Note: This page has been translated by MathWorks. Please click here

To view all translated materals including this page, select Japan from the country navigator on the bottom of this page.

To view all translated materals including this page, select Japan from the country navigator on the bottom of this page.

Thin-plate smoothing spline

`tpaps(x,y) `

tpaps(x,y,p)

[...,p] = tpaps(...)

`tpaps(x,y) `

is the stform
of a thin-plate smoothing spline * f* for the given
data sites

`x(:,j)`

and the given data values `y(:,j)`

.
The `x(:,j)`

must be distinct points in the plane,
the values can be scalars, vectors, matrices, even ND-arrays, and
there must be exactly as many values as there are sites.The thin-plate smoothing spline * f* is the
unique minimizer of the weighted sum

$$pE(f)+(1-p)R(f)$$

$$E(f)={\displaystyle \sum _{j}{\left|y(:,j)-f\left(x(:,j)\right)\right|}^{2}}$$

and * R*(

$$R(f)={{\displaystyle \int (\left|{D}_{1}{D}_{1}f\right|}}^{2}+2{\left|{D}_{1}{D}_{2}f\right|}^{2}+{\left|{D}_{2}{D}_{2}f\right|}^{2})$$

Here, the integral is taken over all of *R*^{2},
|* z*|

`p`

is
chosen so that `(1-p)/p`

equals the average of the
diagonal entries of the matrix `A`

, with ```
A
+ (1-p)/p*eye(n)
```

the coefficient matrix of the linear system
for the `n`

coefficients of the smoothing spline
to be determined. This choice of `p`

is meant to
ensure that we are in between the two extremes, of interpolation (when `p`

is
close to `1`

and the coefficient matrix is essentially `A`

)
and complete smoothing (when `p`

is close to `0`

and
the coefficient matrix is essentially a multiple of the identity matrix).
This should serve as a good first guess for `p`

.`tpaps(x,y,p) `

also inputs
the *smoothing
parameter*, `p`

, a number between 0 and 1. As the smoothing parameter varies from 0 to 1, the
smoothing spline varies, from the least-squares approximation to the
data by a linear polynomial when `p`

is `0`

,
to the thin-plate spline interpolant to the data when `p`

is `1`

.

`[...,p] = tpaps(...) `

also
returns the smoothing parameter actually used.

**Example 1.** The following code
obtains values of a smooth function at 31 randomly chosen sites, adds
some random noise to these values, and then uses `tpaps`

to
recover the underlying exact smooth values. To illustrate how well `tpaps`

does
in this case, the code plots, in addition to the smoothing spline,
the exact values (as black balls) as well as each arrow leading from
a smoothed value to the corresponding noisy value.

rng(23); nxy = 31; xy = 2*(rand(2,nxy)-.5); vals = sum(xy.^2); noisyvals = vals + (rand(size(vals))-.5)/5; st = tpaps(xy,noisyvals); fnplt(st), hold on avals = fnval(st,xy); plot3(xy(1,:),xy(2,:),vals,'wo','markerfacecolor','k') quiver3(xy(1,:),xy(2,:),avals,zeros(1,nxy),zeros(1,nxy), ... noisyvals-avals,'r'), hold off

**Example 2.** The following code
uses an interpolating thin-plate spline to vector-valued
data values to construct a map, from the plane to the plane, that
carries the unit square {* x* : |

n = 64; t = linspace(0,2*pi,n+1); t(end) = []; values = [cos(t); sin(t)]; centers = values./repmat(max(abs(values)),2,1); st = tpaps(centers, values, 1); fnplt(st), axis equal

Note the choice of `1`

for the smoothing parameter
here, to obtain interpolation.

The determination of the smoothing spline involves the solution
of a linear system with as many unknowns as there are data points.
Since the matrix of this linear system is full, the solving can take
a long time even if, as is the case here, an iterative scheme is used
when there are more than 728 data points. The convergence speed of
that iteration is strongly influenced by `p`

, and
is slower the larger `p`

is. So, for large problems,
use interpolation, i.e., `p`

equal to 1,
only if you can afford the time.

Was this topic helpful?