VITS CSE 1

MIDPOINT CIRCLE & ELLIPSE GENERARTING

ALGORITHMS

Circle Generating Algorithm Properties of circle Midpoint circle algorithm

Ellipse-generating algorithm Properties of ellipse Midpoint ellipse algorithm

VITS CSE 2

Circle Generating Algorithm Properties of Circle

Circle is defined as the set of points that are all at a given distance r from a center point (xc,yc)

For any circle point (x,y), the distance relationship is expressed by the Pythagorean theorem in Cartesian coordinate as:

ryc

өө

(x,y)

xc

222 )()( ryyxx cc

VITS CSE 3

Circle Generating Algorithm We could use this equation to calculate the

points on a circle circumference by stepping along x-axis in unit steps from xc–r to xc+r and calculate the corresponding y values as

22 )( xxryy cc

VITS CSE 4

Circle Generating Algorithm The problems:

Involves many computation at each stepi.e., square root and additions /subtractions

Spacing between plotted pixel positions is not uniform

Adjustment: interchanging x & y (step through y values and calculate x values)

Involves many computation too!

VITS CSE 5

Circle Generating Algorithm Another way:

Calculate points along a circular boundary using polar coordinates r and ө

x = xc + r cos ө

y = yc + r sin ө

Using fixed angular step size, a circle is plotted with equally spaced points along the circumference

Problem: trigonometric calculations are still time consuming

VITS CSE 6

Symmetry of Circles (8-way symmetry) Proposed Approach:symmetry of circles

Shape of the circle is similar in each quadrant

i.e. if we determine the curve positions in the 1st quadrant, we can generate the circle section in the 2nd quadrant of the xy plane (the 2 circle sections are symmetric with respect to the y axis)

The circle section in the 3rd and 4th quadrant can be obtained by considering symmetry about the x axis

One step further symmetry between octants

VITS CSE 7

8-way symmetry Circle sections in adjacent octants within 1

quadrant are symmetric with respect to the 45° line dividing the 2 octants

Calculation of a circle point (x, y) in 1 octant yields the circle points for the other 7 octants

450

(y,x)(-y,x)

(x,y)

(x,-y)

(y,-x)(-y,-x)

(-x,-y)

(-x,y)

VITS CSE 8

Midpoint Circle Algorithm As in raster algorithm, we sample at unit

intervals & determine the closest pixel position to the specified circle path at each step

For a given radius, r and screen center position (xc,yc) , we can set up our algorithm to calculate pixel positions around a circle path centered at the coordinate origin (0,0)

Each calculated position (x, y) is moved to its proper screen position by adding xc to x and yc

to y

VITS CSE 9

Midpoint Circle Algorithm Along a circle section from x=0 to x=y in the 1st

quadrant, the slope (m) of the curve varies from 0 to -1.0

i.e. we can take unit steps in the +ve x direction over the octant & use decision parameter to determine which 2 possible positions is vertically closer to the circle path

Positions in the other 7 octants are obtained by symmetry

VITS CSE 10

< 0, if (x,y) is inside the circle boundary

fcirc(x,y) = 0, if (x,y) is on the circle boundary

> 0, if (x,y) is outside the circle boundary

Midpoint Circle Algorithm

To apply the midpoint method, we define a circle function as

fcirc(x,y) = x2 + y2 - r2

The relative positions of any point (x,y) can be determined by checking the sign of the circle function:

VITS CSE 11

Midpoint Circle Algorithm

Consider current position (xk, yk)

next pixel position is either(xk+1, yk) or (xk+1, yk-1)?

xk

yk

xk+1 xk+2

yk-1

yk-1

yk

Circle path

xk

Midpoint

VITS CSE 12

Midpoint Circle Algorithm Our decision parameter is the earlier circle

function evaluated at the mid point between the 2 pixels

< 0: midpoint is inside the circle; plot (xk+1, yk)

+ve: midpoint is outside the circle; plot

(xk+1, yk-1)

Successive decision parameters are obtained using incremental calculation

22

212

21

1

,1

ryx

yxfp

kk

kkk

pk

VITS CSE 13

Midpoint Circle Algorithm To ensure things are as efficient as possible

we can do all of our calculations incrementally

First consider:

or

where yk+1 is either yk or yk-1 depending on

the sign of pk

22

12

111

21]1)1[(

21,1

ryx

yxfp

kk

kkcirck

1)()()1(2 122

11 kkkkkkk yyyyxpp

VITS CSE 14

Midpoint Circle Algorithm

Initial decision parameter is obtained by evaluating the circle function at the start position (x0, y0) = (0,r)

p0 = f(1, r–1/2) = 12 + (r – 1/2)2 – r2

p0 = 5/4 – r

If the radius is specified as an integer, we can simple round p0 to p0 = 1 – r

Then if pk < 0 then the next decision variable is given as:

If pk > 0 then the decision variable is:

12 11 kkk xpp

1212 11 kkkk yxpp

VITS CSE 15

Midpoint Circle Algorithm1. Input radius, r, and circle center (xc, yc), then set the coordinates

for the 1st point on the circumference of a circle centered at the origin as (x0, y0) = (0,r)

2. Calculate initial value of decision parameter:

3. At each xk, starting at k = 0, test the value of pk :

If pk < 0, next point will be (xk+1, yk) and

else, next point will be (xk+1, yk – 1) and 

where 2xk+1 = 2xk + 2 and 2yk+1 = 2yk – 2

4. Determine symmetry points in the other 7 octants.

5. Get the actual point for circle centered at (xc,yc) that is (x+xc, y+yc).

6. Repeat step 3 to 5 until x y.

rp 4

50

,12 11 kkk xpp

.212 111 kkkk yxpp

VITS CSE 16

Try this out!

