Bresenham’s Midpoint Algorithm

CS5600 Computer Graphics

Rich RiesenfeldSpring 2006

Lect

ure

Set

1

Spring 2006 CS 5600 2

Line Characterizations

• Explicit:

• Implicit:

• Constant slope:

• Constant derivative:

Bmxy

0),( cbyaxyxF

kx

y

kxf )(

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Discrete Lines

• Lines vs. Line Segments

• What is a discrete line segment?

–This is a relatively recent problem

–How to generate a discrete line?

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“Good” Discrete Line - 1

• No gaps in adjacent pixels

• Pixels close to ideal line

• Consistent choices; same pixels

in same situations

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“Good” Discrete Line - 2

• Smooth looking

• Even brightness in all orientations

• Same line for as for

• Double pixels stacked up?

PP 10 PP 01

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• Line segment in first octant with

0 < m < 1

• Let us proceed

Restricted Form

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Two Line Equations

• Explicit:

• Implicit:

Define:

Hence,

Bmxy 0),( cbyaxyxF

01

01

xxdx

yydy

Bxdx

dyy

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From previous

We have,

Hence,

Bxdx

dyy

0 Byxdx

dy

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Relating Explicit to Implicit Eq’s

Recall,

Or,

where,

0 Byxdx

dy

0)()()( Bdxydxxdy

0)()()(),( BdxydxxdyyxF

)();();( dxBcdxbdya

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Investigate Sign of FVerify that

Look at extreme values of y

0

),( yxF

above line

below line

on line

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The Picture

x

y

0 ),( yxF

above line

below line

0 ),( yxF

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Key to Bresenham Algorithm

“Reasonable assumptions” have reduced the problem to making a binary choice at each pixel:

(Previous)

NE (next)

E (next)

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Decision Variable d (logical)

Define a logical decision variable d– linear in form– incrementally updated (with addition)– tells us whether to go E or NE

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The Picture

ideal linemidpoint

E

NE

previous

Q

1 pxx

pxx

pyy

1 pyy

M

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The Picture (again)

ideal line

E

NE

previous

Q

),( pp yx

),(2

11 pp yx

)1,1( pp yx

),( 1 pp yx

midpointM

Spring 2006 CS 5600 29

Observe the relationships

• Suppose Q is above M, as before.

• Then , M is below the line

• So, means line is above M,

• Need to move NE, increase y value

0)( MF

0)( MF

Spring 2006 CS 5600 30

The Picture (again)

ideal line

midpoint

E

NE

previous

Q

),( pp yx

),(2

11 pp yx

)1,1( pp yx

),( 1 pp yx

Spring 2006 CS 5600 31

Observe the relationships

• Suppose Q is below M, as before.

• Then F(M) < 0 , implies M is above the

line

• So, F(M) < 0 , means line is below M,

• Need to move to E; don’t increase y

Spring 2006 CS 5600 32

M = Midpoint = :

),(211 pp yx

• Want to evaluate at M

• Will use an incr decision var d

• Let, ),(211 pp yxFd

cybxad pp )((21)1

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How will d be used?

Recall,

Therefore,

cybxad pp )((2

1)1

)(arbitrary 0

line) ideal above(midpoint 0

line) ideal below(midpoint 0

E

E

NE

d

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Case 1: Suppose E is chosen

• Recall

•

• ...

cybxad ppold )((2

1)1

cybxa

yxFd

pp

ppnew

)((

(

2

12

1

)2

),2

1: ; , yE x x y

Spring 2006 CS 5600 35

...

add

cybxa

cybxadd

oldnew

pp

ppoldnew

)(1(

)(2(

2

1

2

1

)

)

Case 1: Suppose E is chosen

Spring 2006 CS 5600 36

Review of Explicit to Implicit

Recall,

Or,

where,

0 Byxdx

dy

0)()()( Bdxydxxdy

0)()()(),( BdxydxxdyyxF

)();();( dxBcdxbdya

Spring 2006 CS 5600 37

Case 1:

.

new oldd d a

increment we add if is chosen.

So, . But remember that

(from line equations).

Hence, ( ) is not evaluated explicitly.

We simply add to update for

E

E

E

E

a

a dy

F M

a d E

Spring 2006 CS 5600 38

Case 2: Suppose NE chosen

Recall

...

1

21)( ( )old p pd a x b y c

3

23

2

2, )

2)

(

( ( )

new p p

p p

d F x y

a x b y c

1 1and, : ; , yNE x x y

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Case 2: Suppose NE

...

badd

cybxa

cybxa

dd

oldnew

pp

pp

oldnew

)(1(

)(2(

2

1

2

3

)

)

Spring 2006 CS 5600 40

Case 2: .badd oldnew

increment that we add if is chosen.

