x = 10, y = 3, rule = B3/S23
3o6bo$obo4b3o$obo6bo!
Have fun.
x = 10, y = 3, rule = B3/S23
3o6bo$obo4b3o$obo6bo!
x = 25, y = 25, rule = B3/S23
b2o$o2bo$o3bo$bo3bo$2bo3bo$3bo3bo$4bo3bo$5bo3bo$6bo3bo$7bo3bo$8bo3bo$
9bo3bo$10bo3bo$11bo3bo$12bo3bo$13bo3bo$14bo3bo$15bo3bo$16bo3bo$17bo3bo
$18bo3bo$19bo3bo$20bo3bo$21bo2bo$22b2o!
Tropylium wrote:This pattern makes (two copies of) a traffic light by creating each blinker separately[/code]
x = 9, y = 7, rule = B3/S23
2bo3b2o$2bo3b2o$3o3b3o2$4bo$4bo$4bo!
x = 5, y = 5, rule = B3/S23
obo$bo2bo$3o$bob2o$ob2o!
x = 8, y = 4, rule = B3/S23
3b4o$2bo$bo5bo$o!
x = 7, y = 7, rule = B3/S23
2o$obo3bo$3o2bo$b4obo$ob4o$2o$2b3obo!
#C N 2G-to-Gs in a spiral absorb 2^N gliders, then kick one back
x = 153, y = 173, rule = B3/S23
101b2o7b2o$101b2o7b2o4$105b2o$105b2o4$98bo$98b3o13b2o$101bo12b2o$95b2o
3b2o6b2o$94bobo10bobo$95bo12bo2$102b2o11b2o$95b2o5b2o11b2o$95b2o2$92bo
$91bobo18b2o$91bobo18b2o$92bo$97b2o$96bobo16b2o$96bo18bobo$95b2o19b2o
4$129b2o$128bo2bo$129b2o$86b2o50bo$76b2o7bobo36bo8b2o2bobo$70b2o4b2o7b
2o37b3o6b2o3b2o$70b2o55bo$81b2o43b2o13b2o$81b2o58bo$59b2o78bobo$59b2o
78b2o10b2o$72b2o60b2o15b2o$72bobo59b2o$73bo$64b2o80b2o$64b2o80b2o$138b
o$76b2o59bobo$59b2o15b2o60b2o$59b2o10b2o78b2o$70bobo78b2o$70bo58b2o$
69b2o13b2o43b2o$84bo55b2o$72b2o3b2o6b3o37b2o7b2o4b2o$72bobo2b2o8bo36bo
bo7b2o$73bo50b2o$81b2o$80bo2bo$81b2o8$111b2o$111b2o3$105b2o$105b2o5$
92b2o$91bobo$93bo11$79b2o$78bobo$80bo11$66b2o$65bobo$67bo11$53b2o$52bo
bo$54bo11$40b2o$39bobo$41bo11$27b2o$26bobo$28bo11$14b2o$13bobo$15bo11$
b2o$obo$2bo!
dvgrn wrote:I can usually resist this kind of thing, but today I noticed that a kickback reaction into one of Guam's snazzy new 2G-to-G converters produces a fairly clean output glider on the same relative lane, so you get a chain reaction.
#C R + blinker = 3190 gens
x = 24, y = 15, rule = B3/S23
2o$b2o$bo12$21b3o!
#C R + blinker = 3319 gens
x = 23, y = 16, rule = B3/S23
2o$b2o$bo11$22bo$22bo$22bo!
x = 21, y = 25, rule = B3/S23
obo$obo$3o20$18b3o$18bobo$18bobo!
x = 22, y = 26, rule = B3/S23
10b2o$10b2o4$2o$bo$bobo11bo$2b2o10bo$14b3o6$4b2o$4b2o2$15b2obo$15b2ob
3o$21bo$5bob2o6b2ob3o$3b3obo4bo3bobo$2bo4bobobobo2bobo$3b3obob2o2bo3bo
$5b2o6b2o!
MikeP wrote:I think of this one as the "Homer Simpson" reflector.
x = 100, y = 33, rule = B3/S23
2bo7b2o$3o7b2o$obo$o$11bo$10bobo$10bobo$11bo27b2o$4b2o33b2o$4bobo$6bo$
6b2o32bo$39bobo$39bobo$40bo27b2o$33b2o33b2o$33bobo$35bo$35b2o32bo$68bo
bo$68bobo$69bo27b2o$62b2o33b2o$62bobo$64bo$64b2o32bo$97bobo$97bobo$98b
o$91b2o$91bobo$93bo$93b2o!
x = 17, y = 13, rule = B3/S238
5bo$4bo$4b3o$3o$o$bo12bobo$15b2o$15bo3$12b3o$14bo$13bo!
DivusIulius wrote:I'm not sure if the gliders come from infinity. How might I verify this?
x = 15, y = 33, rule = B3/S238
13bo$12bo$12b3o3$2bobo$3b2o$3bo16$8b3o$8bo$9bo5$3o$2bo$bo!
x = 21, y = 12, rule = B3/S238
$9bo$8bo$8b3o$4b3o12bo$4bo13bo$5bo12b3o3$17bo$16b2o$16bobo!
dvgrn wrote:I was very surprised to see so many pulsars in the final output, by the way -- until the extra "8" in the rule string caught my eye!
skomick wrote:These gliders come from infinity though, producing the same result
x = 5, y = 25, rule = B3/S23
2o$b2o$bo20$3bo$2b2o$3b2o!
x = 4, y = 6, rule = B3/S23
b2o$o2bo$ob2o$o$b2o$2bo!
x = 6, y = 8, rule = B3/S23
2o$2o5$b3o$3b3o!
x = 5, y = 9, rule = B3/S23
3b2o$bo2bo$bo2bo$bobo4$2o$2o!
x = 9, y = 5, rule = B3/S23
2o$2o3b3o2$5bo2bo$6b3o!
x = 38, y = 24, rule = B3/S23
34bo2bo$33bo$33bo3bo$33b4o3$9b2o$9b3o$8bob2o$8b3o$9bo3$28bo$27b3o$26b
2obo$26b3o$27b2o3$b4o$o3bo$4bo$o2bo!
x = 5, y = 2, rule = B3/S23
3b2o$3obo!
x = 5, y = 2, rule = B3/S23
2bobo$5o!
x = 7, y = 2, rule = B3/S23
2bobobo$7o!
x = 5, y = 3, rule = B3/S23
b4o$b2obo$2o2bo!
x = 4, y = 4, rule = B3/S23
b3o$3bo$2b2o$3o!
x = 13, y = 12, rule = B3/S23
obo$b2o$bo3$12bo$10b2o$11b2o2$8bo$8b2o$7bobo!
x = 21, y = 30, rule = B3/S23
20bo$18b2o$19b2o23$2bo$obo$b2o2bo$5bobo$5b2o!
#C Pi + R = 5548 generations to stabilise
x = 21, y = 23, rule = B3/S23
o$bo$bo$2o17$19b2o$18b2o$19bo!
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