\documentclass[]{article} \usepackage{times} \usepackage{amsmath} \usepackage{hyperref} \usepackage{url} \usepackage{graphicx} %opening \title{} \author{} \begin{document} \maketitle \begin{abstract} \end{abstract} \section{Advanced Encryption Standard} \subsection{Evaluation Criteria for AES} Essentially the cipher's submitted NIST for AES were judged on three broad categories \begin{enumerate} \item Security \begin{enumerate} \item Actual security compared with other submitted algorithms \item Randomness \item Soundness (mathematical) \item Other security factors \end{enumerate} \item Cost \begin{enumerate} \item Licensing \item Computational Efficency \item Memory Requirements \end{enumerate} \item Algorithm and implementation characteristics \begin{enumerate} \item Flexibility in key and block size, wide variety of plateforms and applications, and USE: implemented as a stream cipher, message authentication code, random number generator. \item Hareware and software suitable \item Simplicity \end{enumerate} \end{enumerate} \subsection{The AES Cipher} Figure \ref{fig:aes_diagram} shows a diagram of the AES cipher. Look through the review questions below for a good explanation of each step. \begin{figure}[htb] \centering \includegraphics[width=.65\textwidth]{AES_Diagram.png} \caption{AES cipher diagram.}\label{fig:aes_diagram} \end{figure} \subsection{Review Questions} \textbf{What was the original set of criteria used by NIST to evaluate candidate AES ciphers?} In general they said: \begin{enumerate} \item Security strength equal to or greater than 3DES \item Significantly improved efficiency \item Symmetric block cipher with a block length of 128 bits. \item Support key lengths of 128, 192, and 256 bits. \end{enumerate} \textbf{The specific evaluation criteria:} \begin{enumerate} \item Security: This referes to the effort required to cryptanalyze an algorithm. \item Cost: Practical, Efficient enough to use on high bandwidth links and high speed applications. \item Algorithm and Implementation Characteristics: flexibility, suitability for a variety of hardware and software implementations, simplicity \end{enumerate} \textbf{What was the final set?} \begin{enumerate} \item General Security: \begin{enumerate} \item Software Implementations: Speed \item Hardware Implementations: small hardware size to keep cost down. \item Attacks on Implementations: timing attacks and power attacks. \item Encryption versus decryption: Are they the same... \item Key agility: ability to change keys quickly and efficiently \end{enumerate} \item Other versatility and fexibility: Parameter flexibility (other key and block sizes, change in the number of rounds), \item Implementation Flexibility (optimizing cipher elements for particular environments). \item Potential for instruction-level parallelism:The ability to exploit ILP in processors. \end{enumerate} \textbf{What is the power analysis?} Observing the power used to detect a multiply or add operation or to see if ones or zeros are being written. \textbf{What is the difference between Rijndael and AES?} Rijndael took different blocks sizes of 128, 192, 256. AES only takes 128. \textbf{What is the purpose of the \textit{state} array?} The state array holds the input block that is massaged through each round. \textbf{How is the S-Box constructed?} \begin{enumerate} \item Initialize the $S-Box$ with the byte values in ascending sequence row by row\newline $\left[ \begin{array}{cccc} 00 & 01 & ... & 0F \\ 10 & 11 & ... & 1F \\ \vdots & & \ddots & \\ F0 & F1 & & FF% \end{array}% \right] $\newline Thus any element value in row A element B is 0xAB \item Map each byte in the S-Box to its multiplicative inverse in $GF(2^{8})$ where $00\rightarrow 00$. \item Each byte in the S-Box consists of 8 bits labeled $% (b_{7},b_{6},...,b_{0})$. Apply the following transformation to each bit of each byte:% \[ b_{i}^{\prime}=b_{i}\oplus b_{\left( i+4\right) \operatorname{mod}8}\oplus b_{\left( i+5\right) \operatorname{mod}8}\oplus b_{\left( i+6\right) \operatorname{mod}8}\oplus b_{\left( i+7\right) \operatorname{mod}8}\oplus c_{i} \] where $c_{i}$ is the $i^{th}$ bit of byte $c$ with the value $\left\{ 63\right\} $. That is $\left( c_{7}c_{6}c_{5}c_{4}c_{3}c_{2}c_{1}% c_{0}\right) =\left( 01100011\right) $. \end{enumerate} \textbf{Briefly describe Sub Bytes.