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@cmaes
Created October 7, 2011 07:40
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Small cubic spline implementation in Matlab
function cubic_driver(num_points)
% cubic_driver(n) computes a cubic spline
% interpolant of the runge function
%
% f(x) = 1/(1+25*x^2)
%
% based on n linearly spaced
% points in the interval [-1,1]
runge = @(x) 1./(1+ 25*x.^2);
x = linspace(-1,1,num_points);
y = runge(x);
[s0,s1,s2,s3] = cubic_spline(x',y');
plot_points = 1000;
xx = linspace(-1,1,plot_points);
yy = runge(xx);
plot(xx,yy,'g');
hold on;
plot_cubic_spline(x,s0,s1,s2,s3);
function [s0,s1,s2,s3]=cubic_spline(x,y)
% [s0,s1,s2,s3]=cubic_spline(x,y)
%
% computes the coefficents of a cubic spline
% interpolant through the data points (x,y)
%
% The spline is defined as the piecewise cubic
% polynomial
%
% S(x) = { Sk(x) x(k) <= x <= x(k+1)
% = { 0 otherwise
%
% The cubic polynomial Sk(x) is given by
%
% Sk(x) = sk0 + sk1*(x-x(k)) + sk2*(x-x(k))^2 + sk3*(x-x(k))^3
%
% The coefficents sk0, sk1, sk2, and sk3 for each of the
% polynomials are returned in the vectors s0,s1,s2, and s3
% respectively.
if any(size(x) ~= size(y)) || size(x,2) ~= 1
error('inputs x and y must be column vectors of equal length');
end
n = length(x)
h = x(2:n) - x(1:n-1);
d = (y(2:n) - y(1:n-1))./h;
lower = h(1:end-1);
main = 2*(h(1:end-1) + h(2:end));
upper = h(2:end);
T = spdiags([lower main upper], [-1 0 1], n-2, n-2);
rhs = 6*(d(2:end)-d(1:end-1));
m = T\rhs;
% Use natural boundary conditions where second derivative
% is zero at the endpoints
m = [ 0; m; 0];
s0 = y;
s1 = d - h.*(2*m(1:end-1) + m(2:end))/6;
s2 = m/2;
s3 =(m(2:end)-m(1:end-1))./(6*h);
function plot_cubic_spline(x,s0,s1,s2,s3)
% plot_cubic_spline(x,s0,s1,s2,s3)
%
% plots a cubic spline with break points x
% and coefficents s0, s1, s2, s3
n = length(x);
inner_points = 20;
for i=1:n-1
xx = linspace(x(i),x(i+1),inner_points);
xi = repmat(x(i),1,inner_points);
yy = s0(i) + s1(i)*(xx-xi) + ...
s2(i)*(xx-xi).^2 + s3(i)*(xx - xi).^3;
plot(xx,yy,'b')
plot(x(i),0,'r');
end
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@cmaes
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cmaes commented Oct 7, 2011

A tiny Matlab implementation of cubic spline interpolation, based on work done for the 18.310 class at MIT. This gist was motivated by the answer to this question on stack overflow.

@ouboub
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ouboub commented Feb 18, 2017

Hi
thanks for the code, however when i run it with an example of mine it does not coincide. I found
https://stackoverflow.com/questions/7642921/cubic-spline-program

lower = h(2:end);
main = 2*(h(1:end-1) + h(2:end));
upper = h(1:end-1);

which gives the correct result. Could you please correct here also?

Your code and my calculations do not agree with matlab. here is a simple example
x=[-3 -2 1 4 6]
y=[2 0 3 1 4]
[s0,s1,s2,s3]=cubic_spline(x',y')
comment the length of the vectors is not consistent
s3 s2 s1 s0
0.5044 0 -2.5044 2.0000
-0.2831 1.5132 -0.9912 0
0.2217 -1.0351 0.4430 3.0000
-0.1601 0.9605 0.2193 1.0000

however matlab gives using

clear all
x=[-3 -2 1 4 6]
y=[2 0 3 1 4]
pspl=spline(x,[0 y 0])
[breaks,coefs,l,k,d] = unmkpp(pspl)

 s3           s2            s1       s0
2.0595   -4.0595         0    2.0000

-0.3796 2.1190 -1.9405 0
0.3003 -1.2976 0.5238 3.0000
-0.5387 1.4048 0.8452 1.0000

any ideas why this is so?

thanks

Uwe Brauer

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