So, . But remember that

, and (from line equations).

Hence, ( ) is not evaluated explicitly.

We simply add to update for

NE

NE

NE

NE

a b

a dy b dx

F M

a b d NE

Spring 2006 CS 5600 41

.for update to

i.e., , addsimply wemeans,

and , where,

NEddxdy

ba

dxbdyaba

NE

NE

NE

Case 2: .badd oldnew

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Summary• At each step of the procedure, we

must choose between moving E or NE based on the sign of the

decision variable d• Then update according to

dxdyd

dydd

NENE

EE

where,

or , where,

Spring 2006 CS 5600 43

What is initial value of d ?

• First point is

• First midpoint is

• What is initial midpoint value?

),( 00 yx

),(2

11 00 yx

),(),(2

11

2

11 0000 yxFyxd

Spring 2006 CS 5600 44

0 0 0 0

0 0

0 0

1 11 1)

2 2

2

(2

( , ) ( ( )

, )

b

bF

F x y a x b y c

ax by c a

x y a

What is initial value of d ?

Spring 2006 CS 5600 45

What is initial value of d ?

0 0 0 0Note, ( (, ) 0, since , ) is on line.F x y x y

2

22

11

)(

0),( 00

dx

b

dy

ayxFHence,

Spring 2006 CS 5600 46

What is initial value of d ?

0 0 0 0Note, ( (, ) 0, since , ) is on line.F x y x y

2

22

11

)(

0),( 00

dx

b

dy

ayxFHence,

Spring 2006 CS 5600 47

What Does “2 x ” Do ?

• Has the same 0-set

• Changes the slope of the plane

• Rotates plane about the 0-set line

0)(2),(2 cbyaxyxF

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dxdyyxF

dyyxFdx

)(2),(2

gives, 2by

)(),( gMultiplyin

2

11

22

11

00

00

What is initial value of d ?

Spring 2006 CS 5600 49

2 ( , ) 2( ) 0

So, first value of

2( ) ( )

F x y ax by c

d dy dx

What is initial value of d ?

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More Summary

• Initial value

• Case 1:

• Case 2:

)()(2 dxdy

)(2 where, dydd EE

)}(){(2 where

,

dxdy

dd

NE

NE

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More Summary

Choose

otherwise

0 if

NE

dE

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Example

• Line end points:

• Deltas:

)11,9(),;)8,5(), 1100 (( yxyx

3;4 dydx

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Graph

13

12

11

10

9

8

7

64 5 6 7 8 9 10 11

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Example ( dx = 4; dy = 3 )

• Initial value of 3;4 dydx

NEd

dx-y ,d

2

0246

)()2(2)85(

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Graph

13

12

11

10

9

8

7

64 5 6 7 8 9 10 11

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Example ( dx = 4; dy = 3 )

• Update value of d•

Ed

dy-dx d

NE

022

2)43(2

)(2)9,6(2

so , wasmoveLast

Spring 2006 CS 5600 57

Example ( dx = 4; dy = 3 ) -2

• Update value of d

• Last move was NE, so

2d(6,9) = 2d(y - dy)

= 2(4 - 3) = - 2

d = 2 – 2 = 0 E

Spring 2006 CS 5600 58

Graph

13

12

11

10

9

8

7

64 5 6 7 8 9 10 11

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Example ( dx = 4; dy = 3 )

• Previous move was E

(7,9) 2( )

2(3) 6

0 6 0

d dy

d NE

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Graph

13

12

11

10

9

8

7

64 5 6 7 8 9 10 11

Spring 2006 CS 5600 61

Example ( dx = 4; dy = 3 )

• Previous move was , so

2 (8,10) 2( )

2(3 4) 2

6 2 4

NE

d dy - dx

d NE

Spring 2006 CS 5600 62

Graph

13

12

11

10

9

8

7

64 5 6 7 8 9 10 11

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Graph

13

12

11

10

9

8

7

64 5 6 7 8 9 10 11

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Graph

13

12

11

10

9

8

7

64 5 6 7 8 9 10 11

Spring 2006 CS 5600 65

More Raster Line Issues

• Fat lines with multiple pixel width

• Symmetric lines

• How should end pt geometry look?

• Generating curves, e.g., circles, etc.

• Jaggies, staircase effect, aliasing...

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Pixel Space

12

11

10

9

8

7

6

4 5 6 7 8 9 10

Example

Example

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Bresenham Circles

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The End Bresenham’s Algorithm

Lect

ure

Set

1