_5$.
-   the unit length for the inverse normal which is $\frac{n'}{||n'||} =
-   (\frac{1}{3}, -\frac{2}{3}, \frac{2}{3})$ we know that the perpendicular
-   vector will be $\tan(\theta_t) \times (0, \frac{1}{\sqrt{2}},
-   \frac{1}{\sqrt{2}})$.
 
-   The final step is then adding these vectors together and subtracting the
+   \begin{align}
-   interersection point: $(\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}) +
+   v_4 &= (s - p) + \tan \theta_t \times \|s - p\|_2 \times \frac{v_1}{\|v_1\|_2} \\
-   \tan(\theta_t) \times (0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) - (1, 4,
+   v_4 &= (\r{ (2, 2, 10) - (1, 4, 8) }) + \tan \theta_t \times \|s - p\|_2 \times \frac{v_1}{\|v_1\|_2} \\
-   8)$.
+   v_4 &= \r{ (1, -2, 2) } + \tan \theta_t \times \|\r{(1, -2, 2)}\|_2 \times \frac{v_1}{\|v_1\|_2} \\
+   v_4 &= (1, -2, 2) + \tan \theta_t \times \r{3} \times \frac{v_1}{\|v_1\|_2} \\
+   v_4 &= (1, -2, 2) + \tan \theta_t \times 3 \times \frac{\r{(0, 6, 6)}}{\|\r{(0, 6, 6)}\|_2} \\
+   v_4 &= (1, -2, 2) + \tan \theta_t \times 3 \times \r{\left(0, \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)} \\
+   v_4 &= (1, -2, 2) + \tan \theta_t \times \left(0, \r{\frac{3}{2}\sqrt{2}}, \r{\frac{3}{2}\sqrt{2}} \right) \\
+   v_4 &= (1, -2, 2) + \r{\frac{4}{7}\sqrt{2}} \times \left(0, \frac{3}{2}\sqrt{2}, \frac{3}{2}\sqrt{2} \right) \\
+   v_4 &= (1, -2, 2) + \r{\left(0, \frac{12}{7}, \frac{12}{7} \right)} \\
+   v_4 &= \r{\left(1, \frac{-2}{7}, \frac{26}{7} \right)} \\
+   \end{align}
 
-   \todo all the numbers are wrong
+   This can be normalized into $\boxed{\left(\frac{7}{27}, -\frac{2}{27},
+   \frac{26}{27}\right)}$.
 
 ## Geometric Transformations
 
@@ -380,14 +405,14 @@ author: |
    i) \c{(2 points) Where will the six vertex locations be after such a
    projection, omitting the normalization step?}
 
-   - $[\begin{matrix}-1.732 & -1.732 & -6.928\end{matrix}]$ $\rightarrow$ $[\begin{matrix}-1.0 & -1.0 & 7.0\end{matrix}]$
+   - $[\begin{matrix}- \sqrt{3} & - \sqrt{3} & - 4 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}-1 & -1 & 7\end{matrix}]$
-   - $[\begin{matrix}-1.732 & -1.732 & -3.464\end{matrix}]$ $\rightarrow$ $[\begin{matrix}-1.0 & -1.0 & 5.0\end{matrix}]$
+   - $[\begin{matrix}- \sqrt{3} & \sqrt{3} & - 4 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}-1 & 1 & 7\end{matrix}]$
-   - $[\begin{matrix}-1.732 & 1.732 & -6.928\end{matrix}]$ $\rightarrow$ $[\begin{matrix}-1.0 & 1.0 & 7.0\end{matrix}]$
+   - $[\begin{matrix}\sqrt{3} & - \sqrt{3} & - 4 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}1 & -1 & 7\end{matrix}]$
-   - $[\begin{matrix}-1.732 & 1.732 & -3.464\end{matrix}]$ $\rightarrow$ $[\begin{matrix}-1.0 & 1.0 & 5.0\end{matrix}]$
+   - $[\begin{matrix}\sqrt{3} & \sqrt{3} & - 4 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}1 & 1 & 7\end{matrix}]$
-   - $[\begin{matrix}1.732 & -1.732 & -6.928\end{matrix}]$ $\rightarrow$ $[\begin{matrix}1.0 & -1.0 & 7.0\end{matrix}]$
+   - $[\begin{matrix}- \sqrt{3} & - \sqrt{3} & - 2 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}-1 & -1 & 5\end{matrix}]$
-   - $[\begin{matrix}1.732 & -1.732 & -3.464\end{matrix}]$ $\rightarrow$ $[\begin{matrix}1.0 & -1.0 & 5.0\end{matrix}]$
+   - $[\begin{matrix}- \sqrt{3} & \sqrt{3} & - 2 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}-1 & 1 & 5\end{matrix}]$
-   - $[\begin{matrix}1.732 & 1.732 & -6.928\end{matrix}]$ $\rightarrow$ $[\begin{matrix}1.0 & 1.0 & 7.0\end{matrix}]$
+   - $[\begin{matrix}\sqrt{3} & - \sqrt{3} & - 2 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}1 & -1 & 5\end{matrix}]$
-   - $[\begin{matrix}1.732 & 1.732 & -3.464\end{matrix}]$ $\rightarrow$ $[\begin{matrix}1.0 & 1.0 & 5.0\end{matrix}]$
+   - $[\begin{matrix}\sqrt{3} & \sqrt{3} & - 2 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}1 & 1 & 5\end{matrix}]$
 
    ii) \c{(1 points) Sketch the result, being as accurate as possible and
    labeling the colors of each of the visible faces.}
@@ -396,6 +421,8 @@ author: |
    sides are not visible because the cube is parallel to the axis, and when you
    do an orthographic projection, those faces are lost.
 
+   ![](6aii.jpg){width=40%}
+
    iii) \c{(2 points) Show how you could achieve this transformation using one or
    more matrix multiplication operations. Specify the matrix entries you would
    use, and, if using multiple matrices, the order in which they would be
@@ -405,14 +432,7 @@ author: |
    in this Python script:
 
    ```py
-   def ortho(points):
+   def ortho_transform(left, right, bottom, top, near, far):
-     left = min(map(lambda p: p[0], points))
-     right = max(map(lambda p: p[0], points))
-     bottom = min(map(lambda p: p[1], points))
-     top = max(map(lambda p: p[1], points))
-     near = min(map(lambda p: p[2], points))
-     far = max(map(lambda p: p[2], points))
-
      step_1 = np.array([
        [1, 0, 0, 0],
        [0, 1, 0, 0],
@@ -421,20 +441,20 @@ author: |
      ])
 
      step_2 = np.array([
-       [1, 0, 0, -(left + right) / 2.0],
+       [1, 0, 0, -((left + right) / 2)],
-       [0, 1, 0, -(bottom + top) / 2.0],
+       [0, 1, 0, -((bottom + top) / 2)],
-       [0, 0, 1, -(near + far) / 2.0],
+       [0, 0, 1, -((near + far) / 2)],
        [0, 0, 0, 1],
      ])
 
      step_3 = np.array([
-       [2.0 / (right - left), 0, 0, 0],
+       [(2 / (right - left)), 0, 0, 0],
-       [0, 2.0 / (top - bottom), 0, 0],
+       [0, (2 / (top - bottom)), 0, 0],
-       [0, 0, 2.0 / (far - near), 0],
+       [0, 0, (2 / (far - near)), 0],
        [0, 0, 0, 1],
      ])
 
-     M_ortho = step_3 @ step_2 @ step_1
+     return step_3 @ step_2 @ step_1
    ```
 
    b. \c{Show how you could project the vertices of this cube to the plane $z =
@@ -444,14 +464,45 @@ author: |
    i) \c{(3 points) Where will the six vertex locations be after such a
    projection, omitting the normalization step?}
 
+   - $[\begin{matrix}- \sqrt{3} & - \sqrt{3} & - 4 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}- \frac{4 \sqrt{3}}{3} - 1 & -1 & 7\end{matrix}]$
+   - $[\begin{matrix}- \sqrt{3} & \sqrt{3} & - 4 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}- \frac{4 \sqrt{3}}{3} - 1 & 1 & 7\end{matrix}]$
+   - $[\begin{matrix}\sqrt{3} & - \sqrt{3} & - 4 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}1 - \frac{4 \sqrt{3}}{3} & -1 & 7\end{matrix}]$
+   - $[\begin{matrix}\sqrt{3} & \sqrt{3} & - 4 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}1 - \frac{4 \sqrt{3}}{3} & 1 & 7\end{matrix}]$
+   - $[\begin{matrix}- \sqrt{3} & - \sqrt{3} & - 2 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}- \frac{2 \sqrt{3}}{3} - 1 & -1 & 5\end{matrix}]$
+   - $[\begin{matrix}- \sqrt{3} & \sqrt{3} & - 2 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}- \frac{2 \sqrt{3}}{3} - 1 & 1 & 5\end{matrix}]$
+   - $[\begin{matrix}\sqrt{3} & - \sqrt{3} & - 2 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}1 - \frac{2 \sqrt{3}}{3} & -1 & 5\end{matrix}]$
+   - $[\begin{matrix}\sqrt{3} & \sqrt{3} & - 2 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}1 - \frac{2 \sqrt{3}}{3} & 1 & 5\end{matrix}]$
+
    ii) \c{(2 points) Sketch the result, being as accurate as possible and
    labeling the colors of each of the visible faces.}
 
