Non trivial case of cohen sutherland line clippings, Computer Graphics

Non Trivial Case of cohen sutherland line clippings

Case: assume for the line segment PQ, both the trivial rejection and acceptance tests failed (that is, Case 1 and Case 2 situation do not hold, i.e. the case for l3, l4 and l5 line segments) demonstrated above in Figure. For these non-trivial Cases the algorithm is processed, as given below.

Because, both the bit-codes for the end points P, Q of the line segment cannot be equivalent to 0000.  Assume that the beginning point of the line segment is P those bit-code is not equivalent to 0000. For illustration, for the line segment l5 we decide P to be the bit codes 1001. Here, scan the bit-code of P from the first bit to the 4th bit and determine the position of the bit at that the bit value 1 appears at the first time. For the line segment line 5 it appears at the extremely first position. If the bit value 1 happens at the first position then proceed to intersect the line segment along with the UP edge of the window and allocate the first bit value to its point of intersection like 0. Likewise, if the bit value 1 arises at the second position while scanning the bit values at the initial time then intersect the line segment PQ along with the Down edge of the window. This point of intersection might be labeled as P'. Obviously the line segment PP' is outside the window and hence rejected and the original line segment considered for dipping will be P'Q. The coordinates of P' and its keeping new bit values are calculated. Here, by taking P as P', again we have the new line segment PQ that will again be referred to Case 1 for clipping.

926_non-trivial Case of cohen sutherland line clippings.png

Figure: Line Clipping - Geometrically

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