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dioxin wrote:First, as you're using LOGs and floats it's likely you have numbers stored in floating point format. This will be some representation of the number as:

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sign * Mantisssa  * 2^exponent

Helle wrote:I think, this is fast:

The architectural specification of Java requires IEEE 754 arithmetic, but only a subset of it,

'define a few constants
onethird=0.33333333
onefifth=0.2
oneseventh=0.14285714

k=0.868588964    '2*LOG10(e) for base conversion

'the numer to get the LOG10 of (we assume it's been scaled to between 0.5 and 1.0)
z=0.7500033


'now the calculation
x=(z-1)/(z+1)

x2=x * x
intermediate= x * x2

    answer= x + intermediate * onethird
    intermediate= intermediate * x2
    answer= answer + intermediate * onefifth
    intermediate= intermediate * x2
    answer= answer + intermediate * oneseventh

    answer=answer * k     'convert to BASE10 and scale by 2 as needed by the formula

dioxin wrote:DarkDragon,
I don't know Java but it is very likely to use the same format for it's floats as everywhere else. It would be silly not to as most hardware follows the same floating point standard.

Even if it doesn't, there's no need for a "conversion" to IEE 754. You just need to extract the bits in the mantissa and exponenet from wherever they happen to be.

http://www.javaworld.com/javaworld/jw-1 ... tml?page=2

And here http://www.math.utah.edu/~beebe/software/ieee/ I found this quote:

The architectural specification of Java requires IEEE 754 arithmetic, but only a subset of it,

So it looks like the same IEEE754 spec is used by Java.




Helle,
That's the same basic algorithm I outlined above and you can speed it up in a number of ways.

By extracting the exponent as I mentioned earlier you save that While Z>1 loop with its slow divide each time. This can take a lot of time especially if the initial number is large and DarkDragon did say he wants it fast for large numbers.
As a half-way step, use Z=Z*0.5 instead of the slower Z=Z/2. For large initial Z you'll notice the time saving.

You can also rearrange the arithmetic inside the FOR N loop to reduce the calculations required each loop to less than half of what you're doing. You can get it down to 2 adds and 1 multiply for each iteration of the FOR loop.

You can even do away with the loop completely since 4 iterations gives sufficient accuracy you can just precalculate a lot of the values and assign them as constants.


The entire code, after the scaling, can be reduced to something like this:

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'define a few constants
onethird=0.33333333
onefifth=0.2
oneseventh=0.14285714

k=0.868588964    '2*LOG10(e) for base conversion

'the numer to get the LOG10 of (we assume it's been scaled to between 0.5 and 1.0)
z=0.7500033


'now the calculation
x=(z-1)/(z+1)

x2=x * x
intermediate= x * x2

    answer= x + intermediate * onethird
    intermediate= intermediate * x2
    answer= answer + intermediate * onefifth
    intermediate= intermediate * x2
    answer= answer + intermediate * oneseventh
    intermediate= intermediate * x2

    answer=answer * k     'convert to BASE10 and scale by 2 as needed by the formula

Notice that there is only 1 divide, 9 multiplies and 5 add/subtracts to get accuracy of about 5 decimal digits. That's not bad for a LOG10().

Paul.

dioxin wrote:DarkDragon,
why can't you use pointers? Does Java not have them?

dioxin wrote:DarkDragon,
http://java.sun.com/developer/JDCTechTi ... t1009.html

That certainly suggests to me that unions can be done in Java but I don't do Java so I'll take your word for it.

Another way to approach it would depend on the range of values you need to handle.
If the range is randomly distributed from say 1 to 10,000,000 then, instead of dividing by 2 until the number is small enough you could do some quick checks such as check if it's > 2^20 and if it is then divide by 2^20 and save 19 divides. For randomly distributed data 90% of numbers would be over 10,000,000 so you'd save lots of time that way. e.g.

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if z > 1048576 then     '1048576 = 2^20
    z=z/1048576    'of course, you replace this with the appropriate multiply
    E=E+20
end if

if z > 1024 then                   '1024 = 2^10
    z=z/1024
    E=E+10
end if

if z > 32 then                       '32 = 2^5
    z=z/32
    E=E+5
end if

now use a 32 entry lookup table for the final 5 count rather than looping up to 5 times round a divide by 2 loop.

But this is just speculation as we don't know exactly what your requirements are.

Paul.