三次樣條插值演算法c
㈠ Akima 插值和樣條插值的C語言源代碼，要有注釋。
Akima插值
附帶的圖片為運行結果
#include"stdio.h"
#include"math.h"
#include"interpolation.h"
voidinterpolation_akima(AKINTEPap){
intnum,k,kk,m,l;
doublepio,*mtr,*x,*y,u[5],p,q;
num=ap->n;k=ap->k;
pio=ap->t;mtr=ap->s;
x=ap->x;y=ap->y;
if(num<1){
return;
}
elseif(num==1){
mtr[0]=mtr[4]=y[0];
return;
}
elseif(num==2){
mtr[0]=y[0];
mtr[1]=(y[1]-y[0])/(x[1]-x[0]);
if(k<0)
mtr[4]=(y[0]*(pio-x[1])-y[1]*(pio-x[0]))/(x[0]-x[1]);
return;
}
if(k<0){
if(pio<=x[1])kk=0;
elseif(pio>=x[num-1])kk=num-2;
else{
kk=1;m=num;
while(((kk-m)!=1)&&((kk-m)!=-1)){
l=(kk+m)/2;
if(pio<x[l-1])m=l;
elsekk=l;
}
kk--;
}
}
elsekk=k;
if(kk>=num-1)kk=num-2;
u[2]=(y[kk+1]-y[kk])/(x[kk+1]-x[kk]);
if(num==3){
if(kk==0){
u[3]=(y[2]-y[1])/(x[2]-x[1]);
u[4]=2.0*u[3]-u[2];
u[1]=2.0*u[2]-u[3];
u[0]=2.0*u[1]-u[2];
}
else{
u[1]=(y[1]-y[0])/(x[1]-x[0]);
u[0]=2.0*u[1]-u[2];
u[3]=2.0*u[2]-u[1];
u[4]=2.0*u[3]-u[2];
}
}
else{
if(kk<=1){
u[3]=(y[kk+2]-y[kk+1])/(x[kk+2]-x[kk+1]);
if(kk==1){
u[1]=(y[1]-y[0])/(x[1]-x[0]);
u[0]=2.0*u[1]-u[2];
if(num==4)u[4]=2.0*u[3]-u[2];
elseu[4]=(y[4]-y[3])/(x[4]-x[3]);
}
else{
u[1]=2.0*u[2]-u[3];
u[0]=2.0*u[1]-u[2];
u[4]=(y[3]-y[2])/(x[3]-x[2]);
}
}
elseif(kk>=(num-3)){
u[1]=(y[kk]-y[kk-1])/(x[kk]-x[kk-1]);
if(kk==(num-3)){
u[3]=(y[num-1]-y[num-2])/(x[num-1]-x[num-2]);
u[4]=2.0*u[3]-u[2];
if(num==4)u[0]=2.0*u[1]-u[2];
elseu[0]=(y[kk-1]-y[kk-2])/(x[kk-1]-x[kk-2]);
}
else{
u[3]=2.0*u[2]-u[1];
u[4]=2.0*u[3]-u[2];
u[0]=(y[kk-1]-y[kk-2])/(x[kk-1]-x[kk-2]);
}
}
else{
u[1]=(y[kk]-y[kk-1])/(x[kk]-x[kk-1]);
u[0]=(y[kk-1]-y[kk-2])/(x[kk-1]-x[kk-2]);
u[3]=(y[kk+2]-y[kk+1])/(x[kk+2]-x[kk+1]);
u[4]=(y[kk+3]-y[kk+2])/(x[kk+3]-x[kk+2]);
}
}
mtr[0]=fabs(u[3]-u[2]);
mtr[1]=fabs(u[0]-u[1]);
if((fabs(mtr[0])<0.0000001)&&(fabs(mtr[1])<0.0000001))
p=(u[1]+u[2])/2.0;
elsep=(mtr[0]*u[1]+mtr[1]*u[2])/(mtr[0]+mtr[1]);
mtr[0]=fabs(u[3]-u[4]);
mtr[1]=fabs(u[2]-u[1]);
if((fabs(mtr[0])<0.0000001)&&(fabs(mtr[1])<0.0000001))
q=(u[2]+u[3])/2.0;
elseq=(mtr[0]*u[2]+mtr[1]*u[3])/(mtr[0]+mtr[1]);
mtr[0]=y[kk];
mtr[1]=p;
mtr[3]=x[kk+1]-x[kk];
