Explain the scan-line algorithm, Data Structure & Algorithms

Assignment Help:

Explain the Scan-Line Algorithm

This image-space method for removing hidden surfaces is an extension of the scan-line algorithm for filling polygon interiors. Instead of filling just one surface, we now deal with multiple surfaces. As each scan line is processed, all polygon surfaces intersecting that line are examined to determine which are visible. Across each scan line, depth calculations are made for each overlapping surface to determine which is nearest to the view plane. When the visible surface has been determined, the intensity value for that position is entered into the refresh buffer.

We assume that tables are set up for the various surfaces, which include both an edge table and a polygon table. The edge table contains coordinate end points for each line in the scene, the inverse slope of each line, and pointers into the polygon table to identify the surfaces bounded by each line. The polygon table contains coefficients of the plane equation for each surface, intensity information for the surfaces, and possibly pointers into the edge table. To facilitate the search for surfaces crossing a given scan line, we can set up an active list of edges from information in the edge table. This active list will contain only edges that cross the current scan line, sorted in order of increasing x. In addition, we define a flag for each surface that is set on or off to indicate whether a position along a scan line is inside or outside of the surface. Scan lines are processed from left to right. At the leftmost boundary of a surface, the surface flag is turned on; and at the rightmost boundary, it is turned off.  Figure 3.6 illustrates the scan-line method for locating visible portions of surface for pixel positions along the line. The active list for scan line 1 contains information from the edge table for edges AB and BC, only the flag for surface S1 is on. Therefore, no depth calculations are necessary, and intensity information for surface S1 is entered from the polygon table into the refresh buffer. Similarly, between edges EH and FG, only the flag for surface S2 is on. No other positions along scan line 1 intersect surfaces, so the intensity values in the other areas are set to the background intensity. The background \intensity can be loaded throughout the buffer in an initialization routine. 

For scan lines 2 and 3 in Figure, the active edge list contains edges AD, EH, BC, and FG. Along scan line 2 from edge AD to edge EH, only the flag for surface S1 is on. But between edges EH and BC, the flags for both surfaces are on. In this interval, depth calculations must be made using the plane coefficients for the two surfaces. For this example, the depth of surface S1 is assumed to be less than that of S2, so intensities for surface S1 are loaded into the refresh buffer until boundary BC is encountered. Then the flag for surface S1 goes off, and intensities for surface S2, so intensities for surface S1 are loaded into the refresh buffer until boundary BC is encountered. Then the flag for surface S1 goes off, and intensities for surface S2 are stored until edge FG is passed. 

We can take advantage of coherence along the scan lines as we pass from one scan line to the next. In Figure 3.6, scan line 3 has the same active list of edges as scan line 2. Since no changes have occurred in line intersections, it is unnecessary again to make depth calculations between edges EH and BC. The two surfaces must be in the same orientation as determined on scan line 2, so the intensities for surface S1 can be entered without further calculations.

 

688_data structure.png


Related Discussions:- Explain the scan-line algorithm

Polynomials - represented by using arrays, /* the program accepts two polyn...

/* the program accepts two polynomials as a input & prints the resultant polynomial because of the addition of input polynomials*/ #include void main() { int poly1[6][

Enumerate about the concept of container, Enumerate about the concept of co...

Enumerate about the concept of container A Container can have a size() operation. We can also ask (somewhat redundantly) whether a Container is empty. And even though a Contain

Example of pre order traversal, Example of pre order traversal: Reading of...

Example of pre order traversal: Reading of a book, since we do not read next chapter unless we complete all sections of previous chapter & all its sections. Figure  : Rea

Algorithm for the selection sort, Q. Give the algorithm for the selection s...

Q. Give the algorithm for the selection sort. Describe the behaviours of selection sort when the input given is already sorted.

Interest, I =PR/12 Numbers of years .Interest rate up to 1yrs ...

I =PR/12 Numbers of years .Interest rate up to 1yrs . 5.50 up to 5yrs . 6.50 More than 5 yrs . 6.75 design an algorithm based on the above information

Complexity of an algorithm, An algorithm is a sequence of steps to solve a ...

An algorithm is a sequence of steps to solve a problem; there may be more than one algorithm to solve a problem. The choice of a particular algorithm depends upon following cons

String pattern matching, Document processing is quickly becoming one of the...

Document processing is quickly becoming one of the dominant functions of computers. Computers are utilized to edit, search & transport documents over the Internet, and to display d

Program to manipulate the data structure, Data Structure and Methods: ...

Data Structure and Methods: Build an array structure to accomodate at least 10 elements. Provide routines for the following: An initializer. A routine to populate (

Total impedent of the circuit, an electrical student designed a circuit in...

an electrical student designed a circuit in which the impedence in one part of a series circuit is 2+j8 ohms and the impedent is another part of the circuit is 4-j60 ohm mm program

Illustrate the intervals in mathematics, Illustrate the intervals in mathem...

Illustrate the intervals in mathematics Carrier set of a Range of T is the set of all sets of values v ∈ T such that for some start value s ∈ T and end value e ∈ T, either s ≤

Write Your Message!

Captcha
Free Assignment Quote

Assured A++ Grade

Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd