489 lines
20 KiB
Markdown
489 lines
20 KiB
Markdown
---
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geometry: margin=2cm
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output: pdf_document
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title: Exam 1
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subtitle: CSCI 5607
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date: \today
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author: |
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| Michael Zhang
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| zhan4854@umn.edu $\cdot$ ID: 5289259
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---
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\renewcommand{\c}[1]{\textcolor{gray}{#1}}
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## Image File Formats
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1. \c{(8 points) Suppose you have used your raytracing program to create an
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image in image file type? Other potential image file format options include:
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JPEG, TIFF, and GIF. Discuss the main advantages and disadvantages in
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converting your image from ASCII PPM to these other file types. Try to
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identify at least one strength and one weakness of each option.}
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ASCII PPM has the obvious deficiency of not having a concise representation
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on disk (although P6 solves this problem). On the plus side, it is easy to
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represent it in a way that’s easy to debug, for example if you put double
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line breaks between each row in the image, and added a comment making it easy
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to identify rows in the image quickly.
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JPEG is typically a lossy data format, which means while you can get insanely
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high compression resulting in lower file sizes, you also get compression
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artifacts which ruins the quality of the image. On the other hand, ASCII PPM
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is lossless and preserves the original image at original quality entirely.
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## Color Spaces
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2. \c{(8 points) Your raytracing program generates images using an RGB color
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model. But RGB is not the only color space that artists or programmers use
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when creating or working with digital images. Identify two other color spaces
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that are commonly used in computer graphics or related fields. Describe their
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most important characteristics, and discuss how they are similar to and/or
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different from using an (r, g, b) representation. For each color space, try
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to identify at least one use case where it might be preferred over using an
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RGB representation and explain.}
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Another color space that is typically used for computer graphics is CMYK, and
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YUV.
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CMYK uses the secondary colors cyan, magenta, and yellow, as well as black.
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This is primarily used for printing, because as opposed to when light is
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emitted from a computer screen, reflected light from paper combines
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differently. When applying more ink or combining ink together, the color
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tends to get darker rather than lighter, so lighter base colors are
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preferred. The black is added since it is not possible to create pure black
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otherwise.
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YUV is used more commonly in video formats. It uses luma, and red/blue
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projections of the image rather than RGB. While equally expressive as RGB,
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YUV is more efficient in video since the chroma of an image doesn’t change as
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frequently over video.
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## Color Compositing
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3. \c{"Alpha-blending" or "color compositing" is a method for combining the
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colors of one image with the colors of another. A typical alpha-blending
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function is: $C_{final} = C_{fg} \cdot a_{fg} + C_{bg} \cdot (1 - a_{fg})$
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where $C_{fg}$ is the color of the foreground pixel, $Cbg$ is the color of
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the background pixel, $a_{fg}$ is the opacity of the foreground pixel (where
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$a_{fg} = 1$ represents "fully opaque" and $a_{fg} = 0$ represents "fully
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transparent"), and $a_{bg}$ is assumed to be 1. When $a_{fg} = 0.5$,
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$C_{final} = \frac{C_{fg} + C_{bg}}{2}$ which is equivalent to taking the
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average of the foreground and background colors. Yet, the results of using
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alpha-blending to composite multiple transparent surfaces is different than
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the results of averaging their colors.}
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\c{Consider a pure green image and a pure blue image, each with $\alpha =
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0.5$. If $C_{bg} = (0,0,0,1)$, what color would be obtained by:}
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a) \c{(3 points) compositing the blue image over background, then the green
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image over the result?}
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In this case, compositing the blue $(0, 0, 1, 0.5)$ over the background
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$(0, 0, 0, 1)$ would be $C_{final} = (0, 0, 1) \cdot 0.5 + (0, 0, 0) \cdot
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(1 - 0.5) = (0, 0, 0.5)$.
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Then, compositing green over this gives us: $C_{final} = (0, 1, 0) \cdot
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0.5 + (0, 0, 0.5) \cdot (1 - 0.5) = (0, 0.5, 0) + (0, 0, 0.25) =
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\boxed{(0, 0.5, 0.25)}$.
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b) \c{(2 points) compositing the green image over background, then the blue
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image over the result?}
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The result will be the same, except with the blue and green components
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reversed: $\boxed{(0, 0.25, 0.5)}$.