mtr[2]=(3.0*u[2]-2.0*p-q)/mtr[3];
mtr[3]=(q+p-2.0*u[2])/(mtr[3]*mtr[3]);
if(k<0){
p=pio-x[kk];
mtr[4]=mtr[0]+mtr[1]*p+mtr[2]*p*p+mtr[3]*p*p*p;
}
return;
}
main()
{
doublex[11]={3.0,5.0,8.0,13.0,17.0,25.0,27.0,29.0,31.0,35.0,39.0};
doubley[11]={7.0,10.0,11.0,17.0,23.0,18.0,13.0,6.0,3.0,1.0,0.0};
AKINTEaa={11,x,y,-1,14.0,{0}};
AKINTEab={11,x,y,-1,28.0,{0}};
printf(" ");
interpolation_akima(&aa);
printf("x=%6.3f,f(x)=%e ",aa.t,aa.s[4]);
printf("mtr0=%e,mtr1=%e,mtr2=%e,mtr3=%e ",aa.s[0],aa.s[1],aa.s[2],aa.s[3]);
printf(" ");
interpolation_akima(&ab);
printf("x=%6.3f,f(x)=%e ",ab.t,ab.s[4]);
printf("mtr0=%e,mtr1=%e,mtr2=%e,mtr3=%e ",ab.s[0],ab.s[1],ab.s[2],ab.s[3]);
printf(" ");
}
三次樣條插值的實現
1、程序比較簡單的：
#include<iostream>
#include<iomanip>
usingnamespacestd;
constintMAX=50;
floatx[MAX],y[MAX],h[MAX];
floatc[MAX],a[MAX],fxym[MAX];
floatf(intx1,intx2,intx3){
floata=(y[x3]-y[x2])/(x[x3]-x[x2]);
floatb=(y[x2]-y[x1])/(x[x2]-x[x1]);
return(a-b)/(x[x3]-x[x1]);
}//求差分
voidcal_m(intn){//用追趕法求解出彎矩向量M……
floatB[MAX];
B[0]=c[0]/2;
for(inti=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];
}
voidprintout(intn);
intmain(){
intn,i;charch;
do{
cout<<"Pleaseputinthenumberofthedots:";
cin>>n;
for(i=0;i<=n;i++){
cout<<"PleaseputinX"<<i<<':';
cin>>x[i];//cout<<endl;
cout<<"PleaseputinY"<<i<<':';
cin>>y[i];//cout<<endl;
}
for(i=0;i<n;i++)//求步長
h[i]=x[i+1]-x[i];
cout<<"Please輸入邊界條件 1:已知兩端的一階導數 2:兩端的二階導數已知 默認:自然邊界條件 ";
intt;
floatf0,f1;
cin>>t;
switch(t){
case1:cout<<"PleaseputinY0'Y"<<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;
case2:cout<<"PleaseputinY0"Y"<<n<<"" ";
cin>>f0>>f1;
c[0]=a[n]=0;
fxym[0]=2*f0;fxym[n]=2*f1;
break;
default:cout<<"不可用 ";//待定
};//switch
for(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<<" 輸出三次樣條插值函數： ";
printout(n);
cout<<"Doyoutohaveanthertry?y/n:";
cin>>ch;
}while(ch=='y'||ch=='Y');
return0;
}
voidprintout(intn){
cout<<setprecision(6);
for(inti=0;i<n;i++){
cout<<i+1<<":["<<x[i]<<","<<x[i+1]<<"] "<<" ";
/*
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]<<") ";
cout<<endl;*/
floatt=fxym[i]/(6*h[i]);
if(t>0)cout<<t<<"*("<<x[i+1]<<"-x)^3";
