Hi,
for part 1, I used the properties of the spiral pattern to find the
distance.
The spiral matrix can be viewed as a series of concentric rings with the
last number of a ring being a perfect square.
Calculating the ring number gives me one part of the distance.
Finding the distance to the nearest center of a side gives me the other
part.
|.3 WRAP 1
37 36 35 34 33 32 31
38 17 16 15 14 13 30
39 18 5 4 3 12 29
40 19 6 1 2 11 28
41 20 7 8 *9* 10 27
42 21 22 23 24 *25* 26
43 44 45 46 47 48 *49*
Quick and almost dirty:
dtc NB. distance to center
3 : '(<./|y - csr r)+r=.rn y'
csr NB. side centers
3 : '((_1+t%2)+fsr y)+0 1 2 3*t=. 4%~ nsr y'
fsr NB. first
3 : '(*y)+lsr y-1'
nsr NB. number of elements in ring
3 : '1+(lsr - fsr) y'
lsr NB. last
3 : '2^~1+2*y'
rn NB. ring number
3 : '>.-:<:%:y'
tsr NB. checking for corner cases
3 : '(''ring #'';''first'';''last'';''number'';''centers''),:(,. y);(,. fsr
y);(,. lsr y);(,. nsr y);(csr"0 y)'
tsr (i.9),0, 300+i.2
┌──────┬──────┬──────┬──────┬───────────────────────────┐
│ring #│first │last │number│centers │
├──────┼──────┼──────┼──────┼───────────────────────────┤
│ 0 │ 1│ 1│ 1 │ 0.125 0.375 0.625 0.875│
│ 1 │ 2│ 9│ 8 │ 2 4 6 8│
│ 2 │ 10│ 25│ 16 │ 11 15 19 23│
│ 3 │ 26│ 49│ 24 │ 28 34 40 46│
│ 4 │ 50│ 81│ 32 │ 53 61 69 77│
│ 5 │ 82│ 121│ 40 │ 86 96 106 116│
│ 6 │ 122│ 169│ 48 │ 127 139 151 163│
│ 7 │ 170│ 225│ 56 │ 176 190 204 218│
│ 8 │ 226│ 289│ 64 │ 233 249 265 281│
│ 0 │ 1│ 1│ 1 │ 0.125 0.375 0.625 0.875│
│300 │358802│361201│2400 │359101 359701 360301 360901│
│301 │361202│363609│2408 │361502 362104 362706 363308│
└──────┴──────┴──────┴──────┴───────────────────────────┘
On Sun, Dec 3, 2017 at 9:27 PM, Joe Bogner <[email protected]> wrote:
> Hi Jimmy - I'm also working with WRAP from the phrases. I messed around
> with trying to understand https://oeis.org/A054552 but it seemed more
> complicated than I wanted to undertake. Here's my exploration of my
> solution for part 1
>
> I'm using this currently to locate the x/y coordinate (probably a better
> way?) within a grid
>
> ((#~ *)@:, ,. (I.@:*@:,)) I. (0&=)
>
> ((#~ *)@:, ,. (I.@:*@:,)) I. (0&=) (3 WRAP 0)
> 3 3
>
> (3 WRAP 0)
> 42 43 44 45 46 47 48
> 41 20 21 22 23 24 25
> 40 19 6 7 8 9 26
> 39 18 5 0 1 10 27
> 38 17 4 3 2 11 28
> 37 16 15 14 13 12 29
> 36 35 34 33 32 31 30
>
>
> For my problem -- 347991 -- what's the smallest square grid to satisfy it?
>
> %: 347991
>
> 589.908
>
>
> What argument for wrap is needed?
>
>
> -: %: 347991
>
> 294.954
>
>
> Find my target:
>
>
> ((#~ *)@:, ,. (I.@:*@:,)) I. (347991&=) (1+(295 WRAP 0))
>
> 110 590
>
>
>
>
> Try the smaller just to be sure (returns blank)
>
> ((#~ *)@:, ,. (I.@:*@:,)) I. (347991&=) (1+(294 WRAP 0))
>
>
> Find origin (not really needed but still interesting to me)
>
>
> ((#~ *)@:, ,. (I.@:*@:,)) I. (1&=) (1+(295 WRAP 0))
>
> 295 295
>
>
> Calc manhattan distance
>
>
> (| (110-295)) + (|(590-295))
>
> 480
>
>
> My answer was 480
>
>
> Not particularly pretty but got to the right answer... It needs more work
> to be reusable
>
>
>
>
>
> On Sun, Dec 3, 2017 at 9:01 PM, Jimmy Gauvin <[email protected]>
> wrote:
>
> > Hi,
> >
> > has anybody found something better than a "do. while." to calculate the
> > spiral neighborhood count?
> >
> > Is there some underlying mathematical structure that I am not aware of?
> >
> >
> >
> > Thanks,
> >
> >
> >
> >
> > WRAP is from http://www.jsoftware.com/help/phrases/grids.htm
> >
> > spw is the result of the neighbor calculation
> >
> >
> > (|.2 WRAP 1){spw
> >
> > 147 142 133 122 59
> >
> > 304 5 4 2 57
> >
> > 330 10 1 1 54
> >
> > 351 11 23 25 26
> >
> > 362 747 806 880 931
> > ----------------------------------------------------------------------
> > For information about J forums see http://www.jsoftware.com/forums.htm
> ----------------------------------------------------------------------
> For information about J forums see http://www.jsoftware.com/forums.htm
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