Line Circle Algorithm

8/8/2019 Line Circle Algorithm

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Scan converting lines

circles and ellipses


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Line drawing

Description: Given the specification of the straightline, find the collection of the addressable pixels

which more closely approximates this line

Goal: (Not all of them are acheivable with the discrete

space of a raster device) .Straight line should appear straight.

Lines should start and end accurately.

Lines should have constant brightness.

Lines should be drawn as rapidly as possible.


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ProblemsHow do we determine which pixel to illuminate to

satisfy the above goals?

Vertical, horizontal and lines with slope with +/- 1 are

easy to draw.

Others create problems: Stair casing/ jaggies/aliasing

Quality of the line depends on the location of the

pixels and their brightness.


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Direct solution

Solve y=mx+b, where (0,b) is the y-intercept and m is

the slope.

Go from X0 to X1Calculate round(y) from the equation.

Take an example, b=1 [starting point is (0,1)] and

m=3/5.Then X=1, Y=2= ROUND (8/5)

X=2, Y=2= ROUND (11/5)

X=3, Y=3= ROUND (14/5)

X=4, Y=3= ROUND (17/5)

X=5, Y=4= ROUND (20/5)


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Bresenharms line drawing

algorithm (Midpoint line algorithm)

Mathematical description:

True line

B

A

A = the distance of pixels

which are lying above

the true lineB = the distance of pixels

which are lying below

the true line


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Let us define the decision variable as

di = B A

If di < 0 then it implies that B > A

i:epixel below the true line is closer to the

expected line. If di is greater than or equal

to zero then the pixel above the true line iscloset. Hence, if di 0 increment

both the values of x and y.


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Algorithm

1. Input the two line end points and store the left endpoint in(x0,y0).

2. Load (x0,y0), into the frame buffer, that is, plot the firstpoint.

3. Calculate constants x, y, 2y, and 2y - 2x, and obtainthe starting value for the decision parameter as

d0 = 2y x.

4. At each xk along the line, starting at k=0, perform the

following test:if dk


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11 12 13 14 15 16 17

9

10

11

12

13

14

15

16

6 7 8 9 10

k dk (x ,y)

0

1

2

.

.

So do it..

Problem: Construct a line with

endpoints (6,9) and (17,15).


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Note: This method is well suited for the line having the

slope less than 45. When the magnitude of the slope is lesser than 1, xincremented by one, and the decision variable ismade use to find the incremental value of y.

When the slope is greater than 1, y is incrementedby one and x is incremented based on decisionvariable.

Line is constructed from left to right. For the rightto left appropriate directions need to be reversed.


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DDA line algorithm : Disadvantages

Key disadvantage:

it relies on floating point operations to computepixel positions.

Implications:

Computationally inefficient because floating-

point operations are slow. Round-off errors accumulate, producing incorrect

line drawings (e.g., if m is rounded to 0.9 eventhough it is equal to 0.99, lines of length > 10 will

be drawn inaccurately)


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Circle algorithmStarting with the equation of the circle

We could use this equation to calculate the position of points on

a circle by stepping along the x axis in unit steps from x0-r to

x0+r and calculating the corresponding y values at each

position as :


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This works, but is inefficient coz it involves

multiplication and square root operations. Italso creates large gaps in the circle for values

of x values close to r.


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Exploiting Symmetry Circle can be drawn by

two ways namely :

polynomial method and

trigonometric method.

Take advantage of

symmetries to minimize

cases (slopes) and amount

of curve drawn. We cangenerate all pixel positions

around a circle by

calculating only the points

within the sector from x=0

to x=y

(x, y)

(y, x)(y, -x)

(-x, y)

(-x, -y)

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

(x, -y)


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Bresenhams circle drawing algorithm

1. read the radius of the circle say r

2. d=3-2r (initial decision variable)

x=0, y=r

3. plot the initial four points (x,y),(x,-y), (y,x)and(-y,-x)

4. if (dy stop or else goto step 4 until thecondition is satisfied.


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Draw a circle whose radius is 8 units

and centre is (0,0)

Step1:

Set (h,k) = (0,0)

x=0 and y= 8

d= 3-2r = -13 Step 2:

check , if x>y then stop0>8, which is false.


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Step 3: plot the 8 points,

(x+h, y+k)= (0,8)

(y+h, x+k)=(8,0)

(-y+h, x+k)=(-8,0)(-x +h, y+k)=(0,8)

(-x +h, -y+k)=(0,-8)(-y +h, -x+k)=(-8,0)

(y +h, -x+k)=(8,0)

(x +h, -y+k)=(0,-8)


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Step 4: d= -13

since d8, answer will be no

Again plotting 8 points while taking x=1

x=1, h=0 , k=0, y=8

Points to be plotted are as follows(1,8), (8,1), (-8,1), (-1,8), (-1,-8), (-8,-1), (8,-1), (1,-8)


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Now checking decision parameter,

Since d8Again plot 8 points taking x=2, y=8, h=0,k=0

(2,8), (8,2), (-8,2), (-2,8), (-2,-8),(-8,-2), (8,-2), (2,-8)


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Now d 0 3 0

d=d+4(x-y)+10=3+4(2-8)+10

= -11

x= x+1

= 2+1 = 3

y = y-1= 8-1= 7Now x>y No.

Again plotting 8 points x=3, y=7, h=0, k=0

(3,7), (7,3), (-7,3), (-3,7), (-3,-7), (-7,-3), (7,-3), (3,-7)


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d= -11y which is false

So again computing 8 points with x=4(4,7), (7,4), (-7,4), (-4,7), (-4,-7), (-7,-4), (7,-4),

(4,-7)


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now d 0 i:e 7 >0

d= d+4(x-y)+10

= 7+4(4-8)+10 = 1

x= x+1 = 4+1 = 5y=y-1 =7-1 = 6

Check if x>y which is falseSo again computing 8 points with x=5,y=6

(5,6), (6,5), (-6,5), (-5,6), (-5,-6), (-6,-5), (6,-5),

(5,-6)


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now d 0 i:e d=1

d= d+4(x-y)+10= 1+4(5-6)+10 = 7

x= x+1 = 5+1 = 6

y=y-1 =6-1 = 5

Checking the condition if x>y i:e 6>5

hence we stopSo again computing 8 points with x=5,y=6

(5,6), (6,5), (-6,5), (-5,6), (-5,-6), (-6,-5), (6,-5),(5,-6)


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Line Circle Algorithm

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