Chapter 3

Section

3-11 till page 130

Polygon Rasterization

Some of the material in these slides may have been adapted from University of Virginia, MIT, and Åbo Akademi University

Polygon

● Polygon: A set of points p0, p1, p2, p3,…● Convex Vs Concave

Polygon Vs Triangle

● Triangle is the minimal unit of a polygon■ All polygons can be broken up into triangles■ Triangles are guaranteed to be:

○ Planar○ Convex

General Polygon Rasterization

● We want to fill the inside of a polygon● Two General Approaches

■ Use scan lines■ Start from an interior point

● Use Scan lines■ Used in general purpose libraries for simple shapes

● Starting from interior point■ Interactive programs■ Useful for complex shapes

Scan Line Rasterization

● A polygon is defined by its list of vertices● We want to fill the inside of the polygon● We will use horizontal scan lines

Scan Line Rasterization

● Consider the following polygon:

● How do we know whether a given pixel on the scanline is inside or outside the polygon?

A

B

C

D

E

F

Basic Idea

● Find intersection points● Sort from left to right● Fill the scan line between every pair● Does it work all the time?

C

E

F

D

A

B

AAA

Special Cases

● Both X and Y pass through a polygon vertex● Z passes through a horizontal edge

X

Y

Z

Scan Lines Intersect Vertices

● Both X and Y intersect a polygon vertex● X intersects 4 edges● Y intersects 5 edges

X

Y

Scan Line Intersecting Vertices

● Polygon edges are on the same side of X at point A● Polygon Edges are on opposite side of Y at point C● Y needs special processing (why?)

X

Y

A

C

B

E

D

FG

H

I

1

2,3

4

12,3 4 5

Handling Scan line Intersecting Vertices

● Scan the polygon (counter-)clockwise● If the y-coordinate of endpoints of two consecutive

edges monotonically increase or decrease, the vertex is counted as one intersecting point

● Else count the vertex intersection as two

Scan Line Intersecting Vertices

● X scan line intersect at 1,2 and 3,4● Y scan line intersect at 1,2 and 3,4

X

Y

A

C

B

E

D

FG

H

I

1

2,3

4

12 3 4

Edge Shortening

● Scan in the polygon (counter-)clockwise direction● If the Y-coordinate of two consecutive edges is increasing or decreasing

■ Shorten the Y-coordinate of the edge below the scan line■ i.e. decrease y-coordinate by 1

● This way the scan line intersects one polygon edge only

y+1

y-1

y

Determining Intersection Points

● Given point R, can we determine point B quickly

B

R

● Use Bresenham's Line-Drawing Approach

|m| <1 Vs |m|>1

|m| <1 Vs |m|>1

● For |m|<1, ■ y does NOT get incremented every loop■ x gets incremented every loop ■ Hence we have multiple intersection points per scan line

● For |m| >1, ■ y gets incremented every loop ■ x does not get incremented every loop■ Hence we have one intersection point per scan line

Avoiding |m|<1 problem

● An intersection point at a scan y line may consists of multiple x values

● Let xa be the set of x-values of the intersection with edge a

● Let xb be the set x-values of the intersection with edge b

● For a pair of intersection points xa, xb with two edges■ Fill between the one more than the rightmost x value of

xa and one less than the leftmost x value xb

Example for |m|<1

General Polygon Rasterization

● Basic idea: use a parity test where we fill between every two intersection pointsfor each scanline edgeCnt = 0; for each pixel on scanline (l to r) // IF you move from a edge pixel to

// non-edge pixel if (oldpixel->newpixel crosses edge) edgeCnt ++; // draw the pixel if edgeCnt odd if (edgeCnt % 2) setPixel(pixel);

● Why does this work?● What assumptions are we making?

Faster Polygon Rasterization

● How can we optimize the code?for each scanline edgeCnt = 0; for each pixel on scanline (l to r) if (oldpixel->newpixel crosses edge) edgeCnt ++; // draw the pixel if edgeCnt odd if (edgeCnt % 2) setPixel(pixel);

● Big cost: testing pixels against each edge● Solution: active edge table (AET)

Performance Improvement

● The goal is to compute the intersections more efficiently. Brute force: intersect all the edges with each scanline

■ find the ymin and ymax of each edge and intersect the edge only when it crosses the scanline

■ only calculate the intersection of the edge with the first scan line it intersects

■ calculate dx/dy■ for each additional scanline, calculate the new intersection as x

= x + dx/dy

Data Structure

● Edge table: ■ all edges sorted by their ymin coordinates.■ keep a separate bucket for each scanline■ within each bucket, edges are sorted by increasing x

of the ymin endpoint

Edge Table

x

y

ymin

Active Edge Table (AET)

