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Michael Zhang 2023-05-03 17:47:37 -05:00
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@ -15,7 +15,7 @@ author: |
\newcommand{\now}[1]{\textcolor{blue}{#1}}
\newcommand{\todo}[0]{\textcolor{red}{\textbf{TODO}}}
[ 2 3 8 9 ]
[ 3 8 9 ]
## Reflection and Refraction
@ -95,11 +95,11 @@ author: |
$\frac{v_1}{\|v_1\|_2}$ as our unit vector for $v_5$.
\begin{align}
v_4 &= (s - p) + \tan \theta_t \times \|s - p\|_2 \times \frac{v_1}{\|v_1\|_2} \\
v_4 &= (\r{ (2, 2, 10) - (1, 4, 8) }) + \tan \theta_t \times \|s - p\|_2 \times \frac{v_1}{\|v_1\|_2} \\
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 &= (s - p) + \tan \theta_t \times \|s - p\|\_2 \times \frac{v_1}{\|v_1\|\_2} \\
v_4 &= (\r{ (2, 2, 10) - (1, 4, 8) }) + \tan \theta_t \times \|s - p\|\_2 \times \frac{v_1}{\|v_1\|\_2} \\
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) \\
@ -120,46 +120,59 @@ author: |
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 order we want is (1) do all the rotations, and then (2) translate to the
spot we want. The rotation is done in multiple steps:
There's 2 discrete transformations going on here:
- First we want to make sure the nose of the plane points in the correct
direction in the $xz$ plane (rotating around the $y$ axis). The desired
resulting direction is $(2, y, -2)$, so that means for a nose currently facing
the $+z$ direction which is $(0, y, 1)$, we want to rotate around by around
$-\frac{3}{4}\pi$. The transformation matrix is:
- First, we must rotate the plane. Since we are given orthogonal directions
for the flight and wings, we can just come up with another vector for the
tail direction by doing:
$y' = (2, 1, -2) \times (-2, 2, -1) = (1, 2, 2)$
Now we can construct a 3D rotation matrix:
$$
M_1 = \begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
R =
\begin{bmatrix}
d_x & y_x & w_x & 0 \\
d_y & y_y & w_y & 0 \\
d_z & y_z & w_z & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}
=
\begin{bmatrix}
-2 & 1 & 2 & 0 \\
2 & 2 & 1 & 0 \\
-1 & 2 & -2 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}
$$
- Then we want to rotate the plane vertically, so it's pointing in the right
direction.
- Now we just need to translate this to the position (4, 4, 7). This is easy
with:
$$
T =
\begin{bmatrix}
1 & 0 & 0 & 4 \\
0 & 1 & 0 & 4 \\
0 & 0 & 1 & 7 \\
0 & 0 & 0 & 1 \\
\end{bmatrix}
$$
We can just compose these two matrices together (by doing the rotation first,
then translating!)
$$
TR =
\begin{bmatrix}
1 & 0 & 0 & x \\
0 & 1 & 0 & y \\
0 & 0 & 1 & z \\
0 & 0 & 0 & 1 \\
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & 0 & 4 \\
0 & 1 & 0 & 4 \\
0 & 0 & 1 & 7 \\
0 & 0 & 0 & 1 \\
-2 & 1 & 2 & 4 \\
2 & 2 & 1 & 4 \\
-1 & 2 & -2 & 7 \\
0 & 0 & 0 & 1
\end{bmatrix}
$$
Since the direction of flight was originally $(0, 0, 1)$, we have to
transform it to $(2, 1, -2)$.
3. \c{Consider the earth model shown below, which is defined in object
coordinates with its center at $(0, 0, 0)$, the vertical axis through the
north pole aligned with the direction $(0, 1, 0)$, and a horizontal plane
@ -680,70 +693,222 @@ author: |
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 $\frac{\sqrt{3}}{2}$.
