三次样条插值函数求解例题

文章标题：深度解析Matlab三次样条插值1. 前言在数学和工程领域中，插值是一种常见的数值分析技术，它可以用来估计不连续数据点之间的值。

而三次样条插值作为一种常用的插值方法，在Matlab中有着广泛的应用。

本文将从简单到复杂，由浅入深地解析Matlab中的三次样条插值方法，以便读者更深入地理解这一技术。

2. 三次样条插值概述三次样条插值是一种利用分段三次多项式对数据点进行插值的方法。

在Matlab中，可以使用spline函数来进行三次样条插值。

该函数需要输入数据点的x和y坐标，然后可以根据需要进行插值操作。

3. 三次样条插值的基本原理在进行三次样条插值时，首先需要对数据点进行分段处理，然后在每个分段上构造出一个三次多项式函数。

这些多项式函数需要满足一定的插值条件，如在数据点处函数值相等、一阶导数相等等。

通过这些条件，可以得到一个关于数据点的插值函数。

4. Matlab中的三次样条插值实现在Matlab中，可以使用spline函数来进行三次样条插值。

通过传入数据点的x和y坐标，可以得到一个关于x的插值函数。

spline函数也支持在已知插值函数上进行插值点的求值，这为用户提供了极大的灵活性。

5. 三次样条插值的适用范围和局限性虽然三次样条插值在许多情况下都能够得到较好的插值效果，但也存在一些局限性。

在数据点分布不均匀或有较大噪音的情况下，三次样条插值可能会出现较大的误差。

在实际应用中，需要根据具体情况选择合适的插值方法。

6. 个人观点和总结通过对Matlab中三次样条插值的深度解析，我深刻地理解了这一插值方法的原理和实现方式。

在实际工程应用中，我会根据数据点的情况选择合适的插值方法，以确保得到准确且可靠的结果。

我也意识到插值方法的局限性，这为我在实际工作中的决策提供了重要的参考。

通过以上深度解析，相信读者已经对Matlab中的三次样条插值有了更加全面、深刻和灵活的理解。

在实际应用中，希望读者能够根据具体情况选择合适的插值方法，以提高工作效率和准确性。

/* 三次样条插值计算算法*/#include "math.h "#include "stdio.h "#include "stdlib.h "/*N：已知节点数N+1R：欲求插值点数R+1x，y为给定函数f(x)的节点值{x(i)} (x(i) <x(i+1)) ,以及相应的函数值{f(i)} 0 <=i <=NP0=f(x0)的二阶导数；Pn=f(xn)的二阶导数u:存插值点{u(i)} 0 <=i <=R求得的结果s(ui)放入s[R+1] 0 <=i <=R返回0表示成功，1表示失败*/int SPL(int N,int R,double x[],double y[],double P0,double Pn,double u[],double s[]){/*声明局部变量*/double *h; /*存放步长:{hi} 0 <=i <=N-1 */double *a; /*存放系数矩阵{ai} 1 <=i <=N ；分量0没有利用*/ double *c; /*先存放系数矩阵{ci} 后存放{Bi} 0 <=i <=N-1 */double *g; /*先存放方程组右端项{gi} 后存放求解中间结果{yi} 0 <=i <=N */double *af; /*存放系数矩阵{a(f)i} 1 <=i <=N ；*/double *ba; /*存放中间结果0 <=i <=N-1*/double *m; /*存放方程组的解{m(i)} 0 <=i <=N ；*/int i,k;double p1,p2,p3,p4;/*分配空间*/if(!(h=(double*)malloc(N*sizeof(double)))) exit(1);if(!(a=(double*)malloc((N+1)*sizeof(double)))) exit(1);if(!(c=(double*)malloc(N*sizeof(double)))) exit(1);if(!(g=(double*)malloc((N+1)*sizeof(double)))) exit(1);if(!(af=(double*)malloc((N+1)*sizeof(double)))) exit(1);if(!(ba=(double*)malloc((N)*sizeof(double)))) exit(1);if(!(m=(double*)malloc((N+1)*sizeof(double)))) exit(1);/*第一步：计算方程组的系数*/for(k=0;k <N;k++)h[k]=x[k+1]-x[k];for(k=1;k <N;k++)a[k]=h[k]/(h[k]+h[k-1]);for(k=1;k <N;k++)c[k]=1-a[k];for(k=1;k <N;k++)g[k]=3*(c[k]*(y[k+1]-y[k])/h[k]+a[k]*(y[k]-y[k-1])/h[k-1]); c[0]=a[N]=1;g[0]=3*(y[1]-y[0])/h[0]-P0*h[0]/2;g[N]=3*(y[N]-y[N-1])/h[N-1]+Pn*h[N-1]/2;/*第二步：用追赶法解方程组求{m(i)} */ba[0]=c[0]/2;g[0]=g[0]/2;for(i=1;i <N;i++){af[i]=2-a[i]*ba[i-1];g[i]=(g[i]-a[i]*g[i-1])/af[i];ba[i]=c[i]/af[i];}af[N]=2-a[N]*ba[N-1];g[N]=(g[N]-a[N]*g[N-1])/af[N];m[N]=g[N]; /*P110 公式：6.32*/ for(i=N-1;i> =0;i--)m[i]=g[i]-ba[i]*m[i+1];/*第三步：求值*/for(i=0;i <=R;i++){/*判断u(i)属于哪一个子区间，即确定k */if(u[i] <x[0] || u[i]> x[N]){/*释放空间*/free(h);free(a);free(c);free(g);free(af);free(ba);free(m);return 1;}k=0;while(u[i]> x[k+1])k++;//p1=(h[k]+2*(u[i]-x[k])*pow((u[i]-x[k+1]),2)*y[k])/pow(h[k],3); //p2=(h[k]-2*(u[i]-x[k+1])*pow((u[i]-x[k]),2)*y[k+1])/pow(h[k],3);p1=(h[k]+2*(u[i]-x[k]))*pow((u[i]-x[k+1]),2)*y[k]/pow(h[k],3);p2=(h[k]-2*(u[i]-x[k+1]))*pow((u[i]-x[k]),2)*y[k+1]/pow(h[k],3); p3=(u[i]-x[k])*pow((u[i]-x[k+1]),2)*m[k]/pow(h[k],2);p4=(u[i]-x[k+1])*pow((u[i]-x[k]),2)*m[k+1]/pow(h[k],2);s[i]=p1+p2+p3+p4;}/*释放空间*/free(h);free(a);free(c);free(g);free(af);free(ba);free(m);return 0;}void main(){int N,R;double *x,*y,*u,*s;double P0,Pn;int i;/*验证算法：*/N=7;R=6;/*分配空间*/if(!(x=(double*)malloc((N+1)*sizeof(double)))){printf( "malloc error!\n ");exit(1);}if(!(y=(double*)malloc((N+1)*sizeof(double)))){printf( "malloc error!\n ");exit(1);}if(!(u=(double*)malloc((R+1)*sizeof(double)))){printf( "malloc error!\n ");exit(1);}if(!(s=(double*)malloc((R+1)*sizeof(double)))){printf( "malloc error!\n ");exit(1);}x[0]=0.5;x[1]=0.7;x[2]=0.9;x[3]=1.1;x[4]=1.3;x[5]=1.5;x[6]=1.7;x[7]=1.9;y[0]=0.4794;y[1]=0.6442;y[2]=0.7833;y[3]=0.8912;y[4]=0.9636;y[5]=0.9975;y[6]=0.9917;y[7]=0.9 463;u[0]=0.6;u[1]=0.8;u[2]=1.0;u[3]=1.2;u[4]=1.4;u[5]=1.6;u[6]=1.8;P0=-0.4794;Pn=-0.9463;if(!SPL( N, R, x, y, P0, Pn, u, s)){/*打印结果*/printf( "\nx= ");for(i=0;i <=N;i++)printf( "%8.1f ",x[i]);printf( "\ny= ");for(i=0;i <=N;i++)printf( "%8.4f ",y[i]);printf( "\n\nu= ");for(i=0;i <=R;i++)printf( "%9.2f ",u[i]);printf( "\ns= ");for(i=0;i <=R;i++)printf( "%9.5f ",s[i]);printf( "\nsin= ");for(i=0;i <=R;i++)printf( "%9.5f ",sin(u[i]));}/*释放空间*/free(x);free(y);free(u);free(s);}/* 测试数据来自课本55页例5 《数值分析》清华大学出版社第四版*/ //输入327.7 4.128 4.329 4.130 3.013.0 -4.0//输出输出三次样条插值函数：1: [27.7 , 28]13.07*(x - 28)^3 + 0.22*(x - 27.7)^3+ 14.84*(28 - x) + 14.31*(x - 27.7)2: [28 , 29]0.066*(29 - x)^3 + 0.1383*(x - 28)^3+ 4.234*(29 - x) + 3.962*(x - 28)3: [29 , 30]0.1383*(30 - x)^3 - 1.519*(x - 29)^3+ 3.962*(30 - x) + 4.519*(x - 29)//三次样条插值函数#include<iostream>#include<iomanip>using namespace std;const int MAX = 50;float x[MAX], y[MAX], h[MAX];float c[MAX], a[MAX], fxym[MAX];float f(int x1, int x2, int x3){float a = (y[x3] - y[x2]) / (x[x3] - x[x2]);float b = (y[x2] - y[x1]) / (x[x2] - x[x1]);return (a - b)/(x[x3] - x[x1]);} //求差分void cal_m(int n){ //用追赶法求解出弯矩向量M……float B[MAX];B[0] = c[0] / 2;for(int i = 1; i < n; i++)B[i] = c[i] / (2 - a[i]*B[i-1]);fxym[0] = fxym[0] / 2;for(i = 1; i <= n; i++)fxym[i] = (fxym[i] - a[i]*fxym[i-1]) / (2 - a[i]*B[i-1]);for(i = n-1; i >= 0; i--)fxym[i] = fxym[i] - B[i]*fxym[i+1];}void printout(int n);int main(){int n,i; char ch;do{cout<<"Please put in the number of the dots:";cin>>n;for(i = 0; i <= n; i++){cout<<"Please put in X"<<i<<':';cin>>x[i]; //cout<<endl;cout<<"Please put in Y"<<i<<':';cin>>y[i]; //cout<<endl;}for(i = 0; i < n; i++) //求步长h[i] = x[i+1] - x[i];cout<<"Please 输入边界条件\n 1: 已知两端的一阶导数\n 2:两端的二阶导数已知\n 默认:自然边界条件\n";int t;float f0, f1;cin>>t;switch(t){case 1:cout<<"Please put in Y0\' Y"<<n<<"\'\n";cin>>f0>>f1;c[0] = 1; a[n] = 1;fxym[0] = 6*((y[1] - y[0]) / (x[1] - x[0]) - f0) / h[0];fxym[n] = 6*(f1 - (y[n] - y[n-1]) / (x[n] - x[n-1])) / h[n-1];break;case 2:cout<<"Please put in Y0\" Y"<<n<<"\"\n";cin>>f0>>f1;c[0] = a[n] = 0;fxym[0] = 2*f0; fxym[n] = 2*f1;break;default:cout<<"不可用\n";//待定};//switchfor(i = 1; i < n; i++)fxym[i] = 6 * f(i-1, i, i+1);for(i = 1; i < n; i++){a[i] = h[i-1] / (h[i] + h[i-1]);c[i] = 1 - a[i];}a[n] = h[n-1] / (h[n-1] + h[n]);cal_m(n);cout<<"\n输出三次样条插值函数：\n";printout(n);cout<<"Do you to have anther try ? y/n :";cin>>ch;}while(ch == 'y' || ch == 'Y');return 0;}void printout(int n){cout<<setprecision(6);for(int i = 0; i < n; i++){cout<<i+1<<": ["<<x[i]<<" , "<<x[i+1]<<"]\n"<<"\t";/*cout<<fxym[i]/(6*h[i])<<" * ("<<x[i+1]<<" - x)^3 + "<<<<" * (x - "<<x[i]<<")^3 + "<<(y[i] - fxym[i]*h[i]*h[i]/6)/h[i]<<" * ("<<x[i+1]<<" - x) + "<<(y[i+1] - fxym[i+1]*h[i]*h[i]/6)/h[i]<<"(x - "<<x[i]<<")\n";cout<<endl;*/float t = fxym[i]/(6*h[i]);if(t > 0)cout<<t<<"*("<<x[i+1]<<" - x)^3";else cout<<-t<<"*(x - "<<x[i+1]<<")^3";t = fxym[i+1]/(6*h[i]);if(t > 0)cout<<" + "<<t<<"*(x - "<<x[i]<<")^3";else cout<<" - "<<-t<<"*(x - "<<x[i]<<")^3";cout<<"\n\t";t = (y[i] - fxym[i]*h[i]*h[i]/6)/h[i];if(t > 0)cout<<"+ "<<t<<"*("<<x[i+1]<<" - x)";else cout<<"- "<<-t<<"*("<<x[i+1]<<" - x)";t = (y[i+1] - fxym[i+1]*h[i]*h[i]/6)/h[i];if(t > 0)cout<<" + "<<t<<"*(x - "<<x[i]<<")";else cout<<" - "<<-t<<"*(x - "<<x[i]<<")";cout<<endl<<endl;}cout<<endl;}。

