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My program uses a perfect square grid in all circumstances with Math.sqrt() function. But Math.sqrt() takes double as an argument and returns a double. Can anyone tell me how to obtain the square root in Integer without casting? 

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You can use Guava intMath API to do the square root. But it’s not a wise idea to add dependency on an API just to avoid the cast. In casting, when you give the input as a perfect square int, you’ll get an integer-valued double which is in the int range. So casting doesn’t introduce any rounding error when you try to convert int to double or integer-valued double back to an int. 

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