In this case:
- near = 0.5
- far = 20
- left = bottom = $-\tan 30^\circ \times 0.5 = -\frac{\sqrt{3}}{6}$
- right = top = $\tan 30^\circ \times 0.5 = \frac{\sqrt{3}}{6}$
- right - left = top - bottom = $\frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$
$$
\begin{bmatrix}
\frac{2\times near}{right - left} & 0 & \frac{right + left}{right - left} & 0 \\
0 & \frac{2\times near}{top - bottom} & \frac{top + bottom}{top - bottom} & 0 \\
0 & 0 & -\frac{far + near}{far - near} & -\frac{2\times far\times near}{far - near} \\
0 & 0 & -1 & 0
\frac{2\cdot near}{right-left} & 0 & \frac{right+left}{right-left} & 0 \\
0 & \frac{2\cdot near}{top-bottom} & \frac{top+bottom}{top-bottom} & 0 \\
0 & 0 & -\frac{far+near}{far-near} & -\frac{2\cdot far\cdot near}{far-near} \\
0 & 0 & -1 & 0 \\
\end{bmatrix}
=
\begin{bmatrix}
\sqrt{3} & 0 & 0 & 0 \\
0 & \sqrt{3} & 0 & 0 \\
0 & 0 & -\frac{20.5}{19.5} & \frac{-20}{19.5} \\
0 & 0 & -1 & 0 \\
\end{bmatrix}
$$
$$
= \begin{bmatrix}
\frac{2\times 0.5}{\frac{\sqrt{3}}{2} - (-\frac{\sqrt{3}}{2})} & 0 & \frac{\frac{\sqrt{3}}{2} + (-\frac{\sqrt{3}}{2})}{\frac{\sqrt{3}}{2} - (-\frac{\sqrt{3}}{2})} & 0 \\
0 & \frac{2\times 0.5}{\frac{\sqrt{3}}{2} - (-\frac{\sqrt{3}}{2})} & \frac{\frac{\sqrt{3}}{2} + (-\frac{\sqrt{3}}{2})}{\frac{\sqrt{3}}{2} - (-\frac{\sqrt{3}}{2})} & 0 \\
0 & 0 & -\frac{20 + 0.5}{20 - 0.5} & -\frac{2\times 20\times 0.5}{20 - 0.5} \\
0 & 0 & -1 & 0
\end{bmatrix}
$$
$$
= \boxed{\begin{bmatrix}
\frac{1}{\sqrt{3}} & 0 & 0 & 0 \\
0 & \frac{1}{\sqrt{3}} & 0 & 0 \\
0 & 0 & -\frac{41}{39} & -\frac{40}{39} \\
0 & 0 & -1 & 0
\end{bmatrix}}
$$
\todo the numbers are wrong lmao
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?}
Only the second row changes, since it is the only thing that references
bottom or top. top + bottom is still 0, so the non-diagonal cell doesn't
change, so only top - bottom gets doubled. The result is:
$$
\begin{bmatrix}
\sqrt{3} & 0 & 0 & 0 \\
0 & \r{2}\sqrt{3} & 0 & 0 \\
0 & 0 & -\frac{20.5}{19.5} & \frac{-20}{19.5} \\
0 & 0 & -1 & 0 \\
\end{bmatrix}
$$
c. \c{(3 points) What are the new horizontal and vertical fields of view
after this change has been made?}
The horizontal one doesn't change, so only the vertical one does. Since the
height has increased relative to the near value, the FOV increases to
$120^\circ$.
\c{When the viewing frustum is known to be symmetric, we will have $left =
-right$ and $bottom = -top$. In that case, an alternative definition can be
used for the perspective projection matrix where instead of defining
parameters left, right, top, bottom, the programmer instead specifies a
vertical field of view angle and the aspect ratio of the viewing frustum.}
d. \c{(1 point) What are the entries in $P_{alt}$ when the viewing frustum is
defined by: 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 square aspect ratio?}
Using a 1:1 aspect ratio (since the field of views are the same)
- $\cot(30^\circ) = \sqrt{3}$
$$
\begin{bmatrix}
\cot \left( \frac{\theta_v}{2} \right) & 0 & 0 & 0 \\
0 & \cot \left( \frac{\theta_v}{2} \right) & 0 & 0 \\
0 & 0 & -\frac{far + near}{far - near} & \frac{-2 \cdot far \cdot near}{far - near} \\
0 & 0 & -1 & 0 \\
\end{bmatrix}
=
\begin{bmatrix}
\sqrt{3} & 0 & 0 & 0 \\
0 & \sqrt{3} & 0 & 0 \\
0 & 0 & -\frac{20.5}{19.5} & \frac{-20}{19.5} \\
0 & 0 & -1 & 0 \\
\end{bmatrix}
$$
e. \c{(1 points) Suppose the viewing window is re-sized to be twice as wide as
it is tall. How might you re-define the entries in $P_{alt}$?}
This means the aspect ratio is $\frac{1}{2}$, and the aspect ratio is applied
to the first entry in the matrix.
