From 8147d4ec4a19d304a08fe37c2b0b1b61596abb9c Mon Sep 17 00:00:00 2001 From: Andre Henriques Date: Tue, 7 Nov 2023 21:10:48 +0000 Subject: [PATCH] fix spelling --- cw/cw.tex | 17 +++++++++++++++-- 1 file changed, 15 insertions(+), 2 deletions(-) diff --git a/cw/cw.tex b/cw/cw.tex index b295044..839b2ab 100644 --- a/cw/cw.tex +++ b/cw/cw.tex @@ -219,11 +219,24 @@ which means that - $$(((m^{r_{a\text{ alice}}})^{r_{a\text{ bob}}})^{r_{b\text{ alice}}})^{r_{b\text{ bob}}}) = m (\text{mod } p$$ + $$(((m^{r_{a\text{ alice}}})^{r_{a\text{ bob}}})^{r_{b\text{ alice}}})^{r_{b\text{ bob}}} = m (\text{mod } p)$$ in this case $r_a$ from alice cancels $r_b$ from alice, and $r_a$ from bob cancels $r_b$ from bob \subsubsection*{6.3.2} - To send an encrypted message using this system + To send an encrypted message using this system: + \begin{enumerate} + \item Bob and Alice choose a prime $p$ + \item The sender, let's say Alice, selects $m$ and two random values $r_{a1}$ and $r_{a2}$ such that $(m^{r_{a1}})^{r_{a2}} = m (\text{mod } p)$ + \item Alice then calculates $t1 = m^{r_{a1}} (\text{mod } p)$, Alice sends $t1$ to bob. + \item Bob selects two random values $r_{b1}$ and $r_{b2}$ such that $(m^{r_{b1}})^{r_{b2}} = m (\text{mod } p)$ + \item Bob then calculates $t2 = t1^{r_{b1}} (\text{mod } p)$, Bob sends $t2$ to Alice + \item Alice then calculates $t3 = t2^{r_{a2}} (\text{mod } p)$, this undoes step 3, then Alice sends $t3$ to bob + \item Bob then calculates $m = t3^{r_{b2}} (\text{mod } p)$, this undoes step 5 + \end{enumerate} + \subsubsection*{6.3.3} + Bob and Alice exchange information 4 times, they choose the primes and then 3 exchanges during the encryption process. + While for ElGamal you need to exchange information only twice, once to exchange public keys and the second to exchange the encrypted message + \subsubsection*{6.3.4} \section*{7}