Somehow, the Spaceship needs to lay down a blinker track in front of its main body. We might imagine it shooting spaceships or gliders forward to do it, but how exactly can that work? 17c/45 (0.377c) is inconveniently fast, much too fast for gliders traveling at a mere 0.250c to overtake. Standard c/2 spaceships can overtake it, but we need some way to stop them when they reach the front. This is difficult to do.
The best way that we have found to solve this problem
makes use of the following
reaction, where a glider deletes a LWSS and HWSS, and is
advanced 20 generations in the process:
The reaction is fast, and can be repeated after just 24 generations. This allows a glider to be carried along at speeds approaching 11c/24 (0.458c). This is fast enough to meet our target speed of 0.377c, but there is also a lot of overhead required to change the glider's direction. This overhead slows it down considerably, requiring us to use additional copies of the rection to keep the speed fast enough.
It turns out that a set of 14 of these reactions
will suffice to achieve 17c/45, and return the
glider to its original column, and emit an extra glider
along the way:
The above structure was designed to decay at exactly 17c/45. Its period is 450, and it emits a glider to the left once every period. Enough fuel for two periods is shown. I need a name for this structure, so I'll call it a "helix".
At the top of the Spaceship, we need to build two new blinkers every 45 generations. A helix provides just one glider every 450 generations -- and we need two gliders to crash together to create each blinker. This multiplies out to 40 gliders every 450 generations, so we would need 40 helixes.
Fortunately, we can do a lot better than this. We really
only need one helix, because we can duplicate the gliders
that it emits. The best mechanism that we have found for
this is shown below.
fanout device
This converts a single backward glider into 10 strategically-placed forward gliders, plus another copy of the backward glider. The final 17c/45 Spaceship will use a single helix plus four of these "fanout devices".
Although it takes just three spaceships to duplicate a glider, the spaceships in these fanout devices will still make up the bulk of the population of the 17c/45 Spaceship.
The front of the Spaceship will look something like
this:
There will actually be more space between the center and the fanout devices than the above diagram potrays. There will also need to be a few extra forward spaceships used to reposition the gliders as they travel between the various components.
[Note: The stream of backward gliders at the bottom of the diagram illustrates one way of keeping a p450 structure available -- we will soon need such a thing to help build two p450 backward MWSS streams. The gliders can be hit with a pi or a combination of gliders, and converted to stable objects for later use. There are other, perhaps better, ways to do this, but this is one relatively simple way.]
It may be a good idea to eliminate the extra space
between the helix and the rightmost
fanout device. This could be done by using some extra forward
spaceships to turn the backward gliders into stable objects,
and then back into gliders, something like this:
This huge number of forward spaceships is used to create nothing more than one pair of blinker trails, spaced 32 cells apart. As we move from the front to the back of the Spaceship, these two blinker trails will ultimately be responsible for constructing all of the forward-traveling spaceships, completing the cycle.
What happens if the gliders used to construct the second blinker trail collide with the first blinker trail? The solution is simple: just place one or more pi's on the first blinker trail to reposition it.
In fact, the majority of the pi's used in the Spaceship will be used only to reposition blinker trails. As mentioned earlier, there are 34 possible positions for the blinkers, so positioning the blinkers exactly will require between 0 and 33 pi's (an average of 16.5). [see the appendix]
What if you only need to move the pi's out of the way of an approaching glider, but don't care about their exact position? The phase of the blinker doesn't matter, so there are 17 different possible positions. It turns out that 5 of these 17 positions require no repositioning, 5 require one pi, 5 require two pi's, 1 requires 3 pi's, and 1 requires 4 pi's. So, while the average number of pi's needed is only about 1.3, any particular instance may require up to 4. This is unfortunate, because when building a component, if you play it safe and leave enough room for 4 pi's, you end up with something larger than necessary.
At this point we have two blinker trails. The first thing to
do is to build some more -- lots more. Here is an example of
how to build a third trail if we have just two (separated by 32
cells):
Once we have more than two trails, it becomes easier. Keep in mind that as soon as a new blinker trail is created, we can start putting pi's on it to assist in creating even more trails. A lot of strategy could come into play here -- it's not at all obvious what the optimal order for building the trails would be. It may (or may not) be better to use mostly forward rakes instead of backward rakes, even though they are larger, because the forward glider streams take up less space on the trails that they cross, and because of the glider-to-blinker reaction shown below.
There are three different way to create a blinker from
two gliders:
Also, there is a way to move an existing blinker by
(1,3) cells with a glider. In effect this means a glider
can push a blinker out 1 cell, or pull it in 3 cells.
Here's another reaction that may be useful at this stage.
A forward glider collides with a pi to create a blinker 35
cells away:
35 cells is a convenient distance that's just far enough apart that two neighboring pi's cannot interact. Or instead, the outer blinker can then be hit with a glider to pull it in 3 cells, leaving a 32 cell separation.
[Other reactions like this would be nice, but unfortunately, this one is an anomaly. There are no similar reactions that leave just a blinker more than 18 cells away.]
Continue to build new blinker trails until
the central part of the Spaceship
looks very much like this:
This may seem excessive, but the reasons for so many trails, and for the MWSSs, will be explained. The construction process can be parallelized to some extent, so it shouldn't be as unreasonable as it first appears.
Note that this structure is symmetric. The left half will be responsible for constructing the spaceships on the left, and the right half will construct the spaceships on the right.
The ideal number of (pairs of) trails may be slightly more or less than this, but I think this is pretty close to optimal. There is a point toward the back of the Spaceship where it will probably be best to rearrange the blinker trails (when it comes time to construct the helix), but this should suffice for most of the length of the Spaceship.
We don't want to build any more trails than we need, though. The cost of having too many trails is that we need to put a lot of extra pi's on them to keep moving the blinkers out of the way of the gliders that will need to cross them.
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