From bc5e701a8940fdfb6332cbc494c84e0865874067 Mon Sep 17 00:00:00 2001
From: David Florness <david@florness.com>
Date: Sun, 9 Feb 2020 22:40:52 -0700
Subject: [PATCH] OreSec talk: start section on zero-knowledge

---
 doc/oresec-talk/tallyard.tex | 39 ++++++++++++++++++++++++++++++++++--
 1 file changed, 37 insertions(+), 2 deletions(-)

diff --git a/doc/oresec-talk/tallyard.tex b/doc/oresec-talk/tallyard.tex
index 5d45d04..eb901da 100644
--- a/doc/oresec-talk/tallyard.tex
+++ b/doc/oresec-talk/tallyard.tex
@@ -326,9 +326,44 @@
 \end{frame}
 
 \begin{frame}{The Remedy}
-  The definitive answer:
+  Where I got all of the following material: \pause
 
-  \url{https://vitalik.ca/general/2017/11/09/starks\_part\_1.html}
+  \url{https://vitalik.ca/general/2017/11/09/starks\_part\_1.html} \pause
+
+  The following material was very rushed.
+\end{frame}
+
+\begin{frame}
+  \begin{itemize}
+  \item We want to prove that a given $P_i$ constant term is valid
+    \textit{without} revealing what it is. This is equivalent to checking
+    whether $P_i(0)$ is valid. This is the essence of a Zero-Knowledge Proof.
+    \pause
+  \item Now, let $C(x)$ be a \textit{constraint checking polynomial} that is
+    zero if $x$ is a valid constant and nonzero otherwise. For example, if we
+    assume a valid constant is one that is either a zero or one, we can
+    construct $C(x)$ very simply:
+    \begin{equation*}
+      C(x) = (x - 0) (x - 1) = x^2 - x
+    \end{equation*} \pause
+  \item Now, we can restate the problem as: we need to prove that $C(P(x)) = 0$
+    when $x = 0$.
+  \end{itemize}
+\end{frame}
+
+\begin{frame}
+  \begin{itemize}
+  \item Let $Z(x) = x$ \pause
+  \item It's a known mathematical fact that any polynomial that is zero at $x=0$
+    must be a multiple of $Z$. Therefore, there exists some $D(x)$ such that
+    \begin{equation*}
+      C(P(x)) =  Z(x) \cdot D(x)
+    \end{equation*}
+  \item \pause Before anything, everyone ``commits''~\footnote{see commitment
+      schemes} to their polynomial by creating a merkle tree of the outputs of
+    $P(x)$ and $D(x)$ values and sending the root of the tree to
+    everyone. \pause
+  \end{itemize}
 \end{frame}
 
 \end{document}
\ No newline at end of file
-- 
2.38.4