cs 535 computer graphicsryang/teaching/cs535-2012spr/lectures/...cs 535 computer graphics 2d...

CS 535

Computer Graphics

2D Drawings: Lines, Circle, and Spline

Digital Concepts of Drawing in Raster Arrays

PIXEL is a single array element at (x,y)

- No smaller drawing unit exists

0 1 2

012

P(x,y)

y

Primitive Drawing Operations

From (x1,y1) to (x2,y2)

Equation:

y = mx + b

m = (y2-y1)/(x2-x1)

b = y1 – m*x1

x1 x2

y1

y2

compute -- m, b

for x=x1 to x2, step 1

y = m*x + b

pixel( x, round(y), color)

loop

Drawing Lines

(Digital Differential Analyzer)

[DDA avoids the multiple by slope m]

Equation:

xk+1 = xk + 1

yk+1 = yk + m

if slope |m| > 1

yk+1 = yk + 1

xk+1 = xk + 1/m

compute m (assume |m| <1 )

y = y1, x = x1

pixel(round(x1), round(y1), color)

for x=x1 to x2, step 1

x = x + 1

y = y + m

pixel(x, round(y), color)

End

xk xk+1 xk+2 xk+3

yk

yk+1

yk+2

m>1

M<=1

DDA Algorithm

(Digital Differential Analyzer)

[avoid the float multiple by slope m]

slope m, y – floats

Problems:

1. Necessary to perform float addition

2. Necessary to have float representation

3. Necessary to perform -- round()

Float representations/operations are more expensive than

integer operations. We would like to avoid them if possible.

DDA Algorithm

Uses only integer math to draw a line

Lets assume slope-m is positive and less than 1

we know that: xk+1 = xk + 1

at xk+1 we need to make a decision for next pixel

draw either at:

choice 1 -- (xk + 1, yk )

choice 2 -- (xk + 1, yk + 1)

How do we decide between choice 1 or 2? xk xk+1 xk+2 xk+3

yk

yk+1

yk+2

?

Mid-Point Line Algorithm (Bresenham)

Consider the distance with the actual line (y=mx +b) at the two choices:

d1 = y – yk

= m(xk+1) + b – yk

d2 = (yk+1) – y

= yk + 1 - m(xk+1) -b

xk xk+1

yk

yk+1 d2

d1

Difference between the these two separations is:

d = d1 – d2

if d > 0

move to (yk +1)

else stay at yk

y=mx+b


Still have a “float” operation in calculation of “d”

Lets create a new decision operator by multiplying by x (recall m=y/x )

pk = x (d1 - d2) ~ note pk sign is the same as (d1 - d2)

= 2 y xk - 2 x yk + 2 y + x (2b– 1) [Eq 1]

= 2 y xk - 2 x yk + C

We can use pk as a decision operator – instead of (d1 - d2)


pk+1 = 2 y xk+1 - 2 x yk+1 + C

What is the change from pk to pk +1 ?

pk +1 - pk = 2 y (xk+1 - xk) - 2 x ( yk+1 - yk)

pk +1 = pk + 2 y ( 1 ) - 2 x ( yk+1 - yk) either 1 or 0

so –

if pk < 0 // yk+1 = yk

pk+1 = pk + 2 y

else // yk+1 = yk + 1

pk+1 = pk + 2 y - 2 x

Use [Eq. 1] with x1 and y1 to find:

po = 2 y - x


Compute constants: x, y, 2y, 2y-2x

Calculate: po = 2y - x

plot(x1,y1)

for x=x1 to x2 // let y=y1

if (p < 0)

p = p + 2y

else

y++

p = p + 2y - 2x

end

plot(x,y)

loop

Note -- main loop:

Only integer math.

No float representation,

or operations needed.

Constants: 2dy, 2dx

are integers.


Example:

(20,10) to (30,18)

x = 10, y = 8

(slope 0.8)

p0 = 2y - x = 6

2y = 16 [E]

2y - 2x = -4 [NE]

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

0 6 (21,11) 5 6 (26,15)

1 2 (22,12) 6 2 (27,16)

2 -2 (23,12) 7 -2 (28,16)

3 14 (24,13) 8 14 (29,17)

4 10 (25,14) 9 10 (30,18)

full algorithm -- page 90-91 Hearn

adjusts for slope m>1

re-orders x1,x2,y1,y2 as necessary

http://www.cosc.canterbury.ac.nz/people/mukundan/cogr/LineMP.html


Addressing the Framebuffer

1 2 3 . . . N

Xmax

Logical representation

Physical representation in memory -- 1D Array

(0,0)