elsecout<<-t<<"*(x-"<<x[i+1]<<")^3";
t=fxym[i+1]/(6*h[i]);
if(t>0)cout<<"+"<<t<<"*(x-"<<x[i]<<")^3";
elsecout<<"-"<<-t<<"*(x-"<<x[i]<<")^3";
cout<<" ";
t=(y[i]-fxym[i]*h[i]*h[i]/6)/h[i];
if(t>0)cout<<"+"<<t<<"*("<<x[i+1]<<"-x)";
elsecout<<"-"<<-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]<<")";
elsecout<<"-"<<-t<<"*(x-"<<x[i]<<")";
cout<<endl<<endl;
}
cout<<endl;
}
2、程序比較復雜的：
（程序前面的01.，02.，03.等等為語句編號，實際應用時請一一刪除）01./*=======================================================================*/
02.#include<stdio.h>
03.////////////////////////////////////////////////////////////////////////////////
04.#defineMAXNUM50//定義樣條數據區間個數最多為50個
05.typedefstructSPLINE//定義樣條結構體，用於存儲一條樣條的所有信息
06.{//初始化數據輸入
07.floatx[MAXNUM+1];//存儲樣條上的點的x坐標，最多51個點
08.floaty[MAXNUM+1];//存儲樣條上的點的y坐標，最多51個點
09.unsignedintpoint_num;//存儲樣條上的實際的點的個數
10.floatbegin_k1;//開始點的一階導數信息
11.floatend_k1;//終止點的一階導數信息
12.//floatbegin_k2;//開始點的二階導數信息
13.//floatend_k2;//終止點的二階導數信息
14.//計算所得的樣條函數S(x)
15.floatk1[MAXNUM+1];//所有點的一階導數信息
16.floatk2[MAXNUM+1];//所有點的二階導數信息
17.//51個點之間有50個段，func[]存儲每段的函數系數
18.floata3[MAXNUM],a1[MAXNUM];
19.floatb3[MAXNUM],b1[MAXNUM];
20.//分段函數的形式為Si(x)=a3[i]*{x(i+1)-x}^3+a1[i]*{x(i+1)-x}+
21.//b3[i]*{x-x(i)}^3+b1[i]*{x-x(i)}
22.//xi為x[i]的值，xi_1為x[i+1]的值
23.}SPLINE,*pSPLINE;
24.typedefintRESULT;//返回函數執行的結果狀態，下面為具體的返回選項
25.#ifndefTRUE
26.#defineTRUE1
27.#endif
28.#ifndefFALSE
29.#defineFALSE-1
30.#endif
31.#ifndefNULL
32.#defineNULL0
33.#endif
34.#ifndefERR
35.#defineERR-2
36.#endif
37.//////////////////////////////////////////////////////////////////////////////////
38./*===============================================================================
39.***函數名稱：Spline3()
40.***功能說明：完成三次樣條差值，其中使用追趕法求解M矩陣
41.***入口參數：(pSPLINE)pLine樣條結構體指針pLine中的x[],y[],num,begin_k1,end_k1
42.***出口參數：(pSPLINE)pLine樣條結構體指針pLine中的函數參數
43.***返回參數：返回程序執行結果的狀態TRUEorFALSE
44.================================================================================*/
45.RESULTSpline3(pSPLINEpLine)
46.{
47.floatH[MAXNUM]={0};//小區間的步長
48.floatFi[MAXNUM]={0};//中間量