三次样条插值的求解摘要：分段低次插值虽然解决了高次插值的振荡现象和数值不稳定现象，使得插值多项式具有一致收敛性，保证了插值函数整体的连续性，但在函数插值节点处不能很好地保证光滑性要求，这在某些要求光滑性的工程应用中是不能接受的。

如飞机的机翼一般要求使用流线形设计，以减少空气阻力，还有船体放样等的型值线，往往要求有二阶光滑度（即有二阶连续导数）。

因此，在分段插值的基础上，引进了一种新的插值方法，在保证原方法的收敛性和稳定性的同时，又使得函数具有较高的光滑性的样条插值。

关键字:三转角方程 三弯矩阵方程0. 引言1，三次样条函数定义1：若函数2()[,]S x a b C ∈,且在每个小区间上1,j j x x +⎡⎤⎦⎣上是三次多项式，其中01n a x x x b ⋯=<<<= 是给定节点，则称()s x 是节点01,,,n x x x ⋯上的三次样条函数。

若节点j x 上 给定函数值()j j y f x =(0,1,)j n ⋯= ,且()j j s x y = (0,1,)j n ⋯= （1.1）成立，则称 ()s x 为三次样条差值函数。

从定义知，要求出()s x ，在每个应小区间1[,]j j x x + 上确定4个待定系数，共有 n 个小区间，故应确定4n 个参数，根据()s x 在[,]a b 上二阶导数连续，在节点()1,2,3,,1j x j n ⋯=-处应满足连续性条件(0)(0),j j s x s x -=+ ''(0)(0),j j s x s x -=+''''(0)(0)j j s x s x -=+ （1.2） 共有 3n-3个条件，再加上()s x 满足插值条件（1.1），共有4n-2个条件，因此还需要2个条件才能确定()s x 。

通常可在区间[,]a b 端点0,n a x b x ==上各加一个条件（称边界条件），边界条件可根据实际的问题要求给定。

例1 设)(x f 为定义在[0,3]上的函数，有下列函数值表：且2.0)('0=x f ，1)('3-=x f ，试求区间[0,3]上满足上述条件的三次样条插值函数)(x s本算法求解出的三次样条插值函数将写成三弯矩方程的形式：)()6()()6()(6)(6)(211123131j j jj j j jj j j j jj j jj x x h h M y x x h h M y x x h M x x h M x s --+--+-+-=+++++其中，方程中的系数jj h M 6，jj h M 61+，jj j j h h M y )6(2-，jjj j h h M y )6(211++-将由Matlab代码中的变量Coefs_1、Coefs_2、Coefs_3以及Coefs_4的值求出。

