New upper bound on block sensitivity and certificate complexity in terms of sensitivity

Sensitivity \cite{CD82,CDR86} and block sensitivity \cite{Nisan91} are two important complexity measures of Boolean functions. A longstanding open problem in decision tree complexity, the "Sensitivity versus Block Sensitivity" question, proposed by Nisan and Szegedy \cite{Nisan94} in 1992, is whether these two complexity measures are polynomially related, i.e., whether $bs(f)=O(s(f)^{O(1)})$. We prove an new upper bound on block sensitivity in terms of sensitivity: $bs(f) \leq 2^{s(f)-1} s(f)$. Previously, the best upper bound on block sensitivity was $bs(f) \leq (\frac{e}{\sqrt{2\pi}}) e^{s(f)} \sqrt{s(f)}$ by Kenyon and Kutin \cite{KK}. We also prove that if $\min\{s_0(f),s_1(f)\}$ is a constant, then sensitivity and block sensitivity are linearly related, i.e. $bs(f)=O(s(f))$.Comment: 9 page

Sensitivity versus Certificate Complexity of Boolean Functions

Sensitivity, block sensitivity and certificate complexity are basic complexity measures of Boolean functions. The famous sensitivity conjecture claims that sensitivity is polynomially related to block sensitivity. However, it has been notoriously hard to obtain even exponential bounds. Since block sensitivity is known to be polynomially related to certificate complexity, an equivalent of proving this conjecture would be showing that certificate complexity is polynomially related to sensitivity. Previously, it has been shown that $bs(f) \leq C(f) \leq 2^{s(f)-1} s(f) - (s(f)-1)$. In this work, we give a better upper bound of $bs(f) \leq C(f) \leq \max\left(2^{s(f)-1}\left(s(f)-\frac 1 3\right), s(f)\right)$ using a recent theorem limiting the structure of function graphs. We also examine relations between these measures for functions with small 1-sensitivity $s_1(f)$ and arbitrary 0-sensitivity $s_0(f)$