Clouds with p256 glider guns

[skomick]

Was it Dean Hickerson who created the patterns "life computes pi" and the related pattern "clouds?"

Here are the p256 glider gun versions I created to run fast with Golly's hashlife. I wanted to see more "clouds" faster. I don't know if the p256 version of "life computes pi" actually computes pi, but it's neat to watch in Golly anyway. I've also included the p256 glider gun breeder itself.

p256 Glider Gun Breeder.zip

(190.57 KiB) Downloaded 369 times

The .zip file contains 3 .mc files.

[dvgrn]

Was it Dean Hickerson who created the patterns "life computes pi" and the related pattern "clouds?"

Yes. Here's the original pi-calculating pattern from Hickerson's web page, and the notes on 'Clouds':

http://www.radicaleye.com/DRH/pi.html
http://www.radicaleye.com/DRH/clouds.html

I don't know if the p256 version of "life computes pi" actually computes pi, but it's neat to watch in Golly anyway.

Sure it does! The breeders still travel at c/2, so the various triangular areas are the same shape as in the original pattern. The only significant difference is the population density in those triangular areas, but that just means you have to use 5/8192 instead of 1/90 in the final formula. The breeders and guns have much higher populations than the original p30 ones, of course, but as the area increases their contribution to the total population becomes negligible.

Here are the adjusted comments for the 'life computes pi p256' pattern:

Code: Select all

#C Population in gen t is asymptotic to  5/65536 (pi-2) t^2.
#C Four p120 breeders move north, east, south, and west.
#C The gliders from each p256 gun run into those from the
#C corresponding gun that's 90 degrees clockwise, in a
#C vanish reaction.  Let T(n) be the triangle with vertices
#C (0,0), (0, t/(2n)), and (t/(2(n+2)), 0).  Within the first
#C quadrant, the gliders occupy, with density 5/8192, the
#C triangle T(1), with the triangle T(3) deleted, with the
#C triangle T(5) added back in, with the triangle T(7)
#C deleted, etc.  Since T(n) has area  t^2/(8n(n+2)) =
#C t^2/16 (1/n - 1/(n+2)),  the occupied area in the
#C first quadrant is
#C
#C     t^2/16 ((1 - 1/3) - (1/3 - 1/5) + (1/5 - 1/7) - ...)  
#C   = t^2/16 (1 - 2/3 + 2/5 - 2/7 + ...)  =  t^2/32 (pi-2).
#C
#C  The same is true of the other quadrants, so the
#C  population converges to 4 * 5/8192 * t^2/32 (pi-2) 
#C    =  5/65536 (pi-2) t^2.

[skomick]

I don't know if the p256 version of "life computes pi" actually computes pi, but it's neat to watch in Golly anyway.

Sure it does!

Very nice! Thanks!

[beebop]

The link up top is broken.

[cloudy197]

The link up top is broken.

Try this one.

[beebop]

Thanks for the new link.