+   ![](6bii.jpg){width=40%}
+
+   Excuse the poor sketch but the point is the transformation exposes the green
+   face to the left by doing a shear first.
+
    iii) \c{(4 points) Show how you could achieve this transformation using one
    or more matrix multiplication operations. Specify the matrix entries you
    would use, and, if using multiple matrices, the order in which they would be
    multiplied.}
 
+   It's the same as the ortho, except it uses an extra matrix that is applied
+   before any of the other transformations:
+
+   ```py
+   def oblique_transform(left, right, bottom, top, near, far):
+     step_0 = np.array([
+       [1, 0, (1 / sqrt3), 0],
+       [0, 1, 0, 0],
+       [0, 0, 1, 0],
+       [0, 0, 0, 1],
+     ])
+
+     M_ortho = ortho_transform(left, right, bottom, top, near, far)
+
+     return M_ortho @ step_0
+   ```
+
 7. \c{Consider the simple scene shown in the image below, where two cubes, one
    of height 1 and one of height 2, are both resting on a horizontal groundplane
    ($y = -\frac{1}{2}$), with the smaller cube’s front face aligned with $z =

exam-2/exam2.py

@@ -1,7 +1,12 @@
 import itertools
 import numpy as np
-from sympy import *
 import math
+from sympy import N, Number, Rational, init_printing, latex, simplify, Expr
+from sympy.vector import CoordSys3D, Vector
+
+init_printing()
+
+import sympy
 
 unit = lambda v: v/np.linalg.norm(v)
 
@@ -57,45 +62,65 @@ def print_bmatrix(arr):
     print("\\\\")
 
 def problem_1():
-  p = np.array([1, 4, 8])
+  C = CoordSys3D('C')
-  e = np.array([0, 0, 0])
+  p = 1 * C.i + 4 * C.j + 8 * C.k
-  s = np.array([2, 2, 10])
+  e = 0 * C.i + 0 * C.j + 0 * C.k
+  s = 2 * C.i + 2 * C.j + 10 * C.k
+  radius = 3
 
-  i = p - e
+  v0 = p - e
-  print("incoming", i)
+  print("v0 (incoming ray) =", v0)
-  print("|I| =", np.linalg.norm(i))
+  print(" - |v0| =", v0.magnitude())
-  n = s - p
+  print()
-  print("normal", n)
 
-  n_norm = unit(n)
+  n = p - s
-  print("normal_norm", n_norm)
+  print("n (normal) =", n)
-  cos_theta_i = np.dot(-i, n) / (np.linalg.norm(i) * np.linalg.norm(n))
+  n_norm = n.normalize()
-  print("part a = cos^{-1} of ", cos_theta_i)
+  print(" - n_norm (normalized) =", n_norm)
-  print(np.arccos(cos_theta_i))
+  print()
 
-  proj = n_norm * np.linalg.norm(i) * cos_theta_i
+  cos_theta_i = (-v0).dot(n) / (v0.magnitude() * n.magnitude())
-  print("proj", proj)
+  theta_i = sympy.acos(cos_theta_i)
+  print("[1a] theta_i (angle of incidence) =", theta_i)
+  print()
 
-  p_ = p + proj
+  proj_len = cos_theta_i * v0.magnitude()
-  print("proj point", p_)
+  proj = n_norm * proj_len
+  print("projection of v0 onto n =", proj)
 
-  v2 = p_ - e
+  nx = p + proj
-  print("v2", v2)
+  print("nx (proj point) =", nx)
+  print()
 
-  sin_theta_i = np.sin(np.arccos(cos_theta_i))
+  v1 = nx - e
-  print("sin theta_i =", sin_theta_i)
+  print("v1 (from e to nx) =", v1)
 
-  theta_t = np.arcsin(1.0 / 1.5 * sin_theta_i)
+  z = e + 2 * v1
-  print("approx answer for part d", theta_t)
+  print("z =", z)
 
-  uin = unit(n_norm)
+  v3 = z - p
-  print("unit inv normal", uin)
+  print("v3 =", v3)
+  print()
 
-  uperp = unit(v2)
+  sin_theta_i = sympy.sin(theta_i)
-  print("uperp", uperp)
+  print("sin(theta_i) =", sin_theta_i)
 
