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Typed up notes from 9/11

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\documentclass[12pt,letterpaper,final]{report}
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\usepackage{amsmath}
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\usepackage{amsfonts}
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\usepackage{amssymb}
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\usepackage{amsthm}
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\usepackage[linewidth=1pt]{mdframed}
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\usepackage{fullpage}
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\begin{document}
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\begin{mdframed}
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\center{\Large{\textbf{MATH 314 Fall 2019 - Class Notes}}}
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\center{9/11/2019} %Put the date of the class here!
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\center{Scribe: Christine Adams} %Put Your Name Here!
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\end{mdframed}
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\textbf{\underline{Summary:}} Today's class covered the Hill cipher
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\\
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\textbf{Hill Cipher}\\
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\textbf{m-block size}\\
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\textbf{k-$m \times m$} matrix $mod{26}$}\\
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\textbf{(k has to have an inverse)}\\
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E($\dot{\vec{v}}$)=$\dot{\vec{v}}$k \\
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D($\dot{\vec{c}}$)=$\dot{\vec{c}}$k \\
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\textbf{Inverse of a $2 \times 2 \matrix} \
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(mod{26}) \\
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if k=$ \begin{smallmatrix}a&b\\c&d\end{smallmatrix}$ ($mod{26}$)\\
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\textbf{Then k^{-1}$=(adbc)^{-1} $\begin{smallmatrix}d&-b\\c&a\end{smallmatrix}$ ($mod{26}$)}\\
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\textbf{Determinant has to be a mod 26 value that is odd and not 13}\\
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\\
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\textbf{Gcd (det(k), 26)=1, 1 being the greatest factor they have in common.}\\
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\textbf{This is true for any hill cipher matrix k.}\\
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Find the inverse of
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$$\[
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K=
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\left[ {\begin{array}{cc}
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4 & 1 \\
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3 & 10 \\
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\end{array} } \right]
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\]$$
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\textbf{det(k)=$4\times 10$ -$1\times 3$=37=11 mod {26}}
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K^{-1}=(11)^{-1}=$\begin{smallmatrix}10&-1\\-3&4\end{smallmatrix}$\\
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$$\[19\times
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\left[ {\begin{array}{cc}
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10 & 25 \\
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23 & 4 \\
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\end{array} } \right] mod{26}
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\]$$
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$$\[\equiv
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\left[ {\begin{array}{cc}
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19\times 10 & 19 \times 25 \\
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19\times 23 & 19\times 4 \\
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\end{array} } \right] mod{26}
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\]$$
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=$$\[\equiv
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\left[ {\begin{array}{cc}
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8 & 7 \\
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21 & 24 \\
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\end{array} } \right] \in \textbf{Inverse}
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\]$$
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\textbf{\underline{Chosen Plaintext attack}}\\
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\textbf{Suppose m=2}\\
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\textbf{Pick the plaintext} \enquote{"ba"}\textbf{$<$1,0$>$} \\
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\textbf{E($<1$,0$>$=$<$1,0$>$}
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$\begin{smallmatrix}a&b\\c&d\end{smallmatrix}$ = $<$a,b$>$ or $<$1,0$>$ ( Find the first row of k)}
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\\
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\\
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Encrypt \enquote{"a,b"}- E($<$0,1$>$)=$<$0,1$>$ $\begin{smallmatrix}a&b\\c&d\end{smallmatrix}$
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\\Known plaintext attack\\
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Find the key using linear algebra\\
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Alice sends the ciphertext LIPVPI to Bob \\
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11,19,15,21,15,8 \\
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Eve learns this corresponds to \enquote{"linear"} \\
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11,19,15,21,15,8 \\
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Block size m=2\\
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$<$11,8$>$ k =$<$11,19$>$\\
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$<$13,4$>$k=$<$15,21$>$\\
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$<$0,17$>$k=$<$15,8$>$ $mod{26}$ \\
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\textbf{Matrix equation}
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$$\[ \left[ {\begin{array}{cc}
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11 & 8 \\
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13 & 4 \\
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\end{array} } \right] k=
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\left[{\begin{array}{cc}
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11 & 19 \\
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15 & 21 \\
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\end{array} } \right
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]$$ \\
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Invert the first matrix (find the inverse to get k by itself)
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$$\[
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\left[ {\begin{array}{cc}
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11 & 8 \\
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13 & 4 \\
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\end{array}} \right]^{-1}
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= (44-104)^{-1}
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\left[{\begin{array}{cc}
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4 & -8 \\
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-13 & 11 \\
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\end{array} } \right]
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\]$$
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\textbf{(Even so not invertible)}\\
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\textbf{Try again!\\}
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\textbf{1st and 3rd equation\\}
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$$\[
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\left[ {\begin{array}{cc}
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11 & 8 \\
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0 & 17 \\
