calcyman wrote: Can you have a NW-directed glider from the genesis of the SE glider-producing switch-engine collide with the glider stream from the NW glider-producing switch-engine to obstruct it, thereby dispensing with the pre-block entirely?
I was thinking in those lines for a while now. The arithmetic of the ping pong is kinda tricky. Generally speaking we need
three "full" degree of freedom.
1. Generating the first MWSS (this is mod 723 - the period after which very long stream of "building mechanism" will repeat itself - i.e. moving the switch engine by 723 will emit the same stuff, either the MWSS or the switch engine).
2. The second is making sure the "size" of the "tooth" will be correct, and the MWSS emitting switch engine will continue emitting MWSSs (also mod 723), as it easily can be noticed that even if the first hit will emit MWSS there is no guaranty the second will do the same.
3. The final point, is the emitting of side switch engines, i.e. placing the second switch engine in correct placement (mod 723), will continue emitting the switch engines and make it quadratic.
Here we have three (actually four) degrees of freedom: assuming one of the 10 cells is at (0, 0). We have (x, y) of the second and (x, y) of the trigger. So no luck is needed at this point, only correct calculation and usage of degrees of freedom.
Saying that, some luck is needed, because the first MWSS hits the glider in some limited range of states. This is why the discovery of G + R -> switch engine comes so handy. Some states of MWSS and glider will not reflect a glider, or will cause the switch engine to stuck etc. So the mechanism is pretty fragile.
There might be a lot of different approaches to the ping pong as well. I'm currently implementing the "simplest" one, which uses glider reflected from the MWSS and stream collision to create another "wave". One could use the glider emitted from second switch engine, to reignite the MWSS stream again. I don't see any limitation on that, and it could open some new possibilities, as the arithmetic might be quite different. But one should notice all this nuances are mod 64 and the three degrees of freedom mod 723 are unavoidable.
EDIT The approach you suggested obviously has only two degrees of freedom, the (x, y) of the second 10 cell.
EDIT2 Oh and thx for the compliments.