danny wrote:Wait I'm confused. How does this turn into synthesize a pulsar? Wouldn't there be a heckton of ash at the end? I'm sorry if I'm misunderstanding the thread's contents or anything, I just have lots of questions...
calcyman wrote:Can you please provide the pattern at generation 0 instead of 5000? Thank you!
x = 5383, y = 5268, rule = B3/S23
120bo$121b2o$120b2o2$5377bobo$5377b2o$5378bo80$2806bo$2805bo$2805b3o
25$288bo$289b2o$288b2o2$2783bobo$2783b2o$2784bo39$172bo$173b2o$172b2o
206$227bobo$228b2o$228bo21$248bo$249b2o$248b2o2$3123bobo$3123b2o$3124b
o11$238bo$239b2o$238b2o1949$123bo$124b2o$123b2o18$2652bobo$2652b2o$
154bo2498bo$155bo$153b3o39$185bo$183bobo$184b2o17$2714bo$2713bo$2713b
3o$214bo$215b2o$214b2o5$2198bobo$2199b2o$2199bo96$2752bo$2751bo$2751b
3o67$2786b2o$2786bobo$2786bo36$176bo$176b2o$175bobo$2676b2o$2676bobo$
2676bo18$145b3o$147bo$146bo593$3126b2o$3126bobo$3126bo10$242bo$242b2o$
241bobo$3124b2o$3124bobo$3124bo1585$150b2o$149bobo$151bo61$211bo$211b
2o$210bobo10$2201b3o$2203bo$2202bo$4697b2o$4697bobo$4697bo25$2179b2o$
2180b2o$2179bo67$2749b2o$2748b2o$2750bo$253bo$253b2o$252bobo25$2771b2o
$2771bobo$2771bo25$3o$2bo$bo158$5380b2o$5380bobo$5380bo!
danny wrote:Wait I'm confused. How does this turn into synthesize a pulsar? Wouldn't there be a heckton of ash at the end? I'm sorry if I'm misunderstanding the thread's contents or anything, I just have lots of questions...
dvgrn wrote:Not to worry -- it's not just you, it really is getting painfully confusing! Now is probably a good time for a fresh summary of how to get to a pulsar (or any other constructible object) with these 35 ultra-clever gliders.
I'm hoping I can get close, but someone may need to supply a few corrections.
dvgrn wrote:Step 8: Now it's time to build Painfully Complicated Seed Constellations #1 and #2. When PCSC#1 is triggered, it will produce several Corderships traveling along the various ash tracks in various directions, cleaning up some part of them, or let's just say all of them (I think that will be doable now)... and then crashing into the messy junk at the various GPSE and Sakapuffer construction locations without releasing any additional gliders -- except for one or two heading back toward PCSC#2.
x = 716, y = 639, rule = B3/S23
99bo$88b2o8bobo$88b2o8b2o6$107bo$96b2o8bobo$96b2o8b2o6$115bo$104b2o8bo
bo$104b2o8b2o6$123bo$112b2o8bobo$112b2o8b2o6$131bo$120b2o8bobo$120b2o
8b2o6$139bo$128b2o8bobo$128b2o8b2o6$147bo$136b2o8bobo$136b2o8b2o6$155b
o$144b2o8bobo$144b2o8b2o6$163bo$152b2o8bobo$152b2o8b2o6$171bo$160b2o8b
obo$160b2o8b2o4$196b3o$205bo3b2o$179bo23b2ob5o$168b2o8bobo22b2o3bo$
168b2o8b2o20b2o$199bobob4o$199bo3bo3bo$184bobo20b3o$208b2o$183bo17bo2b
2o11b2o$188bo10b2obob2o11b2o$176b2o7b2obo10bo$176b2o9bo12b4o$201bo4$
217b2o$216b2o4b3o$184b2o31b2o$184b2o32b3o$219bobo$219bo2bo$222bo$220bo
bo$220b3o2$192b2o$192b2o2$219b2o$218bo2bo$206b2o2b3o5b2o$206bobo10b5o$
206bob2o13b2o$202b2ob2obo11b2o2b2o$201b2o2bo14bo2b2o$205b2o14b3o$222bo
6$202b2o12bo$201b2o13bo$201b3o11bo$209bo6b2obo$203bob2obobo5b2ob2o$
203bo2bo2bo7bobo$203bobo59$298b2o$298b2o18$271b3o$271bo$272bo88$354b2o
$354b2o310$154b2o$153bobo$155bo25$125b2o$2o122bobo$b2o123bo587b2o$o
712b2o$715bo!
