Spring 200

CS5600 1

Bresenham’sMidpoint 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

kxy =

ΔΔ

kxf =′ )(

Spring 2006 CS 5600 3

Line Characterizations - 2

• Parametric:

where,

• Intersection of 2 planes

• Shortest path between 2 points

• Convex hull of 2 discrete points

PtPttP 10)()( 1 +−=

PPPP 10 )(;)0( 1 ==

Spring 2006 CS 5600 4

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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Spring 2006 CS 5600 5

“Good” Discrete Line - 1

• No gaps in adjacent pixels

• Pixels close to ideal line

• Consistent choices; same pixels in same situations

Spring 2006 CS 5600 6

“Good” Discrete Line - 2

• Smooth looking

• Even brightness in all orientations

• Same line for as for

• Double pixels stacked up?

PP 10 PP 01

Spring 2006 CS 5600 7

Incremental Fn Eval

• Recall

• Characteristics

–Fast

–Cumulative Error

• Need to define

)()()1( ixixfixf Δ+=+

)( oxf

Spring 2006 CS 5600 8

0 0( , ) (0,0) translatex y• ≠ ⇒

Meeting Bresenham Criteria

0; 1 trivial casesm m• = = ⇒

0 1 flip about -axism x• > > − ⇒

1 flip about m x y• > ⇒ =

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Case 1: Translate to Origin

x

y ), 11( yx

), 00( yx ), 00( yx ′′

), 11( yx ′′

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Case 0: Trivial Situations

• Do not need Bresenham

line horizontal0 ⇒=• m

xym =⇒=• line1

Spring 2006 CS 5600 11

Case 1: Translate to Origin

• Need only consider lines emanating from the origin.

0 0, ) ( yx• Move to the origin

),),

0,0(),

010111

00

((

);(

yyy

y

xxx

x

−−=′

=′

′

′

Spring 2006 CS 5600 12

Case 2: Flip about x-axis

), 00( yx

x

y

), 11( yx), 11( yx ′′

), 00( yx ′′

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Case 2: Flip about x-axis

Flip about -axis ( ) :x y y• ′ = −

0 0 0 0

1 11 1

, ) ( ,

, ) , )

( );( (x y x y

x y x y

−′ ′ =′ ′ = −

1Suppose, 0 ,m• > > −

Spring 2006 CS 5600 14

How do slopes relate?

0

1 0

0

1 0

1

1

; by definition

y y

y y

mx x

mx x

−

−

−

−′ ′

⎫= ⎪⎪⎬⎪′ =⎪⎭

)0

1 0

(1i iSince ,

y ymyy

x x−

− −−− ′′ = =

Spring 2006 CS 5600 15

How do slopes relate?

)0

1 0

1( y ym

x xm m

−

−′ = −

′=−

1010 <′<⇒−>>∴ mm

i.e.,

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Case 3: Flip about line y=x), 11( yx ′′

), 00( yx ′′

y

x

xy =), 11( yx

), 00( yx

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Case 3: Flip about line y=x

, swap and prime them ,

,

y mx Bx yx my B

my x B

= +↔′ ′= +′ ′= −

Spring 2006 CS 5600 18

Case 3: m′=? 1 ,

1 and,

1 0 1

y x Bm

mm

m m

⎛ ⎞′ ′= −⎜ ⎟⎝ ⎠⎛ ⎞′∴ = ⎜ ⎟⎝ ⎠

′> ⇒ < <

Spring 2006 CS 5600 19

• Line segment in first octant with

0 < m < 1

• Let us proceed

Restricted Form

Spring 2006 CS 5600 20

Two Line Equations

• Explicit:• Implicit:

Define:

Hence,

Bmxy +=0),( =++= cbyaxyxF

01

01

xxdxyydy

−=−=

Bxdxdyy +⎟

⎠⎞

⎜⎝⎛=

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Spring 2006 CS 5600 21

From previous

We have,

Hence,

Bxdxdyy +⎟

⎠⎞

⎜⎝⎛=

0=+− Byxdxdy

Spring 2006 CS 5600 22

Relating Explicit to Implicit Eq’s

Recall,

Or,

where,

0=+− Byxdxdy

0)()()( =+−+ Bdxydxxdy

0)()()(),( =+−+=∴ BdxydxxdyyxF

)();();( dxBcdxbdya =−==

Spring 2006 CS 5600 23

Investigate Sign of FVerify that

Look at extreme values of y

⎪⎩

⎪⎨

⎧

−

+=

0

),( yxFabove line

below line

on line

Spring 2006 CS 5600 24

The Picture

x

y

0 ),( >yxF

above line

below line

0 ),( <yxF

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Spring 2006 CS 5600 25

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

Spring 2006 CS 5600 27

The Picture

ideal linemidpoint

E

NE

previous

Q

1+= pxx

pxx =

pyy =

1+= pyy

M

Spring 2006 CS 5600 28

The Picture (again)

ideal line

E

NE

previous

Q

),( pp yx

),(211 ++ pp yx

)1,1( ++ pp yx

),( 1 pp yx +

midpointM

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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

),(211 ++ 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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Spring 2006 CS 5600 33

How will d be used?

