1

CS 430/536Computer Graphics I

Circle Drawing and ClippingWeek 3, Lecture 6

David Breen, William Regli and Maxim Peysakhov

Geometric and Intelligent Computing Laboratory

Department of Computer Science

Drexel Universityhttp://gicl.cs.drexel.edu

2

Outline

• Scan conversion of circles• Clipping circles• Scan conversion of ellipses

3

Scan Conversion of Circles

• Generalization of the line algorithm

• Assumptions:– circle at (0,0)– Fill 1/8 of the circle,

then use symmetry

Using the 8-way symmetry of a circle:

1994 Foley/VanDam/Finer/Huges/Phillips ICG

Not using the 8-way symmetry of a circle

4

Scan Conversion of Circles

• Implicit representation of the circle function:

• Note: < 0 for points inside the circle, and > 0 for points outside the circle

5

Scan Conversion of Circles

• Assume we finished pixel• What pixel to draw next?

(going clockwise)• Note: the slope of the circular

arc is between 0 and –1– Hence, choice is between:

E and SE

• Idea: If the circle passes above the midpoint M, then we go to E next, otherwise we go to SE

1994 Foley/VanDam/Finer/Huges/Phillips ICG

6

Scan Conversion of Circles

• We need a decision variable D:

• If then M is below the arc, hence the E pixel is closer to the line.

• If then M is above the arc, hence the SE pixel is closer to the line.

1994 Foley/VanDam/Finer/Huges/Phillips ICG

7

Case I: When E is next• What increment for computing a new D?• Next midpoint is:

• Hence, increment by:Pics/Math courtesy of Dave Mount @ UMD-CP

8

Case II: When SE is next• What increment for computing a new D?• Next midpoint is:

• Hence, increment by:Pics/Math courtesy of Dave Mount @ UMD-CP

9

Scan Conversion of Circles

• How to compute the initial value of D:

• We start with x = 0 and y = R, so the first midpoint is at x = 1, y = R-1/2:

10

Scan Conversion of Circles

• Converting this to an integer algorithm:– Need only know if D is positive or negative– D & R are integers– Note D is incremented by an integer value– Therefore D + 1/4 is positive only when D

is positive; it is safe to drop the 1/4

• Hence: set the initial D to 1 – R (subtracting 1/4)

11

Circle Scan Conversion Algorithm

• Given radius R and center (0, 0)– First point (0, R)

• Initial decision parameter D = 1- R• While x y

– If (D < 0)• x++; D += 2x + 3;

– else• x++; y--; D += 2(x - y) + 5

– WritePoints(x,y)

12

WritePoints(x,y)

• Writes pixels to the seven other octants

13

Clipping Circles

• Accept/Reject test– Does bounding box of

the circle intersect with clipping box?

• If yes, condition pixel write on clipping box inside/outside test

14

Properties of Ellipses

• “Elongated circle”• For all points on

ellipse, sum of distances to foci is constant

d1+ d2 = const

• If F1=(x1, y1) and   F2=(x2, y2) then

y

x

d1

d2

P=(x,y)F1

F2

€

(x − x1)2 + (y − y1)

2 + (x − x2)2 + (y − y2)2 = const

15

Properties of Ellipses

• Equation simplified if ellipse axis parallel to coordinate axis

• Parametric form

y

x

ry

rx

xc

yc

€

x − xc

rx

⎛

⎝ ⎜

⎞

⎠ ⎟

2

+y − yc

ry

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

2

=1

€

x = xc + rx cosθ

y = yc + ry sinθ

16

Symmetry Considerations

• 4-way symmetry• Unit steps in x until reach

region boundary• Switch to unit steps in y

(x,y)

ry

rx

(x,-y)

(-x,y)

(-x,-y)

Region1

Region 2

Slope = -1

• Step in x while

• Switch to steps in y when

€

f (x,y) = ry2x 2 + rx

2y 2 − rx2ry

2

dy

dx= −

ry2x

rx2y

dy

dx= −1

ry2x = rx

2y

€

ry2x < rx

2y

€

ry2x ≥ rx

2y

17

Midpoint Algorithm (initializing)

