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I am working on a simple linear interpolation program. And I am having some troubles when implementing the algorithm. Assuming there are 12 numbers as a whole, we'll let user input 3 of them(position 0, position 6 and position 12). Then the program will calculate other numbers. Here is a piece of my code to achieve this:

static double[] interpolate(double a, double b){
    double[] array = new double[6];
    for(int i=0;i<6;i++){
        array[i] = a + (i-0) * (b-a)/6;
    }
    return array;
}

static double[] interpolate2(double a, double b){
    double[] array = new double[13];
    for(int i=6;i<=12;i++){
        array[i] = a + (i-6) * (b-a)/6;
    }
    return array;
}

As you can see, I used two functions. But I want to find a universal function to do this job. However, I don't know how to find a common way to represent i-0 and i-6. How to fix it? According to Floating point linear interpolation, I know maybe I should add a formal parameter float f. But I am not quite understand what does float f mean and how to modify my code based on it. Could anyone help me? Thank you.

Community
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turf
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1 Answers1

7

If you want to interpolate intervals to different count of numbers, you can just add the count of output numbers to function parameter. Example:

/***
 * Interpolating method
 * @param start start of the interval
 * @param end end of the interval
 * @param count count of output interpolated numbers
 * @return array of interpolated number with specified count
 */
public static double[] interpolate(double start, double end, int count) {
    if (count < 2) {
        throw new IllegalArgumentException("interpolate: illegal count!");
    }
    double[] array = new double[count + 1];
    for (int i = 0; i <= count; ++ i) {
        array[i] = start + i * (end - start) / count;
    }
    return array;
}

Then you can just call interpolate(0, 6, 6); or interpolate(6, 12, 6); or interpolate(6, 12, 12); or whatever you want.

Luo
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