From 52425244efb4c41f4d36d81950eae52248a51dc9 Mon Sep 17 00:00:00 2001 From: Andre Henriques Date: Tue, 7 Nov 2023 20:50:00 +0000 Subject: [PATCH] fix spelling --- cw/cw.tex | 24 ++++++++++++++++++++++++ 1 file changed, 24 insertions(+) diff --git a/cw/cw.tex b/cw/cw.tex index 27748ff..79944ce 100644 --- a/cw/cw.tex +++ b/cw/cw.tex @@ -201,6 +201,30 @@ a = \frac{s' - s}{h - h'} $$ + \subsection{6.2} + \subsubsection{6.2.1} + To sign a contract $C$ Alice first chooses 2 random values $r$ and $c_2$ then $z$ is callulated $z=g^r\times y_b^{c_2}$. After we have $z$ we can calculate the intermidary value $c$, $c = H(y_a, y_b, C, z)$. After having $c$ we calculate $c_1$, $c_1 = c - c_2$. $c_1$ is then used to callulate $s = r - c1 \times a mod q$. The signature is $(c_1, c_2, s)$ + \subsubsection{6.2.2} + No because Alice only needs Bob's public key which is publicly avaiable + \subsubsection{6.2.3} + The signature is verified if the equation holds + $$c_1 + c_2 = H(y_a, y_b, C, g^s\times y_a^{c_1} \times y_b^{c_2} mod p )$$ + \subsubsection{6.2.4} + No because the signature is generated from multiple public keys and Alice's private key therefore chris will not be able to tell who signed the contract + + \subsection*{6.3} + \subsubsection*{6.3.1} + The encryption works because the numbers that were chosen by Alice and Bo b make this equation work + $$(m^{r_a})^{r_b} = m$$ + + which means that + + $$((((m^{r_{a\text{ alice}}})^{r_{a\text{ bob}}})^{r_{b\text{ alice}}})^{r_{b\text{ bob}}}) = m$$ + + in this case $r_a$ from alice cancels $r_b$ from alice, and $r_a$ from bob cancels $r_b$ from bob + \subsubsection*{6.3.2} + + \section*{7} \subsection*{7.1} $$v1 = (137, 312), v2 = (215, -187)$$