技術標籤：人工智慧

如圖，設計函式f(x) = x^2 - 4*x + 6;

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <math.h>
double f(double x){ // x^2 - 4*x + 6
	return x * x - 4 * x + 6;
}
double getDouleRand(double aL, double aR){
	double tmp = 0;
	do{
		tmp = (rand() % (int)(aR * 10000)) * 1.0 / 10000;
	}while(!(tmp > aL && tmp < aR));
	return tmp;
}
int main(){
	srand(time(0));
	double aL = 0;
	double aR = 4;

	while(aR - aL > 1e-3){
		double a1 = getDouleRand(aL, aR);
		double a2 = getDouleRand(a1, aR);
		printf("%.4lf %.4lf %.4lf %.4lf\n",aL, a1, a2, aR);
		if(f(a1) < f(a2)){
			aR = a2;
		}else{
			aL = a1;
		}
	}
	printf("%.2lf\n", (aL + aR) / 2);
} 
 

可以根據區間消去思想，解決

問題迎刃而解。

如果需要區間縮減率:

double ita0 = aR - aL;
//計算後的aR, aL
double ita1 = aR - aL;
printf("itaK = %.2lf\n", ita1 / ita0);

itaK即為所求