我正在实现用于求解java中的方程的二分法。我最初编码解决方案的预定义多项式方程x ^ 3 + 4x ^ 2-10。现在我正在推广用户输入的任何多项式的解。
我读了相应度数的系数。现在我只需要调整f（）方法，以便我可以计算f（a），f（b）和f（c）。
// BISECTION METHOD IMPLEMENTATION IN JAVA
// This program uses bisection method to solve for x^3 + 4x^2 -10 = 0
package nisarg;
import java.util.Scanner;
public class BetterBisection {
public static void main(String[] args) {
double a, b, c; // a, b and c have the usual meaning
double f_of_a, f_of_b; // f_of_a, f_of_b store values of f(a) and f(b)
// respectively
int highest_degree;
System.out.println("What is the highest degree of your polynomial? ");
Scanner input = new Scanner(System.in);
highest_degree = input.nextInt();
for (int i = highest_degree; i >= 0; i--) {
int coeff_deg_i;
coeff_deg_i = poly_input(i);
// System.out.println(coeff_deg_i);
}
// The following do-while loop keeps asking the user for a and b until
// f(a)f(b) does not become negative
do {
a = input();
b = input();
if (f(a) * f(b) >= 0) {
System.out
.println("Sorry the two numbers are not bracketing the root. Please try again ");
}
} while (f(a) * f(b) >= 0);
f_of_a = f(a);
f_of_b = f(b);
double root = bisectionMethod(f_of_a, f_of_b, a, b);
System.out.println("Root is : " + root);
}
public static double input() { // Reads in the bracketing number i.e a and b
Scanner input = new Scanner(System.in);
System.out.println("Enter a bracketing number");
return (input.nextDouble());
}
public static double f(double num) { // Calculates f(x) given x and returns
// f(x)
final int COEFF_DEG_3 = 1; // Coefficient of x^3
final int COEFF_DEG_2 = 4; // Coefficient of x^2
final int COEFF_DEG_0 = -10; // Coefficient of x^0
return (COEFF_DEG_3 * Math.pow(num, 3) + COEFF_DEG_2 * Math.pow(num, 2) + COEFF_DEG_0
* Math.pow(num, 0));
}
public static double bisectionMethod(double f_of_a, double f_of_b, double a,
double b) { // Does the actual work of evaluating
double c; // the root using the method of bisection.
double f_of_c;
final double TOLERANCE = 0.0001;
while (Math.abs(a - b) > TOLERANCE) {
c = (a + b) / 2;
f_of_c = f(c);
if (f_of_c * f(a) == 0 || f_of_c * f(b) == 0) {
return c;
} else if (f_of_c * f(a) > 0) {
a = c;
} else {
b = c;
}
}
return (a + b) / 2;
}
public static int poly_input(int degree) {
System.out.println("Please enter coefficient for degree " + degree);
Scanner input = new Scanner(System.in);
int coefficient;
coefficient = input.nextInt();
return coefficient;
}
}
答案 0 :(得分：1)
您不能使用循环来定义变量。要么有12个显式变量：
public class global {
public static int coeff_deg_1;
public static int coeff_deg_2;
public static int coeff_deg_3;
// and so on...
}
或者定义一个包含12个元素的数组：
public class global {
public static final int coeff_degs = new int[12];
}
答案 1 :(得分：0)
这是一个递归的：
public static double f(double x) {
return (x*x*x)-4*x-10;
}
public static double RecursiveBisection(Function fct, final double left, final double right, final double tolerance) {
double x = 0;
double dx = 0;
if ( Math.abs(right - left) < tolerance ) // base case
return (left + right) / 2;
else { // recursive case
x = (left + right)/2;
System.out.println("Root obtained: " + x);
dx = right - left;
System.out.println("Estimated error: " + dx);
if ( fct.f(left) * fct.f(x) > 0 ) // on same side
return RecursiveBisection (fct, x, right, tolerance);
else // opposite side
return RecursiveBisection(fct, left, x, tolerance);
}
}
答案 2 :(得分：0)
import java.util.*;
public class Bisection {
private static Scanner input = new Scanner(System.in);
private static List<Double> coefficients = new ArrayList<>();
private static double epsilon = 0.00000001; // precision.
/**
* Input the coefficients of the polynomial
*/
private void inputPolynomial() {
System.out.print("Enter the order of the polynomial: ");
int order = input.nextInt();
System.out.print("Enter the coefficient of terms of the function: ");
while (order >= 0) {
double value = input.nextDouble();
coefficients.add(value);
order--;
}
}
/**
* The function method to compute the value of the function given variable x
*
* @param x
* @return result of inputing variable x
*/
private double function(double x) {
double result = 0.0;
for (int index = 0, order = coefficients.size()-1; index < coefficients.size(); order--, index++) {
result += coefficients.get(index) * (Math.pow(x, order));
}
return result;
}
private void calculateRoot() {
double a, b;
// Ask user for a and b until f(a) * f(b) > 0
do {
System.out.print("Enter the start of the interval: ");
a = input.nextDouble();
System.out.print("Enter the end of the interval: ");
b = input.nextDouble();
if (function(a) * function(b) >= 0) {
System.out.println("Sorry, the root is not within the 2 numbers.\nDo try again.");
}
} while(function(a)*function(b) >= 0);
long startTime = System.nanoTime();
double root = bisectionMethod(a, b);
long timeUsed = System.nanoTime() - startTime;
System.out.printf("\nThe root is: %.6f.\n", root);
System.out.println("Time used is " + timeUsed/1000 + " milliseconds.");
}
private double bisectionMethod(double start, double end) {
double middle = (start + end)/2.0;;
while (Math.abs(start - end) > epsilon) {
//System.out.println("x: " + middle);
if ((function(start) * function(middle) == 0.0) || (function(end) * function(middle) == 0.0)) {
break;
} else if (function(start) * function(middle) > 0.0) {
start = middle;
} else {
end = middle;
}
middle = (start + end)/2.0;
}
return middle;
}
public static void main(String[] args) {
Bisection bisection = new Bisection();
bisection.inputPolynomial();
bisection.calculateRoot();
}
}