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\end{array} } \right]
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k \equiv \left[ {\begin{array}{cc}
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11 & 18\\
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15 & 8 \\
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\end{array} } \right]
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\]$$
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\textbf{Invert this}
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$$\[
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\left[ {\begin{array}{cc}
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11 & 8 \\
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0 & 17 \\
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\end{array} } \right]^{-1}\equiv (11*17-8(0))^{-1}
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\left[ {\begin{array}{cc}
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17 & -8\\
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0 & 11 \\
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\end{array} } \right]
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\equiv 5^{-1} \left[ {\begin{array}{cc}
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17 & -8\\
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0 & 11 \\
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\end{array} } \right]
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\]$$
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$$\[
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\equiv 21 \times
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\left[ {\begin{array}{cc}
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17 & 18\\
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0 & 11 \\
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\end{array} } \right]
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\equiv
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\left[ {\begin{array}{cc}
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19 & 14\\
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0 & 23 \\
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\end{array} } \right]
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\]$$
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\textbf{Multiply both sides of equations on left!}
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$$\[
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\left[ {\begin{array}{cc}
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19 & 14\\
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0 & 23 \\
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\end{array} } \right]
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\left[ {\begin{array}{cc}
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11 & 8\\
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0 & 17 \\
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\end{array} } \right] k\equiv
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\left[ {\begin{array}{cc}
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19 & 14\\
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0 & 23 \\
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\end{array} } \right]
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\left[ {\begin{array}{cc}
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11 & 19\\
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15 & 8 \\
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\end{array} } \right]
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\]$$
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Identity \\
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\\
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$$\[K\equiv
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\left[ {\begin{array}{cc}
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19 & 14\\
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0 & 17 \\
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\end{array} } \right]
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\left[ {\begin{array}{cc}
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11 & 19\\
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15 & 8 \\
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\end{array} } \right]
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\]$$
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$$\[\equiv
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\left[ {\begin{array}{cc}
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19(11)+14(15) & 19(19)+14(8)\\
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23(15) & 23(8) \\
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\end{array} } \right]
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\]$$
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$$\[
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\equiv
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\left[ {\begin{array}{cc}
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1+2 & 23+8\\
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7 & 2 \\
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\end{array} } \right]
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\]$$
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$$\[
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\equiv
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\left[ {\begin{array}{cc}
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3 & 5\\
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7 & 2 \\
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\end{array} } \right]=k
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\]$$
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\textbf{Key matrix has to be invertible }
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\end{document}\documentclass[12pt,letterpaper,final]{report}
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\usepackage{amsmath}
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\usepackage{amsfonts}
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\usepackage{amssymb}
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\usepackage{amsthm}
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\usepackage[linewidth=1pt]{mdframed}
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\usepackage{fullpage}
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\begin{document}
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\begin{mdframed}
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\center{\Large{\textbf{MATH 314 Spring 2018 - Class Notes}}}
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\center{10/14/2015} %Put the date of the class here!
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\center{Scribe: Name} %Put Your Name Here!
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\end{mdframed}
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\textbf{\underline{Summary:}} Insert a short summary of what today's class covered.
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\\
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\textbf{\underline{Notes:}} Include detailed notes from the lecture or class activities. Format your notes nicely using latex such as
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\begin{itemize}
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\item bullets
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\item or
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\end{itemize}
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\begin{enumerate}
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\item lists
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\item of
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\item things
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\end{enumerate}
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or \textbf{other} \underline{formatting} \textit{commands.} Make sure to write $e^{qu}a+i \circ \mathbb{N} s$ in math mode.
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\\
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\textbf{\underline{Examples:}} If including plaintext or ciphertext or other data it is often helpful to write them using \texttt{typewriter text}.
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\end{document}
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