• Similar to circles• The initial value for region 1

• The initial value for region 2

• We have initial values, now we need the increments

1994 Foley/VanDam/Finer/Huges/Phillips ICG

(x,y)

ryRegion

1Region 2

Slope = -1

€

Dinit1 = f (1,ry − 12)

= ry2 − rx

2ry + 14 rx

2

€

Dinit 2 = f (x p + 12 , y p −1)

= ry2(x p + 1

2)2 + rx2(y p −1)2 − rx

2ry2

18

Making a Decision

• Computing the decision variable

• If then M is below the arc, hence the E pixel is closer to the line.

• If then M is above the arc, hence the SE pixel is closer to the line.

1994 Foley/VanDam/Finer/Huges/Phillips ICG

€

D = f (x p +1,y p − 12)

= ry2(x p +1)2 + rx

2(y p − 12)2 − rx

2ry2

19

Computing the Increment

1994 Foley/VanDam/Finer/Huges/Phillips ICG

€

Dnew = f (x p + 2, y p − 12)

= ry2(x p + 2)2 + rx

2(y p − 12)2 − rx

2ry2

= Dold + ry2(2x p + 3)

Dnew = f (x p + 2, y p − 32)

= ry2(x p + 2)2 + rx

2(y p − 32)2 − rx

2ry2

= Dold + ry2(2x p + 3) + rx

2(−2y p + 2)

increment =ry

2(2x p + 3) Dold < 0

ry2(2x p + 3) + rx

2(−2y p + 2) Dold ≥ 0

⎧ ⎨ ⎩

E case

SE case

21

Computing the Increment in 2nd Region

• Decision variable in 2nd region

• If then M is left of the arc, hence the SE pixel is closer to the line.

• If then M is right of the arc, hence the S pixel is closer to the line.

1994 Foley/VanDam/Finer/Huges/Phillips ICG

€

D = f (x p + 12 ,y p −1)

= ry2(x p + 1

2)2 + rx2(y p −1)2 − rx

2ry2

22

Computing the Increment in 2nd Region

1994 Foley/VanDam/Finer/Huges/Phillips ICG

€

Dnew = f (x p + 32 ,y p − 2)

= ry2(x p + 3

2)2 + rx2(y p − 2)2 − rx

2ry2

= Dold + rx2(−2y p + 3) + ry

2(2x p + 2)

Dnew = f (x p + 12 , y p − 2)

= ry2(x p + 1

2)2 + rx2(y p − 2)2 − rx

2ry2

= Dold + rx2(−2y p + 3)

increment =rx

2(−2y p + 3) + ry2(2x p + 2) Dold < 0

rx2(−2y p + 3) Dold ≥ 0

⎧ ⎨ ⎩

SE case

S case

23

Midpoint Algorithm for Ellipses

Set first point to (0,ry)

Set the Decision variable to

Loop (x = x+1)

If D < 0 then pick E and

If D ≥ 0 then pick SE and

Until

2212ykyDDrxr+=++

2221122ykyxkDDrxrry++=++−

22222211222,222ykykyxkxkxrxrxrryryr++=+=−

2222ykxkrxry≥

Set first point to the last computedSet the Decision variable to

Loop (y = y - 1)If D < 0 then pick SE and

If D ≥ 0 then pick S and

Until y < 0Use symmetry to complete the

ellipse

Region 1 Region 2

€

Dinit1 = ry2 − rx

2ry + 14 rx

2

€

D + = ry2(2x p + 3)

€

D + = ry2(2x p + 3) + rx

2(−2y p + 2)

€

2ry2xk ≥ 2rx

2yk

€

Dinit 2 = ry2(x p + 1

2)2 + rx2(y p −1)2 − rx

2ry

€

D + = ry2(2x p + 2) + rx

2(−2y p + 3)

€

D + = rx2(−2y p + 3)