CG iclicker questions 4
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What is happening in the following code that is part of a picking program that we saw (11_4):
if(i==0) gl_FragColor = c[0]; else if(i==1) gl_FragColor = c[1]; else if(i==2) gl_FragColor = c[2]; else if(i==3) gl_FragColor = c[3]; else if(i==4) gl_FragColor = c[4]; else if(i==5) gl_FragColor = c[5]; else if(i==6) gl_FragColor = c[6];
- This assigns to each pixel the number that we'd like returned if the user clicks on that pixel.
- This assigns to each pixel the true color of that polygon.
- This keeps track of how many times the user clicked on that pixel.
- This notes how many vertices that polygon has.
- This notes the shininess exponent for that face.
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Radiosity is better than ray tracing when the scene is all
- diffuse objects
- objects outside the viewing region
- objects that are very close to the viewpoint
- specular objects
- very small objects
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Ray tracing is better than radiosity when the scene is all
- diffuse objects
- objects outside the viewing region
- objects that are very close to the viewpoint
- specular objects
- very small objects
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Firing multiple rays through each pixel handles the problem of
- diffuse objects
- objects outside the viewing region
- objects that are very close to the viewpoint
- specular objects
- very small objects
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Consider a 2D Cartesian cubic Bezier curve with these control points: (0,0), (0,1), (1,1), (1,0). What is the point at t=0? OK to look up the formula.
- (0,0)
- (0,1)
- (1,0)
- (1/2, 3/4)
- (1/2,1)
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What is the point at t=1/2?
- (0,0)
- (0,1)
- (1,0)
- (1/2, 3/4)
- (1/2,1)
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What is the point at t=1?
- (0,0)
- (0,1)
- (1,0)
- (1/2, 3/4)
- (1/2,1)
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If you interpolate a curve through a list of control points instead of approximating a curve near the points, then what happens?
- It will not be possible to join two curves and match the radii of curvature.
- The calculations to compute the curve will take impossibly long.
- The curve will stay within the convex hull of the control points.
- The curve will swing outside the convex hull of the control points.
- This isn't possible for curves of odd degree.
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If you use quadratic Bezier curves, then what happens?
- It will not be possible to join two curves and match the radii of curvature.
- The calculations to compute the curve will take impossibly long.
- The curve will stay within the convex hull of the control points.
- The curve will swing outside the convex hull of the control points.
- This isn't possible for curves of even degree.