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Some more ruminations on directional modifiers:

Betza notation has implied operators for combining modifiers and atoms, which contributes to its compactness. The operations can be joining or intersecting the sets of moves selected by the individual modifiers. E.g. an frW is a Wazir that moves forward OR left (2 moves; (f+r)W = fW + rW = fWrW). OTOH frF would mean a Ferz that moves(along the right-forward diagonal (1 move: (f*r)F), intersecting the set of moves of fF and rF.

So it is not obvious which operation is implied; this depends on the symmetry and orientation of the atom. For 'orthogonal 4-movers' such as W the f, b, l, r modifiers specify disjoint sets of one, so intersecting is pointless, and any combination of modifiers thus implies joining.

In the more general case of an oblique 8-mover these same modifiers specify pairs of moves, though. For historical reasons fsN means the forward moves of the wide Knight, and since s was defined as l+r, (so that fsN = flN + frN), frN would mean the upper-most of the r-pair. By rotation symmetry rvN would mean r(f+b)N = rfN + rbN, and rfN would specify the right member of the forward pair. So rfN and rfN each specify a single, different move. I.e. order of the modifiers matters here, and the combination neither implies joining or intersection: the f and r sets were disjoint. Basically fr and rf (etc.) are one symbol specifying a single direction on oblique atoms. (And in a counter-intuitive way; you would expect the major direction of the jump to be mentioned first!)

This poses a problem when you want to designate a Knight that has only the two forward and two right-most moves (f+r)N, as frN is already taken. Of course you can resort to repeating the atom, (f+r)N = fN + rN = fNrN. The alternative would be to introduce ff etc. as replacement for f in case there was ambiguity. So ffrrN would mean (ff+rr)N = ffN + rrN = fN + rN = fNrN. A bit ugly, and one letter extra. so I guess fNrN would be the preferred form. (And it is one letter shorter anyway.) So ff is not really needed. But fh is still a useful two-letter modifier that would be cumbersome to describe by other means (e.g. fhN = fNfsN).

For diagonal 4-movers, the fr and rf moves collapse into one. So the order no longer matters, and the f,b and r,l sets do intersect. OTOH f and b (as well as l and r) are each other's complement, so that v and s dont't mean anything.

If we want to describe continuation steps of bent leapers, the best way to do it seems in the coordinate system of the previous step. This best exploits symmetry of the moves. E.g. [FW] would view the W second step in 45-degree-rotated coordinates defined by the Ferz move, where it now appears a diagonal mover, and should use the corresponding system to describe its moves. That means that an inplied f ([FW] = [FfW]) now refers to two moves. The reulting two paths are indeed symmetry equivalent, and thus likely to occur together in typical variant pieces. (In fact they describe the Moa.)

By rotating coordinates we can get into situations where we have on-axis 8-movers, though. E.g. when both steps are Knight jumps, continuing in the same direction (as the Nightrider) would be pure f. It is as if orthogonal and diagonal moves are combined in the same atom (like the moves of the sqrt(50) leaper). On such an atom fr (equivalent to rf) is needed to describe the diagonal move, and thus cannot be used to imply joining of the f and r orthogonal move.

Note that in case of the N in the rotated coordinates the fl and fr moves are not perfectly symmetric, as the angle between neighboring oblique rays alternates between two values. In fact such symmetry-breaking can occur for all atom types. After a Knight jump both F and W would be tilted in the new coordinate system, and be neither diagonal nor orthogonal. But the tilting breaks the exact equivalence of moves that a purely diagonal mover suffers, so the simpler system for the orthogonal movers can be applied. It is always clear what f and b mean, however (most outward and inward). But l and r are more tricky, as they would depend on how you take the first N step. If you go from e1 to f3, then going to f4 would be f, and going to g3 would be r. But the symmetry-equivalent path to the latter (which regular pieces would also be able to take, so that it would be very annoying if you would have to specify it separately) would be e1-d3-c3, where d3-c3 is now lW, needing a different notation. Therefore I would propose to assume that the first step of the path was a move in the rf octant (from fW to frF, so to say). Then a n[NrW] would move along all symmetry-equivalent paths of e1-f3-g3, i.e. also e1-d3-c3, e1-f2-f3, e1-c2-c3. So basically the [XYZ...] notation implies [rfXfYfZ...] to specify a path starting in a unique direction, but then includes all 8-fold-symmetry-equivalent versions of that path, unless there are directional modifiers in front of the [XYZ...] (e.g. f[NW]).