Oops, a typo in "Corollary"!

02-splay/splay.tex

@@ -89,7 +89,7 @@ and once negatively. Therefore the right-hand side is equal to $3r_t(x) - 3r_0(x
 as claimed by the theorem.
 \qed
 
-\corr{
+\cor{
 As all ranks are logarithmic, the amortized cost of \alg{Splay} is $\O(\log n)$.
 }
 
@@ -143,7 +143,7 @@ $ \sqrt{\alpha\beta} \le (\alpha + \beta) / 2 $.
 
 As $\log{\alpha+\beta\over 2} = \log(\alpha+\beta) - 1$, the lemma implies:
 
-\corrxn{S}{sum of logarithms}{
+\corxn{S}{sum of logarithms}{
 For positive $\alpha$, $\beta$: $\log \alpha + \log \beta \le 2\log(\alpha+\beta) - 2$.
 }
 

03-abtree/abtree.tex

@@ -100,7 +100,7 @@ $$
 Therefore in each tree, we have $n \le b^h-1$, so $h \ge \log_b (n+1)$.
 \qed
 
-\corr{The height is $\Omega(\log_b n)$ and $\O(\log_a n)$.}
+\cor{The height is $\Omega(\log_b n)$ and $\O(\log_a n)$.}
 
 \subsection{Searching for a~key}
 

04-heaps/heaps.tex

@@ -144,7 +144,7 @@ present in the heap. Since each~$n$ has a~unique binary representation, the
 shape of the heap is fully determined by its size. Still, we have a~lot of
 freedom in location of items in the heap.
 
-\corr{
+\cor{
 A~binomial heap with $n$~items contains $\O(\log n)$ trees, whose ranks and
 heights are $\O(\log n)$. Each node has $\O(\log n)$ children.
 }
@@ -500,7 +500,7 @@ can be written as $F_0 + F_1 + F_2 + \ldots + F_k + 1$. By the previous lemma, t
 is equal to~$F_{k+2}$. This completes the induction step.
 \qed
 
-\corrx{R}{
+\corx{R}{
 The ranks of all nodes are $\O(\log n)$.
 }
 

05-cache/cache.tex

@@ -416,7 +416,7 @@ the same. When OPT starts with non-empty cache, it can save up to $\Copt$ misses
 which is compensated by the extra $\Copt$ term in the statement of the theorem.
 \qed
 
-\corr{
+\cor{
 If we set $\Clru = 2\cdot\Copt$, then $\Tlru \le 2\Topt + \Copt$. So on a~long enough
 access sequence, LRU is $(2+\varepsilon)$-competitive.
 }

06-hash/hash.tex

@@ -78,7 +78,7 @@ By linearity of expectation, we have
 $\E[A] = \E[\sum_i A_i] = \sum_i \E[A_i] \le \sum_i c/m = cn/m$.
 \qed
 
-\corrn{Complexity of hashing with chaining}{
+\corn{Complexity of hashing with chaining}{
 Consider a~hash table with chaining which uses $m$~buckets and a~hash function~$h$
 picked at random from a~$c$-universal family. Suppose that the hash table contains
 items $x_1,\ldots,x_n$.
@@ -338,7 +338,7 @@ So $\Pr[M] \le d/m^2 + cd/mr$. We want to write this as $c' / m^2$, so
 $c' = d + cdm/r = (1 + cm/r)d$.
 \qed
 
-\corr{
+\cor{
 The family ${\cal F}\circ {\cal G}$ is also $(2,c')$-independent for $c' = (c+1)d$.
 }
 
@@ -369,7 +369,7 @@ Of course hashing from a~field to the same field is of little use, so we usually
 consider the family ${\cal P}_k\bmod m$ instead. If we set~$p$ large enough, Lemma~\xx{K}
 guarantees:
 
-\corr{If $p \ge 2km$, the family ${\cal P}_k\bmod m$ is $(k,2)$-independent.}
+\cor{If $p \ge 2km$, the family ${\cal P}_k\bmod m$ is $(k,2)$-independent.}
 
 The downside is that we need time $\Theta(k)$ both to pick a~function from the family
 and to evaluate it for a~single~$x$.
@@ -549,7 +549,7 @@ linear functions. Not only we get a~2-independent family as a~result, but Lemma~
 guarantees that if $p$~is sufficiently large, the big constant from $d$-universality
 disappears.
 
-\corr{Given a~prime~$p$ and the number of buckets~$m$ such that $p \ge 4km$, the
+\cor{Given a~prime~$p$ and the number of buckets~$m$ such that $p \ge 4km$, the
 compound family ${\cal R}\circ {\cal L}$ is $(2,5)$-independent.}
 
 Hash functions of this kind play important role in the \em{Rabin-Karp string search
@@ -788,7 +788,7 @@ is bounded by the probability that one of the 12 blocks
 is critical. By union bound and our estimate on the probability that a~block is critical,
 we get:
 
-\corr{
+\cor{
 Let~$R$ be a~run containing $h(x)$. The probability that
 $\vert R\vert \in [2^{\ell+2}, 2^{\ell+3})$ is at most
 $12 \cdot (\e/4)^{2^\ell / 3}$ = $12\cdot q^{2^\ell}$, where $q = (\e/4)^{1/3} \doteq 0.879$.

07-geom/geom.tex

@@ -119,7 +119,7 @@ Since all paths contain $\O(\log n)$ nodes together, we visit $\O(\log n)$ nodes
 and report $\O(\log n)$ nodes and $\O(\log n)$ subtrees.
 \qed
 
-\corr{
+\cor{
 An~enumeration query is answered in time $\O(\log n + p)$, where~$p$ is the number
 of items reported. If we precompute sizes of all subtrees, a~counting query takes
 $\O(\log n)$ time. Aggregate queries can be answered if we precompute aggregate

08-string/string.tex

@@ -72,7 +72,7 @@ We can easily find the start and end of this range using binary search on the su
 We need $\O(\log |\alpha|)$ steps, each step involves string comparison with~$\alpha$, which
 takes $\O(|\beta|)$ time in the worst case. This makes $\O(|\beta| \log|\alpha|)$ total.
 
-\corr{Using the suffix array for~$\alpha$, we can enumerate all occurrences of a~substring~$\beta$
+\cor{Using the suffix array for~$\alpha$, we can enumerate all occurrences of a~substring~$\beta$
 in time $\O(|\beta| \log |\alpha| + p)$, where $p$~is the number of occurrences reported. Only
 counting the occurrences costs $\O(|\beta| \log |\alpha|)$ time.
 }

tex/adsmac.tex

@@ -755,10 +755,10 @@
 \def\obs{\s{Observation:} \sbody}
 \def\note{\s{Note:} \sbody}
 \def\noten#1{\s{Note (#1):} \sbody}
-\def\corr{\s{Corrolary:} \sbody}
-\def\corrn#1{\s{Corrolary (#1):} \sbody}
-\def\corrx#1{\s{Corrolary #1:} \sbody}
-\def\corrxn#1#2{\s{Corrolary~#1 (#2):} \sbody}
+\def\cor{\s{Corollary:} \sbody}
+\def\corn#1{\s{Corollary (#1):} \sbody}
+\def\corx#1{\s{Corollary #1:} \sbody}
+\def\corxn#1#2{\s{Corollary~#1 (#2):} \sbody}
 \def\nota{\s{Notation:} \sbody}
 \def\example{\s{Example:} \sbody}
 \def\examples{\s{Examples:} \sbody}