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樣條曲線反算公式

%------------------非均勻B樣條擬合MATLAB程式-----------------
clear
k=3;
x=load('data.txt');
[n,m]=size(x);
%-----------弦長引數化--------------------------------------
u(k+n)=0;
for i=1:n-1
 u(k+i+1)=u(k+i)+sqrt((x(i+1,1)-x(i,1))^2+(x(i+1,2)-x(i,2))^2);
end;
L=u(n+k);
for i=1:n
u(k+i)=u(k+i)/L;
end;
for i=1:3
u(k+i+n)=1;
end
%控制多邊線
plot(x(:,1),x(:,2),'o');
hold on
%------------反求n+2個控制點--------------------
%首位重節點v1=v2
%首位與控制多邊形相切
A=zeros(n+2);
A(1,1)=1;A(1,2)=-1;
A(2,2)=1;
A(n+2,n+1)=-1;A(n+2,n+2)=1;
A(n+1,n+1)=1;
for i=3:n
  for j=0:2
    A(i,i+j-1)=Base(i+j-1,k,u,u(i+2));
   end
end
%e:方程右邊.
e=0;
for i=1:m
    e(n+2,i)=0;
end
for i=1:n
    e(i+1,:)=x(i,:);
end
%求出控制點d
d=inv(A)*e;
plot(d(:,1),d(:,2),'g');




hold on
%------------插值並作出樣條曲線-----------------
x=0;y=0;down=0;
for j=1:(n-1)
    uu=(u(j+3)):0.0005:u(j+4);
    for kk=1:length(uu)
       down=down+1;
        x(down)=d(j,1)*Base(j,3,u,uu(kk))+d(j+1,1)*Base(j+1,3,u,uu(kk))+d(j+2,1)*Base(j+2,3,u,uu(kk))+d(j+3,1)*Base(j+3,3,u,uu(kk));
        y(down)=d(j,2)*Base(j,3,u,uu(kk))+d(j+1,2)*Base(j+1,3,u,uu(kk))+d(j+2,2)*Base(j+2,3,u,uu(kk))+d(j+3,2)*Base(j+3,3,u,uu(kk));
    end
end
axis('equal');


plot(x,y,'red');


xlabel('x');ylabel('y');
grid on