Cg Lecture 6

7/31/2019 Cg Lecture 6

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Computer Graphics (CODE: ECS-504)

Mr. Anil kr, Pandey

L6 : Point & Line Drawing


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Introduction

Computer Graphics & Its application

Types of computer graphics

Graphic display : random Scan & Raster Scan display

Frame buffer and video Controller

Points & Lines , line Drawing Algorithm

Circle Generation algorithm

Mid point circle generation algorithm

Parallel version of these algorithm

Unit 1: Point & Line Generation


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Point

Graphics programming packages provide functions todescribe a scene in terms of these basic geometricstructures, referred to as output primitives.

Point plotting is accomplished by converting a singlecoordinate position function furnished by an applicationprogram into appropriate operation for the output devicein use.

Line drawing is accomplished by calculating intermediate

position along the line path between to specified endpoint positions.

An output device is then a random scan display , astraight line can be drawn smoothly form one endpoint tothe other end point.


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line

Must appear straight and continuous

Must interpolate both defining end points

Must have uniform density and intensity Consistent within a line and over all lines

Must be efficient, drawn quickly


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Line

Based on slope-intercept

algorithm from algebra:

y = mx + b

Simple approach: increment x,

solve for y

Floating point arithmetic

required


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Continue..

M =y2-y1 / x2-xy

b = y1-m.x1

for any given x interval x along a line , we can compute the

corresponding y interval y from above equation

y=mx and x = y/m

For lines with slope magnitudes m1 , y can be set proportional to asmall horizontal deflection voltage and corresponding vertical

deflection is then set proportional to x.

If m=1 smooth line.


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DDA(Digital Differential Analyser)

algorithm

its is based on scan conversion line

algorithm based on calculating x or y.

if m < =1 then x=1 and successive y

value yk+1 = yk+m

if m >1 then y=1 and successive x value

xk+1 = xk+1/m


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DDA Algorithm

At each step t , increment tby dx and y by dy

dt

dyyy

dt

dxxx

oldnew

oldnew


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DDA algo.

1. Start.

2. Declare variables x,y,x1,y1,x2,y2,k,dx,dy,s,xi,yi and also declaregdriver=DETECT,gmode.

3. Initialise the graphic mode with the path location in TC folder.

4. Input the two line end-points and store the left end-points in (x1,y1).

5. Load (x1,y1) into the frame buffer;that is,plot the first point.put x=x1,y=y1.

6. Calculate dx=x2-x1 and dy=y2-y1. 7. If abs(dx) > abs(dy), do s=abs(dx).

8. Otherwise s= abs(dy).

9. Then xi=dx/s and yi=dy/s.

10. Start from k=0 and continuing till k


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DDA algorithm

line(int x1, int y1, int x2, int y2)

{

float x,y;

int dx = x2-x1, dy = y2-y1;

int n = max(abs(dx),abs(dy));

float dt = n, dxdt = dx/dt, dydt = dy/dt;

x = x1;

y = y1;

while( n-- ) {

point(round(x),round(y));x += dxdt;

y += dydt;

}

}


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DDA algorithm

Still need a lot of floating point arithmetic.

2 rounds and 2 adds per pixel.

Is there a simpler way ?

Can we use only integer arithmetic ?

Easier to implement in hardware.


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Bresenhams line drawing algorithm

Bresenhams line drawing algorithm

Line drawing algorithm

comparisons

Circle drawing algorithms

A simple technique

The mid-point circle

algorithmThe algorithm was developed by Jack

Bresenham, who we heard about earlier.


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Concept

Move across thex axis in unit intervals and at each step

choose between two differenty coordinates

2 3 4 5

2

4

3

5 For example, from

position (2, 3) we

have to choose

between (3, 3) and

(3, 4)

We would like the

point that is closer to

the original line

(xk,yk)

(xk+1,yk)

(xk+1,yk+1)


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They coordinate on the mathematical line atxk+1 is:

Concept: The Bresenham Line

Algorithm

At sample positionxk+1 the vertical separations from the

mathematical line are labelled dupper and dlower

bxmy k )1(

yyk

yk+1

xk+1

dlower

dupp

er


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So, dupperand dlowerare given as follows:

and:

We can use these to make a simple decision

about which pixel is closer to the

mathematical line

Deriving The Bresenham Line

Algorithm (cont)

klower yyd

kk ybxm

)1(

yyd kupper )1(

bxmy kk )1(1


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This simple decision is based on the difference between thetwo pixel positions:

Lets substitute m with y/xwhere x and

y are the differences between the end-points:


Algorithm (cont)

122)1(2 byxmdd kkupperlower

)122)1(2()(

byxx

yxddx kkupperlower

)12(222 bxyyxxy kk

cyxxy kk 22


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So, a decision parameterpk for the kthstep along a line is given by:

The sign of the decision parameterpk is

the same as that ofdlowerdupper

Ifpk is negative, then we choose the lowerpixel, otherwise we choose the upper pixel

Deriving Algorithm (cont)

cyxxyddxp

kk

upperlowerk

22)(


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Remember coordinate changes occur

along thex axis in unit steps so we can do

everything with integer calculations At step k+1 the decision parameter is given

as:

Subtractingpkfrom this we get:


Algorithm (cont)

cyxxyp kkk 111 22

)(2)(2111 kkkkkk yyxxxypp


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But,xk+1 is the same asxk+1 so:

whereyk+1 - yk is either 0 or 1 depending

on the sign ofpk

The first decision parameter p0 isevaluated at (x0, y0) is given as:

Deriving The Bresenham

Line Algorithm (cont)

)(2211 kkkk yyxypp

xyp 20


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The Bresenham Line Algorithm

BRESENHAMS LINE DRAWING ALGORITHM(for |m| < 1.0)

1. Input the two line end-points, storing the left end-point

in (x0, y0)

2. Plot the point (x0, y0)

3. Calculate the constants x, y, 2y, and (2y - 2x)and get the first value for the decision parameter as:

4. At eachxkalong the line, starting at k = 0, perform the

following test. Ifpk< 0, the next point to plot is

(xk+1, yk) and:

xyp 20

ypp kk 21


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The Bresenham Line Algorithm

(cont)

The algorithm and derivation above assumes slopes are

less than 1. for other slopes we need to adjust the

algorithm slightly

Otherwise, the next point to plot is (xk+1, yk+1) and:

5. Repeat step 4 (x 1) times

xypp kk 221


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Example

Lets have a go at this

Lets plot the line from (20, 10) to (30, 18)

First off calculate all of the constants:

x: 10

y: 8

2y: 16 2y - 2x: -4

Calculate the initial decision parameterp0:

p0= 2yx = 6


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Bresenham Example (cont)

17

16

15

14

13

12

11

10

18

292726252423222120 28 30

k pk (xk+1,yk+1)

0

1

2

3

4

56

7

8

9

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Cg Lecture 6