$$
\begin{bmatrix}
\r{2}\sqrt{3} & 0 & 0 & 0 \\
0 & \sqrt{3} & 0 & 0 \\
0 & 0 & -\frac{20.5}{19.5} & \frac{-20}{19.5} \\
0 & 0 & -1 & 0 \\
\end{bmatrix}
$$
f. \c{(2 points) What would the new horizontal and vertical fields of view be
after this change has been made? How would the image contents differ from
when the window was square?}
The horizontal FOV has doubled, so it goes to $120^\circ$, but the vertical
doesn't change.
g. \c{(1 points) Suppose the viewing window is re-sized to be twice as tall
as it is wide. How might you re-define the entries in $P_{alt}$?}
$$
\begin{bmatrix}
\r{\frac{1}{2}}\sqrt{3} & 0 & 0 & 0 \\
0 & \sqrt{3} & 0 & 0 \\
0 & 0 & -\frac{20.5}{19.5} & \frac{-20}{19.5} \\
0 & 0 & -1 & 0 \\
\end{bmatrix}
$$
So in this case, the $\theta_v$ would be changed, and then we scale the
$\theta_h$ part using the aspect ratio so it stays consistent.
h. \c{(2 points) What would the new horizontal and vertical fields of view be
after this change has been made? How would the image contents differ from
when the window was square?}
Since the horizontal field of view hasn't changed, it remains $60^\circ$. But
the vertical field of view has doubled, which makes it $120^\circ$. More of
the vertical view would be available than if it was just in a square.
i. \c{(1 points) Suppose you wanted the user to be able to see more of the
scene in the vertical direction as the window is made taller. How would you
need to adjust $P_{alt}$ to achieve that result?}
As long as the FOV is increased, more of the scene is available. If for
example, when we changed the height of the window above, the _horizontal_
field of view changed, then we would've _reduced_ the amount of visibility in
the scene.
## 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.}
So for each edge, we have to test if the point is on the inside half of the
line that divides the plane. Here are the (a, b, c) values for each of the
edges:
- Edge 1: (-2, 6, -12)
- Edge 2: (-6, 2, 44)
- Edge 3: (8, -8, 0)
We just shove these numbers into $e(x, y) = ax + by + c$ for each point to
determine if it lies inside the triangle or not. I've written this Python
script to do the detection:
```py
for (i, (p0_, p1_)) in enumerate([(p0, p1), (p1, p2), (p2, p0)]):
a= -(p1_[1] - p0_[1])
b = (p1_[0] - p0_[0])
c = (p1_[1] - p0_[1]) * p0_[0] - (p1_[0] - p0_[0]) * p0_[1]
for x, y in itertools.product(range(3, 12), range(3, 12)):
if (x, y) not in statuses: statuses[x, y] = [None, None, None]
e = a * x + b * y + c
statuses[x, y][i] = e >= 0
```
I then plotted the various numbers to see if they match:
![](9a.jpg){width=40%}
The 3 digit number corresponds to 1 if it's "inside" and 0 if it's not
"inside" for each of the 3 edges. The first digit corresponds to the top
horizontal edge, the second digit corresponds to the right most edge, and the
last digit corresponds to the long diagonal. When all three are 1, the pixel
is officially "inside" the triangle for sure.
There is also edge detection to see if the edge pixels belong to the left or
the top edges. I didn't implement that here but I talk about it below in the
second part b.
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?}
For these three pixels, we can start with $p_4$ and define $a$ and $b$ using
it (going to $p_6$ first to remain in counter-clockwise order).
