JAVA中的拉格朗日插值

Question

我四处搜索了一下，但java中没有任何可用的代码，因此我自己编写了代码，遇到了一些问题。实际上，我是从一个c++源代码中获得这段代码的，并努力将其转换为一个可行的java程序。

http://ganeshtiwaridotcomdotnp.blogspot.sg/2009/12/c-c-code-lagranges-interpolation.html

public static void main(String[] args) {
    int n;
    int i, j;
    int a;
    int x[] = null;
    int f[] = null;
    int sum = 0;
    int mult;
    Scanner input = new Scanner(System.in);
    System.out.println("Enter number of point: ");
    n = input.nextInt();
    System.out.println("Enter value x for calculation: ");
    a = input.nextInt();
    for (i = 0; i < n; i++) {
        System.out.println("Enter all values of x and corresponding functional vale: ");
        x = input.nextInt();
        f = input.nextInt();
    }
    for (i = 0; i <= n - 1; i++) {
        mult = 1;
        for (j = 0; j <= n - 1; j++) {
            if (j != i) {
                mult *= (a - x[j]) / (x[i] - x[j]);
            }
            sum += mult * f[i];
        }
    }
    System.out.println("The estimated value of f(x)= " + sum);
}

Answer 1

对于拉格朗日插值公式-

• 您应该使用双数据类型Array
• 您应该制作具有特定项目数的数组

看看下面的程序，你就会明白了。

    double product, sum = 0;
    Scanner sc = new Scanner(System.in);
    System.out.print("Enter the Number of Terms: ");
    int n = sc.nextInt();
    double[] x = new double[n];
    double[] y = new double[n];
    System.out.println("Enter all the x, y terms: ");
    for (int i = 0; i < n; i++) {
        x[i] = sc.nextDouble();
        y[i] = sc.nextDouble();
    }
    System.out.println("x = {" + Arrays.toString(x) + "}");
    System.out.println("x = {" + Arrays.toString(y) + "}");
    System.out.print("Enter a point to Find it's value: ");
    int xPoint = sc.nextInt();
    // End of inputs
    product = 1;
    // Peforming Arithmatic Operation
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (j != i) {
                product *= (xPoint - x[j]) / (x[i] - x[j]);
            }
        }
        sum += product * y[i];
        product = 1;    // Must set to 1
    }
    System.out.println("The value at point " + xPoint + " is : " + sum);
    // End of the Program
}

}

Answer 2

public static void main(String[] args) {
    System.out.println("Langrages");
        
    Scanner sc = new Scanner(System.in);
    System.out.println("Enter the no of terms: ");
    int n = sc.nextInt();
    
    float[] ax = new float[n];
    System.out.println("Enter values of x");
    for(int i=0 ;i<n;i++) {
        ax[i] = sc.nextFloat();
    }
    
    float[] ay = new float[n];
    System.out.println("Enter values of y");
    for(int i=0 ;i<n;i++) {
        ay[i] = sc.nextFloat();
    }
    
    System.out.println("Enter the value of xn at which you find y");
    int x = sc.nextInt();
    
    float num,den;
    float term =0;
    
    for(int i=0;i<n;i++) {
        num=1;
        den=1;
        for(int j=0;j<n;j++) {
            if(j!=i) {
                num=num*(x-ax[j]);
                den = den*(ax[i]-ax[j]);    
            }
        }
        term += (num/den)*ay[i];
    }
    
    System.out.println(term);
    sc.close();
}   

Answer 3

int x[] = null;
int f[] = null;
...
for (i = 0; i < n; i++) {
    ....
    x = input.nextInt();
    f = input.nextInt();
}

看起来很清楚。只需为x和f:创建数组实例

x = new int[n];
f = new int[n];

之后的某个地方：

n = input.nextInt();

在上述for循环之前，修改for主体：

...
x[i] = input.nextInt();
f[i] = input.nextInt();