一维搜索---斐波拉契法

参考网址：

https://blog.youkuaiyun.com/nwpuwyk/article/details/28470645

https://blog.youkuaiyun.com/w8253497062015/article/details/79534098

原理

代码

JAVA版本

private static void fibonacci(float start, float end, float eps) {  
          
        float Fn_1, Fn;  
        int n;  
        float L1 = end - start;  
        float Ln = eps;  
        n = 2;  
        Fn = F(n);  
        //求要经过几轮区间长度才可以小于eps  
        while(Fn <= L1 / Ln) {  
            n ++;  
            Fn = F(n);  
        }  
        Fn_1 = F(n - 1);  
          
        float t1, t2, f1, f2;  
        float a = Math.min(start, end);  
        float b = Math.max(start, end);  
          
        t1 = b - (b - a) * Fn_1 / Fn;  
        t2 = a + b - t1;  
        f1 = fun(t1);  
        f2 = fun(t2);  
          
        while(b - a >= Ln) {  
            if(f2 < f1) {  
                a = t1;  
                t1 = t2;  
                t2 = a + b - t1;  
                f1 = f2;  
                f2 = fun(t2);  
            } else {  
                b = t2;  
                t2 = t1;  
                t1 = a + b - t2;  
                f2 = f1;  
                f1 = fun(t1);  
            }  
        }  
          
        System.out.println((b + a) / 2);  
    }  
      
    private static float F(int n) {  
          
        if(n == 0 || n == 1) return 1;  
        return F(n - 1) + F(n - 2);  
    }  
      
    private static float fun(float x) {  
        return (float) Math.sin(x);  
    } 

C++版本

#include <stdio.h>
#include <time.h>
#include <math.h>
 
#define f(x)  (-(x + 10*sin(5*x) + 7*cos(4*x))) 
#define Q 0.382
 
float goldsearch(float a, float b, int n)
{
	float x, y;
	for (int i = n; i > 0; i--) {
		float x = a + Q*(b - a);
		float y = a + (1 - Q)*(b - a);
		if (f(x) < f(y)) {
			a = a;
			b = y;
		}
		else {
			a = x;
			b = b;
		}
	}
	if (f(a) < f(b)){ 
		return a; }
	else{
		return b;
	}
}
 
float fibonacci_num(int n)
{
	if (n <= 2) return 1;
	else return fibonacci_num(n - 1) + fibonacci_num(n - 2);
}
 
float fibonacci_search(float a, float b, int n) 
{	
	float q[255] = {0};
	for(int i = n; i >= 0; i--) {
		q[n-i+1] =1- (fibonacci_num(i + 2) / fibonacci_num(i + 3));
	//	printf("q[%d] = %f\n", n - i + 1, q[n - i + 1]);
	}	
	
	float x, y;
	for (int i = 1; i <= n; i++) {
		float x = a + q[i]*(b - a);
		float y = a + (1 - q[i])*(b - a);
		if (f(x) < f(y)) {
			a = a;
			b = y;
		}
		else {
			a = x;
			b = b;
		}
	}
	if (f(a) < f(b)) {
		return a;
	}
	else {
		return b;
	}
 
}
 
double* CallBack_GetTime(float a, float b, int n,float	(*search)(float , float , int ))
{
	double start, end, delta_time, ans[2] = {0};
	start = clock();
	ans[0] = (*search)(a, b, n);
	end = clock();
	delta_time = (end - start)/ CLOCKS_PER_SEC;
	ans[1] = delta_time;
	return  ans;
	
}
 
 
int main()
{
	double *gold_ans = CallBack_GetTime(0, 9, 20, goldsearch);
	printf("黄金分割法——在X = %f 时取得最小值，搜索花费时间为%f\n\n", gold_ans[0],gold_ans[1]);
	double *fibo_ans = CallBack_GetTime(0, 9, 20, fibonacci_search);
	printf("斐波拉契法——在X = %f 时取得最小值，搜索花费时间为%f\n\n", fibo_ans[0],fibo_ans[1]);
	while (1);
	return 0;
}