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\c{When $a_{bg} \ne 1$, the alpha blending function
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becomes slightly more complicated: $a_{final} = a_{fg} + a_{bg} \cdot (1 -
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a_{fg})$; $C_{final} = \frac{C_{fg} \cdot a_{fg} + C_{bg} \cdot a_{bg} \cdot
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(1 - a_{fg})}{a_{final}}$.}
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c) \c{(3 points) What result do you obtain by compositing the blue image over
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the green image, and then compositing that result over the background?}
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First, we want to composite the blue image over the green image, which
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both have an alpha of $0.5$:
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$a_{final} = 0.5 + 0.5 \cdot (1 - 0.5) = 0.75$.
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Then
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\begin{align*}
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C_{final} &= \frac{C_{fg} \cdot a_{fg} + C_{bg} \cdot a_{bg} \cdot (1 -
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a_{fg})}{a_{final}} \\
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&= \frac{(0, 0, 1) \cdot 0.5 + (0, 1, 0) \cdot 0.5 \cdot (1 - 0.5)}{0.75}
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\\
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&= \frac{(0, 0, 0.5) + (0, 0.25, 0)}{0.75} \\
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&= \frac{(0, 0.25, 0.5)}{0.75} \\
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&= (0, 0.\overline{333}, 0.\overline{666})
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\end{align*}
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Now, composite this over the background:
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\begin{align*}
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C_{final} &= C_{fg} \cdot a_{fg} + C_{bg} \cdot (1 - a_{fg}) \\
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&= (0, 0.\overline{333}, 0.\overline{666}) \cdot 0.75 + (0, 0, 0) \cdot (1
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- 0.75) \\
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&= (0, 0.25, 0.5) \cdot 0.75 \\
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&= \boxed{(0, 0.1875, 0.375)}
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\end{align*}
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Alpha is 1 at the end because of blending with the background.
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d) \c{(2 points) How does your answer in (c) differ from what you would get
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by simply averaging the blue and green images, and then superimposing that
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result over the background?}
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Because the alphas are 0.5, the way you blend the images produces a more
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nuanced 0.75 alpha for whichever ones you blend first. So if you were to
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blend into the background first, the background has more of an effect on
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the blended image.
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## Viewing Parameter Specifications
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4. \c{(10 points) Suppose that you are trying to help a marketing manager come
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up with the largest possible number to use to advertise the field of view
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offered by their new head-mounted display device. if the vertical field of
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view is $90^\circ$ and the horizontal field of view is $110^\circ$ what is
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the diagonal field of view?}
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The relationship between FOV angle and the length (whichever way we're
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measuring) is $\tan(\frac{1}{2} \theta) = \frac{L}{2d}$, where $d$ is the
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distance the viewer is from the screen.
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Because the viewer will remain the same distance from the screen while all
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these calculations are being made, we can just ignore it completely by
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setting it to 1. So we have $\tan(\frac{1}{2} \theta) = \frac{L}{2}$.
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Now, we can find $w$ and $h$ by reversing this into $L = 2\tan(\frac{1}{2}
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\theta)$:
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- $w = 2\tan(55^\circ)$
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- $h = 2\tan(45^\circ)$
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The diagonal $D$ can be calculated using Pythagorean's theorem:
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- $D = \sqrt{w^2 + h^2} = 2\sqrt{\tan^2(55^\circ) + \tan^2(45^\circ)} \approx
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3.48$
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```
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In [68]: w = 2 * tan(radians(55))
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In [69]: h = 2 * tan(radians(45))
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In [71]: sqrt(pow(w, 2) + pow(h, 2))
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Out[71]: 3.486893591242196
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```
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Now, we can convert this length back into an angle by turning the original
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equation around: $\theta = 2\tan^{-1}(\frac{1}{2} L)$, which gives us a value
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of around $120^\circ$.
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```
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In [72]: 2 * atan(0.5 * _)
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Out[72]: 2.1000651019312633
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In [73]: degrees(_)
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Out[73]: 120.32486704337242
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```
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## Geometry for Computer Graphics
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5. \c{(4 points) Consider a triangle defined by the vertices: $v_0 = (0, 1, 2),
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v_1 = (4, 1, -2), v_2 = (2, 2, 0)$, specified in counter-clockwise order.
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What is the area of this triangle?}
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The area of a triangle given two of its sides is $\frac{1}{2} |e_1 \times
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e_2|$. In this case, $e_1 = v_1 - v_0 = (4, 0, -4)$, and $e_2 = v_2 - v_0 =
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(2, 1, -2)$.
