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Clipping

Clipping. Clipping is the removal of all objects or part of objects in a modelled scene that are outside the real-world window. Doing this on a pixel-by-pixel basis would be very slow, especially if most objects in the scene are outside the window

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Clipping

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  1. Clipping • Clipping is the removal of all objects or part of objects in a modelled scene that are outside the real-world window. • Doing this on a pixel-by-pixel basis would be very slow, especially if most objects in the scene are outside the window • More practical techniques are necessary to speed up the task window

  2. window Line Clipping • Lines are defined by their endpoints, so it should be possible just to examine these, and not every pixel on the line • Most applications of clipping use a window that is either very large • i.e. nearly the whole scene fits inside, or very small • i.e. most of the scene lies inside the window • Hence, most lines may be either trivially accepted or rejected

  3. window Cohen-Sutherhland • The Cohen-Sutherland line-clipping algorithm is particularly fast for “trivial” cases • i.e. lines completely inside or outside the window • Non-trivial lines such as those that cross a boundary of the window, are clipped by computing the coordinates of the new boundary endpoint of the line where it crosses the edge of the window

  4. xI - xP xQ - xP = yI - yP yQ - yP Computing intersection points (1st case) Q yQ I ymax yI yP P A B ymin xmin xP xI xQ xmax As the triangles PAI and PBQ are similar we can write

  5. xI - xP xQ - xP = yI - yP yQ - yP (ymax - yP)(xQ - xP) xI xP = + yQ - yP yI ymax = Computing intersection points • We have • After replacing yI with ymax and multiplying both sides of this equation by ymax – yP, we get

  6. Computing intersection points • If segment PQ intersects one of the other sides of the rectangle, coordinates are calculated with similar considerations Q’’ ymax Q I’’ I Q’ P P’’ ymin I’ P’ xmin xmax

  7. yI - yP yQ - yP = xI - xP xQ - xP Computing intersections: 2nd case ymax Q yQ yI I P yP A B ymin xP xmin xI xQ xmax As the triangles PAQ and PBI are similar we can write

  8. (xmin - xP)(yQ - yP) yI yP = + xQ - xP xI xmin = yI - yP yQ - yP = xI - xP xQ - xP Computing intersection points • We have • After replacing xI with xmin and multiplying both sides of this equation by xmin – xP, we get

  9. Cohen Sutherland cont. • Having decided that a line is non-trivial, it must be clipped • We “push” each end-point of the line to its “nearest” window boundary 1 2 3 4

  10. 1001 0001 0101 1000 0000 0100 1010 0010 0110 Cohen-Sutherhland • The window edges are assumed to be extended so that the whole picture is divided into 9 regions • Each region is assigned a four-bit code, called an outcode Left Right Below Above

  11. Xmin Xmax 1001 0001 0101 Q Ymax Q` 1000 0000 0100 Ymin P`` P` 1010 0010 0110 P Example • Since P lies to the left of the left rectangle edge, it is replaced with P` • Since P` lies below the lower rectangle edge, it is replaced with P`` • Since Q lies to the right of the rectangle edge, it is replaced with Q` • Line segment P`` Q` (the clipped segment) can now be drawn

  12. Example • Steps 1, 2, 3 loop terminated as follows: • If the four bit code of P and Q are equal to zero • Draw line • If the two four bit codes contain a 1 in the same bit position • P and Q are on the same side of rectangle nothing to be drawn Xmin Xmax 1001 0001 0101 Q Ymax Q` 1000 0000 0100 Ymin P`` P` 1010 0010 0110 P

  13. Sample Java Code • Method for generating outcode for each end of the line is computed, giving codes cP and cQ int clipCode(float x, float y) { return ((x < xmin ? 8 : 0) | (x > xmax ? 4 : 0) | (y < ymin ? 2 : 0) | (y > ymax ? 1 : 0)); } Note: The expression: condition ? value1 : value2 evaluates value1 if condition is true value2 if condition is false

  14. 1001 0001 0101 1000 0000 0100 1010 0010 0110 Sample Java Code • Method for generating outcode for each end of the line is computed, giving codes cP and cQ int clipCode(float x, float y) { return ((x < xmin ? 8 : 0) | (x > xmax ? 4 : 0) | (y < ymin ? 2 : 0) | (y > ymax ? 1 : 0)); } 1000 = 8 0100 = 4 0010 = 2 0001 = 1 x < xminLeft x > xmaxRight y < yminBelow y > ymaxAbove

