Some preliminary information: ZBLL is the state of the cube where only the U layer is left to be solved and the edges are oriented (i.e. there’s a cross formed). This can be considered to be a subgroup of 4!2⋅33/2 = 7776 elements, but a common convention is to somehow quotient out the effect of U turns (aka “adjust U face”, AUF), which can be done by considering the subgroup of order 7776/4 = 1944 with the ULF corner fixed (but possibly misoriented). There are other schemes to reduce the number of states to consider, but this is the simplest that still preserves the group structure. (Note: I don’t think there’s a surjective homomorphism from the whole ZBLL group to this subgroup, but we don’t need that anyway.) The number of “looks” is defined as the number of times the solver has to inspect the cube state to decide what algorithm to use, but note that this is not a very precisely defined term as, technically, blindfold solves are “one-look” but involve very many substeps and possibly visualising the cube state mentally.
The ZBLL group is itself a subgroup of the whole last layer group of index 8, as there are eight possible edge orientations. An earlier search by a SpeedSolving.com forum member Tempus led to a set of eighteen algorithms (up to AUF, mirroring and inversion) such that for every state in the LL group, either this state is in the set or there exist two algs such that their composition gives the desired state.
Such a one-look two-alg system provides a reasonable tradeoff between a fully one-look system and a fully two-look system. It retains the relatively low algorithm count needed for a two-look system (a full one-look LL needs over 1200 algs, compared to only 44 for the two-look system CLL/ELL) while allowing for optimisations in choosing different pairs of algs for the same result. Furthermore, it may even require fewer algs than a two-look system for comparable efficiency. However, memorising the optimal choice of alg for any given state is still necessary (so it’s still 1200+ cases to remember), which may be why two-alg systems have not gained much traction.
ZBLL is a much smaller group in comparison, and unsurprisingly also a lot more tractable. About two weeks ago, I discovered a set of 12 algs such that all combinations of two algs (i.e. the square of the set) is exactly the whole ZBLL group. This was inspired by my laziness to learn all the COLLs and all the 2GLLs, so I just combined the Sune cases from either (5 from COLL, 8 from 2GLL, one alg shared between the two) and realised this was enough to span ZBLL. This combination apparently had the lowest known alg count of any two-alg ZBLL system known then (at least, I did my part in reading all the thousand-odd posts in the concept/method/substep thread on SpeedSolving.com’s forum), and more importantly, it doesn’t even require memorising over a hundred cases to be applied; it’s more like only six cases in addition to all the algs. (I have a post written elaborating on this system which I’ll clean up and publish when I have the time to.)
Yes, had the lowest known alg count. More recently, last weekend I wrote a program to test the ZBLL coverage of any given input alg set in ZBLL, and after some experimentation, I found a set of just ten, six of which were heuristically-chosen “seed” algs that I already knew (Sune, another 2GLL, G perm, Buffy, L3C commutator, Bob), and the remaining four were found with (human) trial and error. Is ten minimal? Probably not, but it took me quite a few hours to even improve on an 11-alg set I’d found earlier. (I think an 8-alg set might be possible, in lieu of counting arguments demonstrating otherwise.) This, of course, loses the low case count of the “Sunes” method of the previous paragraph and three of the four non-seed algs are kind of nasty to execute IRL, so it’s probably useless for speedcubing purposes, but it’d be interesting to know what the theoretical limit is.