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The cross product is (4, 0, 4) (I used numpy to calculate this, see below
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output), which normalizes to $\sqrt{32}$. Half of this is $\boxed{\sqrt{8}}$,
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which is the area.
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```
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In [2]: v0 = np.array([0,1,2])
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In [3]: v1 = np.array([4,1,-2])
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In [4]: v2 = np.array([2,2,0])
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In [5]: e1 = v1 - v0 # (4, 0, -4)
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In [6]: e2 = v2 - v0 # (2, 1, -2)
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In [8]: np.cross(e1, e2)
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Out[8]: array([4, 0, 4])
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```
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## Point-in-Polygon Testing
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6. \c{Consider a triangle defined by $v_0 = (3, 0, -2)$, $v_1 = (2, 4, 0)$, $v_2
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= (-2, 2, 2)$, in counterclockwise order, and a point $p = (2, 1, -1)$.}
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a) \c{(9 points) What are the barycentric coordinates that define the
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location of the point $p$ with respect to the locations of $v_0$, $v_1$,
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and $v_2$?}
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Using $v_0$ as the base, we can write $e_1 = v_1 - v_0 = (-1, 4, 2)$, and
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$e_2 = v_2 - v_0 = (-5, 2, 4)$. We also need $e_p = p - v_0 = (-1, 1, 1)$.
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Now we can proceed to find $\beta$ and $\gamma$.
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Using the above, we get:
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- $d_{11} = e_1 \cdot e_1 = (-1, 4, 2) \cdot (-1, 4, 2) = 1 + 16 + 4 = 21$
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- $d_{12} = e_1 \cdot e_2 = (-1, 4, 2) \cdot (-5, 2, 4) = 5 + 8 + 8 = 21$
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- $d_{22} = e_2 \cdot e_2 = (-5, 2, 4) \cdot (-5, 2, 4) = 25 + 4 + 16 =
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45$
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- $d_{1p} = e_1 \cdot e_p = (-1, 4, 2) \cdot (-1, 1, 1) = 1 + 4 + 2 = 7$
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- $d_{2p} = e_2 \cdot e_p = (-5, 2, 4) \cdot (-1, 1, 1) = 5 + 2 + 4 = 11$
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Plugging this into the formula, we get:
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\begin{align*}
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\begin{bmatrix}
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e_1 \cdot e_1 & e_1 \cdot e_2 \\
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e_1 \cdot e_2 & e_2 \cdot e_2 \\
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\end{bmatrix}
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\begin{bmatrix}
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\beta \\ \gamma
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\end{bmatrix}
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&=
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\begin{bmatrix}
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e_1 \cdot e_p \\ e_2 \cdot e_p
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\end{bmatrix} \\
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\begin{bmatrix} 21 & 21 \\ 21 & 45 \end{bmatrix}
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\begin{bmatrix} \beta \\ \gamma \end{bmatrix}
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&=
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\begin{bmatrix} 7 \\ 11 \end{bmatrix} \\
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\end{align*}
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Checking the determinant $\det \begin{vmatrix} 21 & 21 \\ 21 & 45
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\end{vmatrix} = 21 \times 45 - 21 \times 21 = 504$, which is non-zero.
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This means the matrix is invertible and we can solve for $\beta$ and
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$\gamma$. Now we have:
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- $\beta = (d_{22} d_{1p} - d_{12} d_{2p}) / \det = (45 \times 7 - 21
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\times 11) / 504 = \frac{1}{6}$
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- $\gamma = (d_{11} d_{2p} - d_{12} d_{1p}) / \det = (21 \times 11 - 21
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\times 7) / 504 = \frac{1}{6}$
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Then, $\alpha = 1 - (\beta + \gamma) = \frac{2}{3}$.
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b) \c{(1 point) Does p lie: inside the triangle; outside of the triangle; or
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on the edge of the triangle? (please circle the correct answer)}
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Using these values, we can see that all three values lie between 0 and 1,
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which means the point is definitely inside the triangle.