49.floatU[MAXNUM+1]={0};//中間量
50.floatA[MAXNUM+1]={0};//中間量
51.floatD[MAXNUM+1]={0};//中間量
52.floatM[MAXNUM+1]={0};//M矩陣
53.floatB[MAXNUM+1]={0};//追趕法中間量
54.floatY[MAXNUM+1]={0};//追趕法中間變數
55.inti=0;
56.////////////////////////////////////////計算中間參數
57.if((pLine->point_num<3)||(pLine->point_num>MAXNUM+1))
58.{
59.returnERR;//輸入數據點個數太少或太多
60.}
61.for(i=0;i<=pLine->point_num-2;i++)
62.{//求H[i]
63.H[i]=pLine->x[i+1]-pLine->x[i];
64.Fi[i]=(pLine->y[i+1]-pLine->y[i])/H[i];//求F[x(i),x(i+1)]
65.}
66.for(i=1;i<=pLine->point_num-2;i++)
67.{//求U[i]和A[i]和D[i]
68.U[i]=H[i-1]/(H[i-1]+H[i]);
69.A[i]=H[i]/(H[i-1]+H[i]);
70.D[i]=6*(Fi[i]-Fi[i-1])/(H[i-1]+H[i]);
71.}
72.//若邊界條件為1號條件，則
73.U[i]=1;
74.A[0]=1;
75.D[0]=6*(Fi[0]-pLine->begin_k1)/H[0];
76.D[i]=6*(pLine->end_k1-Fi[i-1])/H[i-1];
77.//若邊界條件為2號條件，則
78.//U[i]=0;
79.//A[0]=0;
80.//D[0]=2*begin_k2;
81.//D[i]=2*end_k2;
82./////////////////////////////////////////追趕法求解M矩陣
83.B[0]=A[0]/2;
84.for(i=1;i<=pLine->point_num-2;i++)
85.{
86.B[i]=A[i]/(2-U[i]*B[i-1]);
87.}
88.Y[0]=D[0]/2;
89.for(i=1;i<=pLine->point_num-1;i++)
90.{
91.Y[i]=(D[i]-U[i]*Y[i-1])/(2-U[i]*B[i-1]);
92.}
93.M[pLine->point_num-1]=Y[pLine->point_num-1];
94.for(i=pLine->point_num-1;i>0;i--)
95.{
96.M[i-1]=Y[i-1]-B[i-1]*M[i];
97.}
98.//////////////////////////////////////////計算方程組最終結果
99.for(i=0;i<=pLine->point_num-2;i++)
100.{
101.pLine->a3[i]=M[i]/(6*H[i]);
102.pLine->a1[i]=(pLine->y[i]-M[i]*H[i]*H[i]/6)/H[i];
103.pLine->b3[i]=M[i+1]/(6*H[i]);
104.pLine->b1[i]=(pLine->y[i+1]-M[i+1]*H[i]*H[i]/6)/H[i];
105.}
106.returnTRUE;
107.}
108.//////////////////////////////////////////////////////////////////////////////////
109.SPLINEline1;
110.pSPLINEpLine1=&line1;
111.//////////////////////////////////////////////////////////////////////////////////
112.main()
113.{
114.line1.x[0]=27.7;
115.line1.x[1]=28;
116.line1.x[2]=29;
117.line1.x[3]=30;
118.line1.y[0]=4.1;
119.line1.y[1]=4.3;
120.line1.y[2]=4.1;
121.line1.y[3]=3.0;
122.line1.point_num=4;
123.line1.begin_k1=3.0;
124.line1.end_k1=-4.0;
125.Spline3(pLine1);
126.return0;
127.}