  15. 1001 0001 0101 1000 0000 0100 1010 0010 0110 Sample Java Code • The expression ((x < xmin ? 8 : 0) | (x > xmax ? 4 : 0) | (y < ymin ? 2 : 0) | (y > ymax ? 1 : 0)); Is an OR of boolean values (bit by bit) Example: 1000 | 0110 = 1110 1st and 2nd bits are mutually exclusive (cannot be both 1 but they can be both 0) Same for 3rd and 4th bits x < xminLeft x > xmaxRight y < yminBelow y > ymaxAbove

  16. 1001 0001 0101 1000 0000 0100 1010 0010 0110 P x < xminLeft x > xmaxRight y < yminBelow y > ymaxAbove Sample Java Code • The expression ((x < xmin ? 8 : 0) | (x > xmax ? 4 : 0) | (y < ymin ? 2 : 0) | (y > ymax ? 1 : 0)); Example: point P 0000 | 0100 | 0010 | 0000 == 0110

  17. void clipLine (Graphics g, float xP, float yP, float xQ, float yQ, float xmin, float ymin, float xmax, float ymax) { int cP = clipCode(xP, yP); // Generate clip code for point P int cQ = clipCode(xQ, yQ); // Generate clip code forpoint Q float dx, dy; while ((cP | cQ) != 0) { if ((cP & cQ) != 0) return; // Both points are outside area so ignore dx = xQ - xP; dy = yQ - yP;

  18. Trivial cases Xmin Xmax CP&CQ==4 (0100) CP&CQ==1 (0001) 1001 0001 0101 Ymax 1000 0000 0100 Ymin 1010 0010 0110 CP&CQ==8 (1000) CP&CQ==2 (0010)

  19. More trivial cases Xmin Xmax 1001 0001 0101 Ymax 1000 0000 0100 Ymin 1010 0010 0110 CP&CQ==0 However PQ does not intersect the window: must run a few steps

  20. More trivial cases (cont.d) Xmin Xmax 1001 0001 0101 Q’ Ymax Now: CP’&CQ’==1 1000 0000 0100 Ymin P’ 1010 0010 0110

  21. if (cP != 0) { if ((cP & 8) == 8) /*i.e.cP & 1000 == 1000 (x<xmin) */ { yP += (xmin-xP) * dy / dx; xP = xmin; } else if ((cP & 4) == 4) { yP += (xmax-xP) * dy / dx; xP = xmax; } else if ((cP & 2) == 2) { xP += (ymin-yP) * dx / dy; yP = ymin; } else if ((cP & 1) == 1) { xP += (ymax-yP) * dx / dy; yP = ymax; } cP = clipCode(xP, yP) } // end outer if

  22. elseif (cQ != 0) { if ((cQ & 8) == 8) { yQ += (xmin-xQ) * dy / dx; xQ = xmin; } elseif ((cQ & 4) == 4) { yQ += (xmax-xQ) * dy / dx; xQ = xmax; } elseif ((cQ & 2) == 2) { xQ += (ymin-yQ) * dx / dy; yQ = ymin; } elseif ((cQ & 1) == 1) { xQ += (ymax-yQ) * dx / dy; yQ = ymax;} cQ = clipCode(xQ, yQ); } // end else if } //end while drawLine(g, xP, yP, xQ, yQ); } // end method

  23. Other Clipping algorithms • The Cohen-Sutherland algorithm requires the window to be a rectangle, with edges aligned with the co-ordinate axes • It is sometimes necessary to clip to any convex polygonal window, e.g. triangular, hexagonal, or rotated. • The Cyrus-Beck, and Liang-Barsky line clippers use the concept of inside/outside half spaces generated by edges (polygons are seen as intersection of half (2D) spaces (planes)

  24. Polygon Clipping • A polygon is usually defined by a sequence of vertices and edges • If the polygons are un-filled, line-clipping techniques are sufficient

  25. Polygon Clipping • However, if the polygons are filled, the process in more complicated • A polygon may be fragmented into several polygons in the clipping process, and the original colour associated with each one

  26. Polygon Clipping Algorithms • The Sutherland-Hodgeman clipping algorithm clips any polygon against a convex clip polygon. • The Weiler-Atherton clipping algorithm will clip any polygon against any clip polygon. The polygons may even have holes

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