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From Friday, April 19th (11:00 PM CDT) through Saturday, April 20th (2:00 PM CDT), 2024, ni.com will undergo system upgrades that may result in temporary service interruption.
We appreciate your patience as we improve our online experience.
04-26-2005 11:52 AM
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07-08-2005 04:49 PM
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07-10-2005 01:47 PM
Thanks!
Yes, I will upload my code, but I since I have no access to my home PC for the next weeks it may take some time. (Actually, I thought that NI would make the code available on the results page, as they had done it in the past).
It is probably not too interesting to look at my code, anyway, since Bruce Ammons solution is so much faster, and I only won since he was not in the judging. Bruce, if you read this: Could you please post your solution in LV 7.0, since I have no access to 7.1 at the moment?
To add to what was said on the results page already:
I used FFT and inverse FFT for the bigint multiply, not the convolution VIs. I also limited the precision of intermediate results, since the Newton iteration typically doubles the number of valid digits in each step and there is no need to calculate more digits than a given iteration step will produce. Starting with an initial value obtained by EXT floating point math, I have typically 14..15 digits of precision initially, which is 3 'digits' in the chosen 10^5 base. One obtains 3*2^i 'digits' after iteration step i. For a 10000 digit decimal number (2000 'digits' in 10^5 base), the square root has 1000 'digits' and should be obtained after 9 iterations (3*2^8<1000, but 3*2^9 > 1000).
I'd really like to see what makes Bruce's solution 15 times faster, since I did not see much where I could speed up my code.
greetings from Taos, New Mexico
Franz
07-14-2005 10:39 AM