an efficient model for line clipping operation

An Efficient Model for

Line Clipping Operation

Kasun Ranga Wijeweera

([email protected])

This Presentation is Based on Following Research Papers

• S. R. Kodituwakku, K. R. Wijeweera, M. A. P. Chamikara (2013), An Efficient Algorithm for Line Clipping in Computer Graphics Programming, Ceylon Journal of Science (Physical Sciences), Volume 17, pp. 1-7.

• S. R. Kodituwakku, K. R. Wijeweera, M. A. P. Chamikara (2012), An Efficient Line Clipping Algorithm for 3D Space, International Journal of Advanced Research in Computer Science and Software Engineering, Volume 2 (Issue 5), pp. 96-101.

What is Clipping ?Any procedure which identifies that portion of a scene which is either inside or outside a region is referred to as a clipping algorithm or clipping

Line Clipping Algorithms• In computer graphics, line clipping is the process of

removing lines or portions of lines outside of a region of interest

• Well known algorithms– Cohen Sutherland 2D Algorithm– Cohen Sutherland 3D Algorithm– Liang Barsky 2D Algorithm– Liang Barsky 3D Algorithm

Clipping Window : Rectangular

x

y

minx maxx

miny

maxy

Clipping Volume : Cuboidal

x

y

minx maxx

miny

maxy

z

minz

maxz

The Scissoring MethodFOR each point ( x, y ) on the line segment

IF ( x, y ) is inside the region

Save ( x, y )

END IF

END FOR

The Scissoring Method(Against a Rectangular Window)

Condition:( x >= minx ) AND ( x <= maxx )

AND

( y >= miny ) AND ( y <= maxy )

The Scissoring Method(Against a Cuboidal Volume)

Condition:( x >= minx ) AND ( x <= maxx )

AND


AND

( z >= minz ) AND ( z <= maxz )

Problem with the Scissoring Method Computational cost =

(Cost to evaluate the condition)

*

(Number of points on the line segment)

Problem:

What will happen in high resolution systems?

Proposed Model • The simplest model ever invented• Faster than well known clipping algorithms• Low memory consumption• Flexible with different shapes of clipping regions• Easy to extend to higher dimensions• A single mathematical model

Proposed Model against a Rectangle• Line segments are divided into four classes • Let ( x1, y1 ) and ( x2, y2 ) be the end points of the line

segment

x1 = x2 y1 = y2 Nature of the line segment

F F Declined

F T Parallel to x-axis

T F Parallel to y-axis

T T A point

Proposed Model against a Rectangle(Separation into Classes)

IF { ( x1 != x2 ) AND ( y1 != y2 ) }

// Declined

ELSE IF ( x1 != x2 )

// Parallel to x-axis

ELSE IF ( y1 != y2 )

// Parallel to y-axis

ELSE

// A point

END IF

Proposed Model against a Rectangle(Declined)

Equation of the line segment

( x – x1) / (x2 – x1 ) = ( y – y1 ) / ( y2 – y1 );

Let’s take

L = x2 – x1;

M = y2 – y1;


Using the parameter t,

( x – x1) / (x2 – x1 ) = ( y – y1 ) / ( y2 – y1 );

( x – x1) / (x2 – x1 ) = t t = ( x – x1) / L;

( y – y1 ) / ( y2 – y1 ) = t t = ( y – y1 ) / M;



( x – x1) / (x2 – x1 ) = ( y – y1 ) / ( y2 – y1 );

( x – x1) / (x2 – x1 ) = t x = x1 + L * t;

( y – y1 ) / ( y2 – y1 ) = t y = y1 + M * t;


Intersection with x = p line,

t = ( p –x1) / L;

x = p;

y = y1 + M * t;


Intersection with y = q line,

t = ( q – y1) / M;

x = x1 + L * t;

y = q;


L = x2 – x1; M = y2 – y1;

FOR each end point ( x, y ) of the line segment

IF ( x < minx )

t = ( minx – x1 ) / L; x = minx; y = y1 + M * t;

ELSE IF ( x > maxx )

t = ( maxx – x1 ) / L; x = maxx; y = y1 + M * t;

IF ( y < miny )

t = ( miny – y1 ) / M; x = x1 + L * t; y = miny;

ELSE IF ( y > maxy )

t = ( maxy – y1 ) / M; x = x1 + L * t; y = maxy;

END IF

END FOR


An important observation after code is executed!

IF line segment is completely outside

Both end points become a single point

ELSE

End points are the end points of clipped line segment

END IF


Example 1: Line segment is completely inside

B

A


Example 2: Line segment is partially inside

B

A

B’B’’


Example 3: Line segment intersects clipping window

B

A

B’

A’

A’’

B’’


Example 4: Line segment is completely outside

B

A

A’

A’’

B’

B’’


How can we prove this result?