import golly as g
n = 64 * int(g.getstring("Enter steps to lengthen or shorten by:"))
x, y, w, h = g.getrect()
g.select([x, y, w // 4, h])
cells = g.getcells(g.getselrect())
g.clear(0)
g.putcells(cells, -n, 0)
g.select([x + w - w // 4, y, w // 4, h])
cells = g.getcells(g.getselrect())
g.clear(0)
g.putcells(cells, n, 0)
g.select([x - n, y, w + 2 * n, h // 4])
cells = g.getcells(g.getselrect())
g.clear(0)
g.putcells(cells, 0, -n)
g.select([x - n, y + h - h // 4, w + 2 * n, h // 4])
cells = g.getcells(g.getselrect())
g.clear(0)
g.putcells(cells, 0, n)
g.select([])
g.fit()
Could you give a link to Elkies' outline?calcyman wrote:If we manage to produce a proof (following Elkies' outline) that every still-life is glider-constructible, then we're left with a finite amount of work to show that every still-life can be constructed in <= 1 glider per cell.
Macbi wrote:Could you give a link to Elkies' outline?calcyman wrote:If we manage to produce a proof (following Elkies' outline) that every still-life is glider-constructible, then we're left with a finite amount of work to show that every still-life can be constructed in <= 1 glider per cell.
On 10th August 1994, Mark Niemiec wrote:Mainly to Harold:
Thanks for the papers on de Bruijn diagrams. Now, a lot of stuff
(at least the basic ideas) are starting to make a little bit of sense.
I am currently working on a related task, attempting to prove that
every still-life (finite or infinite) may be cut along an arbitary
horizontal boundary and everything below the line replaced by a small
finite stabilizer. If this can be proven rigorously (and I believe that it
can), we can prove that all infinite still-lifes can be made finite.
This is done by dividing up the universe into regions
...
aaaaa
aaaaa
aaaaa
bbbbb
ccccc
ddddd
eeeee
eeeee
eeeee
...
where 'a' is the interior of the still-life, is already stable, and has no
effect on the exterior.
'b' is just below the surface, is already stable, and influences row 'c',
'c' is at the surface, and may require stabilization from external row 'd',
'd' consists of cells 'outside' the still-life which we need to add to
stabilize row 'c',
and 'e' consists of anything else we need to stabilize 'd'.
By definition, for every still-life which is cut in this way, for every
sequence
'b' and 'c' in that still-life, there must exist at least one sequence 'd'
(i.e.
the row which occurs just below 'c' if the still-life is not cut).
The idea is to prove that for every possible sequence of 'b' and 'c', some
arbitrary 'd' can be chosen which stabilizes 'c', and is also itself
stabilizable by some arbitary 'e'.
The proof is somewhat similar to examining every edge in a de Bruijn
diagram of a (4,2) automata (the states 0,1,2,3 correspond to no cells,
'b' cell, 'c' cell, and 'b+c' cells being on in any specific column.) to
ensure that such edges preserve the 'c' cell. Wherever this is not
possible, any path leading through that edge must be re-routed through
a parallel edge connecting altered versions of the initial states.
We may replace states 0-3 by new states 4-7 as required (like 0-3
but with an additional 'd' cell.). Since 'd' cells may not always be
placed with impunity, finding optimal alternate paths requires some
trial and error.
So far, I have expanded a fair bit of the tree, and it seems like a
quite possible (although fairly tedious task.) I'lll mail the results
when (if?) I get finished.
A slightly more ambitius task which requires the above as a pre-requisite is
to define all such states at the exterior of a partially-constructed
still-life,
and find all mappings between such states and the corresponding states
in which one more cell is added to the interior:
... ...
aaaaa aaaaa
aaaaa aaaaa
abbbb aabbb
bbccc -> bbbcc
cccdd ccccd
ddddE ddddD
eeeee eeeee
eeeee eeeee
... ...
If cell E is the same as cell D, no translation is needed;
if they differ, a glider construction is required which flips
the state of the single bit (and possibly alters other bits
in 'e'); this will, of course, have no effect on the already-
constructed parts of the still-life 'a', 'b', and 'c'; only
'd' needs to be taken care of by taking care how 'e' is altered.
