## New construction arms

For discussion of specific patterns or specific families of patterns, both newly-discovered and well-known.

### Re: New construction arms

For a macro-spaceship that is its own glider synthesis I'd like to know what initial collision between two 90-degree gliders can most quickly be converted into a clean elbow block (or SPEBOE, I suppose -- Standard Pi-Explodable Block or Equivalent, for anyone who's getting a little rusty on that questionable bit of terminology).

It looks like all the available 90-degree collisions aren't quite lucky enough to make an elbow object in the right place directly. Did I miss a good one somewhere?

I eventually found a traffic-light-making collision that allows a crystal to form --

`x = 76, y = 90, rule = LifeHistory4.A\$4.A.A\$4.2A12\$2A\$A.A\$A21\$23.2A\$23.A.A\$23.A8\$34.A\$33.2A\$33.A.A8\$43.2A\$43.A.A\$43.A8\$54.A\$53.2A\$53.A.A8\$64.A\$63.2A\$63.A.A8\$73.2A\$73.A.A\$73.A!`

-- and we know a crystal can be turned into a standard elbow with enough additional single-channel gliders.

But I suspect I'm missing something like a three- to five-glider collision with a 90-degree glider, just a little beyond my ability to find without a script...?

dvgrn
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Joined: May 17th, 2009, 11:00 pm

### Re: New construction arms

dvgrn wrote:I'd like to know what initial collision between two 90-degree gliders can most quickly be converted into a clean elbow block

Hello. There was a solution lurking in the second code-box of this post:

`x = 52, y = 56, rule = B3/S232bo\$2o\$b2o3\$b2o\$bobo\$bo21\$23b3o\$23bo\$24bo23\$50bo\$49b2o\$49bobo!`

The second glider comes 90 ticks after the first. Is that acceptable or does the first glider need to be a singleton?
chris_c

Posts: 839
Joined: June 28th, 2014, 7:15 am

### Re: New construction arms

chris_c wrote:There was a solution lurking in the second code-box of this post:
...
The second glider comes 90 ticks after the first. Is that acceptable or does the first glider need to be a singleton?

As far as I can see there's no reason to require a singleton. The all-gliders recipe will consist of two diploid chromosomes -- two exact copies of two slightly different recipes. Any stream of gliders that can be duplicated by a memory loop is fair game. Thanks!

When we get to the all-gliders stage, the first copy of the recipe will already be partly used up: the first N gliders in the recipe hit a receding Cordership. Their job is to clean up all the debris and produce that Acceptable Glider, by building a 180-degree one-time turner off to the side if necessary. (Or straight back along the channel -- see below.)

I think it should be fairly easy to extract an elbow block from a crashed Cordership, using the same single-channel lane for every glider. Then we can pull that elbow back and use 0-degree gliders from there to do the cleanup and eventually build the Acceptable Glider.

`x = 175, y = 206, rule = B3/S2332b3o\$31bo2bo\$30bo4bo\$30bo2b3o\$30bo5bo\$31b7o\$37bo\$37bo\$35b2o12b2o\$49b2o3\$34bo\$33bobo\$32b2obo\$32b2o\$33b3o21b2o\$34bo22b2o\$35bo2bo\$32bo6bo\$32bo6bo\$32bo4b3o\$26bo\$25b3o\$24b2ob2o\$23b2obobo\$22b3o2b3o\$23bo6bo\$24b5obo\$25b4ob2o\$2b3o25bo\$bo2bo24bo\$o4bo\$o2b3o\$o5bo\$b7o\$7bo\$7bo\$5b2o12b2o\$19b2o3\$4bo17bo\$3bobo15bo2bobo\$2b2obo14bo2b5o\$2b2o17bobo3bo\$3b3o16bo\$4bo\$5bo2bo\$2bo6bo\$2bo6bo\$2bo4b3o13bo\$22bo\$19b3obo\$19b2o3\$13b3ob2o\$7b2o3b4o\$7b2o3bo5bo\$16bobo\$14b2obo\$16b3o\$13b2o2bo\$14bo2bo\$17bo\$14bo2bo\$15b2o4\$39b3o\$39bo\$40bo20\$62b2o\$62bobo\$62bo41\$104b3o\$104bo\$105bo20\$127b2o\$127bobo\$127bo21\$149b3o\$149bo\$150bo20\$172b2o\$172bobo\$172bo!`

Options
I guess it would work to have either a 180-degree one-time turner triggered by a 90-degree glider that uses up a standard elbow, or a 90-degree one-time turner triggered by a 0-degree glider that uses up the 0-degree elbow. Just have to make sure the next three single-channel gliders can get there in time to meet the Acceptable Glider.

