As mentioned in the book, strong lower bounds on the CNF complexity of graphs could imply impressive lower bounds for depth-$3$ circuits, if we could prove such bounds for graph properties. Of particular interest is the following question: what monotone properties of graphs force their large CNFs? A property of graphs is hereditary if it is preserved under deletion of edges. For example, the property of avoiding some fixed graph $H$ as a subgraph is a hereditary property.
Is there a graph $H$ such that $\AND{G}\geq d^{\Omega(1)}$ holds for every bipartite $H$-free graph $G$ of average degree $d$? In view of \eqref{eq:dim}, this question is equivalent to the following question about the biclique covering number of graphs.
Is there a graph $H$ such that $\bc{\bar{G}}\geq d^{\Omega(1)}$ holds for every bipartite $H$-free graph $G$ of average degree $d$?In 1988, Pudlak, Roedl and Savicky asked whether this holds for $K_{2,2}$-free graphs $G$ (that is, when $H=K_{2,2}$). This was stated as the Research Problem 1.33 in the book (with its equivalent reformulations as Research Problems 4.9 and 11.17). In 2011, a negative answer was given by Katz [2] using probabilistic arguments: there exist bipartite $K_{2,2}$-free graphs $G$ of average degree $d\approx n^{1/10}$ for which $\bc{\bar{G}}=O(\log n)$ holds (see this my comment).
In 2022, Hambardzumyan, Hatami and Ndiaye [1] have found a cute and very simple explicit construction of a bipartite $K_{2,2}$-free graphs $G$ of average degree $d\approx n^{1/3}$ for which $\bc{\bar{G}}=O(\log^4 n)$ holds.
Theorem 1: There are explicit $n \times n$ Boolean matrices $A$ with $\Omega(n^{4/3})$ zero-entries and no $2 \times 2$ all-zero submatrices such that the $1$-entries of $A$ are covered by only $O(\log^4 n)$ all-$1$ submatrices of $A$.By viewing these matrices $A$ as incidence matrices of complements $\bar{G}$ of bipartite graphs $G$, Theorem 1 gives us bipartite $K_{2,2}$-free graphs $G$ of average degree $d\approx n^{1/3}$ for which $\bc{\bar{G}}=O(\log^4 n)$ holds.
Proof Let $n=2m^3$ for a positive integer $m$, and define the sets $ \cP = \cL := [m] \times [2m^2]$. We think of the elements $\ell=(a,b) \in \cL$ as lines $y=ax + b$ in $\RR^2$, and the elements $p=(x,y) \in \cP$ as points in $\RR^2$. A point $(x,y)$ lies on a line $(a,b)$ if $ax+b=y$. Define the Boolean matrix $A$ according to point line incidences: $A[(x,y),(a,b)]=0$ iff the point $(x,y)$ lies on a line $(a,b)$. Observe that $A$ is an $n \times n$ matrix for $n=2m^3$ that satisfies the conditions of Theorem 1:
Let us now construct the desired the covering of $1$-entries of $A$ by all-$1$ submatrices. Let $q$ be a prime number. For every triple $(i,j,k)\in[q]^3$, consider the set $R^{q}_{i,j,k} = \cP^{q}_{i,j,k}\times \cL^{q}_{i,j,k}$ of entries of the matrix $A$ given by the following sets of rows and columns: \begin{align*} \cP^{q}_{i,j,k}&=\{(x,y)\in \cP \colon x \equiv i \ \text{ and } y \not\equiv i j+k \}\\ \cL^{q}_{i,j,k}&=\{(a,b)\in \cL\colon a \equiv j \ \text{ and } \ b \equiv k\}, \end{align*} where $\equiv$ means equality modulo $q$. It is clear that for any choice of $q$ and for any $(i,j,k) \in [q]^3$, the set $R^q_{i,j,k}$ is a $1$-monochromatic rectangle of $A$ because any $[(x,y),(a,b)] \in R^q_{i,j,k}$ satisfies $ax+b \neq y \mod q$ (and, hence, also $ax+b \neq y$). Next, fix $r = \lceil \log_2 (2m^2) \rceil$ and let $Q$ be the set of all prime numbers $q$ smaller than $r$. We claim that the $O(r^4)=O(\log^4 m)=O(\log^4 n)$ all-$1$ submatrices $\{R^{q}_{i,j,k}\colon q\in Q,\ (i,j,k)\in[q]^3\}$ of the matrix $A$ cover all its $1$-entries.
Indeed, if a $1$-entry $A[(x,y),(a,b)]$ is not covered by $S$, then for all $q\in Q$, we have $y=(ax+b)\mod q$. Consequently, $y=(ax+b) \mod \prod_{q\in Q} q$. However, since both $y$ and $ax+b$ are at most $2m^2 \lt \prod_{q\in Q} q$, they can only be equal modulo $\prod_{q\in Q} q$ if they are equal. This implies that $A[(x,y),(a,b)]=0$, so all the $1$-entries of $A$ are covered. $\Box$
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