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$\def\<#1>{\left<#1\right>} \newcommand{\OR}[1]{\mathrm{OR}(#1)} \newcommand{\XOR}[1]{\mathrm{XOR}(#1)} \newcommand{\SUM}[1]{\mathrm{SUM}(#1)} \newcommand{\GEN}{\mathrm{GEN}} \newcommand{\Cov}{c} \newcommand{\Dec}{d} \newcommand{\tr}{\mathrm{tr}} \newcommand{\pr}{\mathrm{rk}_+} \newcommand{\br}{\mathrm{rk}_{\lor}} \newcommand{\Geq}{\succcurlyeq} \newcommand{\Leq}{\preccurlyeq}$

On XOR complexity of direct sum of matrices

Recall that the direct sum of matrices

$A$ and

$B$ is the matrix

$A\boxplus B=\left[\begin{smallmatrix} A & 0 \\ 0 & B\end{smallmatrix}\right]$ . For SUM and OR complexities, we have that (cf., e.g. Thm. 3.18):

$\SUM{A\boxplus B}=\SUM{A}+\SUM{B}\ \ \mbox{ and }\ \ \OR{A\boxplus B}=\OR{A}+\OR{B}\,.$ But what about the XOR-complexity? Does then

$\XOR{A\boxplus B}\geq \XOR{A}+\XOR{B}\qquad (*)$ also hold?

Igor recently observed that (*) does not hold: the complexity of direct sum of matrices may be smaller than the sum of their complexities.

Take a boolean $n\times n$ matrix $A$ such that $\XOR{A}\sim n^2/2\log n$ ; by Thm. 2.2, we know that such matrices exist.

Let $M_n$ be the Kronecker product of the identity matrix $I_n$ with $A$ , that is, $M_1=A\ \ \mbox{ and } \ \ M_{n+1}=\left[\begin{matrix} A & 0 \\ 0 & M_n\end{matrix}\right]$ Held (*), then we would have that $\XOR{M_n}\geq n\cdot \XOR{A}=\Omega(n^3/\log n)$ .

But the outputs of an XOR circuit for $y=M_n\cdot x$ are the entries of the product matrix $Y=A\cdot X$ , where $X$ is the vector $x$ written as an $n\times n$ matrix. By using the Strassen algorithm, all these entries can be computed over $GF(2)$ using only $O(n^{\omega})$ operations, where $\omega < 2.81$ (actually, $\omega < 2.4$ ), which is much smaller than $\Omega(n^3/\log n)$ , a contradiction.

May 22, 2015