A list of edges active for current scanline, sorted in increasing x

y = 9

y = 8

Scan line intersect DE at x=10

Polygon Scan-conversion Algorithm

Construct the Edge Table (ET); Active Edge Table (AET) = null;for y = Ymin to Ymax

Merge-sort ET[y] into AET by x valueFill between pairs of x in AETfor each edge in AET

if edge.ymax = yremove edge from AET

elseedge.x = edge.x + dx/dy

sort AET by x value

end scan_fill

Active Edge Table Algorithm

● Pre-Process:■ Sort all edges by their minimum y-coordinate■ Shorten edges as needed■ Ignore horizontal edges

● Loop■ For each scanline in the screen:

○ Add edges with Ymin Y○ Sort edges in AET by x intersection with the current scan line○ Walk from left to right, setting pixels by parity rule○ Increment scanline○ Retire edges with Ymax < Y

■ Stop when Y > Ymax for last edge

Exercise

C`

C

A

B

D

E

0 1 2 3 4 5 6 7 8 9

7

6

5

4

3

2

1

0

Notes about AET

● At any scan line, AET contains ONLY the edges that the current scan line intersects

● No need to test pixels in the frame buffer● We need to check for Ymin Y for every scan line to add

edges■ Make the check Ymin Y for edges outside the AET■ Because edges are sorted by Ymin, we need to check for the first

edge that is not in the AET● We need to check Ymax < Y for every scan line to retire

edges■ We need to check Ymax < Y for edges in the AET■ If we sort edges in the AET by Ymax

○ We only need to check one edge after every scan line○ Costs O(log2n) for every add/retire from the AET

Determining the interior of a Polygon

● Non-Intersecting polygons are easy

Odd-Even Rule

● We want to determine wither a point P is inside a polygon or not

● Draw a line from P to a point beyond the polygon range

● If the line intersects the polygon in odd number of points then the point is interior

Odd-even Rule

Non-Zero Winding Number

● The number of times polygon edges wind around a point● Interior point have non-zero winding number● Draw a line from P to a distant point without intersecting any vertex and

without being parallel to an edge. ● Consider this line vector u● Walk around the polygon in one direction. ● Consider each edge as vector V in the walk direction● If u intersects any edge V, calculate uxV● If uxV > 1 counter = counter+1● If uxV<1 counter=counter-1● After intersecting all possible edges, if counter != 0 then P is interior

Would both algorithms yield the same result?

Non-Zero Winding Number

A

B

D

E

F

C

Non-Zero Winding Number

A

B

D

E

F

C

Non-Zero Winding Number

A

B

D

E

F

C

What Rule Applied in the Scanline Rasterization

?

What Rule Applied in the Scanline Rasterization

● A interior point P on a scan line bisects the scan line● Hence we have 2 lines from P to two distant point● If P is interior then the each line of the 2 lines will

have odd intersections with each of the two lines● The sum of the two odd numbers is even

What Rule Applied in the Scanline Rasterization

● A interior point P on a scan line bisects the scan line● Hence we have 2 lines from P to two distant point● If P is interior then the each line of the 2 lines will

have odd intersections with each of the two lines● The sum of the two odd numbers is even

AET Algorithm applies odd-even rule

Boundary Fill

● Basic Idea: BoundaryFill(x,y,color)■ Start with an interior point■ Color the starting point■ Recursively Call boundaryFill(x,y, Color) for surrounding points

● Four Connected■ Recursively call BoundaryFill(x,y,Color) for (x+1,y), (x-1,y),

(x,y+1), and (x, y-1)■ What problems can occur because of 4-connected?

● Eight Connected■ Also call the “corners” (x+1,y+1), (x-1, y-1), (x+1, y-1), and (x-1,

y+1)

Boundary Fill

● Problem: stack Overflow leads to program crash● How to solve stack overflow problem?

■ Avoid recursive calls by doing your own stack

Avoiding Program Crash

//4-way floodfill using our own stack routines void floodFill4Stack(int x, int y, int newColor, int oldColor) { if(newColor == oldColor) return; //avoid infinite loop emptyStack(); if(!push(x, y)) return; while(pop(x, y))

{ screenBuffer[x][y] = newColor; if(x + 1 < w && screenBuffer[x + 1][y] == oldColor)

{ if(!push(x + 1, y)) return; } if(x - 1 >= 0 && screenBuffer[x - 1][y] == oldColor)

{ if(!push(x - 1, y)) return; } if(y + 1 < h && screenBuffer[x][y + 1] == oldColor)

{ if(!push(x, y + 1)) return; } if(y - 1 >= 0 && screenBuffer[x][y - 1] == oldColor)

{ if(!push(x, y - 1)) return; } } }