Clipping means identifying portions of a scene that are outside a specified region.

For a 2D graphics the region defining what is to be clipped is called the clip window.

Types of clipping:

1. All-or-none clipping: If any part of object outside clip window then the whole object is rejected.

2. Point clipping: Only keep the points that are inside clip window.

3. Line clipping: Only keep segment of line inside clip window.

4. Polygon clipping: Only keep segment of polygon inside clip window.

5. Text clipping.

Suppose that the clip window is a rectangle. The point P = (x, y) is saved for display if the following are satisfied:

xwmin ≤ x ≤ xwmax

ywmin ≤ y ≤ ywmax

Otherwise, the point will be clipped (not saved for display).

P1= (10,20), P2= (30,50), P3= (60,90), and P4= (130,150). Suppose that the coordinates of the two opposite corners of the clip window are (xwmin, ywmin) = (30, 30) and (xwmax, ywmax) = (130, 110). Which of the above points will be clipped?

P2 and P3 will saved because:

For P2: 30 ≤ 30 ≤ 130 and 30 ≤ 50 ≤ 110.

For P3: 30 ≤ 60 ≤ 130 and 30 ≤ 90 ≤ 110.

For a line segment with endpoints (x1, y1) & (x2, y2) and one or both endpoints outside the clipping window, the parametric representation:

x= x1 + u(x2 – x1) and y= y1 + u(y2 – y1), where 0 ≤ u ≤ 1

Line endpoints P1= (56,10) & P2= (62,16). Suppose that the coordinates of the two opposite corners of the clip window are (xwmin,ywmin)= (60,10) & (xwmax,ywmax)= (70,16). Check the for clipping.

x= 60= 56 + u(62 – 56), u= 4/6. So we deduce the line (56,10)-(62,16) cross the left boundary (x = 60) of the clip window.

x= 70= 56 + u(62 – 56), u= 14/6. So we deduce the line (56,10)-(62,16) is not crossing the right boundary (x = 70) of the clip window.

y= 16= 10 + u(16 – 10), u= 6/6. So we deduce the line (56,10)-(62,16) cross the top boundary (y = 16) of the clip window.

y= 10= 10 + u(16 – 10), u= 0/6. So we deduce the line (56,10)-(62,16) cross the bottom boundary (y = 10) of the clip window.

Step 1 − Assign a region code for each endpoints.

Step 2 − If both endpoints have a region code 0000 then accept this line.

Step 3 − Else, perform the logical AND operation for both region codes.

Step 3.1 − If the result is not 0000, then reject the line.

Step 3.2 − Else you need clipping.

Step 3.2.1 − Choose an endpoint of the line that is outside the window.

Step 3.2.2 − Find the intersection point at the window boundary.

Step 3.2.3 − Replace endpoint with the intersection point and update the region code.

Step 3.2.4 − Repeat step 2 until we find a clipped line either trivially accepted or trivially rejected.

Step 4 − Repeat step 1 for other lines.

Compute the end point Outcodes of A and B;

if (A == 0000 && B == 0000)

the whole line is visible (accept it);

Else if (A AND B != 0000)

the whole line is outside of window (reject it);

else repeat the following {

compute intersection;

clip outside part;

test again;}

Assign a binary 4 bit code to each vertex by comparing it to clip coordinates:

Bit 1 (left): the sign bit of x – xwmin.

Bit 2 (right): the sign bit of xwmax – x.

Bit 3 (below): the sign bit of y – ywmin.

Bit 4 (above): the sign bit of ywmax – y.

Suppose that the coordinates of opposite corners of the clip window are (68,59) & (92,71). Use Cohen-Sutherland clipping line algorithm to find region codes of the line (72,65)-(90,68), and the line (74,93)-(85,100), and the line (60,50)-(120, 93).

Solution:

1. Calculate differences between endpoint coordinates and clipping boundaries.

2. Use the resultant sign bit of each difference calculation to set the corresponding value in the region code.

Line (72,65)-(90,68):

the region code of (72,65) is 0000

the region code of (90,68) is 0000

The line (72-65)-(90,68) is

completely inside

Line (74,93)-(85,100):

the region code of (74,93) is 1000

the region code of (85,100) is 1000

Since 1000 AND 1000 = 1000 (Non Zero), so the line (74,93) – (85,100) lie completely outside the clipping region. So we clip it.

Line (60,50)-(120,95):

the region code of (60,50) is 1010

the region code of (120,95) is 0101

The line is saved because the final region code is 0000.

Suppose that the coordinates of opposite corners of the clip window are (4,2) & (12,8). Use Cohen-Sutherland clipping line algorithm to check the two endpoints (6,10) and (13,15).

the region code of (6,10) is 0001

the region code of (120,95) is 0101

With AND operation the region code is 0001 (Clipped).