Least-squares line fitting: Difference between revisions

Revision as of 19:35, 5 November 2010

@@ Line 9: / Line 9: @@
 brd.suspendUpdate();
 for (i=0;i<100;i++) {
-   yr = 10*(Math.random()-0.5);
+   xr = 10*(Math.random()-0.5);
-   xr = 0*yr+delta*(Math.random()-0.5) +1.0;
+   yr = 0.3*xr+delta*(Math.random()-0.5) +1.0;
    p.push(brd.create('point',[xr, yr], {withLabel:false}));
 }
 brd.unsuspendUpdate();
-// Having constructed the points, we can fit a circle
+// Having constructed the points, we can fit a line described
+// by homogeneous coordinates
 // through the point set, consisting of n points.
-// The (n times 3) matrix consists of
+// The (n times 2) matrix consists of
-//   1, x_1, y_1
+//   x_1, y_1
-//   1, x_2, y_2
+//   x_2, y_2
 //      ...
-//   1, x_n, y_n
+//   x_n, y_n
 // where x_i, y_i is the position of point p_i
-// y is equal to the zero vector.
+// y is equal to the all-minus-one vector.
 var M = [], y= [], MT, B, c, z, n;
 n = p.length;
@@ Line 40: / Line 41: @@
 c = JXG.Math.matVecMult(MT, y);
 z = JXG.Math.Numerics.Gauss(B, c);
-// Finally, we can read from the solution vector z the coordinates [xm, ym] of the center
-// and the radius r and draw the circle.
-//var xm = z[0]*0.5;
-//var ym = z[1]*0.5;
-//var r = Math.sqrt(z[2]+xm*xm+ym*ym);
 brd.create('line',[1, z[0], z[1]]);