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Fixing notation when discussing gcd.

Shit I mixed my variables with knuth's one
Michele Orrù 11 年之前
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共有 1 个文件被更改,包括 5 次插入5 次删除
  1. 5 5
      book/math_prequisites.tex

+ 5 - 5
book/math_prequisites.tex

@@ -25,14 +25,14 @@ protocol, \openssl implemented it with the following signature:
 
 
 The computation proceeds under the well-known Euclidean algorithm, specifically
 The computation proceeds under the well-known Euclidean algorithm, specifically
 the binary variant developed by Josef Stein in 1961 \cite{AOCPv2}. This variant
 the binary variant developed by Josef Stein in 1961 \cite{AOCPv2}. This variant
-exploits some interesting properties of $gcd(u, v)$
+exploits some interesting properties of $gcd(a, b)$
 \begin{itemize}
 \begin{itemize}
   \setlength{\itemsep}{1pt}
   \setlength{\itemsep}{1pt}
   \setlength{\parskip}{0pt}
   \setlength{\parskip}{0pt}
   \setlength{\parsep}{0pt}
   \setlength{\parsep}{0pt}
-  \item if $u,\ v$ are even, then $gcd(u, v) = 2gcd(u/2, v/2)$
-  \item if $u$ is even and $v$ is odd, then $gcd(u, v) = gcd(u/2, v)$
-  \item  $gcd(u, v) = gcd(u-v, v)$, as in the standard Euclid's algorithm
+  \item if $a,\ b$ are even, then $gcd(a, b) = 2gcd(a/2, b/2)$
+  \item if $a$ is even and $b$ is odd, then $gcd(a, b) = gcd(a/2, b)$
+  \item  $gcd(a, b) = gcd(a-b, b)$, as in the standard Euclid's algorithm
   \item the sum of two odd numbers is always even
   \item the sum of two odd numbers is always even
 \end{itemize}
 \end{itemize}
 
 
@@ -256,7 +256,7 @@ explainations, the reference is still \cite{Dijkstra:adop}.
 \end{algorithm}
 \end{algorithm}
 
 
 A fair approxidmation of the magnitude of the Dijkstra algorithm can be studied
 A fair approxidmation of the magnitude of the Dijkstra algorithm can be studied
-by looking at the pseudocode in ~\ref{alg:sqrt:dijkstra_naif}.Exactely as with
+by looking at the pseudocode in ~\ref{alg:sqrt:dijkstra_naif}. Exactely as with
 the dicotomic search case, we split the interval $[a, b]$ in half on each step,
 the dicotomic search case, we split the interval $[a, b]$ in half on each step,
 and choose wether to take the leftmost or the rightmost part. This results in
 and choose wether to take the leftmost or the rightmost part. This results in
 $log(n+1)$ steps. During each iteration, instead, as we have seen in
 $log(n+1)$ steps. During each iteration, instead, as we have seen in