homework: 組合最優化之一維搜索：fibnacci搜索/進退法求極值區間

問題描述：

給定函數$f(x)=x^2-x+2$,設計fibnacci算法，進退算法分別求$f(x)$在區間[0,3]上的極小值區間，要求區間長度不大於原始區間的 $9\%$

Fibnacci算法：

算法設計：

給定區間$[a,b]$以及精度$e$,函數$f(x)$

①預處理出一定數量的斐波那契數列$fib,fib[0]=fib[1]=1,fib[n]=fib[n-1]+fib[n-2]$，然後根據求解區間和求解精度得到計算次數n。因爲我們設定間隔兩個單位區間長度的時候結束，所以$f[n+1] >=2(b-a)/e$
②令$x1 =b-fib[n]/fib[n+1]*(b-a), x2 = a + fib[n]/fib[n+1]*(b-a)$

③判斷$|b -a|$是否小於等於$e$，如果是，則結束計算，得到結果區間$[a,b]$，否則比較$f(x1)$和
$f(x2)$的大小，若$f(x1) > f(x2)$，跳到④，否則跳到⑤。

④令$a = x1, x1 = x2, x2 = a + b - x1$.跳到③

⑤令$b = x2, x2 = x1, x1 = a + b - x2$.跳到③

代碼實現：

#include<bits/stdc++.h>
using namespace std;
int fib[40];
double f(double x){
    return x*(x-1)+2;
}
void init(){
    fib[0] = 1; fib[1] = 1; for(int i = 2; i < 40; ++i) fib[i] = fib[i-1] + fib[i-2];
}
void sol(double l, double r, double eps){
    int s = 2*(r-l)/eps+1;
    int n = 0;
    while(fib[n+1] <= s) ++n;//獲取迭代次數
    double lmid = r - (double)fib[n]/fib[n+1] * (r-l);
    double rmid = l + (double)fib[n]/fib[n+1] * (r-l);
    int ca = 0;
    while(r-l > eps){
        cout<<setprecision(4)<<fixed<<"test"<<++ca<<": l = "<<l<<"    r = "<<r<<"    lmid = "<<lmid<<"    rmid:"<<rmid;
        cout<<"    f(lmid) = "<<f(lmid)<<"    f(rmid) = "<<f(rmid)<<endl<<endl;
        if(f(lmid) > f(rmid)){
            l = lmid;
            lmid = rmid;
            rmid = l + r - lmid;
        }
        else{
            r = rmid;
            rmid = lmid;
            lmid = l + r - rmid;
        }
    }

    cout<<"\n result : \nl:"<<l<<" r:"<<r<<endl;
}
int main()
{
	init();
	double l, r;cout<<"輸入搜索區間："; cin>>l>>r;
	sol(l, r, (r-l)*0.09);
}

進退法：

ps: 這裏寫的進退法經過修改，在找到單峯但是不滿足精度的時候通過減少步長來繼續下降。不知道還算不算進退法:D

算法設計：

給定初始點$x$，精度$e$和函數$f(x)$,初始步長$h_0$

①先確定很小的步數$dx=e/100$,比較$f(x0+dx),f(x)$和$f(x-dx)$來確定下降方向，若$f(x+dx)<f(x)<f(x-dx)$,令$dx=h0$,若$f(x+dx)>f(x)>f(x-dx)$,令$dx = -h0$，否則可確定$[x-dx,x+dx]$爲滿足條件的區間，結束算法。

②令$x1 = x+dx, x2 = x1+dx$.如果$f(x)>f(x1)>f(x2)$，轉③，否則如果$|x2-x|<=e$,已經找到滿足條件單峯區間$[x,x2]$，結束算法。如果$|x2-x|>e$，轉④

③令$x = x1, dx = dx*2$。轉②

④令$dx = dx/2$。轉②

代碼實現：

#include<bits/stdc++.h>
using namespace std;
double f(double x){
    return x*(x-1)+2;
}
void sol(double x0, double eps, double h0){
    double dx = eps/100;
    double x1, x2;
    if(f(x0) < f(x0 + dx) && f(x0) < f(x0 - dx)) {
        cout<<"結果區間爲：["<<x0-dx<<","<<x0+dx<<"]\n";return;
    }
    else if(f(x0) > f(x0 + dx)) dx = h0;
    else dx = -h0;
    int num = 0;
    while(1){
        num++;
        x1 = x0 + dx;  x2 = x1 + dx;
        cout<<setprecision(3)<<fixed<<"x0 = "<<x0<<"\tx1 = "<<x1<<"\tx2 = "<<x2;
        cout<<"\tf(x0) = "<<f(x0)<<"\tf(x1) = "<<f(x1)<<"\tf(x2) = "<<f(x2)<<endl;
        if(f(x0) > f(x1) && f(x1) > f(x2)){
            x0 = x1;
            dx = dx*2.0;
            continue;
        }
        else{
            if(abs(x2-x0) > eps){
                dx = dx/2.0;
            }
            else{
                cout<<"迭代次數:"<<num<<endl;
                cout<<"結果區間爲：["<<min(x0,x2)<<","<<max(x0, x2)<<"]\n";return;
            }
        }
    }
}
int main()
{
	double x0, e, h0;
	cout<<"輸入初始點：";cin>>x0;
    cout<<"輸入需求區間精度：";cin>>e;
    cout<<"輸入初始步長：";cin>>h0;
    sol(x0, e, 0.1);
}