Dividing a BigIntegers to return double

Question

I want to calculate the slope of a line.

public sealed class Point
{
    public System.Numerics.BigInteger x = 0;
    public System.Numerics.BigInteger y = 0;

    public double CalculateSlope (Point point)
    {
        return ((point.Y - this.Y) / (point.X - this.X));
    }
}

I know that BigInteger has a DivRem function that returns the division result plus the remainder but am not sure how to apply it to get a double. The numbers I'm dealing with are far far beyond the range of Int64.MaxValue so the remainder itself could be out of range to calculate by conventional division.

EDIT: Not sure if it helps but I'm dealing with only positive integers (>=1).

IMPORTANT: I only need a few decimal points of precision (5 should be good enough for my purpose).

Answer 1

Get BigRational from Codeplex. Its part of Microsoft's Base Class Library, so it's a work-in-progress for .Net. Once you have that, then do something like:

System.Numerics.BigInteger x = GetDividend() ;
System.Numerics.BigInteger y = GetDivisor() ;

BigRational r     = new BigRational( x , y ) ;
double      value = (double) r ;

Dealing with the inevitable overflow/underflow/loss of precision is, of course, another problem.

Since you can't drop the BigRational library into your code, evidently, the other approach would be to get out the right algorithms book and roll your own...

The easy way, of course, of "rolling one's own" here, since a rational number is represented as the ratio (division) of two integers, is to grab the explicit conversion to double operator from the BigRational class and tweak it to suit. It took me about 15 minutes.

About the only significant modification I made is in how the sign of the result is set when the result is positive or negative zero/infinity. While I was at it, I converted it to a BigInteger extension method for you:

public static class BigIntExtensions
{

  public static double DivideAndReturnDouble( this BigInteger x , BigInteger y )
  {
    // The Double value type represents a double-precision 64-bit number with
    // values ranging from -1.79769313486232e308 to +1.79769313486232e308
    // values that do not fit into this range are returned as +/-Infinity
    if (SafeCastToDouble(x) && SafeCastToDouble(y))
    {
      return (Double) x / (Double)  y;
    }

    // kick it old-school and figure out the sign of the result
    bool isNegativeResult = ( ( x.Sign < 0 && y.Sign > 0 ) || ( x.Sign > 0 && y.Sign < 0 ) ) ;

    // scale the numerator to preseve the fraction part through the integer division
    BigInteger denormalized = (x * s_bnDoublePrecision) / y ;
    if ( denormalized.IsZero )
    {
      return isNegativeResult ? BitConverter.Int64BitsToDouble(unchecked((long)0x8000000000000000)) : 0d; // underflow to -+0
    }

    Double result   = 0              ;
    bool   isDouble = false          ;
    int    scale    = DoubleMaxScale ;

    while ( scale > 0 )
    {
      if (!isDouble)
      {
        if ( SafeCastToDouble(denormalized) )
        {
          result = (Double) denormalized;
          isDouble = true;
        }
        else
        {
          denormalized = denormalized / 10 ;
        }
      }
      result = result / 10 ;
      scale-- ;
    }

    if (!isDouble)
    {
      return isNegativeResult ? Double.NegativeInfinity : Double.PositiveInfinity;
    }
    else
    {
      return result;
    }

  }

  private const           int        DoubleMaxScale      = 308 ;
  private static readonly BigInteger s_bnDoublePrecision = BigInteger.Pow( 10 , DoubleMaxScale ) ;
  private static readonly BigInteger s_bnDoubleMaxValue  = (BigInteger) Double.MaxValue;
  private static readonly BigInteger s_bnDoubleMinValue  = (BigInteger) Double.MinValue;

  private static bool SafeCastToDouble(BigInteger value)
  {
    return s_bnDoubleMinValue <= value && value <= s_bnDoubleMaxValue;
  }

}

Answer 2

The BigRational library has a conversion operator to double.

Also, remember to return infinity as a special case for a vertical line, you'll get a divide by zero exception with your current code. Probably best to just calculate the X1 - X2 first, and return infinity if it's zero, then do the division, to avoid redundant operations.