Explain the steps involved in clipping a line using the midpoint subdivision algorithm. A clipping window ABCD is specified as A(0,0), B(40,0), C(40,40), and D(0,40). Using this al... Explain the steps involved in clipping a line using the midpoint subdivision algorithm. A clipping window ABCD is specified as A(0,0), B(40,0), C(40,40), and D(0,40). Using this algorithm, find the visible portion, if any, of the line segment joining the points P(-10,20) and Q(50,10). Read more

Steps to Solve

1. Midpoint Subdivision Algorithm Explanation

The midpoint subdivision algorithm is a line clipping algorithm that works by recursively dividing a line segment into two halves and testing the midpoint of the segment against the clipping boundaries. If the midpoint is inside the clipping window, we check if the entire line segment is inside or outside the window. If the midpoint is outside the window, we discard the portion of the line that is definitively outside and recursively apply the algorithm to the remaining portion. This process continues until the visible portion of the line is found or it is determined that the line is entirely outside the clipping window.

2. Define Clipping Window and Line Segment

   The vertices of the clipping window ABCD are A(10, 10), B(60, 10), C(60, 50), and D(10, 50). The line segment to be clipped is defined by its endpoints P(20, 60) and Q(80, 20).

3. Determine the Outcodes

   The outcode for a point indicates its location relative to the clipping window. The outcode is a 4-bit binary code, where each bit represents a region outside the window. The bits correspond to (Top, Bottom, Right, Left). A '1' indicates the point is outside that boundary, and a '0' indicates it is inside or on the boundary. The boundaries of the clipping window are:

   • Left: $x = 10$
   • Right: $x = 60$
   • Bottom: $y = 10$
   • Top: $y = 50$

   Now, we calculate the outcodes for P(20, 60) and Q(80, 20):

   For P(20, 60):

   • Top: $y > 50$, so bit = 1
   • Bottom: $y < 10$ is false, so bit = 0
   • Right: $x > 60$ is false, so bit = 0
   • Left: $x < 10$ is false, so bit = 0 Thus, the outcode for P is 1000.

   For Q(80, 20):

   • Top: $y > 50$ is false, so bit = 0
   • Bottom: $y < 10$ is false, so bit = 0
   • Right: $x > 60$, so bit = 1
   • Left: $x < 10$ is false, so bit = 0 Thus, the outcode for Q is 0010.

4. Check for Trivial Accept/Reject

   Trivial Accept: Both outcodes are 0000. The line is entirely inside the window. Trivial Reject: The bitwise AND of the outcodes is not 0000. The line is entirely outside the window.

   In our case, the outcodes are 1000 and 0010. Their bitwise AND is 0000. So no trivial reject or accept.

5. Midpoint Subdivision Calculate the midpoint M of line segment PQ: $M = (\frac{x_P + x_Q}{2}, \frac{y_P + y_Q}{2}) = (\frac{20 + 80}{2}, \frac{60 + 20}{2}) = (50, 40)$

   Calculate the outcode for M(50, 40):

   • Top: $y > 50$ is false, so bit = 0
   • Bottom: $y < 10$ is false, so bit = 0
   • Right: $x > 60$ is false, so bit = 0
   • Left: $x < 10$ is false, so bit = 0 Thus, the outcode for M is 0000.

6. Recursive Clipping Since M is inside the clipping window, consider the segments PM and MQ.

   • Segment PM: Outcodes are P(1000) and M(0000). Since P is outside, clip PM. Find the midpoint $M_1$ of PM: $M_1 = (\frac{20 + 50}{2}, \frac{60 + 40}{2}) = (35, 50)$. Outcode of $M_1$ is 0000 since it lies right on the top boundary.

   Since $M_1$ is inside, consider the segment $PM_1$. Find the midpoint $M_2$ of $PM_1$: $M_2 = (\frac{20 + 35}{2}, \frac{60 + 50}{2}) = (27.5, 55)$. The outcode of $M_2$ is 1000.

   Since $M_2$ is outside, find the midpoint $M_3$ of $M_1M_2$: $M_3 = (\frac{35 + 27.5}{2}, \frac{50 + 55}{2}) = (31.25, 52.5)$. The outcode of $M_3$ is 1000.

   Since $M_3$ is outside, find the midpoint $M_4$ of $M_1M_3$: $M_4 = (\frac{35 + 31.25}{2}, \frac{50 + 52.5}{2}) = (33.125, 51.25)$. The outcode of $M_4$ is 1000.

   Since $M_4$ is outside, find the midpoint $M_5$ of $M_1M_4$: $M_5 = (\frac{35 + 33.125}{2}, \frac{50 + 51.25}{2}) = (34.0625, 50.625)$. The outcode of $M_5$ is 1000.

   After continuing this bissection a few more times, this converges to the point (35, 50) which is $M_1$.

   Therefore, the intersection point is such that $y=50$. Find the x coordinates of the point of intersection between PQ and the line $y = 50$. The equation of line PQ is: $\frac{y - 60}{x - 20} = \frac{20 - 60}{80 - 20}$ which gives $\frac{y-60}{x-20} = \frac{-40}{60} = -\frac{2}{3}$ $3(y-60) = -2(x-20)$ $3y - 180 = -2x + 40$ $2x + 3y = 220$

   Plugging in y=50: $2x + 3(50) = 220$ $2x + 150 = 220$ $2x = 70$ $x = 35$ So, the intersection point is (35, 50).

   • Segment MQ: Outcodes are M(0000) and Q(0010). Since Q is outside, clip MQ. Find the intersection location of $x=60$. $2x + 3y = 220$ $2 * 60 + 3y = 220$ $120 + 3y = 220$ $3y = 100$ $y = \frac{100}{3} \approx 33.33$ So, the intersection point is (60, 33.33).

7. Visible Portion

   The visible portion of the line segment PQ is the segment between (35, 50) and (60, 33.33).