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# spmak

Put together spline in B-form

## Syntax

```spmak(knots,coefs) spmak(knots,coefs,sizec) spmak sp = spmak(knots,coeffs) ```

## Description

The command `spmak(...)` puts together a spline function in B-form, from minimal information, with the rest inferred from the input. `fnbrk` returns all the parts of the completed description. In this way, the actual data structure used for the storage of this form is easily modified without any effect on the various `fn...` commands that use this construct.

`spmak(knots,coefs) ` returns the B-form of the spline specified by the knot information in `knots` and the coefficient information in `coefs`.

The action taken by `spmak` depends on whether the function is univariate or multivariate, as indicated by `knots` being a sequence or a cell array. For the description, let `sizec` be `size(coefs)`.

If `knots` is a sequence (required to be non-decreasing), then the spline is taken to be univariate, and its order `k` is taken to be` length(knots)-sizec(end)`. This means that each `column' `coefs(:,j)` of `coefs` is taken to be a B-spline coefficient of the spline, hence the spline is taken to be `sizec(1:end-1)`-valued. The basic interval of the B-form is [`knots(1)` .. `knots(end)`].

Knot multiplicity is held to be ≤ `k`. This means that the coefficient `coefs(:,j)` is simply ignored in case the corresponding B-spline has only one distinct knot, i.e., in case `knots(j)` equals `knots(j+k)`.

If `knots` is a cell array, of length `m`, then the spline is taken to be `m`-variate, and `coefs` must be an (`r+m`)-dimensional array, – except when the spline is to be scalar-valued, in which case, in contrast to the univariate case, `coefs` is permitted to be an` m`-dimensional array, but `sizec` is reset by

```sizec = [1, sizec]; r = 1; ```

The spline is `sizec(1:r)`-valued. This means the output of the spline is an array with `r` dimensions, e.g., if `sizec(1:2) = [2, 3]` then the output of the spline is a 2-by-3 matrix.

The spline is `sizec(1:r)`-valued, the `i`th entry of the` m`-vector `k` is computed as `length(knots{i}) `-` sizec(r+i)`, `i=1:m`, and the `i`th entry of the cell array of basic intervals is set to `[knots{i}(1), knots{i}(end)]`.

`spmak(knots,coefs,sizec) ` lets you supply the intended size of the array `coefs`. Assuming that `coefs` is correctly sized, this is of concern only in the rare case that `coefs` has one or more trailing singleton dimensions. For, MATLAB® suppresses trailing singleton dimensions, hence, without this explicit specification of the intended size of `coefs`, `spmak` would interpret `coefs` incorrectly.

`spmak ` prompts you for `knots` and `coefs`.

`sp = spmak(knots,coeffs)` returns the spline `sp`.

## Examples

`spmak(1:6,0:2)` constructs a spline function with basic interval [1. .6], with 6 knots and 3 coefficients, hence of order 6 - 3 = 3.

`spmak(t,1)` provides the B-spline B(·|t) in B-form.

The coefficients may be `d`-vectors (e.g., 2-vectors or 3-vectors), in which case the resulting spline is a curve or surface (in R2 or R3).

If the intent is to construct a 2-vector-valued bivariate polynomial on the rectangle [–1..1] × [0..1], linear in the first variable and constant in the second, say

``` coefs = zeros([2 2 1]); coefs(:,:,1) = [1 0;0 1]; ```

then the straightforward

```sp = spmak({[-1 -1 1 1],[0 1]},coefs); ```

will result in the error message ```'There should be no more knots than coefficients'```, because the trailing singleton dimension of `coefs` will not be perceived by `spmak`, while proper use of that third argument, as in

```sp = spmak({[-1 -1 1 1],[0 1]},coefs,[2 2 1]); ```

will succeed. Replacing here `[2 2 1]` by `size(coefs)` would not work.

See the example “Intro to B-form” for other examples.

## Diagnostics

There will be an error return if the proposed knot sequence fails to be nondecreasing, or if the coefficient array is empty, or if there are not more knots than there are coefficients. If the spline is to be multivariate, then this last diagnostic may be due to trailing singleton dimensions in `coefs`.

## Support

#### Machine Learning Challenges: Choosing the Best Classification Model and Avoiding Overfitting

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