Minor fixes.

diff --git a/notation.tex b/notation.tex
index f51e7462a7f1e667195a602cecce9ed16b8ea8ae..6e34746ed83b560285ebc60f437b79a0b4698dc8 100644 (file)
--- a/notation.tex
+++ b/notation.tex
@@ -44,8 +44,8 @@
 \n{$W$}{word size of the RAM \[wordsize]}
 \n{$\(x)$}{number~$x\in{\bb N}$ written in binary \[bitnota]}
 \n{$\(x)_b$}{$\(x)$ zero-padded to exactly $b$ bits \[bitnota]}
 \n{$W$}{word size of the RAM \[wordsize]}
 \n{$\(x)$}{number~$x\in{\bb N}$ written in binary \[bitnota]}
 \n{$\(x)_b$}{$\(x)$ zero-padded to exactly $b$ bits \[bitnota]}

-\n{$x[i]$}{when $x$ is a~number: the value of the $i$-th bit of~$x$ \[bitnota]}
-\n{$\pi[i]$}{when $\pi$ is a~sequence: the $i$-th element of~$x$, starting with $x[1]$ \[brackets]}
+\n{$x[i]$}{when $x\in{\bb N}$: the value of the $i$-th bit of~$x$ \[bitnota]}
+\n{$\pi[i]$}{when $\pi$ is a~sequence: the $i$-th element of~$\pi$, starting with $\pi[1]$ \[brackets]}

 \n{$\sigma^k$}{the string~$\sigma$ repeated $k$~times \[bitnota]}
 \n{$\0$, $\1$}{bits in a~bit string \[bitnota]}
 \n{$\equiv$}{congruence modulo a~given number}
 \n{$\sigma^k$}{the string~$\sigma$ repeated $k$~times \[bitnota]}
 \n{$\0$, $\1$}{bits in a~bit string \[bitnota]}
 \n{$\equiv$}{congruence modulo a~given number}

diff --git a/rank.tex b/rank.tex
index 1bbd3c03b5dc9e2e3542ff84f0a5b15f8796ddc0..ede335fe16d05c88a00aead4ac9a985b15e88a60 100644 (file)
--- a/rank.tex
+++ b/rank.tex
@@ -143,7 +143,7 @@ or operations on the data structure.
 \qed
 
 \example
 \qed
 
 \example

-If store~$A$ in an~ordinary array, we have insertion and deletion in constant time,
+If we store~$A$ in an~ordinary array, we have insertion and deletion in constant time,

 but ranking and unranking in~$\O(n)$, so $t(n)=\O(n)$ and the algorithm is quadratic.
 Binary search trees give $t(n)=\O(\log n)$. The data structure of Dietz \cite{dietz:oal}
 improves it to $t(n)=O(\log n/\log \log n)$. In fact, all these variants are equivalent
 but ranking and unranking in~$\O(n)$, so $t(n)=\O(n)$ and the algorithm is quadratic.
 Binary search trees give $t(n)=\O(\log n)$. The data structure of Dietz \cite{dietz:oal}
 improves it to $t(n)=O(\log n/\log \log n)$. In fact, all these variants are equivalent