ধন্যবাদ, কাভেহ, প্রুফ জটিলতার অধ্যায়গুলি দেখার ইচ্ছা করার জন্য!
Concerning Robin's question, first that note AC0AC0 contains functions requiring formulas (and even circuits) of size nknk for any constant kk. This follows, say, from a simple fact that AC0AC0 contains all DNFs with constantly long monomials. Thus, AC0AC0 contains at least exp(nk)exp(nk) distinct functions, for any kk. On the other hand, we have at most about exp(tlogn)exp(tlogn) functions computable by formulas of size tt.
I shortly discussed the issue of getting explicit lower bounds of n2n2 or larger with Igor Sergeev (from Moscow university). One possibility could be to use Andreev's method, but applied to some another, easier computable function instead of Parity. That is, consider a function of nn variables of the form F(X)=f(g(X1),…,g(Xb))F(X)=f(g(X1),…,g(Xb)) where b=lognb=logn and gg is a function in AC0AC0 of n/bn/b variables; ff is some most complex function of bb variables (mere existence of ff is enough). We only need that the function gg cannot be "killed" in the following sense: if we fix all but kk variables in XX, then it must be possible to fix all but one of the remaining variables of gg so that the obtained subfunction of gg is a single variable. Then applying Andreev's argument and using Hastad's result that the shrinking constant is at least 22 (not just 3/23/2 as earlier shown by Sybbotovskaya), the resulting lower bound for F(X)F(X) will be about n3/k2n3/k2. Of course, we know that every function in AC0AC0 can be killed by fixing all but n1/dn1/d variables, for some constant d≥2d≥2. But to get a n2n2 lower bound it would be enough to find an explicit function in AC0AC0 which cannot be killed by fixing all but, say, n1/2n1/2 variables. One should search for such a function in depth larger than two.
Actually, for the function F(X)F(X) as above, one can obtain lower bounds about n2/lognn2/logn via simple greedy argument, no Nechiporuk, no Subbotovskaya and no random restrictions! For this, it is merely enough that the "inner function" g(Y) is non-trivial (depends on all its n/bn/b variables). Moreover, the bound holds for any basis of constant fanin-gates, not just for De Morgan formulas.
Proof: Given a formula for F(X)F(X) with ss leaves, select in each block XiXi a variable which appears the smallest number of times as a leaf. Then set all remaining variables to the corresponding constants so that each g(Xi)g(Xi) turns to a variable or its negation. The obtained formula will then be at least n/bn/b times smaller than the original formula. Thus, ss is at least n/b=n/lognn/b=n/logn times the formula size 2b/logb=n/loglogn2b/logb=n/loglogn of ff, that is, s≥n2−o(1)s≥n2−o(1). Q.E.D.
To get n2n2 or more, one has to incorporate Subbotovskaya-Hastad shrinking effect under random restrictions. A possible candidate could be some version of Sipser's function used by Hastad to show that depth-(d+1)(d+1) circuits are more powerful than those of depth dd.