Recall,

Therefore,

cybxad pp ++++= )((21)1

⎪⎩

⎪⎨

⎧

⇒=⇒<⇒>

=)(arbitrary 0

line) ideal above(midpoint 0line) ideal below(midpoint 0

EENE

d

Spring 2006 CS 5600 34

Case 1: Suppose E is chosen• Recall

•

• ...

cybxad ppold ++++= )((21)1

cybxa

yxFd

pp

ppnew

++++=

++=∴)((

(

21

21

)2

),2

1: ; , yE x x y⇒ ← + ←

Spring 2006 CS 5600 35

...

add

cybxa

cybxadd

oldnew

pp

ppoldnew

+=

⎟⎠⎞⎜

⎝⎛ ++++−

⎟⎠⎞⎜

⎝⎛ ++++=−

)(1(

)(2(

21

21

)

)

Case 1: Suppose E is chosen

Spring 2006 CS 5600 36

Review of Explicit to ImplicitRecall,

Or,

where,

0=+− Byxdxdy

0)()()( =+−+ Bdxydxxdy

0)()()(),( =+−+=∴ BdxydxxdyyxF

)();();( dxBcdxbdya =−==

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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

Ea

a dyF M

a d E

Δ ≡Δ ==

Δ =

Spring 2006 CS 5600 38

Case 2: Suppose NE chosen

Recall

...

12

1)( ( )old p pd a x b y c= + + + +

32

32

2, )

2)

(

( ( )

new p p

p p

d F x y

a x b y c

= + +

= + + + +

∴1 1and, : ; , yNE x x y⇒ ← + ← +

Spring 2006 CS 5600 39

Case 2: Suppose NE

...

badd

cybxa

cybxa

dd

oldnew

pp

pp

oldnew

++=

⎟⎠⎞⎜

⎝⎛ ++++−

⎟⎠⎞⎜

⎝⎛ ++++=

=−

)(1(

)(2(

21

23

)

)

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

NEa b

a dy b dxF M

a b d NE

Δ ≡Δ = +

= = −

Δ = +

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Spring 2006 CS 5600 41

.for update to

i.e., , addsimply wemeans,

and , where,

NEddxdy

ba

dxbdyaba

NE

NE

NE

−=Δ

+=Δ

−==+=Δ

Case 2: .badd oldnew ++=

Spring 2006 CS 5600 42

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

⎪⎩

⎪⎨⎧

−=ΔΔ+

=ΔΔ+←

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),(

21

1 00 ++ yx

),(),(21

121

1 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 ?

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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

221

1

)(

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

221

1

)(

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

Spring 2006 CS 5600 48

dxdyyxF

dyyxF dx

−=++

⎟⎠⎞

⎜⎝⎛−=++

)(2),(2

gives, 2by

)(),( gMultiplyin

21

1

221

1

00

00

What is initial value of d ?

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Spring 2006 CS 5600 49

2 ( , ) 2( ) 0So, first value of 2( ) ( )

F x y ax by c

d dy dx

= + + =

= −

What is initial value of d ?

Spring 2006 CS 5600 50

More Summary

• Initial value

• Case 1:

• Case 2:

)()(2 dxdy −

)(2 where, dydd EE =ΔΔ+←

)}(){(2 where ,

dxdydd

NE

NE

−=ΔΔ+←

Spring 2006 CS 5600 51

More Summary

Choose⎪⎩

⎪⎨⎧ ≤

otherwise

0 if

NE

dE

Spring 2006 CS 5600 52

Example

• Line end points:

• Deltas:

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

3;4 == dydx

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Graph1312 11 10

9 8 7 6

4 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

⇒=>=−=

=

20246)()2(2)85(

Spring 2006 CS 5600 55

Graph1312 11 10

9 8 7 6

4 5 6 7 8 9 10 11

Spring 2006 CS 5600 56

Example (dx=4; dy=3 )

• Update value of d•

Ed

dy-dx dNE

⇒=−=−=−=

=

0222)43(2

)(2)9,6(2so , wasmoveLast

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

• Update value of d• Last move was NE, so

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

= 2(4 - 3) = - 2d = 2 – 2 = 0 ⇒ E

Spring 2006 CS 5600 58

Graph1312 11 10

9 8 7 6

4 5 6 7 8 9 10 11

Spring 2006 CS 5600 59

Example (dx=4; dy=3 )

• Previous move was E

(7,9) 2( )2(3) 60 6 0

d dy

d NE

== == + > ⇒

Spring 2006 CS 5600 60

Graph1312 11 10

9 8 7 6

4 5 6 7 8 9 10 11

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Example (dx=4; dy=3 )

• Previous move was , so2 (8,10) 2( )

2(3 4) 26 2 4

N Ed dy - dx

d N E

== − = −= − = ⇒

Spring 2006 CS 5600 62

Graph1312 11 10

9 8 7 6

4 5 6 7 8 9 10 11

Spring 2006 CS 5600 63

Graph1312 11 10

9 8 7 6

4 5 6 7 8 9 10 11

Spring 2006 CS 5600 64

Graph1312 11 10

9 8 7 6

4 5 6 7 8 9 10 11

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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...

Spring 2006 CS 5600 66

Pixel Space

1211109876

4 5 6 7 8 9 10

Example Example

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Bresenham Circles

Spring 2006 CS 5600 70

The End Bresenham’s Algorithm

Lect

ure

Set

1