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## Smooth Shading
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7. \c{(6 points) Consider a triangle defined by $v_0 = (1, 1, -2)$, $v_1 = (-1,
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0, 1)$, $v_2 = (2, -1, -1)$, in counter-clockwise order, where the associated
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normal directions are: $vn_0 = (\frac{2}{3}, -\frac{2}{3}, \frac{1}{3})$,
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$vn_1 = (-\frac{2}{3}, \frac{1}{3}, \frac{2}{3})$, $vn_2 = (-\frac{1}{3},
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\frac{2}{3}, \frac{2}{3})$. If the barycentric coordinates $\alpha = 0.3$,
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$\beta = 0.6$, $\gamma = 0.1$ describe the location of $p$ with respect to
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$v_0$, $v_1$, $v_2$, what is the smooth-shading surface normal direction at
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$p$ (interpolated from the vertex normals) that should be used when computing
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the Phong illumination at $p$?}
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This is simply applying each of the barycentric coordinates (which is really
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just a weight of how close $p$ is to a particular vertex) to the
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corresponding vertex, and summing them. The corresponding vertex is the one
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_opposite_ from the side that the coordinate defines.
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\begin{align*}
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n &= \alpha \cdot vn_0 + \beta \cdot vn_1 + \gamma \cdot vn_2 \\
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&= 0.3 \cdot vn_0 + 0.6 \cdot vn_1 + 0.1 \cdot vn_2 \\
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&= 0.3 \cdot \left(\frac{2}{3}, -\frac{2}{3}, \frac{1}{3} \right) + 0.6
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\cdot \left(-\frac{2}{3}, \frac{1}{3}, \frac{2}{3} \right) + 0.1 \cdot
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\left(-\frac{1}{3}, \frac{2}{3}, \frac{2}{3} \right) \\
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&= (0.2, -0.2, 0.1) + (-0.4, 0.2, 0.4) + \left(-\frac{0.1}{3},
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\frac{0.2}{3}, \frac{0.2}{3} \right) \\
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&= \boxed{(-0.2\overline{333}, 0.0\overline{666}, 0.5\overline{666})}
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\end{align*}
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## Phong Illumination
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8. \c{(10 points) Suppose you are rendering a simple scene that contains a
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single sphere of radius 3 centered at (0, 0, 0) and illuminated by a single
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white point light source located at (0, 4, 2). Use the Blinn-Phong
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illumination model to determine what color a viewer at (6, –2, 3) will
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perceive at the point $p = (2, 2, 1)$ if the material properties of the
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sphere are $k_a = 0.2$, $k_d = 0.6$, $k_s = 0.3$, $n=2$, $O_{d\lambda} = (1,
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0, 0)$ and $O_{s\lambda} = (1, 1, 0)$. Please provide the $(r, g, b)$ color
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definition and also name the color that this value corresponds to.}
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First, we have to find some vectors needed for the Blinn-Phong equation:
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- For the normal $\vec{N}$, we can just subtract the sphere center from the
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intersection point, yielding $\vec{N} = (2, 2, 1)$.
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- For the vector to the viewer $\vec{V}$, we can subtract the intersection
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point from the viewer location, yielding $\vec{V} = (6, -2, 3) - (2, 2,
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1) = (4, -4, 2)$.
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- For the light direction $\vec{L}$, subtract the intersection point from the
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light $\vec{L} = (0, 4, 2) - (2, 2, 1) = (-2, 2, 1)$.
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- The halfway direction is just halfway between the light and the viewer,
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which is $\vec{H} = (\vec{L} + \vec{V}) / 2 = ((-2, 2, 1) + (4, -4, 2)) /
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2 = (2, -2, 3) / 2 = (1, -1, 1.5)$.
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First, substitute the values given into the Blinn-Phong equation:
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\begin{align*}
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I_\lambda &= k_a O_{d\lambda} + k_d O_{d\lambda} (\vec{N} \cdot \vec{L}) +
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k_s O_{s\lambda} (\vec{N} \cdot \vec{H})^n \\
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I_\lambda &= 0.2 \cdot (1, 0, 0) + 0.6 \cdot (1, 0, 0) ((2, 2, 1) \cdot
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(-2, 2, 1)) + 0.3 \cdot (1, 1, 0) ((2, 2, 1) \cdot (1, -1, 1.5))^2 \\
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I_\lambda &= (0.2, 0, 0) + (0.6, 0, 0) \cdot (-4 + 4 + 1) + (0.3, 0.3, 0)
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(2 - 2 + 1.5)^2 \\
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I_\lambda &= (0.2, 0, 0) + (0.6, 0, 0) + (0.3, 0.3, 0) \cdot 2.25 \\
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I_\lambda &= (0.8, 0, 0) + (0.675, 0.675, 0) \\
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I_\lambda &= (1.475, 0.675, 0) \\
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\end{align*}
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We clamp to 1, so the result is $\boxed{(1, 0.675, 0)}$. This is an
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orange-ish brown color.
