Talk:Total circles area: Difference between revisions
→Integral solution?
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main()</lang>
::: Both examples have exactly one unit circle, and the regular method's error is about 8 times smaller and runs 10 times faster. Both being more or less uniform sampling, there's no inherent reason one should be more accurate than the other, but regular sampling is just going to be faster due to simplicity. Unless you can provide good reasoning or meaningful error estimate, it's very inapproriate to claim some method is better at it than others. --[[User:Ledrug|Ledrug]] 08:30, 15 October 2012 (UTC)
::Hi Ledrug, are you saying that the results quoted may be correct for the programs as given on the task page, but are either:
::* Not due to the use Van der Corput?
::* Or peculiar to the example circles chosen?
I must admit that I I only trawled the results stated for the other programs and did not do any extensive reruns with different circles as you have done. I just extracted accuracy vs points. --[[User:Paddy3118|Paddy3118]] 08:56, 15 October 2012 (UTC)
== C89 vs C99 ==
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There are a few concerns about this algorithm: firstly, it's fast only if there are few circles. Its complexity is maybe O(N^3) with N = number of circles, while normal scanline method is probably O(N * n) or less, with n = number of scanlines. Secondly, step 4 needs to be accurate; a small precision error there may cause an arc to remain or be removed by mistake, with disastrous consequences. Also, it's difficult to estimate the error in the final result. The scanline or Monte Carlo methods have errors mostly due to statistics, while this method's error is due to floating point precision loss, which is a very different can of worms. --[[User:Ledrug|Ledrug]] 01:24, 15 October 2012 (UTC)
== Integral solution? ==
Does this solution with the [[wp:Heaviside step function|Heaviside step function]] (<math>H</math>) work?
:<math>\int_{(x,y)\in\mathbb{R}^2} (\prod_{i=1}^N H((r^{(i)})^2 - (x-c_x^{(i)})^2 - (y-c_y^{(i)})^2)) dx dy</math>
If it does maybe then we can write <math>H(x) = \lim_{k\to\infty}\frac{1}{1+e^{-2kx}}</math> and hope for things to simplify somehow?
--[[User:Grondilu|Grondilu]] ([[User talk:Grondilu|talk]]) 12:51, 3 October 2023 (UTC)
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