以下为Matlab 代码：%=============================% 本段代码解决作业题的例1%============================= clear all clc% 自变量x 与因变量y ，两个边界条件的取值 IndVar = [0, 1, 2, 3]; DepVar = [0, 0.5, 2, 1.5];LeftBoun = 0.2; RightBoun = -1;% 区间长度向量，其各元素为自变量各段的长度 h = zeros(1, length(IndVar) - 1); for i = 1 : length(IndVar) - 1h(i) = IndVar(i + 1) - IndVar(i); end% 为向量μ赋值mu = zeros(1, length(h));for i = 1 : length(mu) - 1mu(i) = h(i) / (h(i) + h(i + 1));endmu(i + 1) = 1;% 为向量λ赋值lambda = zeros(1, length(h));lambda(1) = 1;for i = 2 : length(lambda)lambda(i) = h(i) / (h(i - 1) + h(i));end% 为向量d赋值d = zeros(1, length(h) + 1);d(1) = 6 * ( (DepV ar(2) - DepVar(1) ) / ( IndVar(2) - IndVar(1) ) - LeftBoun) / h(1);for i = 2 : length(h)a = ( DepVar(i) - DepVar(i - 1) ) / ( IndVar(i) - IndVar(i - 1) );b = ( DepVar(i + 1) - DepVar(i) ) / ( IndVar(i + 1) - IndVar(i) );c = (b - a) / ( IndVar(i + 1) - IndVar(i - 1) );d(i) = 6 * c;endd(i + 1) = 6 *( RightBoun - ( DepVar(i + 1) - DepVar(i) ) / ( IndVar(i + 1) - IndVar(i) ) ) / h(i);% 为矩阵A赋值% 将主对角线上的元素全部置为2A = zeros( length(d), length(d) );for i = 1 : length(d)A(i, i) = 2;end% 将向量λ的各元素赋给主对角线右侧第一条对角线for i = 1 : length(d) - 1A(i, i + 1) = lambda(i);end% 将向量d的各元素赋给主对角线左侧第一条对角线for i = 1 : length(d) - 1A(i + 1, i) = mu(i);end% 求解向量MM =A \ d';% 求解每一段曲线的函数表达式for i = 1 : length(h)Coefs_1 = M(i) / (6 * h(i));Part_1 = conv( Coefs_1, ...conv( [-1, IndVar(i + 1)], ...conv( [-1, IndVar(i + 1)], [-1, IndVar(i + 1)] ) ) );S_1 = polyval (Part_1, [IndVar(i) : 0.01 : IndVar(i + 1)]);Coefs_2 = M(i + 1)/(6 * h(i));Part_2 = conv( Coefs_2, ...conv( [1, -IndVar(i)], ...conv( [1, -IndVar(i)], [1, -IndVar(i)] ) ) );S_2 = polyval (Part_2, [IndVar(i) : 0.01 : IndVar(i + 1)]);Coefs_3 = (DepVar(i) - M(i) * h(i)^2 / 6) / h(i);Part_3 = conv(Coefs_3, [-1, IndVar(i + 1)]);S_3 = polyval (Part_3, [IndVar(i) : 0.01 : IndVar(i + 1)]);Coefs_4 = (DepVar(i + 1) - M(i + 1) * h(i)^2 / 6) / h(i);Part_4 = conv(Coefs_4, [1, -IndVar(i)]);S_4 = polyval (Part_4, [IndVar(i) : 0.01 : IndVar(i + 1)]);S = S_1 + S_2 + S_3 + S_4;plot ([IndVar(i) : 0.01 : IndVar(i + 1)], S, 'LineWidth', 1.25)% 在样条插值曲线的相应位置标注该段曲线的函数表达式text(i - 1, polyval(Part_1, 3), ...['\itS', num2str(i), '(x)=', num2str(Coefs_1), '(', num2str( IndVar(i + 1) ), '-x)^{3}+', ...num2str(Coefs_2), '(x-', num2str( IndVar(i) ), ')^{3}+', num2str(Coefs_3), ...'(', num2str( IndVar(i + 1) ), '-x)+', num2str(Coefs_4), '(x-', num2str( IndVar(i) ), ')'], ...'FontName', 'Times New Roman', 'FontSize', 14)hold onend% 过x=1和x=2两个横轴点作垂线%line([1, 1], [2.5, -0.5], 'LineStyle', '--');line([2, 2], [2.5, -0.5], 'LineStyle', '--');% 为x轴和y轴添加标注xlabel( '\itx', 'FontName', 'Times New Roman', ...'FontSize', 14, 'FontWeight', 'bold');ylabel( '\its(x)', 'FontName', 'Times New Roman', ...'Rotation', 0, 'FontSize', 14, 'FontWeight', 'bold');最终，三次样条插值函数s(x)表达式为：[][][]⎪⎩⎪⎨⎧∈-+-+-+--∈-+-+---∈+-++--=.3,2,)2(44.1)3(62.2)2(06.0)3(62.0,2,1,)1(62.2)2(08.0)1(62.0)2(42.0,1,0,08.0)1(06.042.0)1(06.0)(333333x x x x x x x x x x x x x x x x s曲线的图像如图所示：例2 已知函数值表：试求在区间[1,5]上满足上述函数表所给出的插值条件的三次自然样条插值函数)(x s本算法求解出的三次样条插值函数将写成三弯矩方程的形式：)()6()()6()(6)(6)(211123131j j jj j j jj j j j jj j jj x x h h M y x x h h M y x x h M x x h M x s --+--+-+-=+++++其中，方程中的系数jj h M 6，jj h M 61+，jj j j h h M y )6(2-，jjj j h h M y )6(211++-将由Matlab代码中的变量Coefs_1、Coefs_2、Coefs_3以及Coefs_4的值求出。

