Contesting for Idiot du Jour

Pi Digital sean at pidigital.co.uk
Mon Sep 7 05:04:31 EDT 2020


Sure. Draw a circle on a 10x10, 20x20, 100x100, etc grid. Only Whole pixels get counted (Pass or 1 in digital binary). Depending on the methodology, either 1) only those within the circle line (complete) or 2) those on the line itself  and within it (incomplete). 

In your example, a 200x200 circle has a resolution where it’s practically negligible from regular mathematics. However, the resolution at 100x100 and lower starts to flat wildly away from it. If you are measuring using collision and it’s accounting for antialiased pixels it can become even more diverse from standard math as it does not ‘see’ it in percentages of visible, only on or off. 

So, the difference between measuring only inside of a 200x200 and outside of the 100x100 will throw off considerably any ordinary calculations you might expect. Even a 400x400 PixelWise ‘circle’. 

Look up Gauss’ Circle Problem. The same chap we get the name for Gaussian blur from. 

> On 7 Sep 2020, at 05:26, Roger Guay via use-livecode <use-livecode at lists.runrev.com> wrote:
> 
> I’m sorry, I don’t understand your terminology. Could you please elaborate? 
> 
> Thanks,
> Roger
> 
>> On Sep 6, 2020, at 10:54 AM, Pi Digital via use-livecode <use-livecode at lists.runrev.com> wrote:
>> 
>> Pixel math:
>> 
>> Counting incomplete pixels within a circle outline (%Pass)(%Fail):
>> 10x10 = 88 (88%)(12%)
>> 20x20 = 344 (86%)(14%)
>> 100x100 = 8012 (80%)(20%)
>> 
>> Counting complete pixels:
>> 10x10 = 48 (48%)(52%)
>> 20x20 = 276 (69%)(31%)
>> 100x100 = 7444 (74.4%)(26%)
>> 
>> Your conclusion here: _________________________
>> 
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