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We like to think about a scene as mathematical primitives in a world-space.

This scene is then rendered into the frame buffer. This allows a logical separation of the world from the view of that world.

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In a language such as OpenGL a programmer can generate a 2D line segment in world-space with code like the following: In a language such as OpenGL polygons are very restricted to improve speed: To generate the outline of a triangular 2D polygon in world-space using OpenGL a programmer can write code like the following: Note that large complex objects are often reduced down to a large number of triangles i.

How tekstverwerkung line segments tekstverwerkng polygons in world-space converted into illuminated pixels on the screen?

First these coordinates in world-space must be converted to coordinates in the viewport ie pixel coordinates in the frame buffer. This may involve the conversion from a 2D world to a 2D frame buffer which we will study in a couple weeksor the reduction from a 3D world to a 2D frame buffer which we will study a couple weeks later.

Then these coordinates in the viewport must be used to draw lines and polygons made up of individual pixels rasterization. This is the topic we will discuss now.

Most of the algorithms in Computer Graphics will follow the same pattern below. There is the simple braindead algorithm that works, cnoversie is too slow. Then that algorithm is repeatedly refined, making it more complicated to understand, but much faster for the computer to implement.

## Scan conversion

So we need more complex algorithms which use simpler operations to decrease the speed. The basic improvement of this algorithm over the purely braindead one is that instead of calculating Y for each X from the equation of a line one multiplication and one addition tfkstverwerking, we will calculate it from the previous Y by just adding a fixed constant one addition only.

This works because the delta change in X from one column to the next is known to be exactly 1. This algorithm follows sczn the previous one, and further takes advantage of the observation that for line slopes between 0 and 1, the change in Y from one column to the next will be either 0 or 1.

This algorithm requires no rounding, no floating point numbers, and no multiplications. Now we need to compute what Midpoint is for the next iteration.

Assuming integral endpoints for the line segment if not then make them integral starting at the leftmost edge of the line: The full algorithm is given in C in the white book as figure 3. Calculate which pixels will be highlighted using the midpoint algorithm to draw a line from 5,8 to 9,