常见插值算法--拉格朗⽇插值、三次卷积插值、三次样条插值、兰克索斯插值写在前⾯本⽂简单介绍了⼏种常见的插值算法并附带了相应的python代码，本⽂公式使⽤latex编写，如有错误欢迎评论指出，如果谁知道如何修改latex字号也欢迎留⾔关于⼀维、⼆维和多维插值三次卷积插值、拉格朗⽇两点插值（线性插值）、兰克索斯插值在⼆维插值时改变x和y⽅向的计算顺序不影响最终结果，这三个也是图像缩放插值时常⽤的插值算法，⽽其他插值在改变计算顺序时会产⽣明显差异，多维的情况笔者没有尝试，读者可以⾃⾏尝试或推导最近邻插值法（Nearest Neighbour Interpolation）在待求像素的四邻像素中，将距离待求像素最近的像素值赋给待求像素p_{11}p_{12}pp_{21}p_{22}python代码1def NN_interpolation(srcImg, dstH, dstW):2 scrH, scrW, _ = srcImg.shape3 dstImg = np.zeros((dstH, dstW, 3), dtype=np.uint8)4for i in range(dstH - 1):5for j in range(dstW - 1):6 scrX = round(i * (scrH / dstH))7 scrY = round(j * (scrW / dstW))8 dstImg[i, j] = srcImg[scrX, scrY]9return dstImg拉格朗⽇插值（Lagrange Interpolation）拉格朗⽇插值法需要找到k个p_i(x)函数，使得每个函数分别在在x_i处取值为1，其余点取值为0，则y_ip_i(x)可以保证在x_i处取值为y_i，在其余点取值为0，因此L_k(x)能恰好经过所有点，这样的多项式被称为拉格朗⽇插值多项式，记为L_k(x)=\sum_{i=1}^ky_ip_i(x)p_i(x)=\prod_{j \neq i}^{1 \leq j \leq k}\frac{x-x_j}{x_i-x_j}以四点即三次图像插值为例，因为横坐标间隔为1，则设四个点横坐标为-1、0、1和2，可得p_1(x)、p_2(x)、p_3(x)和p_4(x)假设y_1、y_2、y_3和y_4分别为1、2、-1、4，则可得拉格朗⽇函数如下图所⽰，待插值点横坐标范围为[0,1]在K=2时在k=2时，也被称为线性插值通⽤公式p_1=\frac{x-x_2}{x_1-x_2}p_2=\frac{x-x_1}{x_2-x_1}\begin{align} L_2x &= p_1y_1+p_2y_2 \nonumber \\ &= \frac{x-x_2}{x_1-x_2}y_1 + \frac{x-x_1}{x_2-x_1}y_2 \nonumber \end{align}图像插值像素分布如图所⽰p_{11}p_{12}pp_{21}p_{22}即当x_{i+1}=x_i+1时，设p与p_{11}的横纵坐标差分别为dx和dy\begin{align} L_2x &= \frac{x-x_2}{x_1-x_2}y_1 + \frac{x-x_1}{x_2-x_1}y_2 \nonumber \\ &= (x_2-x)y_1+(x-x_1)y_2 \nonumber \\ &= (1-dx)y_1+dxy_2 \nonumber \\ &= (y_2-y_1)dx+y_1 \nonumber \end{align}L_2'x=y_2-y_1在K=3时通⽤公式p_1=\frac{x-x_2}{x_1-x_2}\frac{x-x_3}{x_1-x_3}p_2=\frac{x-x_1}{x_2-x_1}\frac{x-x_3}{x_2-x_3}p_3=\frac{x-x_1}{x_3-x_1}\frac{x-x_2}{x_3-x_2}\begin{align} L_3x &= p_1y_1+p_2y_2+p_3y_3 \nonumber \\ &= \frac{x-x_2}{x_1-x_2}\frac{x-x_3}{x_1-x_3}y_1+\frac{x-x_1}{x_2-x_1}\frac{x-x_3}{x_2-x_3}y_2+\frac{x-x_1}{x_3-x_1}\frac{x-x_2}{x_3-x_2}y_3 \nonumber \end{align}图像插值像素分布如图所⽰p_{11}p_{12}p_{13}p_{21}p_{22}p_{23}pp_{31}p_{32}p_{33}即当x_{i+1}=x_i+1时，设p与p_{11}的横纵坐标差分别为dx和dy\begin{align} L_3x &= \frac{x-x_2}{x_1-x_2}\frac{x-x_3}{x_1-x_3}y_1 + \frac{x-x_1}{x_2-x_1}\frac{x-x_3}{x_2-x_3}y_2 + \frac{x-x_1}{x_3-x_1}\frac{x-x_2}{x_3-x_2}y_3 \nonumber \\ &= \frac{-dx(1-dx)}{(-1)\cdot(-2)}y_1 + \frac{-(1+dx)(1-dx)}{1\cdot(-1)}y_2 + \frac{(1+dx)dx}{2\cdot 1}y_3 \nonumber \\ &= (\frac{1}{2}d^2x-\frac{1}{2}dx)y_1 - (d^2x-1)y_2 + (\frac{1}{2}d^2x+\frac{1}{2}dx)y_3 \nonumber \\ &= d^2x(\frac{1}{2}y_1-y_2+\frac{1}{2}y_3)+dx(-\frac{1}{2}y_1+\frac{1}{2}y_3)+y_2 \nonumber \end{align}L_3'x=dx(y_1-2y_2+y_3)+(\frac{1}{2}y_3-\frac{1}{2}y_1)在K=4时通⽤公式p_1=\frac{x-x_2}{x_1-x_2}\frac{x-x_3}{x_1-x_3}\frac{x-x_4}{x_1-x_4}p_2=\frac{x-x_1}{x_2-x_1}\frac{x-x_3}{x_2-x_3}\frac{x-x_4}{x_2-x_4}p_3=\frac{x-x_1}{x_3-x_1}\frac{x-x_2}{x_3-x_2}\frac{x-x_4}{x_3-x_4}p_4=\frac{x-x_1}{x_4-x_1}\frac{x-x_2}{x_4-x_2}\frac{x-x_3}{x_4-x_3}\begin{align} L_4x &= p_1y_1+p_2y_2+p_3y_3+p_4y_4 \nonumber \\ &= \frac{x-x_2}{x_1-x_2}\frac{x-x_3}{x_1-x_3}\frac{x-x_4}{x_1-x_4}y_1 + \frac{x-x_1}{x_2-x_1}\frac{x-x_3} {x_2-x_3}\frac{x-x_4}{x_2-x_4}y_2 + \frac{x-x_1}{x_3-x_1}\frac{x-x_2}{x_3-x_2}\frac{x-x_4}{x_3-x_4}y_3 + \frac{x-x_1}{x_4-x_1}\frac{x-x_2}{x_4-x_2}\frac{x-x_3}{x_4-x_3}y_4\nonumber \end{align}图像插值p_{11}p_{12}p_{13}p_{14}p_{21}p_{22}p_{23}p_{24}pp_{31}p_{32}p_{33}p_{34}p_{41}p_{42}p_{43}p_{44}即当x_{i+1}=x_i+1时，设p与p_{11}的横纵坐标差分别为dx和dy\begin{align} L_4x &= \frac{x-x_2}{x_1-x_2}\frac{x-x_3}{x_1-x_3}\frac{x-x_4}{x_1-x_4}y_1 + \frac{x-x_1}{x_2-x_1}\frac{x-x_3}{x_2-x_3}\frac{x-x_4}{x_2-x_4}y_2 + \frac{x-x_1}{x_3-x_1}\frac{x-x_2}{x_3-x_2}\frac{x-x_4}{x_3-x_4}y_3 + \frac{x-x_1}{x_4-x_1}\frac{x-x_2}{x_4-x_2}\frac{x-x_3}{x_4-x_3}y_4 \nonumber \\ &= \frac{dx[-(1-dx)][-(2-dx)]}{(-1)\cdot(-2)\cdot(-3)}y_1 + \frac{(1+dx)[-(1-dx)][-(2-dx)]}{1\cdot(-1)\cdot(-2)}y_2 + \frac{(1+dx)dx[-(2-dx)]}{2\cdot 1\cdot(-1)}y_3 + \frac{(1+dx)dx[-(1-dx)]}{3\cdot 2\cdot 1}y_4 \nonumber \\ &= \frac{d^3x-3d^2x+2dx}{-6}y1 + \frac{d^3x-2d^2x-dx+2}{2}y_2 + \frac{d^3x-d^2x-2dx}{-2}y_3 + \frac{d^3x-dx}{6}y_4 \nonumber \\ &= d^3x(-\frac{1}{6}y_1+\frac{1}{2}y_2-\frac{1} {2}y_3+\frac{1}{6}y_4)+d^2x(\frac{1}{2}y_1-y_2+\frac{1}{2}y_3)+dx(-\frac{1}{3}y_1-\frac{1}{2}y_2+y_3-\frac{1}{6}y_4)+y_2 \nonumber \end{align}\begin{align} L_4'x &= d^2x(-\frac{1}{2}y_1+\frac{3}{2}y_2-\frac{3}{2}y_3+\frac{1}{2}y_4)+dx(y_1-2y_2+y_3)+(-\frac{1}{3}y_1-\frac{1}{2}y_2+y_3-\frac{1}{6}y_4) \nonumber \\ &= -[\frac{1}{2}d^2x(y_1-3y_2+3y_3-y_4)-dx(y_1-2y_2+y_3)+\frac{1}{6}(2y_1+3y_2-6y_3+y_4)] \nonumber \end{align}python代码插值核计算的时候乘法和加减法计算的顺序不同可能会导致结果存在细微的差异，读者可以⾃⾏研究⼀下1class BiLagrangeInterpolation:2 @staticmethod3def LagrangeInterpolation2(x, y1, y2):4 f1 = 1 - x5 f2 = x6 result = y1 * f1 + y2 * f27return result89 @staticmethod10def LagrangeInterpolation3(x, y1, y2, y3):11 f1 = (x ** 2 - x) / 2.012 f2 = 1 - x ** 213 f3 = (x ** 2 + x) / 2.014 result = y1 * f1 + y2 * f2 + y3 * f315return result1617 @staticmethod18def LagrangeInterpolation4(x, y1, y2, y3, y4):19 f1 = - (x ** 3 - 3 * x ** 2 + 2 * x) / 6.020 f2 = (x ** 3 - 2 * x ** 2 - x + 2) / 2.021 f3 = - (x ** 3 - x ** 2 - 2 * x) / 2.022 f4 = (x ** 3 - x) / 6.023 result = y1 * f1 + y2 * f2 + y3 * f3 + y4 * f424return result2526def biLag2_2(self, srcImg, dstH, dstW):27 dstH, dstW = int(dstH), int(dstW)28 srcH, srcW, _ = srcImg.shape29 srcImg = np.pad(srcImg, ((1, 1), (1, 1), (0, 0)), 'edge')30 dstImg = np.zeros((dstH, dstW, 3), dtype=np.uint8)31for dstY in range(dstH):32for dstX in range(dstW):33for channel in [0, 1, 2]:34# p11 p1235# p36# p21 p2237# 储存为 p(y, x)38 p = [dstY * srcH / dstH, dstX * srcW / dstW]39 p11 = [math.floor(p[0]), math.floor(p[1])]40 p12 = [p11[0], p11[1] + 1]4142 p21 = [p11[0] + 1, p11[1]]43 p22 = [p21[0], p12[1]]4445 diff_y, diff_x = p[0] - p11[0], p[1] - p11[1]46 r1 = grangeInterpolation2(diff_x, srcImg[p11[0], p11[1], channel], srcImg[p12[0], p12[1], channel])47 r2 = grangeInterpolation2(diff_x, srcImg[p21[0], p21[1], channel], srcImg[p22[0], p22[1], channel])4849 c = grangeInterpolation2(diff_y, r1, r2)5051 dstImg[dstY, dstX, channel] = np.clip(c, 0, 255)52return dstImg5354def biLag3_3(self, srcImg, dstH, dstW):55 dstH, dstW = int(dstH), int(dstW)56 srcH, srcW, _ = srcImg.shape57 srcImg = np.pad(srcImg, ((1, 1), (1, 1), (0, 0)), 'edge')58 dstImg = np.zeros((dstH, dstW, 3), dtype=np.uint8)59for dstY in range(dstH):60for dstX in range(dstW):61for channel in [0, 1, 2]:62# p11 p12 p1363#64# p21 p22 p2365# p66# p31 p32 p3367# 储存为 p(y, x)68 p = [dstY * srcH / dstH, dstX * srcW / dstW]69 p22 = [math.floor(p[0]), math.floor(p[1])]70 p21 = [p22[0], p22[1] - 1]71 p23 = [p22[0], p22[1] + 1]7273 p11 = [p21[0] - 1, p21[1]]74 p12 = [p11[0], p22[1]]75 p13 = [p11[0], p23[1]]7677 p31 = [p21[0] + 1, p21[1]]78 p32 = [p31[0], p22[1]]79 p33 = [p31[0], p23[1]]8081 diff_y, diff_x = p[0] - p22[0], p[1] - p22[1]82 r1 = grangeInterpolation3(diff_x, srcImg[p11[0], p11[1], channel], srcImg[p12[0], p12[1], channel], srcImg[p13[0], p13[1], channel])83 r2 = grangeInterpolation3(diff_x, srcImg[p21[0], p21[1], channel], srcImg[p22[0], p22[1], channel], srcImg[p23[0], p23[1], channel])84 r3 = grangeInterpolation3(diff_x, srcImg[p31[0], p31[1], channel], srcImg[p32[0], p32[1], channel], srcImg[p33[0], p33[1], channel]) 8586 c = grangeInterpolation3(diff_y, r1, r2, r3)8788 dstImg[dstY, dstX, channel] = np.clip(c, 0, 255)89return dstImg9091def biLag4_4(self, srcImg, dstH, dstW):92 dstH, dstW = int(dstH), int(dstW)93 srcH, srcW, _ = srcImg.shape94 srcImg = np.pad(srcImg, ((1, 2), (1, 2), (0, 0)), 'edge')95 dstImg = np.zeros((dstH, dstW, 3), dtype=np.uint8)96for dstY in range(dstH):97for dstX in range(dstW):98for channel in [0, 1, 2]:99# p11 p12 p13 p14100#101# p21 p22 p23 p24102# p103# p31 p32 p33 p34104#105# p41 p42 p43 p44106# 储存为 p(y, x)107 p = [dstY * srcH / dstH, dstX * srcW / dstW]108 p22 = [math.floor(p[0]), math.floor(p[1])]109 p21 = [p22[0], p22[1] - 1]110 p23 = [p22[0], p22[1] + 1]111 p24 = [p22[0], p22[1] + 2]112113 p11 = [p21[0] - 1, p21[1]]114 p12 = [p11[0], p22[1]]115 p13 = [p11[0], p23[1]]116 p14 = [p11[0], p24[1]]117118 p31 = [p21[0] + 1, p21[1]]119 p32 = [p31[0], p22[1]]120 p33 = [p31[0], p23[1]]121 p34 = [p31[0], p24[1]]122123 p41 = [p21[0] + 2, p21[1]]124 p42 = [p41[0], p22[1]]125 p43 = [p41[0], p23[1]]126 p44 = [p41[0], p24[1]]127128 diff_y, diff_x = p[0] - p22[0], p[1] - p22[1]129 r1 = grangeInterpolation4(diff_x, srcImg[p11[0], p11[1], channel], srcImg[p12[0], p12[1], channel], srcImg[p13[0], p13[1], channel], srcImg[p14[0], p14[1], channel]) 130 r2 = grangeInterpolation4(diff_x, srcImg[p21[0], p21[1], channel], srcImg[p22[0], p22[1], channel], srcImg[p23[0], p23[1], channel], srcImg[p24[0], p24[1], channel]) 131 r3 = grangeInterpolation4(diff_x, srcImg[p31[0], p31[1], channel], srcImg[p32[0], p32[1], channel], srcImg[p33[0], p33[1], channel], srcImg[p34[0], p34[1], channel]) 132 r4 = grangeInterpolation4(diff_x, srcImg[p41[0], p41[1], channel], srcImg[p42[0], p42[1], channel], srcImg[p43[0], p43[1], channel], srcImg[p44[0], p44[1], channel]) 133134 c = grangeInterpolation4(diff_y, r1, r2, r3, r4)135136 dstImg[dstY, dstX, channel] = np.clip(c, 0, 255)137return dstImg三次卷积插值法（Cubic Convolution Interpolation）使⽤上图中的卷积核进⾏加权平均计算，卷积核为u(s)，四个等距（距离为1）的采样点记为x_0、x_1、x_2和x_3，采样数值记为y_0、y_1、y_2和y_3，且保证四个点均在[-2,2]区间上，计算得到g(x)，假设y_1、y_2、y_3和y_4分别为1、2、-1、4，则可得三次卷积插值函数如下图所⽰，待插值点横坐标范围为[0,1]公式推导设u(s)=\begin{cases} A_1|s|^3+B_1|s|^2+C_1|s|+D_1, &0<|s|<1 \\ A_2|s|^3+B_2|s|^2+C_2|s|+D_2, &1<|s|<2 \\ 1, &s=0 \\ 0, &otherwise \end{cases}\because函数在s=0,1,2处连续\therefore\begin{cases} 1=u(0^+)=D_1 \\ 0=u(1^-)=A_1+B_1+C_1+D_1 \\ 0=u(1^+)=A_2+B_2+C_2+D_2 \\ 0=u(2^-)=8A_2+4B_2+2C_2+D_2 \end{cases} (1)\because函数在s=0,1,2处导函数连续\therefore\begin{cases} u'(0^-)=u'(0+) \\ u'(1^-)=u'(1+) \\ u'(2^-)=u'(2+)\end{cases} \Rightarrow \begin{cases} -C_1=C_1 \\ 3A_1+2B_1+C_1=3A_2+2B_2+C_2\\ 12A_2+4B_2+C+2=0 \end{cases} ~~~~ (2)联⽴⽅程组(1)(2)，设A_2=a，解得\begin{cases} A_1=a+2 \\ B_1=-(a+3) \\ C_1=0 \\ D_1=1 \\ A_2=a \\ B_2=-5a \\ C_2=8a \\ D_2=-4a \end{cases}\Rightarrow u(s)=\begin{cases} (a+2)|s|^3-(a+3)|s|^2+1, &0<|s|<1 \\ A_2|s|^3+B_2|s|^2+C_2|s|+D_2, &1<|s|<2\\ 1, &s=0 \\ 0, &otherwise \end{cases}\because g(x)=\sum_kC_ku(s+j-k), ~~~~k=j-1,j, j+1,j+2且0<s<1⼜\because \begin{cases}\begin{align} u(s+1)&=as^3-2as^2+as \nonumber \\ u(s)&=(a+2)s^3-(a+3)s^2+1 \nonumber \\ u(s-1)&=-(a+2)s^3+(2a+3)s^2-as \nonumber \\ u(s-2)&=-as^3+as^2 \nonumber \end{align}\end{cases}\begin{align} \therefore g(x) &= C_{j-1}u(s+1)+C_{j}u(s)+C_{j+1}u(s-1)+C_{j+2}u(s-2) \nonumber \\ &= C_{j-1}(as^3-2as^2+as)+C_j[(a+2)s^3-(a+3)s^2+1]+C_{j+1}[-(a+2)s^3+ (2a+3)s^2-as]+C_{j+2}[-a^3+as^2] \nonumber \\ &= s^3[aC_{j-1}+(a+2)C_j-(a+2)C_{j+1}-aC_{j+2}]+s^2[-2aC_{j-1}-(a+3)C_j+(2a+3)C_{j+1}+aC_{j+2}]+s[aC_{j-1}-aC_{j+1}]+C_j \nonumber \end{align} ~~(3)f在x_j处泰勒展开得到f(x)=f(x_j)+f'(x_j)(x-x_j)+\frac{1}{2}f''(x_j)(x-x_j)^2+\cdots\therefore \begin{cases} f(x_{j+1})=f(x_j)+f'(x_j)(x_{j+1}-x_j)+\frac{1}{2}f''(x_j)(x_{j+1}-x_j)^2+\cdots \\ f(x_{j+2})=f(x_j)+f'(x_j)(x_{j+2}-x_j)+\frac{1}{2}f''(x_j)(x_{j+2}-x_j)^2+\cdots \\ f(x_{j-1})=f(x_j)+f'(x_j)(x_{j-1}-x_j)+\frac{1}{2}f''(x_j)(x_{j-1}-x_j)^2+\cdots \end{cases}令x_{j+1}-x_j=h\because x_{i+1}=x_i+1\therefore x_{j+2}-x_j=2h，x_{j-1}-x_j=-h\therefore \begin{cases} f(x_{j+2})=f(x_j)+2f'(x_j)h+2f''(x_j)h^2+\cdots \\ f(x_{j+1})=f(x_j)+f'(x_j)h+\frac{1}{2}f''(x_j)h^2+\cdots \\ f(x_{j-1})=f(x_j)-f'(x_j)h+\frac{1}{2}f''(x_j)h^2+\cdots \end{cases}\therefore \begin{cases} c_{j-1}=f(x_j)-f'(x_j)h+\frac{1}{2}f''(x_j)h^2+o(h^3) \\ c_j=f(x_j) \\ c_{j+1}=f(x_j)+f'(x_j)h+\frac{1}{2}f''(x_j)h^2+o(h^3)\\ c_{j+2}=f(x_j)+2f'(x_j)h+2f''(x_j)h^2+o(h^3) \end{cases} ~~ (4)将(4)代⼊(3)，得g(x)=-(2a+1)[2hf'(x_j)+h^2f''(x_j)]s^3+[(6a+3)hf'(x_j)+\frac{4a+3}{2}h^2f''(x_j)]s^2-2ahf'(x_j)s+f(x_j)+o(h^3)\because h=1,s=x-x_J\therefore sh=x-x_j\begin{align}\therefore f(x)&= f(x_j)+f'(x_j)(x-x_j)+\frac{1}{2}f''(x_j)(x-x_j)^2+o(h^3) \nonumber \\ &= f(x_j)+f'(x_j)sh+\frac{1}{2}f''(x_j)s^2h^2+o(h^3) \nonumber \end{align}\therefore f(x)-g(x)=(2a+1)[2hf'(x_j)+h^2f''(x_j)]s^3-(2a+1)[3hf'(x_j)+h^2f''(x_j)]s^2+[(2a+1)hf'(x_j)]s+o(h^3)\because 期望f(x)-g(x)趋于0\therefore 2a+1=0 \Rightarrow a=-\frac{1}{2}\therefore u(s)=\begin{cases} \frac{3}{2}|s|^3-\frac{5}{2}|s|^2+1, &0<|s|<1 \\ -\frac{1}{2}|s|^3+\frac{5}{2}|s|^2-4|s|+2, &1<|s|<2 \\ 1, &s=0 \\ 0, &otherwise \end{cases}\therefore g(s)=s^3[-\frac{1}{2}c_{j-1}+\frac{3}{2}c_j-\frac{3}{2}c_{j+1}+\frac{1}{2}c_{j+2}]+s^2[c_{j-1}-\frac{5}{2}c_j+2c_{j+1}-\frac{1}{2}c_{j+2}]+s[-\frac{1}{2}c_{j-1}+\frac{1} {2}c_{j+1}]+c_j图像插值p_{11}p_{12}p_{13}p_{14}p_{21}p_{22}p_{23}p_{24}pp_{31}p_{32}p_{33}p_{34}p_{41}p_{42}p_{43}p_{44}python代码1class BiCubicConvInterpolation:2 @staticmethod3def CubicConvInterpolation1(p0, p1, p2, p3, s):4# ⽤g（s）公式计算，已经将四个u（s）计算完毕并整理5# as^3 + bs^2 + cs + d6 a = 0.5 * (-p0 + 3.0 * p1 - 3.0 * p2 + p3)7 b = 0.5 * (2.0 * p0 - 5.0 * p1 + 4.0 * p2 - p3)8 c = 0.5 * (-p0 + p2)9 d = p110return d + s * (c + s * (b + s * a))1112 @staticmethod13def CubicConvInterpolation2(s):14# ⽤u（s）公式计算15 s = abs(s)16if s <= 1:17return 1.5 * s ** 3 - 2.5 * s ** 2 + 118elif s <= 2:19return -0.5 * s ** 3 + 2.5 * s ** 2 - 4 * s + 220else:21return 02223def biCubic1(self, srcImg, dstH, dstW):24# p11 p12 p13 p1425#26# p21 p22 p23 p2427# p28# p31 p32 p33 p3429#30# p41 p42 p43 p4431 dstH, dstW = int(dstH), int(dstW)32 scrH, scrW, _ = srcImg.shape33 srcImg = np.pad(srcImg, ((1, 1), (1, 1), (0, 0)), 'edge')34 dstImg = np.zeros((dstH, dstW, 