(x,y)

(Xmax , Ymax )

1D memory as a 2D Framebuffer

Indexing 1D memory

Coordinates (x,y) [or (i,j)] (assume array is unsigned char)

1 byte pixel (grayscale)

frameBuffer[ y * Xmax + x ] = grayscale or index color

4 byte pixel (RGBA) frameBuffer[ (y*Xmax+x)*4 + 0 ] = R

frameBuffer[ (y*Xmax+x)*4 + 1 ] = G

frameBuffer[ (y*Xmax+x)*4 + 2 ] = B

frameBuffer[ (y*Xmax+x)*4 + 3 ] = A

CIRCLE

2D-Drawings

Representing A Circle

Explicit representation:

Implicit representation:

Parametric representation:

Equation for a circle:

Given: center = (xc, yc) and radius r

(x – xc)2 + (y – yc)

2 = r2

y = yc ± r2 – (xc– x)

2

Not an ideal approach:

Non-uniform spacing during evaluation

causes noticeable artifacts

Also, we have to evaluate ^2, and sqrt()

Figure 3-13

xc

yc

r

Drawing Circles

Parametric Polar Representation for circle:

x = xc + r cos

y = yc + r sin

Allows a uniform spacing via the variable

step size is often set to be 1/r.

We can render adjacent points

via lines (to avoid gaps)

xc

yc

step

step

Drawing Circles

Only need to compute

1 octant

We can reduce the number of calculation:

Exploit Symmetry of Circle

Need only to compute (0,45) or (90,45)

Drawing Circles

The Circle Midpoint Algorithm

),( pp yx

Key idea: at each

step, decide which

pixel (E or SE) is

closest to the circle.

Choose that pixel

and draw it.

Goals:

Implicit formulation

Finite differences (incremental)

Integer arithmetic

M

SE

E

The Implicit Circle Equation

If M lies inside the circle, draw pixel E

If M lies outside the circle, draw pixel SE

),( pp yx

M

SE

E

Implicit Formulation

0),( 222 ryxyxF222 )( ryx Notice that

22

222

222 )(

xry

xry

ryx

2),( ryxF means

0),( yxF

0),( yxF

points outside circle

points inside circle

The Midpoint Test

),1(21 pp yxM

The sign of F(M) gives the

answer as to which pixel,

E or SE, to draw

),( pp yx

M

SE

E

Formulating an Algorithm

Let d be a decision variable which makes the midpoint test.

Then the test to decide which pixel to draw is just

if (d < 0)

then

// the midpoint is inside the circle

// the circle passes closer to the upper pixel

draw the E pixel

else

// the midpoint is outside the circle

// the circle passes closer to the lower pixel

draw the SE pixel


MidpointCircle (radius, x_center, y_center)

x = 0; y = radius;

CirclePoints (x_center, y_center, x, y);

while (y > x) {

if (d < 0) // the midpoint is inside the circle

// the circle is closer to E pixel

x = x + 1;


else // the midpoint is outside the circle

// the circle is closer to SE pixel

x = x + 1; y = y - 1;


end


CirclePoints (a, b, x, y)

DrawPoint (a + x, b + y);

DrawPoint (a + x, b - y);

DrawPoint (a - x, b + y);

DrawPoint (a - x, b - y);

DrawPoint (a + y, b + x);

DrawPoint (a + y, b - x);

DrawPoint (a - y, b + x);

DrawPoint (a - y, b - x);

end

Using Polar Coordinates

x = xc + rx cos

y = yc + ry sin

Implicit

f(x,y)ellipse = rx2 (x – xc)

2 + ry2 (y - yc)

2 - rx2ry

2

Exploit Symmetry

xc

ry

rx yc

(-x,y) (x,y)

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

Drawing Ellipses

Bresenham approach is typically used for low-level

scan conversion of Lines, Circles & Ellipses.

In principle, any algebraic curve can be expressed

in the Bresenham style.

1. Measure error to actual curve (make decision)

2. Find integer form of error

3. Find incremental update of error criteria

In practice, curve beyond conics are drawn as

a series of short Bresenham straight lines.

Other Algebraic Surfaces

SPLINES

2D-Drawings

2D Freeform Curves or Splines

1

2

3

4

5

6 7

A typical free form curve designed to

pass through or near a sequence of

points.

Splines are parameterized curves that

generalize linear interpolation of straight

lines to higher powers of interpolation

parameter.

