[jira] [Created] (NUMBERS-120) Major loss of precision in BigFraction.doubleValue() and BigFraction.floatValue()

["Heinrich Bohne (JIRA)" <ji...@apache.org>]

Heinrich Bohne created NUMBERS-120:
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             Summary: Major loss of precision in BigFraction.doubleValue() and BigFraction.floatValue()
                 Key: NUMBERS-120
                 URL: https://issues.apache.org/jira/browse/NUMBERS-120
             Project: Commons Numbers
          Issue Type: Bug
          Components: fraction
    Affects Versions: 1.0
            Reporter: Heinrich Bohne


The method {{BigFraction.doubleValue()}} calculates the double value of fractions with numerators or denominators that, when converted to a {{double}}, round up to {{Double.POSITIVE_INFINITY}}, by right-shifting both the numerator and denominator synchronously until both numbers fit into 1023 bits. Apart from the fact that the maximum number of bits an integer representable as a finite {{double}} can have is 1024 (an unbiased exponent of 1023, which is the largest possible unbiased exponent of a {{double}} number, means 1.xxxx ⋅ 2^1023^, which amounts to 1024 bits), this way of converting the fraction to a {{double}} is incredibly wasteful with precision if the numerator and denominator have a different bit length, because the smaller of the two numbers will be truncated beyond what is necessary to represent it as a finite {{double}}. Here is an extreme example:

The smallest integer that rounds up to {{Double.POSITIVE_INFINITY}} when converted to a {{double}} is 2^1024^ - 2^970^. This is because {{Double.MAX_VALUE}} as an integer is a 1024-bit number with the most significant 53 bits set to 1 and all other 971 bits set to 0. If the 970 least significant bits are changed in any way, the number will still round down to {{Double.MAX_VALUE}} as long as the 971st bit remains 0, but as soon as the 971st bit is set to 1, the number will round up to {{Double.POSITIVE_INFINITY}}.

The smallest possible denominator greater than 1 where a single right-shift will cause a loss of precision is 3. 2^1024^ - 2^970^ is divisible by 3, so in order to create an irreducible fraction, let's add 1 to it:

(2^1024^ - 2^970^ + 1) / 3 ≈ 5.992310449541053 ⋅ 10^307^ (which can be verified with {{BigDecimal}}, or, more easily, with [this online tool|https://www.wolframalpha.com/input/?i=(2%5E1024+-+2%5E970+%2B+1)+%2F+3]. However, the current implementation of BigFraction.doubleValue() returns 8.98846567431158 ⋅ 10^307^, which differs from the correct result by a relative error of 50%! The same problem applies to the method {{BigFraction.floatValue()}}.

This can be prevented by truncating the numerator and denominator separately, so that for each of the two numbers, the maximum possible precision is retained.



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