\begin{eqnarray*} \frac{(q-1)(\lfloor \log_q(D-1)\rfloor +a)}2 +2 & & \text{if $q$ is even and $r+a$ is odd,} \\ \frac{(q-1)(\lfloor \log_q(D-1)\rfloor+a+1)}2 +2 & & \text{otherwise.} \end{eqnarray*}
Here $q$ is the base field size, $D$ the degree of the HFE polynomial, $r=\lfloor \log_q(D-1)\rfloor +1$ and $a$ is the number of removed equations (Minus number).
This allows us to present an estimate of the complexity of breaking the HFE
Challenge 2: \vskip .1in \begin{itemize} \item the complexity to break the HFE Challenge 2 directly using algebraic solvers is about $2^{96}$. \end{itemize}
Category / Keywords: public-key cryptography / multivariate, degree of regularity Date: received 21 Oct 2011 Contact author: jintai ding at gmail com Available formats: PDF | BibTeX Citation Version: 20111025:170048 (All versions of this report) Discussion forum: Show discussion | Start new discussion