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2D CGA - Midtest 2014

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COMPUTER GRAPHICS AND ANIMATION MID TEST 150 minutes. Open book. No electronic devices allowed. Answer the questions below. You can score more than 100, but the maximum score is capped to 100. There are a lot of questions and you might not have enough time to answer all of them. Start by answering the questions that you think are easy to answer. The questions are not too hard to answer; if you paid attention during class, tried doing the review questions and exercises at the end of each PPT file, and did the homeworks without mindlessly copying them from someone else, you should not have too much trouble answering the questions. But should you become frustrated during the test, feel free to create your own handmade graphics on your answer sheet. The basics Explain the difference between Computer Graphics, Image Processing, and Image Interpretation. Give an example for each. (10 points) Answer Computer graphics is a field that focused on
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HW 2D CGA - PPT 5 - Clipping 2

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Question 1 Solution: We need to declare:  $$A=\begin{bmatrix}2\\7\end{bmatrix}, B=\begin{bmatrix}10\\2\end{bmatrix}, C=\begin{bmatrix}7\\4\end{bmatrix}$$ And then find \(\overrightarrow {AB}\) and \(\vec N\): $$\overrightarrow {AB}=\begin{bmatrix}10-2\\2-7\end{bmatrix}=\begin{bmatrix}8\\-5\end{bmatrix}$$ Here are some basic rule you need to know to easily define \(\vec N\): $$if \ \vec N \ is \ to \ the \ left \ of \ any \ vector \rightarrow \vec N=\begin{bmatrix}-dy\\dx\end{bmatrix}$$  $$if \ \vec N \ is \ to \ the \ right \ of \ any \ vector \rightarrow \vec N=\begin{bmatrix}dy\\-dx\end{bmatrix}$$ From the rule above, we can see that because \(\vec N\) is facing to the left (up as stated in the question) of \(\overrightarrow {AB}\), we can conclude as stated below: $$\vec N=\begin{bmatrix}-dy\\dx\end{bmatrix}=\begin{bmatrix}5\\8\end{bmatrix}$$ Variable \(dx\) and \(dy\) is taken from \(\overrightarrow {AB}\). After that the question requires us to find \(\o
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HW 2D CGA - PPT 8 - Polygon Fill

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Question Solution: First of all, we need to know the element to be inserted into SET and AEL. The element must contain \(y_{max}\), \(x \ of \ y_{min}\), \(dx\), and \(dy\). Also carry but only in AEL, carry doesn't need to be inserted into SET. With that said, let's begin with the first polygon. Polygon 1 A(1, 6), B(5,4), C(3,1) From these 3 vertices we need to find the edges element containing \(y_{max}\), \(x \ of \ y_{min}\), \(dx\), and \(dy\). Let's take a look at edge AB , the \(y_{max}\) is 6, the \(x \ of \ y_{min}\) is 5, the \(dx\) is 4, and the \(dy\) is -2. Now if we found \(dy < 0\) we need to make it positive, by multiplying \(dx\) and \(dy\) with -1, so we got \(dx\) = -4 and \(dy\) = 2. The same rule applies with edges BC and CA. After we've found all of the elements then, we put it into the SET just like below. SET y Edges \(y_{max}\) 6 5 4 \( \rightarrow AB \) 6 5 -4 2 3 2 \(y_{min}\) 1 \( \rightarrow
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