uniform knot vector for splines

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SA-W il 12 Apr 2024 alle 13:33

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Modificato: Bruno Luong il 13 Apr 2024 alle 13:31

Suppose we have uniform interpolation points, e.g

x = [1 2 3 4 5 6]
x = 1x6
     1     2     3     4     5     6
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The functions for knot generation produce knots of the form

t = aptknt(x,5)
t = 1x11
    1.0000    1.0000    1.0000    1.0000    1.0000    3.5000    6.0000    6.0000    6.0000    6.0000    6.0000
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where the first and last knot is repeated k times, but the breaks are no longer uniform. Similar for other knot functions like optknt.

I would like to carry over the property of equal distance between the interp. points to the knots. The knot vector

t = 1 1 1 2 3 4 5 6 6 6 6

satisfies this, as well as the Schoenberg-Whitney conditions. I arbitrarily repeated the first knot 3 times and the last knot four times.

Is such a knot vector plausible and why does MATLAB not offer functions for uniform knot sequences?

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Answer 1

Bruno Luong il 12 Apr 2024 alle 15:45

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Modificato: Bruno Luong il 12 Apr 2024 alle 15:55

Apri in MATLAB Online

Your knot sequence seems NOT to be suitable for interpolation. The interpolation matrix is singular.

x = linspace(1,6,6);

xi = linspace(min(x),max(x),61);

k = 5;

k1 = floor(k/2);

k2 = k-k1;

tSAW = [repelem(x(1),1,k1) x repelem(x(end),1,k2)]

tSAW = 1x11

1 1 1 2 3 4 5 6 6 6 6

t = aptknt(x,k);

for p=0:k-1

y = x.^p;

sSAW = spapi(tSAW,x,y); % Not workingn coefs are NaN

s = spapi(t,x,y);

subplot(3,2,p+1);

yiSAW = fnval(sSAW, xi);

yi = fnval(s, xi);

plot(x, y, 'ro', xi, yiSAW, 'b+', xi, yi, 'g-') % Blue curve not plotted since yiSAW are NaN

end

Warning: Matrix is singular, close to singular or badly scaled. Results may be inaccurate. RCOND = NaN.

It seems Schoenberg-Whitney conditions are violated despite what you have claimed.

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Bruno Luong il 12 Apr 2024 alle 17:18

Modificato: Bruno Luong il 12 Apr 2024 alle 17:20

Apri in MATLAB Online

You need to revise the spline theory, or make correct logical dediction.

k = 5;

x = linspace(1,6,6);

k1 = floor(k/2);

k2 = k-k1;

tSAW = [repelem(x(1),1,k1) x repelem(x(end),1,k2)]

tSAW = 1x11

1 1 1 2 3 4 5 6 6 6 6

n = length(tSAW) - k;

figure

hold on

% Interpolation matrix at station x

M = zeros(n);

,for i = 1:n

si = spmak(tSAW, accumarray(i, 1, [n 1])');

fnplt(si);

M(:,i) = fnval(si,x);

end

% All basis functions is 0 at first and last knots/station

% First and last rows of M contain 0s

M

M = 6x6

0 0 0 0 0 0 0.5139 0.3194 0.0417 0 0 0 0.0556 0.4444 0.4583 0.0417 0 0 0 0.0417 0.4583 0.4444 0.0556 0 0 0 0.0417 0.3194 0.5139 0.1250 0 0 0 0 0 0

% Singularity

cond(M)

ans = Inf

rank(M)

ans = 4

Bruno Luong il 13 Apr 2024 alle 6:24

Modificato: Bruno Luong il 13 Apr 2024 alle 8:28

Apri in MATLAB Online

@SA-W knots(i) < x(i) < knots(i+k), i=1:length(tau)

This Schoenberg Whitney (SW) conditions are violated twice.

