On DNF approximators for monotone Boolean functions (Eric Blais)
In this talk, Eric Blais presented joint work with Johan Håstad, Rocco Servedio and Li-Yang Tan, concerned with Boolean functions. More precisely, "the simplest representations of functions": DNF (Disjunctive Normal Form) formulas.For a little bit of background, recall that a Boolean function f : {0,1}n →{0,1} defined on the hypercube [2] is a DNF if it can be written as
A nice property of DNF formulas is that they are arguably amongst the simplest of all representations of Boolean functions; while formulas of depth ≥ 3 are a nightmare, DNFs have been extensively studied, and by now "Everything is known about them". Well, almost everything.
Indeed, amidst other facts, we have that
Theorem 3 (Korshunov '81, Kuznetsov '83). A random Boolean function can be computed by DNFs of size Θ(2n∕log n) (and
requires that much).
So... are we done yet? The mere presence of Eric in the auditorium
was a clear hint that all was not settled. And as it turns out, if the
picture is well understood for exact computation
of Boolean functions by DNFs, what about approximate
representation of a function? That is, what about the size required to
approximate a Boolean function by a DNF, if one allows error
ε (as a fraction of the inputs).This leads to the notion of DNF approximator complexity; and here again some results – much more recent results:
Theorem 4 (Blais–Tan ’13). Every Boolean function can be approximated by a DNF of size
O(2n∕log n). Furthermore, our all friend PARITYn only needs DNF size O(2(1-2ε)n).
That’s way better than 2n-1. So, again – are we done here? And, again... not quite. This brings us to the main point of the paper, namely: what about
monotone functions? Can they be computed more efficiently?
Approximated more efficiently? (Recall that a Boolean function f is
monotone if x ≼ y implies f(x) ≤ f(y), where ≼
is the coordinate-wise partial order on bit-strings.)As a starter: no.
Theorem 5 (Folklore) Every monotone Boolean function can be computed by a DNF of size O(2n∕n1/2) (using subcubes rooted on each min-term); and again, this is tight for PARITY.
Furthermore, and quite intuitively, using negations does not buy you
anything to compute a monotone function (and why should it, indeed?):
Theorem 6 (Quine ’54). To compute monotone Boolean functions, monotone DNFs are the best amongst DNFs.
Not surprising, I suppose? Well... it’s a whole new game when one
(one, again!) asks only for approximations; and that’s the gist of the
paper presented here. First of all, drastic
savings in the size of the formulas!
Theorem 7 (Blais–Hastad–Servedio–Tan ’14). Every monotone Boolean function can be approximated
by a DNF of size O(2n∕2Ω(n1/2) ).
Eric gave a high-level view of the proof: again, it works by considering the subcubes rooted on each min-term, but in two steps:- Regularity lemma: the world would be much simpler if all subcubes were rooted on the same level of the hypercube; so first, reduce it to this case (writing f = f1 + .....+ fk, each fi has this property)
- then, approximate independently each fi, using a probabilistic argument (via random sampling), to prove there exists a good approximator for all fi’s, and then stitching them together.
So, approximation does buy us a lot. But clearly, using negations shouldn’t, should it? Why would allowing non-monotone DNF’s to approximate monotone functions ever help? (Hint: it does.) (Yep.)
Theorem 8 (Blais–Hastad–Servedio–Tan ’14). For every n, there exists εn and
f : {0,1}6n →{0,1} such that
(Take that, intuition!)- f can be εn-approximated by DNFs of size O(n);
- any monotone DNF εn-approximating f must have size Ω(n2).
The upshot: exact computation and approximate computation have intrinsically very different properties!
Eric then concluded with an open question: namely, how to improve/better understand the gap between approximating functions with monotone DNF vs. approximating them with general DNF’s (the currently known gap in the size being quite huge – almost exponential). Additionally, how to get a separation as in the mind-boggling theorem above, but changing the quantifiers – that is, for a constant ε independent of n?
Also, can someone fix my intuition? I think it’s broken.
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