Splines

),,,( 1111

zyxx

),,,( 2222

zyxx

21

121

)1(

)()(

txxt

xxtxtx

21 )1( and )0( xxxx

start :

end :

where :

Jagged Lines :

Linear interpolation (1st order Spline)

x 1

x 2

x 3

x 4

x 6

x 7 t=0

t=1 x 5

G0: geometric continuity: the endpoints coincide

22

2

)1()1(

)(

CttBttA

ctbtatx

but 2nd derivative is choppy

matches derivatives

Quadratic interpolation

G1: tangents have the same slope

32

3223 )1()1()1()(

ttt

tDttCttBtAtx

Cubic Spline & Interpolation

C1: first derivative on both curves match at join point

The "speed" is the same before and after

Examples:

• Natural Cubic Spline

• Hermit

• Beizer

• B-spline

Representing Curves

Cubic curve ( to maintain C1 continuity)

Matrix Representation

y

y

y

y

x

x

x

x

d

c

b

a

d

c

b

a

tttyx ]1[][ 23

yyyy

xxxx

dtctbtay

dtctbtax

23

23

Solving for Coefficients

Continuity – C0, C1, etc.

The Gradient of a Cubic Curve

y

y

y

y

x

x

x

x

dt

dy

dtdx

d

c

b

a

d

c

b

a

tt ]0123[][ 12

yyyy

xxxx

dtctbtay

dtctbtax

23

23

023

023

2

2

yyydt

dy

xxxdtdx

ctbta

ctbta

Matrix notion:

The Hermite Specification as a Matrix

Equation

y

y

y

y

x

x

x

x

dt

dy

dt

dy

dt

dx

dt

dx

d

c

b

a

d

c

b

a

y

y

x

x

0123

0100

1111

1000

2

1

2

1

2

1

2

1

Solving the Hermite Coefficients

y

y

y

y

x

x

x

x

dt

dy

dt

dy

dt

dx

dt

dx

d

c

b

a

d

c

b

a

y

y

x

x

2

1

2

1

2

1

2

1

0001

0100

1233

1122

MHermite GHermite

Cubic Hermite Spline Equation:

HermiteHermite

23 ]1ttt[][ GMyx

Another Way to Think About Splines

Cubic Hermite Spline Equation

After reordering multiplications

HermiteHermite

23 ]1ttt[][ GMyx

24132211

2

1

2

1

23

23

23

23

)()()()(

2

32

132

][

ptfptfptfptf

p

p

p

p

tt

ttt

tt

tt

yx

T

Hermite Blending Functions

Beizer Spline

p1

p2

p3

p4

Anchor Point

Anchor Point Control Point

Control Point

)(3

)(3

432

121

ppp

ppp

Converting from Beizer to Hermite

Since

Substituting gives:

4

3

2

1

4

3

2

1

2

1

2

1

3300

0033

1000

0001

2

1

2

1

y

y

y

y

x

x

x

x

y

y

x

x

dt

dy

dt

dy

dt

dx

dt

dx

4

3

2

1

4

3

2

1

Hermite

y

y

y

y

x

x

x

x

TM

d

c

b

a

d

c

b

a

y

y

y

y

x

x

x

x

T

0001

0033

0363

1331

MBeizer

Blending Functions for Beizer Splines

0 t

1 B1 B4

B2 B3

3

4

2

3

2

2

3

1

3)1(

3)1(

)1(

tB

ttB

ttB

tB

Bezier Curve Properties

Blending functions always

•Sum to 1 (partition of unity)

•Are non-negative

•Satisfy 14321 BBBB

Curve always

•Passes inside the convex hull of the control points

•Goes through the two endpoints

•Is tangent to the control polygon at the endpoints

Convex Hull Property

Drawing Spline Curves

for t between 0 and 1 gives all the points on the curve:

0 1/6 1/3 1/2 2/3 5/6 1

xxxx

xxxx

PPPP

PPPPx

481

383

283

187

4

3

21

3212

21

2

2

21

21

1

3

21

21 )1(3)1(3)1()(

)](),([ tytx

Displaying Cubic Curves

•Divide interval [0, 1] into n even pieces

•Evaluate the curve at each value of t within the interval

•Draw line segments connecting the points

What is the cost of displaying curves this way?

An incremental approach possible too.

xxxx PtPttPttPttx 4

3

3

2

2

2

1

3 )1(3)1(3)1()(

Summary

Continuity of free-form curves (G0, G1, C0, C1,

C2…)

Cubic Curves (Splines)

Hermite

Matrix representation

Blending

Beizer

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