By considering all the possible test cases

1 2 3

4 5 6

7 8 9


• One end point can be in 9 regions• Therefore both the end points can be in 9*9 (= 81)

ways• Now the problem can be divided into 81 classes• Each class can be further divided into subclasses

based on dissimilarities• Since the result is true for all the test cases then the

result is true for all the possible cases


Example 1: Both end points are in the region 1

BA

B’A’

Due to “minx”:Subclass 1

A’’

B’’



B

A

B’A’Due to “minx”:Subclass 2



B

A

B’A’

Due to “minx”:Subclass 3

A’’B’’


Example 2: End points are in the regions 2 and 4B

A

Subclass 1

A’

A’’B’


Example 2: End points are in the regions 2 and 4

B

A

Subclass 2 A’

B’


An important observation after code is executed!

IF line segment is completely outside

Both end points become a single point

ELSE

End points are the end points of clipped line segment

END IF

Now we have proved the result !


IF{ [ abs ( L ) < 1 ] AND [ abs ( M ) < 1 ] }

// Ignore

ELSE

FOR each end point (x, y) of the line segment

x = FLOOR ( x + 0.5 );

y = FLOOR ( y + 0.5 );

END FOR

Line ( x1, y1, x2, y2 );

END IF

Proposed Model against a Rectangle(Precision Error)

Example 1:

1 / 3 = 0.33333…

• But in computer it is stored as 0.333333• Therefore checking the exact equality causes a

problem

Proposed Model against a Rectangle(Precision Error)

Example 2:

• Mathematically; 1 / 3 = 1 - ( 2 / 3 )• In computer

1 / 3 = 0.333333

2 / 3 = 0.666666

1 - ( 2 / 3 ) = 0.333334 Not equal

Proposed Model against a Rectangle(Dealing with Precision Error)

• Seek approximate equality instead of exact equality

( x1 = x2 ) AND ( y1 = y2)

is replaced with

[ abs ( L ) < 1 ] AND [ abs ( M ) < 1 ]

Proposed Model against a Rectangle(Integer Inputs to Line Drawing Algorithms)

• Efficient line drawing algorithms deal only with integers– Example: Bresenham line drawing algorithm

• If floating point inputs were provided then they would be truncated

• Truncation may badly affect the quality of graphics– Example: ( 3.9, 4.0 ) ( 3, 4 )

• Therefore proper approximation should be done when converting into integers

Proposed Model against a Rectangle(Better Approximation)

p = FLOOR ( p + 0.5 )

Example:

10 10.59.5 119


IF { ( x1 != x2 ) AND ( y1 != y2 ) }

// Declined





ELSE

// A point

END IF

Proposed Model against a Rectangle(Parallel to x-axis)

IF { ( y1 >= miny ) AND ( y1 =< maxy ) }

FOR each end point (x, y) of the line segment

IF ( x < minx )

x = minx;


x = maxx;

END IF

END FOR

IF ( x1 != x2 )

Line ( x1, y1, x2, y2 );

END IF

END IF



BA


Example 2: Line segment is partially inside

BA B’


Example 3: Line segment intersects clipping window

BA B’A’


Example 4: Line segment is completely outside 1

BA B’A’



A B


IF { ( x1 != x2 ) AND ( y1 != y2 ) }

// Declined





ELSE

// A point

END IF

Proposed Model against a Rectangle(Parallel to y-axis)

IF { ( x1 >= minx ) AND ( x1 =< maxx ) }


IF ( y < miny )

y = miny;


y = maxy;

END IF

END FOR

IF ( y1 != y2 )

Line ( x1, y1, x2, y2 );

END IF

END IF



B

A


Example 2: Line segment is partially insideB

A

B’


Example 3: Line segment intersects clipping windowB

A

B’

A’



B

A

B’A’



B

A


IF { ( x1 != x2 ) AND ( y1 != y2 ) }

// Declined





ELSE

// A point

END IF

Proposed Model against a Rectangle(A Point)

IF <condition>

PutPixel ( x, y );

END IF

Condition:

( x >= minx ) AND ( x <= maxx )

AND


Similar to Scissoring Method !