In the majority of cases in which 'e' is empty, this is not
likely to be a problem. However, if the local area of 'e'
contains a convoluted stabilizer, mucking with an inner cell
could be quite a problem. (On the other hand, such convoluted
stabilizers are usually forced by one obnoxious cell; if
that cell is the one which needs to be flipped, most of
the stabilizer could be removed as well.)
If every such state-transition can be provided with a glider
synthesis, we will have a proof that every still-life can be
constructed from gliders, one bit at a time.
(But don't hold your breath; this could involve thousands of
cases, and even though Dave Buckingham is the best person to
generate the syntheses, I am sure he is not enough of a
masochist to try to find them all!) Perhaps some kind of
automated search will be required here.
Has anyone else researched this area? (or determined that it is
either too trivial or intractible to bother?)
-- MDN
chris_c wrote:Excellent. Thanks for finding that and for clarifying those other issues. The 35 glider synthesis was pretty easy to obtain from the 43 glider version. Just 6 GPSE's now and the minimum population is 143 after 5000 generations:
calcyman wrote:I've just realised that this implies the existence of a 143-cell sawtooth. The smallest explicitly-constructed sawtooth has 177 cells, by comparison.
dvgrn wrote:calcyman wrote:I've just realised that this implies the existence of a 143-cell sawtooth. The smallest explicitly-constructed sawtooth has 177 cells, by comparison.
Seems like it would be a fairly complicated implication, though. Just the fact that a 143-cell pattern eventually constructs, let's say, a 177-cell sawtooth pattern, doesn't mean that the 143-cell pattern actually exhibits sawtooth-ish behavior starting from T=0. It will start acting like a sawtooth after umpteen bajillion ticks, but I'm not sure that counts.
Don't we have predecessors of sawtooth 177 that have fewer than 177 cells, for example (e.g., put a block instead of a boat in the boat-bit positions, and then take a cell out of all the blocks)?
To be a proper sawtooth, doesn't it have to return to 143 cells at ever-increasing intervals? How would that work in an RCT context, exactly?
calcyman wrote:dvgrn wrote:To be a proper sawtooth, doesn't it have to return to 143 cells at ever-increasing intervals? How would that work in an RCT context, exactly?
It synthesises those 35 gliders, except with greater separation, such that there's a new NOP on the end of the tape each time.
dvgrn wrote:-- Ouch, this stuff still hurts to think about... maybe mostly because I really like to be able to build working examples of things, and even the simplest "simple RCT" design is just hopeless.
calcyman wrote:quincunx*
...
* it's so useful this word exists for the 5-spot configuration on the face of a die or domino.
x = 418, y = 345, rule = LifeHistory
265.2B$265.3B$265.4B$266.4B$267.4B$268.4B$265.B3.4B$265.2B3.4B$265.3B
3.4B$265.4B3.4B$266.4B3.4B$267.4B3.4B$268.4B3.4B$269.4B3.4B$270.4B3.
4B$271.4B3.BABA$272.4B3.B2AB$273.4B3.A3B$274.4B3.4B35.2C$275.4B3.4B
34.2C$276.4B3.4B$277.4B3.4B$278.4B3.4B$279.4B3.4B$280.4B3.4B$281.4B3.
4B$282.4B3.4B$283.4B3.4B$284.4B3.4B$285.4B3.4B$286.4B3.4B$287.4B3.4B$
288.4B3.4B$289.4B3.4B$290.4B3.4B$291.4B3.4B$292.4B3.4B$293.4B3.4B$
294.4B3.4B$295.4B3.4B$296.4B3.4B$297.4B3.4B18.2A$298.4B3.4B17.2A16.2A
$299.4B3.4B12.2D20.2A$300.4B3.4B12.2D$301.4B3.4B10.D$302.4B3.4B$303.
4B3.BABA6.2AB$304.4B3.B2AB4.B2A2B$305.4B3.A3B3.5B.2B$306.4B3.4B2.9B$
307.4B3.4B.12B$308.4B3.17B7.2A$309.4B3.17B6.2A$310.4B2.16B$311.19B$
311.19B5.2A$310.16B2A4B3.2A$310.16B2A6B26.2A$311.23B26.2A$312.23B$
313.22B19.2A$313.22B19.2A$316.20B$316.20B$317.19B14.2A$317.21B.5B6.2A
16.2A$318.18B2A8B22.2A$318.17BA2BA9B$319.17B2A11B$320.30B$321.11BA18B
$322.9BABA18B$323.8BABA19B$324.8BA21B$325.30B$326.19B2A9B7.2A$327.17B
A2BA8B7.2A$329.16B2A8B$329.4B.19B$330.3B2.6BA11B5.3B$331.B4.4BABA14B.