Then we use the checks to determine if the $a$ and $b$ values satisfy the
conditions for being left or top edges:
```py
p4 = (6, 4)
p5 = (7, 7)
p6 = (10, 8)
for (i, (p0_, p1_)) in enumerate([(p4, p6), (p6, p5), (p5, p4)]):
a= -(p1_[1] - p0_[1])
b = (p1_[0] - p0_[0])
c = (p1_[1] - p0_[1]) * p0_[0] - (p1_[0] - p0_[0]) * p0_[1]
print(a, b, c, end=" ")
if a == 0 and b < 0: print("top")
elif a > 0: print("left")
else: print()
```
This tells us that the $p_6 \rightarrow p_5$ and the $p_5 \rightarrow p_4$
edges are both left edges. If you graph this on the grid, this is accurate.
This means for those particular edges, the points that lie exactly on the
edge will be considered "inside" and for others, it will not.
Edge detection can be done by subtracting the point from the normal and
seeing if the resulting vector is normal or not.
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

170
exam-2/exam2.py
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@ -1,14 +1,37 @@
import itertools
import numpy as np
import math
from sympy import N, Number, Rational, init_printing, latex, simplify, Expr
from sympy import N, Matrix, Number, Rational, init_printing, latex, simplify, Expr
from sympy.vector import CoordSys3D, Vector
from PIL import Image, ImageDraw
init_printing()
import sympy
C = CoordSys3D('C')
unit = lambda v: v/np.linalg.norm(v)
vec = lambda a, b, c: a * C.i + b * C.j + c * C.k
def ap(matrix, vector):
vector_ = np.r_[vector, [1]]
trans_ = matrix @ vector_
trans = trans_[:3]
return trans
def ap2(matrix, vector):
c = vector.components
vector_ = vector.to_matrix(C).col_join(Matrix([[1]]))
trans_ = matrix @ vector_
return trans_[0] * C.i + trans_[1] * C.j + trans_[2] * C.k
def pv(vector):
c = vector.components
x = c.get(C.i, 0)
y = c.get(C.j, 0)
z = c.get(C.k, 0)
return (x, y, z)
def perspective_matrix(vfov, width, height, left, right, bottom, top, near, far):
aspect = width / height
@ -62,7 +85,6 @@ def print_bmatrix(arr):
print("\\\\")
def problem_1():
C = CoordSys3D('C')
p = 1 * C.i + 4 * C.j + 8 * C.k
e = 0 * C.i + 0 * C.j + 0 * C.k
s = 2 * C.i + 2 * C.j + 10 * C.k
@ -302,10 +324,9 @@ def problem_6():
near = min(map(lambda p: p[2], points))
far = max(map(lambda p: p[2], points))
M_this = M(left, right, bottom, top, near, far)
for point in points:
point_ = np.r_[point, [1]]
trans_ = M(left, right, bottom, top, near, far) @ point_
trans = trans_[:3]
trans = ap(M_this, point)
v = np.vectorize(lambda x: float(x.evalf()))
l = np.vectorize(lambda x: latex(simplify(x)))
point = l(point)
@ -330,12 +351,139 @@ def problem_6():
calculate(oblique_transform, points)
def problem_9():
p0 = (3, 3)
p1 = (9, 5)
p2 = (11, 11)
statuses = {}
for (i, (p0_, p1_)) in enumerate([(p0, p1), (p1, p2), (p2, p0)]):
a= -(p1_[1] - p0_[1])
b = (p1_[0] - p0_[0])
c = (p1_[1] - p0_[1]) * p0_[0] - (p1_[0] - p0_[0]) * p0_[1]
for x, y in itertools.product(range(3, 12), range(3, 12)):
if (x, y) not in statuses: statuses[x, y] = [None, None, None]
e = a * x + b * y + c
statuses[x, y][i] = e >= 0
CELL_SIZE = 30
im = Image.new("RGB", (9 * CELL_SIZE, 9 * CELL_SIZE))