} SubBytes: Uses the S-box described above to perform a byte-by-byte substitution of the state (or input) block as show in Figure~\ref{fig:subbytes} \begin{figure} \centering \includegraphics{AES_SubBytes.png} \caption{S-Box substitution}\label{fig:subbytes} \end{figure} In the decryption algorithm an Inverse-S-Box is used. $S:EA \rightarrow 87$ and $S^{-1}:87 \rightarrow EA$. \textbf{Briefly describe ShiftRow Transformation.} To perform the ShiftRow transformation, we take the state and ''left circular shift'' row 0 by 0 byts, 1 by 1 byte, row 2 by 2 bytes, and row 3 by 3 bytes. To perform the inverse we use right shifts instead of left shifts. \textbf{How many bytes in ''State'' are affected by Shift Rows?} 12 Bytes \textbf{Briefly describe MixColumns.} MixColumns operates on each column individually and is defined by the following matrix multiplication on state:% \[ \left[ \begin{array} [c]{cccc}% 02 & 03 & 01 & 01\\ 01 & 02 & 03 & 01\\ 01 & 01 & 02 & 03\\ 03 & 01 & 01 & 02 \end{array} \right] \left[ \begin{array} [c]{cccc}% S_{0,0} & S_{0,1} & S_{0,2} & S_{0,3}\\ S_{1,0} & S_{1,1} & S_{1,2} & S_{1,3}\\ S_{2,0} & S_{2,1} & S_{2,2} & S_{2,3}\\ S_{3,0} & S_{3,1} & S_{3,2} & S_{3,3}% \end{array} \right] =% \begin{array} [c]{cccc}% S_{0,0}^{\prime} & S_{0,1}^{\prime} & S_{0,2}^{\prime} & S_{0,3}^{\prime}\\ S_{1,0}^{\prime} & S_{1,1}^{\prime} & S_{1,2}^{\prime} & S_{1,3}^{\prime}\\ S_{2,0}^{\prime} & S_{2,1}^{\prime} & S_{2,2}^{\prime} & S_{2,3}^{\prime}\\ S_{3,0}^{\prime} & S_{3,1}^{\prime} & S_{3,2}^{\prime} & S_{3,3}^{\prime}% \end{array} \] In the matrix multiplication we must remember that we are doing multiplication in $G\left( 2^{8}\right) $. We do multiplication as follows:% \begin{align*} 01\ast S_{i,j} & =S_{i,j}\\ 02\ast S_{i,j} & =\left\{ \begin{array} [c]{cc}% (b_{6}b_{5}b_{4}b_{3}b_{2}b_{1}b_{0}0) & if~~b_{7}=0\\ (b_{6}b_{5}b_{4}b_{3}b_{2}b_{1}b_{0}0)\oplus(00011011) & if~~b_{7}=1 \end{array} \right. \\ 03\ast S_{i,j} & =\left\{ \begin{array} [c]{cc}% (b_{6}b_{5}b_{4}b_{3}b_{2}b_{1}b_{0}0)\oplus(b_{7}b_{6}b_{5}b_{4}b_{3}% b_{2}b_{1}b_{0}) & if~~b_{7}=0\\ (b_{6}b_{5}b_{4}b_{3}b_{2}b_{1}b_{0}0)\oplus(00011011)\oplus(b_{7}b_{6}% b_{5}b_{4}b_{3}b_{2}b_{1}b_{0}) & if~~b_{7}=1 \end{array} \right. \end{align*} The inverse matrix is even uglier because it contains elements such as $0x$ where $x \geq 9$. \textbf{Briefly describe Add Round Key.} Recall that the $key$ is $4-32$ bit words. and that the key block is arranged \[ k=\left[ \begin{array} [c]{cccc}% w_{0} & w_{1} & w_{2} & w_{3}% \end{array} \right] \] where each word is a column of 32 bits. to write this as a square we just break the 32 bits into 8 bit bytes per row. Then we can just $\oplus$ the state with the key to get the next state:% \[ \left[ \begin{array} [c]{cccc}% S_{0,0} & S_{0,1} & S_{0,2} & S_{0,3}\\ S_{1,0} & S_{1,1} & S_{1,2} & S_{1,3}\\ S_{2,0} & S_{2,1} & S_{2,2} & S_{2,3}\\ S_{3,0} & S_{3,1} & S_{3,2} & S_{3,3}% \end{array} \right] \oplus\left[ \begin{array} [c]{cccc}% w_{0,0} & w_{1,0} & w_{2,0} & w_{3,0}\\ w_{0,1} & w_{1,1} & w_{2,1} & w_{3,1}\\ w_{0,2} & w_{1,2} & w_{2,2} & w_{3,2}\\ w_{0,3} & w_{1,3} & w_{2,3} & w_{3,3}% \end{array} \right] \] \textbf{Breifly describe the key expansion algorithm.} \begin{tabbing} We start with a 16 byte (128 bit) key and perform the following: \smallskip \= KeyExpansion(byte key[16], word w[44]) \\ \> { \\ \> \= word temp \\ \> \> for (i=0; i<4; i++) w[i] = (key[4*i], key[4*i+1], key[4*i+2], key[4*i+3]); \\ \> \> \= for (i=4; i<44; i++) \\ \> \> \> { \\ \> \> \> \= temp = w[i-1]; \\ \> \> \> \> if (i mod 4 = 0) temp = SubWord(RotWord(temp)) XOR Rcon[i/4]; \\ \> \> \> \> w[i] = w[i-4] XOR temp \\ \> \> \> } \\ \> \> RotWord(word x) performs a left circular rotation by 1 word. \\ \> \> SubWord(word x) uses the S-Box as a lookup table to perform a substitution of each byte in the word. \\ \> } \\ \end{tabbing} \textbf{What is the difference between SubBytes and SubWord?} SubBytes performs takes a byte and performs the substitution using the S-Box. SubWord takes a word (4 bytes) and performs SubBytes on each byte in place. \textbf{What is the difference between ShiftRows and RotWord?} Nothing really except that Shift Rows really does shift a row, and the words are stored in a column which for RotWord we can view as a row. \textbf{What is the difference between the AES decryption algorithm and the equivalent inverse cipher?} Because Round 10 is different than the other rounds you can not just reverse the process. Plus you must use inverse S-box which is not the same as the original S-Box. similarly the SubBytes and MixCols are not there own inverse, thus the decryption can not be the same as the encryption. \end{document}