3B$336.4BABA18B23.2A$338.3BA17B25.2A$341.19B$344.11B2A3B18.2A$344.10B
A2BAB19.2A$345.10B2A4B$346.15B$347.B.2BA9B13.2A$349.BABA9B12.2A16.2A$
350.ABA9B30.2A$350.BA12B2$364.2A$364.2A5$387.2A$387.2A2$230.2A$231.2A
150.2A$230.A152.2A$408.2A$408.2A2$402.2A$380.2A20.2A$380.2A2$210.2A
186.2A$211.2A185.2A16.2A$210.A205.2A3$388.2A$388.2A5$411.2A$411.2A3$
407.2A$407.2A5$404.2A$404.2A7$412.2A$412.2A13$201.A$201.2A$200.A.A4$
167.A$167.2A$166.A.A12$191.A$191.2A$190.A.A8$147.2A$148.2A$147.A8$
127.2A$128.2A$127.A22$141.2A$142.2A$141.A17$118.A$118.2A$117.A.A4$84.
A$84.2A$83.A.A12$108.A$108.2A$107.A.A8$64.2A$65.2A$64.A8$44.2A$45.2A$
44.A22$58.2A$59.2A$58.A17$35.A$35.2A$34.A.A4$.A$.2A$A.A12$25.A$25.2A$
24.A.A!
x = 442, y = 135, rule = LifeHistory
27$46.A119.A119.A119.A$45.A.A117.A.A117.A.A117.A.A$45.A.A117.A.A117.A
.A117.A.A$46.A119.A119.A119.A2$41.2A7.2A109.2A7.2A109.2A7.2A109.2A7.
2A$40.A2.A5.A2.A107.A2.A5.A2.A107.A2.A5.A2.A107.A2.A5.A2.A$41.2A7.2A
102.2A5.2A7.2A102.2A5.2A7.2A101.2A6.2A7.2A$35.2A116.A.A117.A.A116.A.A
$34.A.A9.A108.A10.A108.A10.A107.A11.A$36.A8.A.A117.A.A117.A.A117.A.A$
45.A.A117.A.A117.A.A117.A.A$46.A119.A119.A119.A$148.2A118.2A$147.A.A
117.A.A$149.A119.A3$29.2A8.2A$28.A.A7.A.A101.2A$30.A9.A100.A.A$143.A
118.2A$261.A.A$263.A8$20.2A$19.A.A231.2A$21.A230.A.A$254.A4$374.2A$
373.A.A$375.A4$129.2A$128.A.A$130.A17$339.2A$338.A.A$340.A10$341.2A$
340.A.A$342.A!
calcyman wrote:I don't believe anyone finished off the proof that we can convert monochromatic monophase gliders into arbitrary slow-salvos. Fortunately, we can. Firstly, use the following technique to move a block as far away as necessary... and then coax it into a honeyfarm of the correct parity* to use Chris Cain's 4-glider recipes from the HBK project...
In an email, calcyman wrote:... we don't need a BABC design or sliding-block memory or similar in order to get a record-breaking knightship: we just make the cleanup salvos leave a block at the back of each of the four diagonal arms, and then slow-salvo-manipulate these into the four (fixed!) tips of the quincunx seed.
x = 195, y = 330, rule = B3/S23
obo$b2o$bo30$32bobo$33b2o$33bo6$33bobo$34b2o$34bo22$64bobo$65b2o$65bo
38$97bobo$98b2o$98bo22$128bobo$129b2o$129bo30$160bobo$161b2o$161bo30$
192bobo$193b2o$193bo5$193b2o$192bobo$194bo62$129b2o$128bobo$130bo62$
65b2o$64bobo$66bo!
x = 35, y = 42, rule = B3/S23
obo$b2o$bo14$9bobo$10b2o$10bo14$32bobo$33b2o$33bo5$33b2o$32bobo$34bo!
dvgrn wrote:The salvo heading southeast isn't exactly a slow salvo, since the gliders heading northeast depend on their exact timing.
dvgrn wrote:I'm not entirely sure I'm understanding the suggestion correctly, though.
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