With

 k = 5;
% station vector is 
x = [1 2 3 4 5 6];
% Knot vector is 
tSAW = [1 1 1, 2 3 4 5, 6 6 6 6];

for i = 1:

knot(i) = tSAW(1) = 1
tau(i) = x(1) = 1
So knots(i) is NOT < tau(i)

for i = 6:

knot(i+k) = tSAW(11) = 6
tau(i) =x(6) = 6
So tau(i) is NOT < knots(i+k)

% Small code to check SW condition
if SWtest(x, tSAW)
    fprintf('Schoenberg Whitney  test pass\n');
else
    fprintf('Schoenberg Whitney  test fails\n');
end
Schoenberg Whitney  test fails
function OK = SWtest(x, t)
x = sort(x); t = sort(t);
n = length(x);
k = length(t)-n;
OK = true;
for i = 1:n
    xi = x(i);
    tl = t(i);
    multiplicity = nnz(t == tl);
    if multiplicity >= k
        test_i = tl <= xi;
    else
        test_i = tl < xi;
    end
    tr = t(i+k);
    multiplicity = nnz(t == tr);
    if multiplicity >= k
        test_i = test_i && xi <= tr;
    else
        test_i = test_i && xi < tr;
    end
    OK = OK && test_i;
end
end

Note that all the basis functions N_i,5 for i=1,2...6 are continuous since the (extreme) knots are repeats less than k=5 times in your case.

All of them vanish outside the support, meaning for x < 1 or x > 6 So all of them must = 0 at x = 0 and x = 6 (since they are C0). This violate SW original condition of N_i,5 (x(i)) must be > 0. This is nother way to prove Schoenberg Whitney conditions are violated.

That explains aslo why the interpolation matrix M has 2 null rows and is singular therefore not suitable for interpolation, and spapi won't work with such knot sequence.

Bruno Luong il 13 Apr 2024 alle 9:01

Modificato: Bruno Luong il 13 Apr 2024 alle 13:31

Apri in MATLAB Online

SWtest.m

"That said, there is no way to have uniform knot breaks on the interpolation domain?"

What is the purpose of it? What if your x is non unform to start with, an assumption you never spell out.

You could chose

x = 1:6;

k = 5;

n = length(x);

dt =(max(x)-min(x))/(n-k+1);

teq = min(x) + dt*(-k+1:n)

teq = 1x11

-9.0000 -6.5000 -4.0000 -1.5000 1.0000 3.5000 6.0000 8.5000 11.0000 13.5000 16.0000

<mw-icon class=""></mw-icon>

OK = SWtest(x, teq)

OK = logical

1

figure

hold on

for i = 1:n

si = spmak(teq, accumarray(i, 1, [n 1])');

fnplt(si);

end

xlim([min(x) max(x)]);

or

dx = mean(diff(x));

dt = dx;

a = (n+k-1)/2;

teqx = mean(x) + dt*(-a:a)

teqx = 1x11

-1.5000 -0.5000 0.5000 1.5000 2.5000 3.5000 4.5000 5.5000 6.5000 7.5000 8.5000

<mw-icon class=""></mw-icon>

OK = SWtest(x, teqx)

OK = logical

1

figure

hold on

for i = 1:n

si = spmak(teqx, accumarray(i, 1, [n 1])');

fnplt(si);

end

xlim([min(x) max(x)]);

SA-W il 13 Apr 2024 alle 10:50

Modificato: Bruno Luong il 13 Apr 2024 alle 11:06

What is the purpose of it? What if your x is non unform to start with, an assumption you never spell out.

I know that my interpolant is barely sampled at some regions and I can circumvent this with the positioning of interpolation points x. For instance, when I do csape(x,y), the interpolation segments are determined by x, whereas for spapi and friends, the knot vector determiens the interpolation segments and I only have semi-control.

Anyway, if x is uniform, the knot vector aptknt(x,k) is uniform as well except at the boundary. So i can live with that. I was just wondering if there are alternatives and you provided some.

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