# Finding formula?

Abu Nene
Ranch Hand
Posts: 56
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:

calculate(3, 4, 4) return 0
calculate(1, 5, 5) return 0
calculate(2, 4, 2) return -2
calculate(0, 5, 3) return -2
calculate(4, 6, 4) return -2
calculate(0, 3, 5) return 1

Any idea what is the best way to find out the formula? Do pardon me if this sound more like a math problem rather then Java. Please advise thanks.

Ulf Dittmer
Rancher
Posts: 42968
73
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.

Wouter Oet
Saloon Keeper
Posts: 2700
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

Greenhorn
Posts: 2
You need to tell what logic/algorithm, you want to calculate in the method..

Abu Nene
Ranch Hand
Posts: 56
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
Rancher
Posts: 42968
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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.

Jesper de Jong
Java Cowboy
Saloon Keeper
Posts: 15354
39
This might be a bit silly, but here's a 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.

Henry Wong
author
Marshal
Posts: 21190
80

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.

Henry

David Newton
Author
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Posts: 12617
I used a partial differential equation and it seems to work.

Ernest Friedman-Hill
author and iconoclast
Marshal
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35
A lookup table would work too!

fred rosenberger
lowercase baba
Bartender
Posts: 12143
30
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.

Rob Spoor
Sheriff
Posts: 20546
56
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.

Rob Spoor
Sheriff
Posts: 20546
56
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.

fred rosenberger
lowercase baba
Bartender
Posts: 12143
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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>

Sha Jar
Greenhorn
Posts: 23
Abu Nene wrote:find out the formula.

Do you have any idea of what kind of formula we're talking about?

Or to put it differently. What kind of black box is between the 3 input values and the output value? What's the nature of the black box?

Rob Spoor
Sheriff
Posts: 20546
56
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?