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.

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.

Mistakes makes human being better..

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 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.

Henry

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.

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.

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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.

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ax + by + cz = <return value>

There are only two hard things in computer science: cache invalidation, naming things, and off-by-one errors

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?

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?

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