RFR of 8032027: Add BigInteger square root methods
brian.burkhalter at oracle.com
Mon Oct 5 19:00:31 UTC 2015
Andrew / Joe,
Thank you for the much appreciated comments.
On Oct 3, 2015, at 11:45 AM, joe darcy <joe.darcy at oracle.com> wrote:
> For an initial implementation, I think it is acceptable to use a simple algorithm that is clearly correct. That algorithm can then be replaced with faster ones once adequate tests are built up (with the original implementation perhaps retiring to the regression tests area where it can be used for comparison purposes).
This was in fact my intention for the initial implementation: something simple that is numerically accurate.
On Oct 3, 2015, at 2:38 AM, Andrew Haley <aph at redhat.com> wrote:
> On 02/10/15 21:41, Brian Burkhalter wrote:
>> Any suggestions as to the improvement of the approach concerning
>> memory use or any other aspect of the algorithm would be
>> appreciated, as would be suggestions regarding the test.
> This algorithm doesn't look like the best to me because it's got this
> slow division in the middle. If we calculate 1/sqrt(x) instead we can
> use a version of Newton's iteration which requires no division.
> Starting with an approximation y = 1/sqrt(x) using the first few digits,
> y = y * (3 - x*y^2)
> ... and fix it all up with a single iteration of Heron's algorithm at the
I will investigate the foregoing and Louis suggestion for this pass of the implementation. Thanks for the most welcome ideas.
> But even better is Karatsuba Square Root:
> I guess it all depends on how much work we think this deserves. In
> general the core algorithms in BigInteger are among the best, and it
> would be nice to continue this tradition.
The Karatsuba Square Root approach by Zimmerman was in fact my original intention and you will find a link to the same work already in the issue on file.
On Oct 3, 2015, at 11:58 AM, joe darcy <joe.darcy at oracle.com> wrote:
> Here is a trick to consider for future performance tuning: all large floating-point numbers are integers. Once the size of the positive exponent exceeds the number of bits of precision, the value must be an integer. For double, that means values greater than 2^54 are integers and the full double exponent range goes out to 2^1023.
> Consequently, if a BigInteger is less than 2^1023 and the spread of 1 bits in the BigInteger is contained within a double , the result could be directly calculated as
> without needing any iterations on the BigInteger side. The bit spread could be able to be calculated from something like
> bi.bitLenth() - bi.getLowestSetBit()
Good observation: I’ll look into this as well.
Thanks for all for the ideas.
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