Of course that part is all a very arbitrary song-and-dance to make it so there are nothing but gliders in the spaceship at some point. Otherwise it would be much simpler to just make the Cordership debris into an elbow, and restart the construction cycle from there.

-- Come to think of it, a recipe that converts an elbow into a clean 180-degree one-time turner would also work, if the turner sends back a glider on the lane shown here. That seems likely to be the cheapest option -- no need to build anything complicated off to the side.

I can easily compile a working Cordership-to-180-degree-turner conversion recipe manually, using 0-degree gliders. But a direct conversion is bound to be cheaper. Might something like that be within reach of a one-off search with an existing search script?

dvgrn
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Joined: May 17th, 2009, 11:00 pm

### Re: New construction arms

In May of 2017, simeks wrote:
simsim314 wrote:I'm also looking for 180 degrees similar to your posts but with destruction of the initial block ...

If you need to save a few gliders I could search for a direct recipe with a 180° glider that leaves nothing behind, but then it would be helpful to know which lane you need the 180° glider on.

I cobbled together a cleaner long-distance elbow duplicator for the new HashLife-friendly Demonoid:

`#C clean long distance elbow duplicator -- 46 gliders#C The arbitrary long wait is '16278'.  Can be adjusted up or down#C by 8N to move the final near block location by N.#C cleanleftparen=[0, 109, 91, 94, 91, 91, 96, 92, 107, 91, 93, 143, 149, 91, 133, 135, 93, 90, 92, 135, 108, 92, 16278, 94, 91, 158, 91, 90, 90, 90, 91, 116, 91, 137, 91, 90, 91, 90, 90, 149, 200, 90, 90, 154, 90, 91]x = 5245, y = 5242, rule = B3/S232o\$2o\$5b3o\$5bo\$6bo25\$33b2o\$32b2o\$34bo21\$55b3o\$55bo\$56bo21\$79b2o\$79bobo\$79bo21\$102b2o\$101b2o\$103bo21\$124b3o\$124bo\$125bo22\$148b3o\$148bo\$149bo21\$171b3o\$171bo\$172bo24\$199bo\$198b2o\$198bobo21\$221b2o\$221bobo\$221bo21\$245bo\$244b2o\$244bobo34\$280b2o\$280bobo\$280bo35\$318bo\$317b2o\$317bobo21\$340b2o\$340bobo\$340bo31\$374bo\$373b2o\$373bobo32\$407b2o\$407bobo\$407bo21\$431bo\$430b2o\$430bobo21\$453b2o\$452b2o\$454bo21\$476b2o\$475b2o\$477bo32\$509b3o\$509bo\$510bo25\$536b3o\$536bo\$537bo21\$559b3o\$559bo\$560bo4067\$4629b2o\$4629bobo\$4629bo22\$4652b3o\$4652bo\$4653bo20\$4676bo\$4675b2o\$4675bobo38\$4715b2o\$4714b2o\$4716bo21\$4737b3o\$4737bo\$4738bo20\$4760b2o\$4760bobo\$4760bo21\$4782b3o\$4782bo\$4783bo20\$4805b2o\$4805bobo\$4805bo21\$4828b2o\$4827b2o\$4829bo27\$4857b2o\$4856b2o\$4858bo21\$4879b3o\$4879bo\$4880bo32\$4914b2o\$4913b2o\$4915bo21\$4936b3o\$4936bo\$4937bo20\$4959b2o\$4959bobo\$4959bo21\$4982b2o\$4981b2o\$4983bo20\$5005bo\$5004b2o\$5004bobo21\$5027b2o\$5026b2o\$5028bo35\$5064b2o\$5064bobo\$5064bo48\$5114b2o\$5114bobo\$5114bo21\$5136b3o\$5136bo\$5137bo20\$5159b2o\$5159bobo\$5159bo37\$5197b3o\$5197bo\$5198bo20\$5220b2o\$5220bobo\$5220bo21\$5243b2o\$5242b2o\$5244bo!`

However, there's another use case where you don't really want the elbow duplicated, you just want it moved quickly. For that I'd need a destructive recipe for a return glider on a lane 3hd from the single-channel lane.

Is there any chance of a destructive 3hd return glider recipe showing up magically when I wish for it, or is it time for me to go see if I can get some search code working again? At this point I only have embarrassingly vague memories of what's available in this department. I can fake a solution by appending a 0-degree recipe on a near-zero lane, but that seems unnecessarily expensive.

(Really the 3hd offset isn't a requirement. It could be any offset that can be cheaply hit with another single-channel glider and cheaply converted back into a clean elbow. Most likely 46 gliders is painfully suboptimal already.)

dvgrn
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Joined: May 17th, 2009, 11:00 pm