Question

I’m dealing with the BigInteger class with numbers in the order of 2 raised to the power 10,000,000.

The BigInteger Log function is now the most expensive function in my algorithm and I am desperately looking for an alternative.

Since I only need the integral part of the log, I came across this answer which seems brilliant in terms of speed but for some reason I am not getting accurate values. I do not care about the decimal part but I do need to get an accurate integral part whether the value is floored or ceiled as long as I know which.

Here is the function I implemented:

public static double LogBase2 (System.Numerics.BigInteger number)
{
    return (LogBase2(number.ToByteArray()));
}

public static double LogBase2 (byte [] bytes)
{
    // Corrected based on [ronalchn's] answer.
    return (System.Math.Log(bytes [bytes.Length - 1], 2) + ((bytes.Length - 1) * 8));
}

The values are now incredibly accurate except for corner cases. The values 7 to 7.99999, 15 to 15.9999, 23 to 23.9999 31 to 31.9999, etc. return -Infinity. The numbers seem to revolve around byte boundaries. Any idea what’s going on here?

Example:

LogBase2(                    1081210289) = 30.009999999993600 != 30.000000000000000
LogBase2(                    1088730701) = 30.019999999613300 != 30.000000000000000
LogBase2(                    2132649894) = 30.989999999389400 != 30.988684686772200
LogBase2(                    2147483648) = 31.000000000000000 != -Infinity
LogBase2(                    2162420578) = 31.009999999993600 != -Infinity
LogBase2(                    4235837212) = 31.979999999984800 != -Infinity
LogBase2(                    4265299789) = 31.989999999727700 != -Infinity
LogBase2(                    4294967296) = 32.000000000000000 != 32.000000000000000
LogBase2(                    4324841156) = 32.009999999993600 != 32.000000000000000
LogBase2(                  545958373094) = 38.989999999997200 != 38.988684686772200
LogBase2(                  549755813887) = 38.999999999997400 != 38.988684686772200
LogBase2(                  553579667970) = 39.009999999998800 != -Infinity
LogBase2(                  557430119061) = 39.019999999998900 != -Infinity
LogBase2(                  561307352157) = 39.029999999998300 != -Infinity
LogBase2(                  565211553542) = 39.039999999997900 != -Infinity
LogBase2(                  569142910795) = 39.049999999997200 != -Infinity
LogBase2(                 1084374326282) = 39.979999999998100 != -Infinity
LogBase2(                 1091916746189) = 39.989999999998500 != -Infinity
LogBase2(                 1099511627775) = 39.999999999998700 != -Infinity

Answer 1

Try this:

public static int LogBase2(byte[] bytes)
{
    if (bytes[bytes.Length - 1] >= 128) return -1; // -ve bigint (invalid - cannot take log of -ve number)
    int log = 0;
    while ((bytes[bytes.Length - 1]>>log)>0) log++;
    return log + bytes.Length*8-9;
}

The reason for the most significant byte being 0 is because the BigInteger is a signed integer. When the most significant bit of the high-order byte is 1, an extra byte is tacked on to represent the sign bit of 0 for positive integers.

Also changed from using the System.Math.Log function because if you only want the rounded value, it is much faster to use bit operations.

---

If you have Microsoft Solver Foundation (download at http://msdn.microsoft.com/en-us/devlabs/hh145003.aspx), then you can use the BitCount() function:

public static double LogBase2(Microsoft.SolverFoundation.Common.BigInteger number)
{
    return number.BitCount;
}

---

Or you can use the java library. Add a reference to the vjslib library (found in the .NET tab – this is the J# implementation of the java library).

You can now add “using java.math” in your code.

java.math.BigInteger has a bitLength() function