andr3/ComputerSecurity
Archived
1
0
Fork 0
Code Issues Pull Requests Packages Projects Releases 2 Wiki Activity

fix spelling
All checks were successful
continuous-integration/drone/push Build is passing
Details

Browse Source
This commit is contained in:
Andre Henriques 2023-11-07 20:53:38 +00:00
parent 52425244ef
commit ec37a83922
1 changed files with 8 additions and 7 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

15
cw/cw.tex
View File

@ -201,28 +201,29 @@
a = \frac{s' - s}{h - h'} a = \frac{s' - s}{h - h'}
$$ $$
\subsection{6.2} \subsection*{6.2}
\subsubsection{6.2.1} \subsubsection*{6.2.1}
To sign a contract $C$ Alice first chooses 2 random values $r$ and $c_2$ then $z$ is callulated $z=g^r\times y_b^{c_2}$. After we have $z$ we can calculate the intermidary value $c$, $c = H(y_a, y_b, C, z)$. After having $c$ we calculate $c_1$, $c_1 = c - c_2$. $c_1$ is then used to callulate $s = r - c1 \times a mod q$. The signature is $(c_1, c_2, s)$ To sign a contract $C$ Alice first chooses 2 random values $r$ and $c_2$ then $z$ is callulated $z=g^r\times y_b^{c_2}$. After we have $z$ we can calculate the intermidary value $c$, $c = H(y_a, y_b, C, z)$. After having $c$ we calculate $c_1$, $c_1 = c - c_2$. $c_1$ is then used to callulate $s = r - c1 \times a mod q$. The signature is $(c_1, c_2, s)$
\subsubsection{6.2.2} \subsubsection*{6.2.2}
No because Alice only needs Bob's public key which is publicly avaiable No because Alice only needs Bob's public key which is publicly avaiable
\subsubsection{6.2.3} \subsubsection*{6.2.3}
The signature is verified if the equation holds The signature is verified if the equation holds
$$c_1 + c_2 = H(y_a, y_b, C, g^s\times y_a^{c_1} \times y_b^{c_2} mod p )$$ $$c_1 + c_2 = H(y_a, y_b, C, g^s\times y_a^{c_1} \times y_b^{c_2} mod p )$$
\subsubsection{6.2.4} \subsubsection*{6.2.4}
No because the signature is generated from multiple public keys and Alice's private key therefore chris will not be able to tell who signed the contract No because the signature is generated from multiple public keys and Alice's private key therefore chris will not be able to tell who signed the contract
\subsection*{6.3} \subsection*{6.3}
\subsubsection*{6.3.1} \subsubsection*{6.3.1}
The encryption works because the numbers that were chosen by Alice and Bo b make this equation work The encryption works because the numbers that were chosen by Alice and Bo b make this equation work
$$(m^{r_a})^{r_b} = m$$ $$(m^{r_a})^{r_b} = m (\text{mod } p)$$
which means that which means that
$$((((m^{r_{a\text{ alice}}})^{r_{a\text{ bob}}})^{r_{b\text{ alice}}})^{r_{b\text{ bob}}}) = m$$ $$(((m^{r_{a\text{ alice}}})^{r_{a\text{ bob}}})^{r_{b\text{ alice}}})^{r_{b\text{ bob}}}) = m (\text{mod } p$$
in this case $r_a$ from alice cancels $r_b$ from alice, and $r_a$ from bob cancels $r_b$ from bob in this case $r_a$ from alice cancels $r_b$ from alice, and $r_a$ from bob cancels $r_b$ from bob
\subsubsection*{6.3.2} \subsubsection*{6.3.2}
To send an encrypted message using this system
\section*{7} \section*{7}