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Math Help - Cohen-Sutherland Line Clipping Problem

  1. #1
    Junior Member
    Apr 2013

    Cohen-Sutherland Line Clipping Problem


    I'm trying to learn how to do the Cohen-Sutherland Line Clipping algorithm by hand. I have an example 2D space below:

    Cohen-Sutherland Line Clipping Problem-cohen-sutherland.png

    Now if I understand correctly, the goal is to clip everything outside of the rectangle correct? The problem I'm having is the math approach. For example the segment AD. I know the code for A and D is 0000 and 1001 respectively. Now the logical and of the two endpoints needs to not be 0 to be rejected which means if the logical and is 1 then then segment is accepted? So the logical and of AD is 0 so trivial accept correct? I have a feeling it should be rejected because the point D is outside of the viewport (rectangle).
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  2. #2
    Super Member
    Dec 2012
    Athens, OH, USA

    Re: Cohen-Sutherland Line Clipping Problem

    At the very least, you need psuedo code for the algorithm. Rather than provide such, I just excerpted a Java implementation from one of my graphics programs. I hope you can read Java or at least some "C like" language. There are some subtle points in the algorithm. For example, in computing a clipped point, a division by 0 is never performed. First is a diagram of a window and a line to be clipped. Following is the Java code.

      /* Following is an implementation of the Cohen Sutherland clip algorithm against
         a window in the "world" with lower left coordinates (wx0,wy0) and upper right
         corner (wx1,xy1).  If the line is completely outside the window, just return;
         Otherwise, draw the clipped line.
      public void drawLine(Graphics2D g2,double x1,double y1,double x2,double y2) {
        int c1,c2,c;
        double x,y;
        // code for point P1=(x1,y1)
        c1=x1<wx0 ? 1 : x1>wx1 ? 2 : 0;
        if (y1<wy0) c1 |=4;
        else if (y1>wy1) c1 |=8;
        // code for P2=(x2,y2)
        c2=x2<wx0 ? 1 : x2>wx1 ? 2 : 0;
        if (y2<wy0) c2 |=4;
        else if (y2>wy1) c2 |=8;
        while ((c1|c2)!=0) {
          if ((c1&c2) != 0) return; // both points completely outside window
          c=(c1!=0) ? c1 : c2;
          if ((c&1)!=0 || (c&2)!=0) { // point to left or right of window
            x=((c&1)!=0) ? wx0 : wx1;
          else { // point above or below wintdow
            y=((c&4)!=0) ? wy0 : wy1;
          if (c==c1) {
            x1=x; y1=y;
            c1=(x1<wx0) ? 1 : (x1>wx1) ? 2 : 0;
            if (y1<wy0) c1 |=4;
            else if (y1>wy1) c1 |=8;
          else {
            x2=x; y2=y;
            c2=(x2<wx0) ? 1 : (x2>wx1) ? 2 : 0;
            if (y2<wy0) c2 |=4;
            else if (y2>wy1) c2 |=8;
        // if we get to here, a line needs to be drawn
        // finally draw the line in Java: code omitted
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