Every end-to-end joining knot (ie 'bend') has 4 corresponding eye knots.
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Note the 4 corresponding eye knots which are derived from the Butterfly bend
I was going to let the first statement stand,
taking it on "e2e KNOT"
(vs. *tangle* associated with the knot
(i.e., the cookie cutter boundary
within which is the entangled cordage
--here, 2 pieces, cannonically 1-2 & A-B--
and the 4 ends emerging awaiting some
assigned Loading Profile);
in which case we'd limit SParts to being either 1 or A.)
BUT if you're counting as you do that EEL Butterfly
with the 2nd statement, you've moved to taking ALL ends
and all possible eye knots :: that => 8.
1 v 2+A
1 v 2+B
2 v 1+A
2 v 1+B
... similarly with A & B qua SParts.
Yes, for a symmetric knot such as SmitHunter's,
there are essential repeats (you might exclaim
about "cheirality"; I won't). But here we nicely
have an asymmetric tangle, so a full set of 8!
But for me, this is an association of the *tangle*,
and that between eye & end-2-end knots I will
limit, and then have another idea altogether,
as I've previously explained (where one ties
one piece to a bight qua single joining strand
and then figures how to *fuse* parts for making
a single-strand eye knot.
As for tying the OP's knot qua "Competition knot",
that is not this exterior-loaded one but the other,
INterior-loaded version, but both sort of taking the
*offset* aspect into their corresponding eye knot.
.:. Interesting knot & discussion!
--dl*
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