1), dtype=np.uint8)35for dstY in range(dstH):36for dstX in range(dstW):37for channel in [0]:38 y = dstY * scrH / dstH39 x = dstX * scrW / dstW40 y1 = math.floor(y)41 x1 = math.floor(x)4243 array = []44for i in [-1, 0, 1, 2]:45 temp = self.CubicConvInterpolation1(srcImg[y1 + i, x1 - 1, channel],46 srcImg[y1 + i, x1, channel],47 srcImg[y1 + i, x1 + 1, channel],48 srcImg[y1 + i, x1 + 2, channel],49 x - x1)50 array.append(temp)5152 temp = self.CubicConvInterpolation1(array[0], array[1], array[2], array[3], y - y1)53 dstImg[dstY, dstX, channel] = np.clip(temp, 0, 255)5455return dstImg5657def biCubic2(self, srcImg, dstH, dstW):58# p11 p12 p13 p1459#60# p21 p22 p23 p2461# p62# p31 p32 p33 p3463#64# p41 p42 p43 p4465 dstH, dstW = int(dstH), int(dstW)66 scrH, scrW, _ = srcImg.shape67 srcImg = np.pad(srcImg, ((1, 1), (1, 1), (0, 0)), 'edge')68 dstImg = np.zeros((dstH, dstW, 3), dtype=np.uint8)69for dstY in range(dstH):70for dstX in range(dstW):71for channel in [0, 1, 2]:72 y = dstY * scrH / dstH73 x = dstX * scrW / dstW74 y1 = math.floor(y)75 x1 = math.floor(x)7677 array = []78for i in [-1, 0, 1, 2]:79 temp = 080for j in [-1, 0, 1, 2]:81 temp += srcImg[y1 + i, x1 + j, channel] * self.CubicConvInterpolation2(x - (x1 + j))82 array.append(temp)8384 temp = 085for i in [-1, 0, 1, 2]:86 temp += array[i + 1] * self.CubicConvInterpolation2(y - (y1 + i))87 dstImg[dstY, dstX, channel] = np.clip(temp, 0, 255)8889return dstImg三次样条插值在n-1个区间上寻找n-1个三次曲线，使其满⾜相邻曲线在端点处值相等、⼀阶导数相等，⼆阶导数相等，在加以边界条件后可得每个曲线的⽅程，然后沿x轴依次偏移对应的距离即可得到插值结果，如仅需要特定范围内的结果，则可以⼤幅减少计算量公式推导设S_i(x)=a_i+b_i(x-x_i)+c_i(x-x_i)^2+d_i(x-x_i)^3, ~~~~i=0,1,...,n-1则 \begin{cases} S_i'(x)=b_i+2c_i(x-x_i)+3d_i(x-x_i)^2\\ S_i''(x)=2c_i+6d_i(x-x_i)\\ S_i'''(x)=6d_i\\ \end{cases} ~~~~i=0,1,...,n-1设h_i(x)=x_{i+1}-x_i，可得\begin{cases} S_i(x)=a_i+b_ih_i+c_ih_i^2+d_ih_i^3\\ S_i'(x)=b_i+2c_ih_i+3d_ih_i^2\\ S_i''(x)=2c_i+6d_ih_i\\ S_i'''(x)=6d_i\\ \end{cases} ~~~~i=0,1,...,n-1\because S_i(x)过点(x_i,y_i)\therefore S_i(x)=a_i=y+i, ~~~~i=0,1,...,n-1 ~~~~~~(1)\because S_i(x)与S_{i+1}(x)在X_{i+1}处相等\therefore S_i(x_{i+1})=S_{i+1}(x_{i+1})\Rightarrow a_i+b_ih_i+c_ih_i^2+d_ih_i^3=y_{i+1}, ~~~~i=0,1,...,n-2~~~~~~(2)\because S_i'(x)与S_{i+1}'(x)在X_{i+1}处相等\therefore S_i'(x)-S_{i+1}'(x)=0\Rightarrow b_i+2c_ih_i+3d_ih_i^2-b_{i+1}=0~~~~~~(3)\because S_i''(x)与S_{i+1}''(x)在X_{i+1}处相等\therefore S_i''(x)-S_{i+1}''(x)=0\Rightarrow 2c_i+6d_ih_i-2c_{i+1}=0, ~~~~i=0,1,...,n-2~~~~~~(4)设m_i=S_i(x_i)=2c_i，即c_i=\frac{1}{2}m_i, ~~~~i=0,1,...,n-1~~~~~~(5)将(5)代⼊(4)，得2c_i+6d_ih_i-2c_{i+1}=0\Rightarrow m_i+6h_id_i-m_{i+1}=0\Rightarrow d_i=\frac{m_{i+1}-m_i}{6h_i}, ~~~~i=0,1,...,n-2~~~~~~(6)将(1)(5)(6)代⼊(2)，得\begin{align} &a_i+b_ih_i+c_ih_i^2+d_ih_i^3=y_{i+1} \nonumber \\ \Rightarrow&y_i+b_ih_i+\frac{1}{2}m_ih_i^2+\frac{m_{i+1}-m_i}{6h_i}h_i^3=y_{i+1} \nonumber \\\Rightarrow&b_i=\frac{y_{i+1}-y_i}{h_i}-\frac{1}{2}m_ih_i-\frac{1}{6}(m_{i+1}-m_i)h_i \nonumber \\ \Rightarrow&b_i=\frac{y_{i+1}-y_i}{h_i}-\frac{1}{3}m_ih_i-\frac{1}{6}m_{i+1}h_i, ~~~~i=0,1,...,n-2~~~~~~(7) \nonumber \end{align}将(5)(6)(7)代⼊(3)，得\begin{align} &\frac{y_{i+1}-y{i}}{h_i}-\frac{1}{3}m_ih_i-\frac{1}{6}m_{i+1}h_i+2\cdot\frac{1}{2}m_ih_i+3\frac{m_{i+1}-m_i}{6h_i}h_i^2-(\frac{y_{i+2}-y_{i+1}}{h_{i+1}}-\frac{1}{3}m_{i+1}h_{i+1}-\frac{1}{6}m_{i+2}h_{i+1})=0 \nonumber \\ \Rightarrow&\frac{y_{i+1}-y{i}}{h_i}-\frac{1}{3}m_ih_i-\frac{1}{6}m_{i+1}h_i+m_ih_i+\frac{1}{2}(m_{i+1}-m_i)h_i-\frac{y_{i+2}-y_{i+1}}{h_{i+1}}+\frac{1}{3}m_{i+1}h_{i+1}+\frac{1}{6}m_{i+2}h_{i+1}=0 \nonumber \\ \Rightarrow&m_ih_i(-\frac{1}{3}+1-\frac{1}{2})+m_{i+1}h_i(-\frac{1}{6}+\frac{1} {2})+\frac{1}{3}m_{i+1}h_{i+1}+\frac{1}{6}m_{i+2}h_{i+1}=\frac{y_{i+2}-y_{i+1}}{h_{i+1}}-\frac{y_{i+1}-y_{i}}{h_{i}} \nonumber \\ \Rightarrow&\frac{1}{6}(m_ih_i+2m_{i+1}h_i+2m_{i+1}h_{i+1}+m_{i+2}h_{i+1})=\frac{y_{i+2}-y_{i+1}}{h_{i+1}}-\frac{y_{i+1}-y_{i}}{h_{i}} \nonumber \\ \Rightarrow&m_ih_i+2m_{i+1}(h_i+h_{i+1})+m_{i+2}h_{i+1}=6(\frac{y_{i+2}-y_{i+1}}{h_{i+1}}-\frac{y_{i+1}-y_{i}}{h_{i}}), ~~~~i=0,1,...,n-2~~~~~~(8) \nonumber \end{align}由(8)可知i=0,1,...,n-2，则有m_0,m_1,...,m_n，需要两个额外条件⽅程组才有解⾃然边界(Natural)m_0=0，m_n=0\begin{bmatrix} \tiny 1 & 0 & 0 & 0 & 0 & \cdots & 0\\ h_0 & 2(h_0+h_1) & h_1 & 0 & 0 & \cdots & 0\\ 0 & h_1 & 2(h_1+h_2) & h_2 & 0 & \cdots & 0\\ 0 & 0 & h_2 & 2(h_2+h_3) & h_3 & \cdots & 0\\ \vdots& & & \ddots & \ddots & \ddots & \vdots\\ 0 & \cdots & & & h_{n-2} & 2(h_{n-2}+h_{n-1}) & h_{n-1}\\ 0 & \cdots & & & 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} m_0\\m_1\\m_2\\m_3\\\vdots\\m_{n-1}\\m_n \end{bmatrix}=6\begin{bmatrix} 0\\ \frac{y_2-y_1}{h_1}-\frac{y_1-y_0}{h_0}\\ \frac{y_3-y_2}{h_2}-\frac{y_2-y_1}{h_1}\\ \frac{y_4-y_3}{h_3}-\frac{y_3-y_2}{h_2}\\ \vdots\\ \frac{y_n-y_{n-1}}{h_{n-1}}-\frac{y_{n-1}-y_{n-2}}{h_{n-2}}\\ 0 \end{bmatrix}固定边界(Clamped)\begin{align} &\begin{cases} S_0'(x_0)=A\\ S_{n-1}'(x_n)=B \end{cases} \nonumber \\ \Rightarrow&\begin{cases} b_0=A\\ b_{n-1}+2c_{n-1}h_{n-1}+3d_{n-1}h_{n-1}^2=B\end{cases} \nonumber \\ \Rightarrow&\begin{cases} A=\frac{y_1-y_0}{h_0}-\frac{h_0}{2}m_0-\frac{h_0}{6}(m_1-m_0)\\ B=\frac{y_n-y_{n-1}}{h_{n-1}}-\frac{1}{3}m_{n-1}h_{n-1}+m_{n-1}h_{n-1}+\frac{1}{2}m_nh_{n-1}-\frac{1}{2}m_{n-1}h_{n-1} \end{cases} \nonumber \\ \Rightarrow&\begin{cases} 2h_0m_0+h_0m_1=6(\frac{y_1-y_0}{h_0}-A)\\ h_{n-1}m_{n-1}+2h_{n-1}m_{n}=6(B-\frac{y_n-y_{n-1}}{h_{n-1}}) \end{cases} \nonumber \\ \end{align}\begin{bmatrix} 2 & 1 & 0 & 0 & 0 & \cdots & 0\\ h_0 & 2(h_0+h_1) & h_1 & 0 & 0 & \cdots & 0\\ 0 & h_1 & 2(h_1+h_2) & h_2 & 0 & \cdots & 0\\ 0 & 0 & h_2 & 2(h_2+h_3) & h_3 & \cdots & 0\\ \vdots& & & \ddots & \ddots & \ddots & \vdots\\ 0 & \cdots & & & h_{n-2} & 2(h_{n-2}+h_{n-1}) & h_{n-1}\\ 0 & \cdots & & & 0 & 1 & 2 \end{bmatrix}\begin{bmatrix} m_0\\m_1\\m_2\\m_3\\\vdots\\m_{n-1}\\m_n \end{bmatrix}=6\begin{bmatrix} \frac{y_1-y_0}{h_0}-A\\ \frac{y_2-y_1}{h_1}-\frac{y_1-y_0}{h_0}\\ \frac{y_3-y_2}{h_2}-\frac{y_2-y_1}{h_1}\\ \frac{y_4-y_3}{h_3}-\frac{y_3-y_2}{h_2}\\ \vdots\\\frac{y_n-y_{n-1}}{h_{n-1}}-\frac{y_{n-1}-y_{n-2}}{h_{n-2}}\\ B-\frac{y_n-y_{n-1}}{h_{n-1}} \end{bmatrix}⾮节点边界(Not-A-Knot)\begin{align} &\begin{cases} S_0'''(x_1)=S_1'''(x_1)\\ S_{n-2}'''(x_{n-1})=S_{n-1}'''(x_{n-1}) \end{cases} \nonumber \\ \Rightarrow&\begin{cases} 6\cdot\frac{m_1-m_0}{6h_0}=6\cdot\frac{m_2-m_1}{6h_1}\\ 6\cdot\frac{m_{n-1}-m_{n-2}}{6h_{n-2}}=6\cdot\frac{m_n-m_{n-1}}{6h_{n-1}} \end{cases} \nonumber \\ \Rightarrow&\begin{cases} h_1(m_1-m_0)=h_0(m_2-m_1)\\ h_{n-1}(m_{n-1}-m_{n-2})=h_{n-2}(m_n-m_{n-1}) \end{cases} \nonumber \\ \Rightarrow&\begin{cases} -h_1m_0+(h_1+h_0)m_1-h_0m_2=0\\ -h_{n-1}m_{n-2}+(h_{n-1}+h_{n-2})m_{n-1}-h_{n-2}m_n=0 \end{cases} \nonumber \\ \end{align}\begin{bmatrix} -h_1 & h_0+h_1 & -h_0 & 0 & 0 & \cdots & 0\\ h_0 & 2(h_0+h_1) & h_1 & 0 & 0 & \cdots & 0\\ 0 & h_1 & 2(h_1+h_2) & h_2 & 0 & \cdots & 0\\ 0 & 0 & h_2 &2(h_2+h_3) & h_3 & \cdots & 0\\ \vdots& & & \ddots & \ddots & \ddots & \vdots\\ 0 & \cdots & & & h_{n-2} & 2(h_{n-2}+h_{n-1}) & h_{n-1}\\ 0 & \cdots & & & -h_{n-1} & h_{n-1}+h_{n-2} & -h_{n-2} \end{bmatrix}\begin{bmatrix} m_0\\m_1\\m_2\\m_3\\\vdots\\m_{n-1}\\m_n \end{bmatrix}=6\begin{bmatrix} 0\\ \frac{y_2-y_1}{h_1}-\frac{y_1-y_0}{h_0}\\ \frac{y_3-y_2}{h_2}-\frac{y_2-y_1}{h_1}\\ \frac{y_4-y_3}{h_3}-\frac{y_3-y_2}{h_2}\\ \vdots\\ \frac{y_n-y_{n-1}}{h_{n-1}}-\frac{y_{n-1}-y_{n-2}}{h_{n-2}}\\ 0 \end{bmatrix}在n=4时通⽤公式⾃然边界\begin{bmatrix} 1 & 0 & 0 & 0 \\ h_0 & 2(h_0+h_1) & h_1 & 0 \\ 0 & h_1 & 2(h_1+h_2) & h_2 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}\begin{bmatrix} m_0\\m_1\\m_2\\m_3 \end{bmatrix}=6\begin{bmatrix} 0\\ \frac{y_2-y_1}{h_1}-\frac{y_1-y_0}{h_0}\\ \frac{y_3-y_2}{h_2}-\frac{y_2-y_1}{h_1}\\ 0 \end{bmatrix}固定边界\begin{bmatrix} 2 & 1 & 0 & 0 \\ h_0 & 2(h_0+h_1) & h_1 & 0 \\ 0 & h_1 & 2(h_1+h_2) & h_2 \\ 0 & 0 & 1 & 2 \\ \end{bmatrix}\begin{bmatrix} m_0\\m_1\\m_2\\m_3 \end{bmatrix}=6\begin{bmatrix} \frac{y_1-y_0}{h_0}-A\\ \frac{y_2-y_1}{h_1}-\frac{y_1-y_0}{h_0}\\ \frac{y_3-y_2}{h_2}-\frac{y_2-y_1}{h_1}\\ B-\frac{y_3-y_2}{h_2} \end{bmatrix}⾮节点边界\begin{bmatrix} -h_1 & h_0+h_1 & -h_0 & 0 \\ h_0 & 2(h_0+h_1) & h_1 & 0 \\ 0 & h_1 & 2(h_1+h_2) & h_2 \\ 0 & -h_2 & h_1+h_2 & -h_1 \\ \end{bmatrix}\begin{bmatrix} m_0\\m_1\\m_2\\m_3 \end{bmatrix}=6\begin{bmatrix} 0\\ \frac{y_2-y_1}{h_1}-\frac{y_1-y_0}{h_0}\\ \frac{y_3-y_2}{h_2}-\frac{y_2-y_1}{h_1}\\ 0 \end{bmatrix}图像插值x_{i+1}-x_i=1 \Rightarrow h_i(x)=1在n=4时，即四个点时如下所⽰p_{11}p_{12}p_{13}p_{14}p_{21}p_{22}p_{23}p_{24}pp_{31}p_{32}p_{33}p_{34}p_{41}p_{42}p_{43}p_{44}⾃然边界（可⽤TDMA或化简计算）\begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 4 & 1 & 0 \\ 0 & 1 & 4 & 1 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}\begin{bmatrix} m_0\\m_1\\m_2\\m_3 \end{bmatrix}=6\begin{bmatrix} 0\\ y_0+y_2-2y_1\\ y_1+y_3-2y_2\\ 0 \end{bmatrix}固定边界（只能⽤TDMA计算）\begin{bmatrix} 2 & 1 & 0 & 0 \\ 1 & 4 & 1 & 0 \\ 0 & 1 & 4 & 1 \\ 0 & 0 & 1 & 2 \\ \end{bmatrix}\begin{bmatrix} m_0\\m_1\\m_2\\m_3 \end{bmatrix}=6\begin{bmatrix} y_1-y_0-A\\ y_0+y_2-2y_1\\ y_1+y_3-2y_2\\ y_2-y_3+B \end{bmatrix}⾮节点边界（只能化简计算）\begin{bmatrix} -1 & 2 & -1 & 0 \\ 1 & 4 & 1 & 0 \\ 0 & 1 & 4 & 1 \\ 0 & -1 & 2 & -1 \\ \end{bmatrix}\begin{bmatrix} m_0\\m_1\\m_2\\m_3 \end{bmatrix}=6\begin{bmatrix} 0\\ y_0+y_2-2y_1\\ y_1+y_3-2y_2\\ 0 \end{bmatrix}python代码1class BiSplineInterpolation:2 @staticmethod3 def TDMA(a, b, c, d):4 n = len(d)56 c[0] = c[0] / b[0]7 d[0] = d[0] / b[0]89for i in range(1, n):10 coef = 1.0 / (b[i] - a[i] * c[i - 1])11 c[i] = coef * c[i]12 d[i] = coef * (d[i] - a[i] * d[i - 1])1314for i in range(n - 2, -1, -1):15 d[i] = d[i] - c[i] * d[i + 1]1617return d1819 @staticmethod20 def Simplified_Natural4(y1, y2, y3, y4):21 # 四点⾃然边界化简公式22 d1 = y1 + y3 - 2 * y223 d2 = y2 + y4 - 2 * y32425 k0 = 026 k1 = (4 * d1 - d2) * 0.427 k2 = (4 * d2 - d1) * 0.428 k3 = 02930return [k0, k1, k2, k3]3132 @staticmethod33 def Simplified_Not_A_Knot4(y1, y2, y3, y4):34 # 四点⾮节点边界化简公式35 d1 = y1 + y3 - 2 * y236 d2 = y2 + y4 - 2 * y33738 k0 = 2 * d1 - d239 k1 = d140 k2 = d241 k3 = 2 * d2 - d14243return [k0, k1, k2, k3]4445 # TDMA矩阵说明46 # a0 和 c3 没有实际意义，占位⽤47 # a0 [b0 c0 00 ] [x0] [d0]48 # [a1 b1 c1 0 ] [x1] = [d1]49 # [0 a2 b2 c2] [x2] [d2]50 # [00 a3 b3] c3 [x3] [d3]5152 def SplineInterpolationNatural4(self, x, y1, y2, y3, y4):53 # ⽤TDMA计算54 # matrix_a = [0, 1, 1, 0]55 # matrix_b = [1, 4, 4, 1]56 # matrix_c = [0, 1, 1, 0]57 # matrix_d = [0, 6 * (y1 + y3 - 2 * y2), 6 * (y2 + y4 - 2 * y3), 0]58 # matrix_x = self.TDMA(matrix_a, matrix_b, matrix_c, matrix_d)5960 # 化简计算61 matrix_x = self.Simplified_Natural4(y1, y2, y3, y4)6263 a = y264 b = y3 - y2 - matrix_x[1] / 3.0 - matrix_x[2] / 6.065 c = matrix_x[1] / 2.066 d = (matrix_x[2] - matrix_x[1]) / 6.06768 s = a + b * x + c * x * x + d * x * x * x69return s7071 def SplineInterpolationClamped4(self, x, y1, y2, y3, y4):72 # 仅有TDMA计算，⽆法化简73 A, B = 1, 17475 matrix_a = [0, 1, 1, 1]76 matrix_b = [2, 4, 4, 2]77 matrix_c = [1, 1, 1, 0]78 matrix_d = [6 * (y2 - y1 - A), 6 * (y1 + y3 - 2 * y2), 6 * (y2 + y4 - 2 * y3), 6 * (B - y4 + y3)]79 matrix_x = self.TDMA(matrix_a, matrix_b, matrix_c, matrix_d)8081 a = y282 b = y3 - y2 - matrix_x[1] / 3.0 - matrix_x[2] / 6.083 c = matrix_x[1] / 2.084 d = (matrix_x[2] - matrix_x[1]) / 6.08586 s = a + b * x + c * x * x + d * x * x * x87return s8889 def SplineInterpolationNotAKnot4(self, x, y1, y2, y3, y4):90 # ⽆法使⽤TDMA计算91 matrix_x = self.Simplified_Not_A_Knot4(y1, y2, y3, y4)9293 a = y294 b = y3 - y2 - matrix_x[1] / 3.0 - matrix_x[2] / 6.095 c = matrix_x[1] / 2.096 d = (matrix_x[2] - matrix_x[1]) / 6.09798 s = a + b * x + c * x * x + d * x * x * x99return s100101 def biSpline4(self, srcImg, dstH, dstW):102 dstH, dstW = int(dstH), int(dstW)103 srcH, srcW, _ = srcImg.shape104 srcImg = np.pad(srcImg, ((1, 2), (1, 2), (0, 0)), 'edge')105 dstImg = np.zeros((dstH, dstW, 3), dtype=np.uint8)106for dstY in range(dstH):107for dstX in range(dstW):108for channel in [0, 1, 2]:109 # p11 p12 p13 p14110 #111 # p21 p22 p23 p24112 # p113 # p31 p32 p33 p34114 #115 # p41 p42 p43 p44116 # 储存为 p(y, x)117 p = [dstY * srcH / dstH, dstX * srcW / dstW]118 p22 = [math.floor(p[0]), math.floor(p[1])]119 p21 = [p22[0], p22[1] - 1]120 p23 = [p22[0], p22[1] + 1]121 p24 = [p22[0], p22[1] + 2]122123 p11 = [p21[0] - 1, p21[1]]124 p12 = [p11[0], p22[1]]125 p13 = [p11[0], p23[1]]126 p14 = [p11[0], p24[1]]127128 p31 = [p21[0] + 1, p21[1]]129 p32 = [p31[0], p22[1]]130 p33 = [p31[0], p23[1]]131 p34 = [p31[0], p24[1]]132133 p41 = [p21[0] + 2, p21[1]]134 p42 = [p41[0], p22[1]]135 p43 = [p41[0], p23[1]]136 p44 = [p41[0], p24[1]]137138 diff_y, diff_x = p[0] - p22[0], p[1] - p22[1]139 r1 = self.SplineInterpolationNatural4(diff_x, srcImg[p11[0], p11[1], channel], srcImg[p12[0], p12[1], channel], srcImg[p13[0], p13[1], channel], srcImg[p14[0], p14[1], channel]) 140 r2 = self.SplineInterpolationNatural4(diff_x, srcImg[p21[0], p21[1], channel], srcImg[p22[0], p22[1], channel], srcImg[p23[0], p23[1], channel], srcImg[p24[0], p24[1], channel]) 141 r3 = self.SplineInterpolationNatural4(diff_x, srcImg[p31[0], p31[1], channel], srcImg[p32[0], p32[1], channel], srcImg[p33[0], p33[1], channel], srcImg[p34[0], p34[1], channel]) 142 r4 = self.SplineInterpolationNatural4(diff_x, srcImg[p41[0], p41[1], channel], srcImg[p42[0], p42[1], channel], srcImg[p43[0], p43[1], channel], srcImg[p44[0], p44[1], channel]) 143144 c = self.SplineInterpolationNatural4(diff_y, r1, r2, r3, r4)145146 dstImg[dstY, dstX, channel] = np.clip(c, 0, 255)。

