Hi guys, in the Java application that I'm working with, I need to create a method which takes in 3 int parameters, calculate and return a int. Problem is I need to find out the formula. I've some input data and return result data as follows:

There's an unlimited number of formulas that produce this output given those inputs. Without more information on how inputs and outputs relate to one another this can't be solved.

I would contact the author because there are a million and one solutions. Are you sure that you don't have the necessary code/documentation?
Another option would be decompiling with javap -c className

"Any fool can write code that a computer can understand. Good programmers write code that humans can understand." --- Martin Fowler
Please correct my English.

You need to tell what logic/algorithm, you want to calculate in the method..

Mistakes makes human being better..

Abu Nene
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The logic have not been created. How I get the data currently is getting the 3rd parameter minus the 2nd parameter but it'll not satisfy "calculate(0, 3, 5) return 1".

The logic is a mathematical formula which satisfy the all given inputs together with the output. Is there something like a reserve math or something?

Ulf Dittmer
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Ulf Dittmer wrote:There's an unlimited number of formulas that produce this output given those inputs.

If the choice of mathematical operators in the formula isn't limited (to, say, basic arithmetic), then there simply is no single solution.

If this is the only information you've got, you're never going to find a good solution. Suppose you put in three other numbers, then how are you ever going to know what the correct answer is? It's impossible.

In the field of statistics, I believe this is done via regression analysis -- although admittedly, it is done as "best fit", and not exact. In the field of graphics, there is "curve fitting", which may be able to be applied.

Maybe a google of "regression analysis" and / or "curve fitting" can get you started.

you could define an infinite number of formulas that 'solve' the 6 data points you have. As EFH said, a simple table would work:

if (input = 3,4,4) return 0
else if (input = 1,5,5) return 0
else if (input = 2,4,3) return -2
else if (input = 0,5,3) return -2
else if (input = 4,6,4) return -2
else if (input = 0,3,5) return 1

//at this point, I could add 0 to 10,000 more "else if" lines defining any 3-digit combination I want
//and an optional 'else' line

If you're looking for a mathematical formula, again, you have 6 data points in a 3-d space. If you are not limited to some kind of surface/shape, there are an infinite number of solutions.

If you are only using addition and multiplication, and no if-statements, you get a lot of different variables in your function:
3 * a + 4 * b + 4 * c + d == 0 (1)
1 * a + 5 * b + 5 * c + d == 0 (2)
2 * a + 4 * b + 2 * c + d == -2 (3)
0 * a + 5 * b + 3 * c + d == -2 (4)
4 * a + 6 * b + 4 * c + d == -2 (5)
0 * a + 3 * b + 5 * c + d -- 1 (6)

By combining all these you can get the solution for a, b, c and d. For instance, combining (1) and (2):
3 * a + 4 * b + 4 * c + d == 1 * a + 5 * b + 5 * c + d
=== {subtract 1 * a, 4 * b, 4 * c and d on each side}
2 * a == b + c (7)

You can then combine (3) and (7), by replacing the 2 * a with b + c in (3):
b + c + 4 * b + 2 * c + d == -2
===
5 * b + 3 * c + d == -2 (8)

If you have enough statements you can eventually work out the values of a, b, c and d. Note that the more variables you have, the more statements you need. 6 may not be enough for 4 variables.

Ok, I've worked it out, and unless I've made a mistake then there are no a, b, c and d that match these equations. Consider:
c == 1 and c == 1/2 so there is contradiction. If the result of (6) was 2 instead of 1 then a == 0, b == -1, c == -1 and d == 0 would be the solution.

Of course this doesn't prevent you using more complex functions like using powers of any of the numbers. All I've shown is that a simple linear function is not possible.

If you're talking about straight algebra, you need one formula per variable, so 6 is more than enough. in fact, here three are too many. Three would do it (the three inputs are the co-efficients of x, y and z in

fred rosenberger wrote:If you're talking about straight algebra, you need one formula per variable, so 6 is more than enough. in fact, here three are too many. Three would do it (the three inputs are the co-efficients of x, y and z in

ax + by + cz = <return value>

Wouldn't you need 4, with the "+ d" I specified as correction for the return value?