CODEWITHSUNDEEP
javamath
user3168961
user3168961

Reputation: 375

Math in Java when precision is lost

The below algorithm works to identify a factor of a small number but fails completely when using a large one such as 7534534523.0

double result = 7; // 7534534523.0;

double divisor = 1;

for (int i = 2; i < result; i++){

   double r = result / (double)i;

   if (Math.floor(r) == r){
      divisor = i;
      break;
   }
}


System.out.println(result + "/" + divisor + "=" + (result/divisor));

The number 7534534523.0 divided by 2 on a calculator can give a decimal part or round it (losing the 0.5). How can I perform such a check on large numbers? Do I have to use BigDecimal for this? Or is there another way?

Upvotes: 3

Views: 107

Answers (3)

Aaditya Gavandalkar
Aaditya Gavandalkar

Reputation: 824

BigDecimal is the way to move ahead for preserving high precision in numbers.

DO NOT do not use constructor BigDecimal(double val) as the rounding is performed and the output is not always same. The same is mentioned in the implementation as well. According to it:

The results of this constructor can be somewhat unpredictable. One might assume that writing new BigDecimal(0.1) in Java creates a BigDecimal which is exactly equal to 0.1 (an unscaled value of 1, with a scale of 1), but it is actually equal to 0.1000000000000000055511151231257827021181583404541015625. This is because 0.1 cannot be represented exactly as a double (or, for that matter, as a binary fraction of any finite length). Thus, the value that is being passed in to the constructor is not exactly equal to 0.1, appearances notwithstanding.

ALWAYS try to use constructor BigDecimal(String val) as it preserves precision and gives same output each time.

Upvotes: 0

Erwin Bolwidt
Erwin Bolwidt

Reputation: 31269

I don't see what the problem is in your code. It works exactly like it should. When I run your code I get this output:

7.534534523E9/77359.0=97397.0

That may have confused you, but its perfectly fine. It's just using scientific notation, but there is nothing wrong with that.

7.534534523E9 = 7.534534523 * 109 = 7,534,534,523

If you want to see it in normal notation, you can use System.out.format to print the result:

System.out.format("%.0f/%.0f=%.0f\n", result, divisor, result / divisor);

Shows:

7534534523/77359=97397

But you don't need double or BigDecimal to check if a number is divisible by another number. You can use the modulo operator on integral types to check if one number is divisible by another. As long as your numbers fit in a long, this works, otherwise you can move on to a BigInteger:

long result = 7534534523L;
long divisor = 1;
for (int i = 2; i < result; i++) {
    if (result % i == 0) {
        divisor = i;
        break;
    }
}
System.out.println(result + "/" + divisor + "=" + (result / divisor));

Upvotes: 1

Rockstar
Rockstar

Reputation: 2288

If your goal is to represent a number with exactly n significant figures to the right of the decimal, BigDecimal is the class to use.

Immutable, arbitrary-precision signed decimal numbers. A BigDecimal consists of an arbitrary precision integer unscaled value and a 32-bit integer scale. If zero or positive, the scale is the number of digits to the right of the decimal point. If negative, the unscaled value of the number is multiplied by ten to the power of the negation of the scale. The value of the number represented by the BigDecimal is therefore (unscaledValue × 10-scale).

Additionally, you can have a better control over scale manipulation, rounding and format conversion.

Upvotes: 3

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