128.//////////////////////////////////////////////////////////////////////////////////
㈡ 三次樣條插值計算步驟
三次樣條插值在實際中有著廣泛的應用，在計算機上也容易實現。下面介紹用計算機求取三樣條插值函數S（x）的演算法步驟：
（1）輸入初始節點離散數據xi，yi（i＝0，1，…，n）；
（2）依據式（6－46），計算hi＝xi－xi－1，λi和Ri（i＝1，…，n－1）；
（3）根據實際問題，從式（6－49）、式（6－51）和式（6－53）中選擇一類對應的邊界條件，求取v0，w0，u0，R0，un，vn，wn，Rn；
（4）根據形成的方程組（6－54）的特點，選用追趕法、高斯法等解方程組，求出Mi（i＝0，1，2，…，n）；
（5）依據式（6－41）、式（6－42），計算插值點的三樣條插值函數值和該點的導數值。
㈢ C語言實現三次樣條插值的子函數
void SPL(int n, double *x, double *y, int ni, double *xi, double *yi)； 是你所要。
已知 n 個點 x,y; x 必須已按順序排好。要插值 ni 點，橫坐標 xi[], 輸出 yi[]。
程序里用double 型，保證計算精度。
SPL調用現成的程序。
現成的程序很多。端點處理方法不同，結果會有不同。想同matlab比較，你需 嘗試 調用 spline（）函數 時，令 end1 為 1, 設 slope1 的值，令 end2 為 1 設 slope2 的值。
#include <stdio.h>
#include <math.h>
int spline (int n, int end1, int end2,
double slope1, double slope2,
double x[], double y[],
double b[], double c[], double d[],
int *iflag)
{
int nm1, ib, i, ascend;
double t;
nm1 = n - 1;
*iflag = 0;
if (n < 2)
{ /* no possible interpolation */
*iflag = 1;
goto LeaveSpline;
}
ascend = 1;
for (i = 1; i < n; ++i) if (x[i] <= x[i-1]) ascend = 0;
if (!ascend)
{
*iflag = 2;
goto LeaveSpline;
}
if (n >= 3)
{
d[0] = x[1] - x[0];
c[1] = (y[1] - y[0]) / d[0];
for (i = 1; i < nm1; ++i)
{
d[i] = x[i+1] - x[i];
b[i] = 2.0 * (d[i-1] + d[i]);
c[i+1] = (y[i+1] - y[i]) / d[i];
c[i] = c[i+1] - c[i];
}
/* ---- Default End conditions */
b[0] = -d[0];
b[nm1] = -d[n-2];
c[0] = 0.0;
c[nm1] = 0.0;
if (n != 3)
{
c[0] = c[2] / (x[3] - x[1]) - c[1] / (x[2] - x[0]);
c[nm1] = c[n-2] / (x[nm1] - x[n-3]) - c[n-3] / (x[n-2] - x[n-4]);
c[0] = c[0] * d[0] * d[0] / (x[3] - x[0]);
c[nm1] = -c[nm1] * d[n-2] * d[n-2] / (x[nm1] - x[n-4]);
}
/* Alternative end conditions -- known slopes */
if (end1 == 1)
{
b[0] = 2.0 * (x[1] - x[0]);
c[0] = (y[1] - y[0]) / (x[1] - x[0]) - slope1;
}
if (end2 == 1)
{
b[nm1] = 2.0 * (x[nm1] - x[n-2]);
c[nm1] = slope2 - (y[nm1] - y[n-2]) / (x[nm1] - x[n-2]);
}
/* Forward elimination */
for (i = 1; i < n; ++i)
{
t = d[i-1] / b[i-1];
b[i] = b[i] - t * d[i-1];
c[i] = c[i] - t * c[i-1];
}
/* Back substitution */
c[nm1] = c[nm1] / b[nm1];
for (ib = 0; ib < nm1; ++ib)