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## Ray-Object Intersection
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9. \c{(14 points) The set of all points on the surface of an infinite,
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circularly symmetric, one-sheet hyperboloid, centered at the origin and
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aligned with the z axis, can be defined by the implicit equation:
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$\frac{x^2}{r^2} + \frac{y^2}{r^2} - \frac{z^2}{c^2} = 1$ where $r$ and $c$
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are parameters that control the radius and curvature of the hyperboloid, and
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the set of all points $(x, y, z)$ for which the implicit equation is true lie
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on the surface of the hyperboloid. Let $r = 2$ and $c = 1$, to obtain the
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surface shown at the right.}
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\c{Consider a viewing ray that starts at $p_0 = (-2, 1, -1)$ and travels in
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the direction $d = (0, 1, 1)$. Does this ray intersect this surface? If so,
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at what point(s)?}
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{width=300px}
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Starting by substituting in the ray equation, we would get:
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\begin{align*}
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\frac{(x_0 + tx_d)^2}{r^2} +
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\frac{(y_0 + ty_d)^2}{r^2} -
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\frac{(z_0 + tz_d)^2}{c^2} &= 1 \\
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\frac{x_0^2 + 2tx_0x_d + t^2x_d^2}{r^2} +
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\frac{y_0^2 + 2ty_0y_d + t^2y_d^2}{r^2} -
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\frac{z_0^2 + 2tz_0z_d + t^2z_d^2}{c^2} &= 1 \\
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\left(
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\frac{x_d^2}{r^2} + \frac{y_d^2}{r^2} - \frac{z_d^2}{c^2}
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\right) t^2 +
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2 \left(
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\frac{x_0x_d}{r^2} + \frac{y_0y_d}{r^2} - \frac{z_0z_d}{c^2}
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\right) t +
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\left(
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\frac{x_0^2}{r^2} + \frac{y_0^2}{r^2} - \frac{z_0^2}{c^2} - 1
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\right) &= 0
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\end{align*}
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Then, substituting in the values for $r$, $c$, and the ray parameters gives:
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\begin{align*}
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\left(
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\frac{0^2}{2^2} + \frac{1^2}{2^2} - \frac{1^2}{1^2}
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\right) t^2 +
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2 \left(
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\frac{-2 \cdot 0}{2^2} + \frac{1 \cdot 1}{2^2} - \frac{-1 \cdot 1}{1^2}
|
||
\right) t +
|
||
\left(
|
||
\frac{(-2)^2}{2^2} + \frac{1^2}{2^2} - \frac{(-1)^2}{1^2} - 1
|
||
\right) = 0 \\
|
||
\left( 0 + \frac{1}{4} - 1 \right) t^2 +
|
||
2 \left( \frac{1}{4} + 1 \right) t +
|
||
\left( 1 + \frac{1}{4} - 1 - 1 \right) = 0 \\
|
||
-\frac{3}{4} t^2 + \frac{5}{2} t -\frac{3}{4} = 0
|
||
\end{align*}
|
||
|
||
Now we can just use quadratic equation with $A = -\frac{3}{4}$, $B =
|
||
\frac{5}{2}$, and $C = -\frac{3}{4}$. Quickly checking the determinant $B^2 -
|
||
4AC = \frac{5}{2}^2 - 4 \cdot (-\frac{3}{4})^2 = \frac{25}{4} - \frac{9}{4} =
|
||
\frac{16}{4} = 4$, which is positive so there are \boxed{\textrm{two
|
||
solutions}}.
