Room for Shutao TANG and J. M.

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J. M.

03:08

I had the main chatroom in mind, but this is fine… :)

Shutao TANG

03:25

Sorry, I cannot log in this room, so I restart my PC

Now I construct a matrix as follow:

n = 40;
RandomSeed[10];
pts =
RandomInteger[{1, 10}, {n, 2}];
knots = {Range[0, 1, 1/n][[-4 ;; -2]] - 1, Range[0, 1, 1/n],
Range[0, 1, 1/n][[2 ;; 4]] + 1} // Flatten;
paras1 = Range[0, 1, 1/n] // Most // N;
mat =
Outer[BSplineBasis[{3, knots}, #2, #1] &, paras1,
Range[0, (n - 1) + 3]];
{a, b} = Dimensions[mat];
mat = Take[mat, a, a] + PadRight[Take[mat, a, a - b], {a, a}];
LinearSolve[mat]

I want to interpolate a sets of points with a closed B-spline curve

J. M.

And you're asking if that configuration will always yield a well-conditioned matrix.

Did you have an example where it became ill-conditioned?

Shutao TANG

Yes, I discovered that the mat always well-conditioned. So I would like to know why. Namely, how to prove it

testConditionNumer[n_] :=
Module[{pts, knots, paras, a, b, mat},
RandomSeed[10];
pts = RandomInteger[{1, 10}, {n, 2}];
With[{inter = Range[0, 1, 1/n]},
knots = {inter[[-4 ;; -2]] - 1, inter, inter[[2 ;; 4]] + 1} //
Flatten
];
paras = Range[0, 1, 1/n] // Most // N;
mat = Outer[BSplineBasis[{3, knots}, #2, #1] &, paras,
Range[0, (n - 1) + 3]]; {a, b} = Dimensions[mat];
mat = Take[mat, a, a] + PadRight[Take[mat, a, a - b], {a, a}];
LinearSolve[mat]
]

In the function testConditionNumer, 3 is the degree of B-spline, and 3 is odd

J. M.

03:44

I think that's a coincidence more than anything.

If memory serves, de Boor had papers on this. It's related to the fact that the banded matrices involved are diagonally dominant and those matrices can then be proven to be well-conditioned.

Shutao TANG

For 5 points, the mat shown as below

case for n=20

J. M.

Yes, I think it's the diagonal dominance. Your system can be permuted to a "periodic banded" system by reordering the variables, and then the diagonal dominance can be shown.

Shutao TANG

I had deboor's book A practical guide to spline. However I didn't find the message about it. Maybe I missed something

J. M.

Either that or his papers, I don't have my books with me now.

In that cubic case, the system can be permuted to a periodic tridiagonal system where the diagonal elements are 2/3 and the off-diagonals and corners are 1/6. That is definitely diagonally dominant.

Shutao TANG

OK, now it's time for lunch now. Let us continue this discussion this afternoon. thanks

also sorry for my poor English:)

J. M.

03:54

Or tonight. I'm not sure about my time today. You can leave questions here, and I'll try to answer when I see them.

1 hour later…

Shutao TANG

05:00

I see:)