Given a circle radius r=10, demonstrate the midpoint circle algorithm by determining positions along the circle octant in the 1st quadrant

p0 = ?

(x0,y0) = ?

k pk (xk+1,yk+1) 2xk+1 2yk+1

0

1

2

3

4

5

6

VITS CSE 17

Plot pixel positions

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2

1

0

0 1 2 3 4 5 6 7 8 9 10

VITS CSE 18

Try this again!

Given a circle with r=8, calculate each pixel positions along the circumference at the 2nd quadrant

VITS CSE 19

Ellipse Generating Algorithm Ellipse – an elongated circle.

A modified circle whose radius varies from a maximum value in one direction to a minimum value in the perpendicular direction.

A precise definition in terms of distance from any point on the ellipse to two fixed position, called the foci of the ellipse.

The sum of these two distances is the same value for all points on the ellipse.

P=(x,y)

F2

F1d2

d1

VITS CSE 20

Ellipse Generating Algorithm

If the distance of the two focus positions from any point P=(x, y) on the ellipse are labeled d1 and d2, the general equation of an ellipse:

d1 + d2 = constant

Expressing distance d1 and d2 in terms of the focal coordinates F1=(x1, y1) and F2=(x2, y2), we have constant2

22

22

12

1 yyxxyyxx

VITS CSE 21

Ellipse Generating Algorithm

However, we will only consider ‘standard’ ellipse:

1

22

y

c

x

c

r

yy

r

xxry

xc

rx

yc

VITS CSE 22

2-way symmetry

An ellipse only has a 2-way symmetry.

(x,y)

(x,-y)

(-x,y)

(-x,-y)

rx

ry

VITS CSE 23

Equation of an ellipse Consider an ellipse centered at the origin,

(xc,yc)=(0,0):

Ellipse function

…and its properties: fellipse(x,y) < 0 if (x,y) is inside the ellipse

fellipse(x,y) = 0 if (x,y) is on the ellipse

fellipse(x,y) > 0 if (x,y) is outside the ellipse

222222, yxxye rryrxryxf

1

22

yx r

y

r

x

VITS CSE 24

Midpoint Ellipse Algorithm Ellipse is different from circle. Similar approach with circle, different in

sampling direction. Slope of ellipse is: dy/dx = - 2ry

2x/2rx2y

Region 1: Sampling is at x direction Choose between (xk+1, yk), or (xk+1,

yk-1)

Move of region 1 if 2r2yx >= 2r2

xy Region 2:

Sampling is at y direction Choose between (xk, yk-1), or (xk+1,

yk-1)

Slope = -1

Region 1

Region 2

ry

rx

VITS CSE 25

Decision parameters

Region 1: Decision parameter  

p1k < 0: midpoint is inside choose pixel (xk+1, yk) 

p1k >= 0: midpoint is outside/on choose pixel (xk+1, yk-1)

21,11 kkek yxfp

Region 2:• Decision parameter 

p2k <= 0:

• midpoint is inside/on

• choose pixel (xk+1, yk-1)

 p2k > 0:

• midpoint is outside

• choose pixel (xk, yk-1)

1,2 21 kkek yxfp

VITS CSE 26

Midpoint Ellipse Algorithm1. Input rx, ry and ellipse center (xc, yc). Obtain the first point on an ellipse

centered on the origin (x0, y0) = (0, ry).

2. Calculate initial value for decision parameter in region 1 as:

3. At each xk in region 1, starting from k = 0, test p1k :

  If p1k < 0, next point (xk+1, yk) and

else, next point (xk+1, yk-1) and

  with 2ry2xk+1 = 2ry

2xk + 2ry2, 2rx

2yk+1 = 2rx2yk – 2rx

2

and repeat step 1 to 3 until 2ry2x 2rx

2y 

2

4122

01 xyxy rrrrp

21

21 211 ykykk rxrpp

21

21

21 2211 ykxkykk ryrxrpp

VITS CSE 27

Midpoint Ellipse Algorithm4. Initial value for decision parameter in region 2:

where (x0, y0) is the last position calculate in region 1

5. At each yk in region 2, starting from k = 0, test p2k:

  If p2k > 0, next point is (xk, yk-1) and

else, next point is (xk+1, yk-1) and

  continue until y=0 

21

21 222 xkxkk ryrpp

2220

22

21

02

0 12 yxxy rryrxrp

21

21

21 2222 xkxkykk ryrxrpp

VITS CSE 28

Midpoint Ellipse Algorithm

6. For both region determine symmetry points in the other 3 quadrants

7. Move each calculated pixel position (x,y) onto the elliptical path centered on (xc,yc) and plot the coordinate values

x=x + xc, y =y + yc

 

VITS CSE 29

Try this out!

Given input ellipse parameter rx=8 and ry=6, determine pixel positions along the ellipse path in the first quadrant using the midpoint ellipse algorithm

2ry2x = ?

2rx2y = 2rx

2ry = ?

p10 = ?

k p1k (xk+1,yk+

1)2ry

2 xk+1 2 rx2yk+1

0 -332 (1,6) 72 768

1 -224 (2,6) 144 768

2 -44 (3,6) 216 768

3 208 (4,5) 288 640

4 -108 (5,5) 360 640

5 288 (6,4) 432 52

6 244 (7,3) 504 384

k p2k (xk+1,yk+

1)2rv

2 xk+1 2 rx2yk+1

0 -151 (8,2) 576 256

1 233 (8,1) 576 128

2 745 (8,0) -- --

We now move out from R1 since2ry2 x> 2 rx

2y

For R2,initial point is(xo,yo)=(7,3)& p2o=f(7+1/2,2)=-151

VITS CSE 30

Plot pixel positions

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1

0

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