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Known that NQ=P where N is the Bspline basis function, Q is the control point that we wish to locate, P is the data points given ( the curve must pass through it) is there any way to modify the "N" such that it is rationalized with desired weight so that we can interpolate a NURBS curve through it?

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I've tried to modify the Basis function matrix and tried to Rationalize it by adding weight and dividing. The result is the curve didn't pass through data points.

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    $\begingroup$ I don't see why this question is specific to the software Mathematica. I vote to move this to http://scicomp.stackexchange.com/. $\endgroup$ – halirutan Feb 10 '15 at 7:46
  • $\begingroup$ My apologies, I've edited the question. It originally is related with mathematica. I just didn't upload my code. $\endgroup$ – Keith Lim Feb 10 '15 at 7:56
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    $\begingroup$ You should post your actual code, not images. You can go to the help centre and read more about proper code formatting $\endgroup$ – Sektor Feb 10 '15 at 9:57
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The procedure for doing a weighted B-spline interpolation is not too different from the unweighted case. I'll use the same point set in the docs, and add a weight vector that gives higher weight to the second and fifth points:

pts = {{1., 2.}, {0., 1.}, {2., 0.}, {2., 2.}, {3., 3.}, {5., 2.}};
wts = {1, 4, 1, 1, 4, 1};

Here again is Lee's algorithm, which I'll use this time to perform chord-length parametrization:

parametrizeCurve[pts_ /; MatrixQ[pts, NumericQ], a : (_?NumericQ) : 1/2] := 
       FoldList[Plus, 0, Normalize[(Norm /@ Differences[pts])^a, Total]];
tvals = parametrizeCurve[pts, 1];

Set up the knots for a cubic NURBS, and the basis function matrix:

m = 3;
knots = Join[ConstantArray[0, m + 1], MovingAverage[ArrayPad[tvals, -1], m],
             ConstantArray[1, m + 1]];
bas = Table[BSplineBasis[{m, knots}, j - 1, tvals[[i]]],
            {i, Length[pts]}, {j, Length[pts]}];

Up to this point, the proceedings have been exactly the same as with the unweighted case. For comparison purposes, let's generate the unweighted control points:

cn = LinearSolve[bas, pts];

Now, spot the differences between the previous snippet and the following snippet for the weighted control points:

cw = LinearSolve[bas, bas.wts pts]/wts;

Now, let's show the points along with the unweighted (red) curve and the weighted (blue) curve:

Graphics[{{Red, BSplineCurve[cn, SplineDegree -> m, SplineKnots -> knots]},
          {Blue, BSplineCurve[cw, SplineDegree -> m, SplineKnots -> knots, 
                              SplineWeights -> wts]},
          {Green, AbsolutePointSize[6], Point[pts]}},
         PlotRange -> {{-1, 6}, {-1, 13/4}}]

unweighted and weighted B-spline

Note the sharper turns in the blue curve. You should now be able to see why weighted interpolations are in fact not that commonly done.

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  • $\begingroup$ Dear J.M. I would like to ask you a question. For the built-in BSplineBasis, what is the default knots in the BSplineBasis[d,n,x]? For instance, By the result of BSplineBasis[3,1,x] // PiecewiseExpand, I guess the default knots is knots ={1/4,1/4,1/4,1/4,2/4,3/4,4/4,5/4,5/4,5/4,5/4}. However, when I execute the code BSplineBasis[{3,knots}},1,x] // PiecewiseExpand, which gives a different result from BSplineBasis[3,1,x] // PiecewiseExpand .Namely, my guess is wrong. THX. $\endgroup$ – xyz Jul 29 '15 at 8:20
  • $\begingroup$ @Shutao, clamped uniform knots over $[0,1]$ are used by default, if memory serves. $\endgroup$ – J. M. will be back soon Jul 29 '15 at 8:37
  • $\begingroup$ According @J. M. to your description, is it the defaut knots={0,0,0,0,1/4,2/4,3/4,1,1,1,1}? But the result of BSplineBasis[3,1,x] // PiecewiseExpand is different from BSplineBasis[{3,knots}},1,x] // PiecewiseExpand. Maybe I have a wrong understanding:-) $\endgroup$ – xyz Jul 29 '15 at 9:19
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    $\begingroup$ I do not have access to a computer right now to check. Let me get back to you when I can do so. $\endgroup$ – J. M. will be back soon Jul 29 '15 at 9:46
  • $\begingroup$ OK, @J. M. THX a lot :) $\endgroup$ – xyz Jul 29 '15 at 10:01

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