draw = ImageDraw.Draw(im)
in_color = (180, 255, 180)
out_color = (255, 180, 180)
for (x, y), status in statuses.items():
color = in_color if all(status) else out_color
sx, sy = x - 3, y - 3
ex, ey = sx + 1, sy + 1
draw.rectangle([
(sx * CELL_SIZE, sy * CELL_SIZE),
(ex * CELL_SIZE, ey * CELL_SIZE),
], color)
text = "".join(map(lambda s: "1" if s else "0", status))
_, _, w, h = draw.textbbox((0, 0), text)
draw.text(
(sx * CELL_SIZE + (CELL_SIZE - w) / 2.0, sy * CELL_SIZE + (CELL_SIZE - h) / 2.0),
text,
"black"
)
im.save("9a.jpg")
p4 = (6, 4)
p5 = (7, 7)
p6 = (10, 8)
for (i, (p0_, p1_)) in enumerate([(p4, p6), (p6, p5), (p5, p4)]):
a= -(p1_[1] - p0_[1])
b = (p1_[0] - p0_[0])
c = (p1_[1] - p0_[1]) * p0_[0] - (p1_[0] - p0_[0]) * p0_[1]
print(a, b, c, end=" ")
if a == 0 and b < 0: print("top")
elif a > 0: print("left")
else: print()
pass
print("\nPROBLEM 8 -------------------------"); problem_8()
print("\nPROBLEM 5 -------------------------"); problem_5()
def problem_2():
y_axis_angle = 3 * sympy.pi / 4
cos_t = -2
sin_t = 2
step_1 = Matrix([
[cos_t, 0, sin_t, 0],
[0, 1, 0, 0],
[-sin_t, 0, cos_t, 0],
[0, 0, 0, 1],
])
sqrt2 = sympy.sqrt(2)
cos_t = 2 * sqrt2
sin_t = 1
step_2 = Matrix([
[1, 0, 0, 0],
[0, cos_t, -sin_t, 0],
[0, sin_t, cos_t, 0],
[0, 0, 0, 1],
])
up_dir = vec(0, 1, 0)
nose_dir = vec(0, 0, 1)
left_wing_dir = vec(1, 0, 0)
right_wing_dir = vec(+1, 0, 0)
def apply(m):
print("- nose (z):", pv(ap2(m, nose_dir)))
print("- up (y):", pv(ap2(m, up_dir)))
print("- leftwing (+x):", pv(ap2(m, left_wing_dir)))
print("- rightwing (-x):", pv(ap2(m, right_wing_dir)))
print("step 1")
apply(step_1)
print()
print("step 2")
apply(step_2)
print()
print("step 2 @ step 1")
apply(step_2 @ step_1)
print()
print("step 1 @ step 2")
apply(step_1 @ step_2)
print()
print("SHIET")
print(vec(2, 1, -2).cross(vec(-2, 2, -1)))
R = Matrix([
[-2, 2, -1, 0],
[1, 2, 2, 0],
[2, 1, -2, 0],
[0, 0, 0, 1],
]).transpose()
print(R)
apply(R)
print()
T = Matrix([
[1, 0, 0, 4],
[0, 1, 0, 4],
[0, 0, 1, 7],
[0, 0, 0, 1],
])
print("T @ R")
print(T @ R)
# print("\nPROBLEM 8 -------------------------"); problem_8()
# print("\nPROBLEM 5 -------------------------"); problem_5()
# print("\nPROBLEM 9 -------------------------"); problem_9()
# print("\nPROBLEM 7 -------------------------"); problem_7()
# print("\nPROBLEM 4 -------------------------"); problem_4()
# print("\nPROBLEM 6 -------------------------"); problem_6()
# print("\nPROBLEM 1 -------------------------"); problem_1()
print("\nPROBLEM 2 -------------------------"); problem_2()
print("\nPROBLEM 9 -------------------------"); problem_9()
print("\nPROBLEM 7 -------------------------"); problem_7()
print("\nPROBLEM 4 -------------------------"); problem_4()
print("\nPROBLEM 6 -------------------------"); problem_6()
print("\nPROBLEM 1 -------------------------"); problem_1()

3
flake.nix
View file

@ -43,7 +43,8 @@
zip
zathura
(python310.withPackages (p: with p; [ ipython numpy scipy sympy ]))
(python310.withPackages
(p: with p; [ ipython numpy scipy sympy pillow ]))
]) ++ (with toolchain; [
cargo
rustc