|
||
|
||
To get the points themselves, we solve for $t$:
|
||
|
||
\begin{align*}
|
||
t &= \frac{-\frac{5}{2} \pm \sqrt{4}}{2 \cdot -\frac{3}{4}} \\
|
||
t &= \frac{-\frac{5}{2} \pm 2}{-\frac{3}{2}} \\
|
||
t &= \frac{-5 \pm 4}{-3} \\
|
||
t &\in \{\frac{1}{3}, 3\}
|
||
\end{align*}
|
||
|
||
Now, just substitute the $t$ back into the ray equation $p = p_0 + dt$ to get
|
||
|
||
\begin{align*}
|
||
p_0 &= (-2, 1, -1) + \frac{1}{3} (0, 1, 1) \\
|
||
&= (-2, 1, -1) + (0, \frac{1}{3}, \frac{1}{3}) \\
|
||
&= (-2, \frac{4}{3}, -\frac{2}{3})
|
||
\end{align*}
|
||
|
||
\begin{align*}
|
||
p_1 &= (-2, 1, -1) + 3 (0, 1, 1) \\
|
||
&= (-2, 1, -1) + (0, 3, 3) \\
|
||
&= (-2, 4, 2)
|
||
\end{align*}
|
||
|
||
The ray intersects the hyperboloid at \boxed{(-2, \frac{4}{3}, -\frac{2}{3})
|
||
\textrm{and} (-2, 4, 2)}.
|
||
|
||
## Texture Mapping
|
||
|
||
10. \c{(10 points) Suppose that a texture image has been mapped onto the surface
|
||
of a truncated cylinder such that the values of $u$ increase from 0 to 1 as
|
||
the values of $\theta$ increase from $0$ to $2\pi$, and the values of $v$
|
||
decrease from 1 to 0 as the position along the cylinder’s height increases
|
||
from its base to its top. Consider a cylinder of radius 2 and height 4,
|
||
whose base is centered at $(0, 0, 0)$. What is the texture coordinate $(u,
|
||
v)$ at the point $p = (-\sqrt{2}, \sqrt{2}, 1)$?}
|
||
|
||
The point $(-\sqrt{2}, \sqrt{2})$ occurs at $\theta = \frac{3}{4}\pi$, so
|
||
when mapped to $u$, this corresponds to a value of $u = \frac{3}{8}$.
|
||
|
||
The height of 1 occurs halfway on the cylinder, which corresponds to a $v =
|
||
0.5$.
|
||
|
||
The coordinate is $\boxed{(\frac{3}{8}, 0.5)}$.
|
||
|
||
11. \c{Consider a texture image that is 1024 pixels wide and 512 pixels tall. If
|
||
bi-linear interpolation is used to retrieve a color from this image
|
||
corresponding to the texture coordinate (u, v) = (0.6, 0.2):}
|
||
|
||
a) \c{(4 points) From which four pixels $[i, j]$ in the image will the
|
||
texture colors be retrieved?}
|
||
|
||
Multiplying it out, we get the exact coordinates (0.6 * 1024, 0.2 * 512)
|
||
= (614.4, 102.4). This means, we will be sampling from:
|
||
|
||
- (614, 102)
|
||
- (614, 103)
|
||
- (615, 102)
|
||
- (615, 103)
|
||
|
||
b) \c{(6 points) Suppose that the colors in the texture image have been
|
||
defined to be white when $i$ and $j$ are both even numbers, red when $i$
|
||
is even and $j$ is odd, green when $i$ is odd and $j$ is even, and blue
|
||
when both $i$ and $j$ are odd. What color will be returned from the
|
||
texture lookup, after the retrieved colors have been appropriately
|
||
combined using bilinear interpolation?}
|
||
|
||
Using this information, the colors corresponding to each of the above
|
||
are:
|
||
|
||
- (614, 102): white
|
||
- (614, 103): red
|
||
- (615, 102): green
|
||
- (615, 103): blue
|
||
|
||
Then, we need to weigh these according to how close the point we want to
|
||
observe is from those four corners:
|
||
|
||
- (614, 102): 0.6 * 0.6 * (1, 1, 1) = 0.36 * (1, 1, 1) = (0.36, 0.36,
|
||
0.36)
|
||
- (614, 103): 0.6 * 0.4 * (1, 0, 0) = 0.24 * (1, 0, 0) = (0.24, 0, 0)
|
||
- (615, 102): 0.4 * 0.6 * (0, 1, 0) = 0.24 * (0, 1, 0) = (0, 0.24, 0)
|
||
- (615, 103): 0.4 * 0.4 * (0, 0, 1) = 0.16 * (0, 0, 1) = (0, 0, 0.16)
|
||
|
||
Taking the sum, we get (0.36 + 0.24, 0.36 + 0.24, 0.36 + 0.16) =
|
||
\boxed{(0.6, 0.6, 0.52)}.
|