MATLAB 程序设计期中考查在许多问题中，通常根据实验、观测或经验得到的函数表或离散点上的信息，去研究分析函数的有关特性。

其中插值法是一种最基本的方法，以下给出最基本的插值问题——三次样条插值的基本提法：对插值区间[]b a ,进行划分：b x x x a n ≤<⋯⋯<<≤10，函数()x f y =在节点i x 上的值()()n i x f y i i ⋯⋯==,2,1,0，并且如果函数()x S 在每个小区间[]1,+i i x x 上是三次多项式，于[]b a ,上有二阶连续导数，则称()x S 是[]b a ,上的三次样条函数，如果()x S 在节点i x 上还满足条件()()n i y x S i i ⋯⋯==,1,0则称()x S 为三次样条插值函数。

三次样条插值问题提法：对[]b a ,上给定的数表如下.求一个分段三次多项式函数()x S 满足插值条件()()n i y x S i i ⋯⋯==,1,0 式，并在插值区间[]b a ,上有二阶连续导数。

这就需要推导三次样条插值公式：记()x f '在节点i x 处的值为()i i m x f ='(n i ⋯⋯=,1,0)(这不是给定插值问题数表中的已知值)。

在每个小区间[]1,+i i x x 利用三次Hermite 插值公式，得三次插值公式：()()()()1111+++++++=i i i i i i i i i m m x y x y x x S ββαα，[]1,+∈i i x x x 。

为了得到这个公式需要n 4个条件：(1).非端点处的界点有n 2个；(2).一阶导数连续有1-n 个条件；(3).二阶导数连续有1-n 个条件，其中边界条件：○1()()n n m x S m x S ='=' 00 ○2()()αα=''=''n x S x S 00 ○3()()()()16500403βααβαα=''+'=''+'n n x S x S x S x S○4n y y =0 ()()()()000000-''=+''-'=+'n nx S x S x S x S 其中：()⎩⎨⎧=≠=j i j i x j i,1,0α ()0='j i x α ()0=j i x β 且(1,0,=j i )。