Finding a "correct" value that is unknown prior to a calculation

Question

The problem is as follows i have a large or small number (can be either one) and i need to tweak this number and put it through a caluclation. Given the result of the calulation it has to mach a certain value on the 5th decimal at least.

So i need to make a method that takes this starting value, tries to increase or decrease it given what the current result is until i get the correct result. I have made some atempts with no success.

Here is an example that won't woork at all but it hints towards what i mean... (this is just a small scale test case)

   public class Test {
        public static void main(String[]args)
        {
            double ran = 100 + (int)(Math.random() * 100000.999999999);
            int count = 0;              
            double tmpPay = 3666.545;
            double top = tmpPay;
            double low = 0;             

            while ( tmpPay != ran )
            {
                if ( tmpPay > ran)
                {
                    if( low == 0)
                    {
                            tmpPay = top / 2;
                            top = tmpPay;
                    }
                    else
                    {
                        tmpPay = tmpPay + ((top - low) / 2);
                        top = tmpPay;
                    }        
                }           

                if (tmpPay  < ran)
                {
                    tmpPay = top * 1.5;
                    low = top;
                    top = tmpPay;                   
                }
            }
            System.out.println(" VAlue of RAN: " +ran + "----VALUE OF tmpPay: " + tmpPay + "---------- COUNTER: " + count);         
}

Example 2 mabey a more clear description. This is my solution now..

guessingValue = firstImput;

while (amortization > tmpPV)
{
    guessingValue -= (decimal)1;
    //guessingVlue -- > blackbox
    amortization = blackboxResults;
}
 while (amortization < tmpPV)
{
    guessingValue += (decimal)0.00001;
    //guessingVlue -- > blackbox
    amortization = blackboxResults;
}

}

Not really a binary search. I need to find what the in parameter should be for a calculation to get a specific answer. Think of the calculation itself as a black box. — SomeRandomName, Oct 29 '13 at 12:30
Comparing doubles using standard java operators ==,<,> etc. is not a good idea. You can find why here http://stackoverflow.com/questions/3832592/test-for-floating-point-equality-fe-floating-point-equality — Luke, Oct 29 '13 at 12:32
If the calculation is truly a black box, it's impossible - consider `if (x == 0.3433) return 1; else return 0;`. It's impossible to know when the function would return `1` here. — Bernhard Barker, Oct 29 '13 at 12:36
If trying to run the program, it will easily be in infinite loop, since the while condition could hardly equal by double values comparison. e.g. There are 2 values, double value1 = 3666.545; double value2 = 3666.545002; value1 == value2 is false. Even this kind of values are not equal. You'd better define a range for deviation. Then, e.g, if |value1 - value2| <0.0002, then break the while condition and print the random num information. — Mengjun, Oct 29 '13 at 12:48

score 1 · Answer 1 · answered Oct 29 '13 at 12:43

1

One way would be to define your problem as a local optimization task and use an local optimizer (for example Brent's method or Nelder Mead Simplex from Apache commons).

Your goal function here would be the distance between the desired value and what you get from your black box.

answered Oct 29 '13 at 12:43

dratewka

2,104
14
15

Mabey that is a solution i can try and specify my problem even further and add another example in the question. ANd if this applies could u hint at how to implement this? – SomeRandomName Oct 29 '13 at 13:18

score 1 · Answer 2 · answered Oct 29 '13 at 12:45

1

If I understand correctly, you have a function g(x) and a value K, you want to find x0 such that g(x0) = K. This is equivalent to find the roots of the function f(x) = g(x) - K, because f(x0) == f(x0) - K == K - K == 0.

A simple algorithm would be Newton's method.

answered Oct 29 '13 at 12:45

Marco Pantaleoni

2,529
15
14

Thank's for the answer! – SomeRandomName Nov 01 '13 at 07:31

score 1 · Answer 3 · answered Oct 29 '13 at 13:00

If trying to run the program, it will easily be in infinite loop, since the while condition (for double values comparison) could hardly equal. e.g. There are 2 values as follows:

double value1 = 3666.545;

double value2 = 3666.54500001;

value1 == value2 is false.

Even this kind of values are not equal.

You'd better define a range for deviation.

e.g, if |value1 - value2| < 0.005, then break the while condition and print the random num information.

score 1 · Accepted Answer · answered Oct 29 '13 at 13:32

As I already mentioned in the comment above you should not compare doubles using build-in operators. This is the main reason why your code is not working. The second one is that in else clause tmpPay = tmpPay + ((top-low) /2); instead of tmpPay = tmpPay - ((top-low) /2 );

Complete fixed code is below :

public class Test {
    private static final double EPSILON = 0.00001;
    public static boolean isEqual( double a, double b){

        return (Math.abs(a - b) < EPSILON);

    }


    public static void main(String[]args)
    {

        double ran = 100 + (int)(Math.random() * 100000.999999999);
        int count = 0;              
        double tmpPay = 3666.545;
        double top = tmpPay;
        double low = 0;             

        while ( !isEqual(tmpPay, ran))
        {
            if ( tmpPay > ran)
            {
                if( isEqual(low, 0.0))
                {
                        tmpPay = top / 2;
                        top = tmpPay;
                }
                else
                {
                    tmpPay = tmpPay - ((top - low) / 2);
                    top = tmpPay;
                }        
            }           

            if (tmpPay  < ran)
            {
                tmpPay = top * 1.5;
                low = top;
                top = tmpPay;                   
            }
            System.out.println("RAN:"+ran+" tmpPay:"+tmpPay+" top:"+top+" low:"+low+" counter:"+count);
            count++;
        }
        System.out.println(" VAlue of RAN: " +ran + "----VALUE OF tmpPay: " + tmpPay + "---------- COUNTER: " + count);




    }
}

Thank you, good to hear i was not COMPLEATLY off target. This will be a good example to build a solution around. — SomeRandomName, Oct 29 '13 at 13:46

Finding a "correct" value that is unknown prior to a calculation

4 Answers4