A question to Raul please: I really liked your simplified approach to plot
against X^4 + y^4 <= 1 using:
require ‘viewmat’
viewmat 1>: +&|/~ ((i:300)%200)^4
But I noticed that since this is to the 4th power, why do you invoke +&| ?
I see the same result using:
viewmat 1>: +/~ ((i:300)%200)^4
Thanks, Rob
> On 17 Aug 2022, at 8:37 am, Elijah Stone <elro...@elronnd.net> wrote:
>
> And the visualisation part (mostly just getting a screenful of normalised
> coordinates to apply to):
>
> ' #' {~ 0 >: sdsquircle"1] 25 50 %"1~ _25 _50 +"1] ($ (#: i.@(*/))) 51 101
>
> Ratio of 50 to 100 just accounts for aspect ratio; should be 1:1 if working
> with square pixels.
>
> Also, sdsquircle _does_ seem to actually be a signed distance function, if
> you halve it. I don't think this is true in general, but relies on some
> specific properties of squares and circles (convexity?).
>
> On Mon, 15 Aug 2022, Elijah Stone wrote:
>
>> Was about to go to bed, but this tickled my imagination, so I will say a
>> little something.
>>
>> Rather than calculus, I would do this as an implicit curve.
>>
>> Start off with the signed distance function of a circle:
>>
>> sdc=. ]: -~ +/&.:*:
>>
>> Then that of a box, taken from
>> https://iquilezles.org/articles/distfunctions2d/:
>>
>> sdb=. (+/&.:*:@:(0&>.) + 0 <. >./)@:(]: -~ |)
>>
>> Both of these are adverbs, taking a radius as an operand and then a vector
>> coordinate as an argument. (The latter can also be given a vector, in which
>> case it calculates the distance to an n-dimensional rectangle, but that is
>> irrelevant here.)
>>
>> Then, we are looking for the case when the distance to the box is equal to
>> the distance to the circle. Such points will be inside the square, but
>> outside the circle, so the distance functions of those shapes will have
>> opposite sign, and we can just add them together:
>>
>> sdsquircle=. 1 sdc + 1 sdb
>>
>> sdsquircle is not, strictly speaking, a signed distance function. However,
>> like a signed distance function, it has a value of 0 when applied to a
>> coordinate on the squircle; and it is negative inside, and positive outside.
>> So it is trivial to render the shape from it.
>>
>> I can expand further on this tomorrow, but I really must be getting to bed
>> now.
>>
>> -E
>>
>> On Mon, 15 Aug 2022, Richard Donovan wrote:
>>
>>> Hi
>>>
>>> I want to construct and plot a Squircle in J.
>>>
>>> There is a lengthy article in Wikipedia but in simple language I want my
>> Squircle to be defined as the continuous line between a unit circle and the
>> unit square that encloses it such that every point on the Squircle is the
>> mean of the nearest points of the circle and the square.
>>>
>>> Thus, the mean is zero at the four points where the circle and the square
>> touch, and a maximum of (-: @ <: @ %:2) at the four corners of the square.
>>>
>>> Each intermediate point between 0 degrees and 90 degrees will be somewhere
>> in the middle.
>>>
>>> I suspect the calculation of the intermediate points is a calculus
>> function?
>>>
>>> Has anyone a good idea for performing that calculation? Could the J
>> function “ plot “ then draw the Squircle?
>>>
>>> Thanks
>>>
>>> Richard Donovan
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