IF { ( x1 != x2 ) AND ( y1 != y2 ) }

// Declined





ELSE

// A point

END IF

Proposed Model against a Rectangle(Experimental Comparison)

• Programming Language:

C++;• Computer:

Intel(R) Pentium(R) Dual CPU;

E2180 @ 2.00 GHz;

2.00 GHz, 0.98 GB RAM;• IDE Details:

Turbo C++;

Version 3.0;

Copyright(c) 1990, 1992 by Borland International, Inc;


Method:• Clip window with values minx = miny = 100 and maxx =

maxy = 300 was used• Random points were generated in the range 0 - 399

– randomize() function

• Those random points were used to generate random line segments

• Number of clock cycles taken by each algorithm to clip 100000000 random line segments were counted– clock() function

Proposed Model against a Rectangle(Experimental Results)

Step Cohen-Sutherland

Liang-Barsky

Proposed Algorithm

1 2596 2452 2296

2 2593 2452 2296

3 2594 2452 2296

4 2595 2452 2296

5 2596 2451 2296

6 2593 2451 2296

7 2593 2452 2296

8 2592 2451 2296

9 2593 2452 2296

10 2594 2451 2296


Average Ratio =

Clock cycles for traditional algorithm

Clock cycles for proposed algorithm


Average Ratios

Cohen Sutherland: Proposed = 1.1297

Liang Barsky: Proposed = 1.0677

The proposed algorithm is 1.13 times faster than

Cohen Sutherland and 1.07 times faster than

Liang Barsky

Proposed Model against a Cuboid(Separation into Classes)

• Let ( x1, y1, z1 ) and ( x2, y2, z2 ) be the end points of the line segment

IF { ( x1 != x2 ) AND ( y1 != y2 ) AND ( z1 != z2 ) }

// Declined

ELSE

// Simply a 2D clipping problem

END IF

Proposed Model against a Cuboid(Declined)

Equation of the line segment

( x – x1) / (x2 – x1 ) = ( y – y1 ) / ( y2 – y1 ) = ( z – z1) / ( z2 – z1 );

Let’s take

L = x2 – x1;

M = y2 – y1;

N = z2 – z1;



( x – x1) / (x2 – x1 ) = ( y – y1 ) / ( y2 – y1 ) = ( z – z1) / ( z2 – z1 );

( x – x1) / (x2 – x1 ) = t t = ( x – x1) / L;

( y – y1 ) / ( y2 – y1 ) = t t = ( y – y1 ) / M;

( z – z1) / ( z2 – z1 ) = t t = ( z – z1) / N;



( x – x1) / (x2 – x1 ) = ( y – y1 ) / ( y2 – y1 ) = ( z – z1) / ( z2 – z1 );

( x – x1) / (x2 – x1 ) = t x = x1 + L * t;

( y – y1 ) / ( y2 – y1 ) = t y = y1 + M * t;

( z – z1) / ( z2 – z1 ) = t z = z1 + N * t;


Intersection with x = p plane,

t = ( p –x1) / L;

x = p;

y = y1 + M * t;

z = z1 + N * t;


Intersection with y = q plane,

t = ( q – y1) / M;

x = x1 + L * t;

y = q;

z = z1 + N * t;


Intersection with z = r plane,

t = ( r –z1) / N;

x = x1 + L * t;

y = y1 + M * t;

z = r;


L = x1 – x2; M = y1 – y2; N = z1 – z2;


IF ( x < minx )

t = ( minx –x1) / L; x = minx; y = y1 + M * t; z = z1 + N * t;


t = ( maxx –x1) / L; x = maxx; y = y1 + M * t; z = z1 + N * t;

IF ( y < miny )

t = ( miny – y1) / M; x = x1 + L * t; y = miny; z = z1 + N * t;


t = ( maxy – y1) / M; x = x1 + L * t; y = maxy; z = z1 + N * t;

IF (z < minz )

t = ( minz –z1) / N; x = x1 + L * t; y = y1 + M * t; z = minz;

ELSE IF ( z > maxz )

t = ( maxz –z1) / N; x = x1 + L * t; y = y1 + M * t; z = maxz;

END IF

END FOR


IF { [ abs ( L ) < 1 ] AND [ abs ( M ) < 1 ] AND [ abs ( N ) < 1 ] }

// Ignore

ELSE

FOR each end point ( x, y, z ) of the line segment

x = FLOOR ( x + 0.5 );

y = FLOOR ( y + 0.5 );

z = FLOOR ( z + 0.5 );

END FOR

Line ( x1, y1, z1, x2, y2, z2 );

END IF

Proposed Model against a Cuboid(Proof)

• Consider all the possible test cases• Proof is done as we did for a rectangular clipping

window

Proposed Model against a Cuboid(Experimental Comparison)

Average Ratios

Cohen Sutherland: Proposed = 1.1268

Liang Barsky: Proposed = 1.0651

The proposed algorithm is 1.13 times faster than

Cohen Sutherland and 1.07 times faster than

Liang Barsky

Any Questions?

Thank You !