Worked it out, but it isn't pretty. Major edit 10/26 for improved explanation. Hopefully none of it is flat out wrong this time.
There are lots of equivalent ways to pose the problem; the most practical one at this stage is as follows:
Given a helix that is aligned with the insertion stage, what needs to be changed for it to be viable at the top? Being viable at the top requires meeting 3 constraints: the heisenburn succeeding, the LoM collision succeeding, and the horizontal offset between the topmost and second rakes being correct.
Degrees of Freedom
What can we change? We can move the trigger glider back in increments of (19, 19). We can add a recipe glider assuming there is some deletion step of the recipe that can be made longer, which moves the trigger glider a minimum distance of (53, 53). We can also move the topmost gliders in increments of (1,-1), which causes them to meet the helix later but with the same impact parameter. This moves the NE rebound gliders by (-1,-1), as shown below.
Part 1: Heisenburn
Code: Select all
x = 33, y = 53, rule = LifeHistory
The heisenburn is a matter of satisfying an impact parameter with a NW glider. This means if we satisfy the heisenburn, the helix itself may then move any multiple of (1, -1), as well as its own translational symmetry of (2,18), with the heisenburn unaffected. The only thing we can actually do though is stall the helix creation, which is a vertical shift of some multiple of (0,2). A shift of (0, -20) is equivalent to (2, -2) via the symmetry of the helix, so there are 10 cosets and we can reach them all as 19 and 53 are both coprime to 10. Once aligned with the heisenburn, our realizable freedom is to shift the helix in increments of (2, -2).
Part 2: NE rake matchup
In this analysis we have been in the reference frame of the construction effort, so the LoM initial target is fixed, as are the rakes that produce it. Matching the NE gliders is a 2D constraint (mod (34,18), which is the translation between gliders in a x2 NE rake). As we showed above we can move these gliders in increments of (1, 1) without moving the helix. To get the other dimension of freedom, we need to move the helix some number of cells according to the freedom identified above, which moves everything in increments of (2, -2).
It is possible to be unable to meet this constraint, as these shifts all satisfy dx - dy = 4N, and half the time the circumstances might require dx - dy = 4N + 2. Assume though that we can shift everything to satisfy the NE gliders. What freedoms remain to use in matching the 3rd criteria for viability?
Part 3: Top rake horizontal offset
We matched the NE gliders modulo the (34, 18) symmetry of the NE stream. If we perform the following convoluted combination:
Translate the helix by (16, -16) via stalling the insertion glider by (80, 80)
Translate the top of the helix by (2, 18) by adding one unit cell
Move the NE gliders by (16, 16), and the topmost gliders the corresponding (-16, 16)
Then the net displacement of the NE gliders is (34, 18), exactly their symmetry. The net displacement of the NW gliders is (2, 18). The horizontal offset between the topmost and second rakes requires they be skewed by 4 mod 19; their vertical separation is irrelevant. Thus, the freedom identified above is a shift of 2 horizontal cells, sufficient to meet the 3rd and final criterion.
We have everything now in terms of how much to shift the trigger glider modulo various parameters. Perform unit shifts to satisfy the heisenburn, then shifts mod (10, 10) to match the NE rake, then shifts mod (80, 80) to achieve the correct horizontal offset, modulo a total overall shift of (1520, 1520) which impacts nothing of interest.
Since 1520 is divisible by 19, the (53, 53) shift incurred by adding a recipe glider is generally necessary for the last part. However, we might have plenty of different recipes to choose from, all of which proceed via a splitter/rephaser. Depending on the properties of the rephaser portion of this, the number of required extra recipe gliders can change. I think I found a way to express that rather simply.
To thaw the LWSS, a glider needs to be provided which is advanced by one phase relative to the rakes and is of the same color. When a splitter provides that glider, it adjusts the incoming glider by moving it 2Y cells vertically and advancing it 1 + 4K generations. Y and K can be positive or negative, and only K is important to the result. Splitters may have different values of K.
Assuming the construction proceeds via the 2fd offset honeyfarm, we will call the rake that hits that honeyfarm the "first" rake. The eventual trigger glider is the Nth rake. In the earlier attachment, N = 33.
To have the correct horizontal match between the trigger rake and the LWSS, there is a restriction on N and K. I believe after redundant checks that the restriction is N = (8 - 5K) mod 19. That the modulus is 19 makes sense, since adding 19 to either N or K leaves the problem unchanged, and that the coefficients are 1 and 5 makes sense as adding 4 to K is equivalent to adding 1 to N.
In the earlier attachment K = -2, meaning we need N = 37, so we need to burn 4 more gliders. Perhaps there is a way to set up the same splitter to delete the problematic block when the recipe is 4 G longer, but I haven't found it.
Certain values of K are more reasonable than others. If K = 0 for example, we need a recipe of length 27 or 46, and 27 is too tight to expect the 3SL constellation and a clean splitter. I have found a number of recipes in the 32-36 glider range that nearly achieve the goal, so 46 would be needlessly wasteful (remembering that every rake fired also means a lot of downstream cleanup..)
That one other problem we swept under the rug
Earlier I said there was a 50% chance we can't match the NE gliders, because of requiring dx - dy = 4N + 2. Consider two splitters, A and B, where the K value of A is one greater than that of B, and which activate the same LWSS seed. If A is used, then the helix is 2 cells lower than if B is used, so dx - dy is changed by 2.
Consider instead two identical LWSS seeds built (1, -1) relative to one another. My adaptation to chris_c's program sought out a way to construct the seeds in one of the 5 hive-compatible positions, some of which had relative dx - dy of 4N + 2. For two of these seeds, their resulting helices also have dx - dy changed by 2.
Thus, depending on where the seed is ultimately placed, there is an additional constraint on the parity of K. The route I have been investigating requires K be even, but if the other background search I have been running shows some promise for a different positioning of the seed then an odd K might be necessary. It's just important to know this constraint also exists and how to meet it.
Physics: sophistication from simplicity.