{
i = n - ib - 2;
c[i] = (c[i] - d[i] * c[i+1]) / b[i];
}
b[nm1] = (y[nm1] - y[n-2]) / d[n-2] + d[n-2] * (c[n-2] + 2.0 * c[nm1]);
for (i = 0; i < nm1; ++i)
{
b[i] = (y[i+1] - y[i]) / d[i] - d[i] * (c[i+1] + 2.0 * c[i]);
d[i] = (c[i+1] - c[i]) / d[i];
c[i] = 3.0 * c[i];
}
c[nm1] = 3.0 * c[nm1];
d[nm1] = d[n-2];
}
else
{
b[0] = (y[1] - y[0]) / (x[1] - x[0]);
c[0] = 0.0;
d[0] = 0.0;
b[1] = b[0];
c[1] = 0.0;
d[1] = 0.0;
}
LeaveSpline:
return 0;
}
double seval (int n, double u,
double x[], double y[],
double b[], double c[], double d[],
int *last)
{
int i, j, k;
double w;
i = *last;
if (i >= n-1) i = 0;
if (i < 0) i = 0;
if ((x[i] > u) || (x[i+1] < u))
{
i = 0;
j = n;
do
{
k = (i + j) / 2;
if (u < x[k]) j = k;
if (u >= x[k]) i = k;
}
while (j > i+1);
}
*last = i;
w = u - x[i];
w = y[i] + w * (b[i] + w * (c[i] + w * d[i]));
return (w);
}
void SPL(int n, double *x, double *y, int ni, double *xi, double *yi)
{
double *b, *c, *d;
int iflag,last,i;
b = (double *) malloc(sizeof(double) * n);
c = (double *)malloc(sizeof(double) * n);
d = (double *)malloc(sizeof(double) * n);
if (!d) { printf("no enough memory for b,c,d\n");}
else {
spline (n,0,0,0,0,x,y,b,c,d,&iflag);
if (iflag==0) printf("I got coef b,c,d now\n"); else printf("x not in order or other error\n");
for (i=0;i<ni;i++) yi[i] = seval(ni,xi[i],x,y,b,c,d,&last);
free(b);free(c);free(d);
};
}
main(){
double x[6]={0.,1.,2.,3.,4.,5};
double y[6]={0.,0.5,2.0,1.6,0.5,0.0};
double u[8]={0.5,1,1.5,2,2.5,3,3.5,4};
double s[8];
int i;
SPL(6, x,y, 8, u, s);
for (i=0;i<8;i++) printf("%lf %lf \n",u[i],s[i]);
return 0;
}
㈣ 涓夋℃牱鏉℃彃鍊
璁維(x)婊¤凍鏍鋒湰鐐硅佹眰錛屽垯鍙闇鍦ㄦ瘡涓瀛愬尯闂碵 ]涓婄『瀹1涓涓夋″氶」寮忥紝鍋囪句負錛
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S(x) 鍦 [ ]錛坖=1,2,⋯,n-1錛変笂鏄涓夋″氶」寮忥紝浜庢槸S"(x)鍦╗ ]涓婃槸涓嬈″氶」寮忥紝鍋囪維"(x) 鍦╗ ]錛坖=1,2,⋯,n-1錛変袱絝鐐逛笂鐨勫煎凡鐭ワ紝璁
鍏朵腑
瀵 榪涜屼袱嬈＄Н鍒嗗彲寰楋細
浠ヤ笂鏄鍦 涓婃眰寰楃殑 鍚岀悊鍙奼 錛屽皢 鍚屾椂浠ｅ叆涓や釜鍑芥暟鑱旂珛鏂圭▼錛屽彲浠ユ眰寰
灝 錛
奼傚煎悗寰楋細
鍚岀悊鍒嗗埆鍐欏嚭 ,鑱旂珛絳夊紡,綆鍖栧悗鍙寰楋細
鍦╩atlab瀹炵幇鏃舵敞鎰忥細n涓鐐癸紝n-1鏉＄嚎錛屼互涓婄煩闃墊槸鐢辯浉閭葷殑涓ゆ潯綰跨殑寰鍒嗘柟紼嬭仈絝嬭屾潵錛堜竴闃惰繛緇錛夛紝鍥犳ゆ柟紼嬫諱釜鏁板噺灝戜簡1錛岀煩闃典腑鏈塶-2涓鏂圭▼銆 鍙﹀栵紝鐢╩atlab瀹炵幇鏃墮渶瑕佹敞鎰忥紝matlab涓涓嬫爣浠1寮濮嬶紝鍏朵粬璇璦涓嬫爣鍙鑳戒粠0寮濮嬨