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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 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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HW 2D CGA - PPT 6 - Polygon Clipping

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Question 1 Solution: Sutherland-Hodgman Method We will be solving this problem with visual clipping so there will be no calculations. Below is the graph of the polygon before any clipping is done: Polygon before clipping First step is to clip at the top edge of the clipping window to get rid of any excess polygon above the window. The given polygon is polygon P with it's members p1, p2, p3, p4, p5, p6, p7, p8, and p9 . Below is the first step we take to clip the top edge: First clipping at TOP EDGE Firstly we check vector p1p2 on the graph, both of the start point and end point are inside of top edge so we will be taking the ending point as the new point of q1. We'll do that for all vectors of polygon P until the last vector p9p1, the start point is outside and end point is inside so we will be taking the intersection and end point as q8 and q9. From the first clipping we get a new Polygon Q with it's members q1, q2, q3, q4, q5, q6, q7, q8, and q9 in ...
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