# Euclidean distance returning strange results

im writing a program to compare two images against each other based on color and them using euclidean distance algorithm, however when i run it and pass two images i get one distance and then when i pass the same images and the other i get a completely different set of results.

- is this normal or should the answers be the same?

The statement I use to calculate the Euclidean distance is:

```
distance = (int) Math.sqrt( (rgb1.getR()-rgb2.getR())^2
+ (rgb1.getG()-rgb2.getG())^2
+ (rgb1.getB()-rgb2.getB())^2
);
```

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Looking at the code you posted it looks like your RGB values. However, the operator is `^`

not a power operator, but XOR (exclusive-OR) is a bitwise operation. Therefore, to calculate the squares correctly, use regular multiplication - for example, use a temporary variable `int deltaR = rgb1.getR()-rgb2.getR();`

, and then write in the formula `deltaR*deltaR`

instead of the operator `^`

. Your RGB values will probably be between 0 and 255, so there shouldn't be any overflow issues. Alternatively, you can use `Math.pow(rgb1.getR()-rgb2.getR(),2)`

etc. In the formula.

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To square numbers in Java, use `Math.pow(x, 2)`

, or even easier `x * x`

. The expression `x ^ 2`

has no square `x`

, instead XORs `x`

with `2`

.

In your code:

```
int diffR = rgb1.getR() - rgb2.getR();
int diffG = rgb1.getG() - rgb2.getG();
int diffB = rgb1.getB() - rgb2.getB();
int distance = (int) Math.sqrt(diffR*diffR + diffG*diffG + diffB*diffB);
```

... Although I'm not entirely sure about your algorithm, this is a different problem.

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As people said, you can use `Math.pow(x, 2)`

for squaring. Just from personal experience, if you are going to call this function a lot, it would be better to write the multiplication yourself, i.e. `Math.sqrt((deltaX * deltaX) + (deltaY * deltaY) + (deltaZ * deltaZ));`

It may sound ugly, but if you profile both forms of code, you will see that the calls are `Math.pow`

much slower than simple multiplications. Obviously there is nothing to do with the challenge `Math.sqrt`

.

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