-  answer_1e = uin + math.tan(theta_t) * uperp - p
+  theta_t = sympy.asin(Rational(2, 3) * sin_theta_i)
-  print("1e answer", unit(answer_1e))
+  print("[1d] theta_t (angle of transmission) =", theta_t)
+  print(" - approx =", theta_t.evalf())
+  print()
+
+  v5_len = sympy.tan(theta_t) * (s - p).magnitude()
+  print("v5 len =", v5_len)
+  v4 = (s - p) + v1.normalize() * v5_len
+  print("v4 =", v4)
+  print(" - [1e] normalized =", v4.normalize())
+  print()
+
+  print("s - p =", s - p)
+  print("|s - p| =", (s - p).magnitude())
+  print("|v1| =", v1.normalize())
+  print("tan(theta_t) =", sympy.tan(theta_t))
 
 def problem_4():
   camera_pos = np.array([2, 3, 5])
@@ -229,14 +254,11 @@ def problem_7():
   solve(camera_pos, angle)
 
 def problem_6():
-  def calculate(points):
+  # sqrt3 = math.sqrt(3)
-    left = min(map(lambda p: p[0], points))
+  sqrt3 = sympy.sqrt(3)
-    right = max(map(lambda p: p[0], points))
+  width = 2 * sqrt3
-    bottom = min(map(lambda p: p[1], points))
-    top = max(map(lambda p: p[1], points))
-    near = min(map(lambda p: p[2], points))
-    far = max(map(lambda p: p[2], points))
 
+  def ortho_transform(left, right, bottom, top, near, far):
     step_1 = np.array([
       [1, 0, 0, 0],
       [0, 1, 0, 0],
@@ -245,46 +267,75 @@ def problem_6():
     ])
 
     step_2 = np.array([
-      [1, 0, 0, -(left + right) / 2.0],
+      [1, 0, 0, -((left + right) / 2)],
-      [0, 1, 0, -(bottom + top) / 2.0],
+      [0, 1, 0, -((bottom + top) / 2)],
-      [0, 0, 1, -(near + far) / 2.0],
+      [0, 0, 1, -((near + far) / 2)],
       [0, 0, 0, 1],
     ])
 
     step_3 = np.array([
-      [2.0 / (right - left), 0, 0, 0],
+      [(2 / (right - left)), 0, 0, 0],
-      [0, 2.0 / (top - bottom), 0, 0],
+      [0, (2 / (top - bottom)), 0, 0],
-      [0, 0, 2.0 / (far - near), 0],
+      [0, 0, (2 / (far - near)), 0],
       [0, 0, 0, 1],
     ])
 
-    M_ortho = step_3 @ step_2 @ step_1
+    return step_3 @ step_2 @ step_1
+
+  def oblique_transform(left, right, bottom, top, near, far):
+    step_0 = np.array([
+      [1, 0, (1 / sqrt3), 0],
+      [0, 1, 0, 0],
+      [0, 0, 1, 0],
+      [0, 0, 0, 1],
+    ])
+
+    M_ortho = ortho_transform(left, right, bottom, top, near, far)
+
+    return M_ortho @ step_0
+
+  def calculate(M, points):
+    left = min(map(lambda p: p[0], points))
+    right = max(map(lambda p: p[0], points))
+    bottom = min(map(lambda p: p[1], points))
+    top = max(map(lambda p: p[1], points))
+    near = min(map(lambda p: p[2], points))
+    far = max(map(lambda p: p[2], points))
 
     for point in points:
       point_ = np.r_[point, [1]]
-      trans_ = M_ortho @ point_
+      trans_ = M(left, right, bottom, top, near, far) @ point_
       trans = trans_[:3]
-      print("-", print_trans(np.around(point, 3), np.around(trans, 3)))
+      v = np.vectorize(lambda x: float(x.evalf()))
+      l = np.vectorize(lambda x: latex(simplify(x)))
+      point = l(point)
+      trans = l(trans)
+      print("-", print_trans(point, trans))
 
-  sqrt3 = math.sqrt(3)
+  cube_center = np.array([0, 0, -3 * sqrt3])
-  width = 2.0 * sqrt3
-  cube_center = np.array([0, 0, -3.0 * sqrt3])
-  print("HELLOSU")
 
   points = []
-  for (dx, dy, dz) in itertools.product([-1, 1], [-1, 1], [-1, 1]):
+  for (dz, dx, dy) in itertools.product([-1, 1], [-1, 1], [-1, 1]):
-    point = cube_center + np.array([dx * width / 2, dy * width / 2, dz * width / 2])
+    point = cube_center + np.array([
+      dx * width / 2,
+      dy * width / 2,
+      dz * width / 2,
+    ])
     points.append(point)
 
-  calculate(points)
+  print("Helosu Ortho")
+  calculate(ortho_transform, points)
+
+  print("Helosu Oblique")
+  calculate(oblique_transform, points)
 
 def problem_9():
   pass
 
 print("\nPROBLEM 8 -------------------------"); problem_8()
-print("\nPROBLEM 1 -------------------------"); problem_1()
 print("\nPROBLEM 5 -------------------------"); problem_5()
 print("\nPROBLEM 9 -------------------------"); problem_9()
 print("\nPROBLEM 7 -------------------------"); problem_7()
-print("\nPROBLEM 6 -------------------------"); problem_6()
 print("\nPROBLEM 4 -------------------------"); problem_4()
+print("\nPROBLEM 6 -------------------------"); problem_6()
+print("\nPROBLEM 1 -------------------------"); problem_1()