From 5ce685aacfdf8ad58c4773e9a79f2f1213260725 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?Ondra=20Mi=C4=8Dka=20=40=20miles-teg?=
 <mitch.ondra@gmail.com>
Date: Thu, 9 Sep 2021 11:50:39 +0200
Subject: [PATCH] Dynamization: typos

---
 vk-dynamic/dynamic.tex | 8 ++++----
 1 file changed, 4 insertions(+), 4 deletions(-)

diff --git a/vk-dynamic/dynamic.tex b/vk-dynamic/dynamic.tex
index 2fed7da..9b99bc6 100644
--- a/vk-dynamic/dynamic.tex
+++ b/vk-dynamic/dynamic.tex
@@ -196,7 +196,7 @@ such that $\forall A, B \subseteq U_X$, $A \cap B = \emptyset$ and $\forall q
 
 \list{o}
 \: Let $X \subseteq {\cal U}$. Is $q \in X$? This is a classic search problem
-where universes $U_Q, U_R$ are both set ${\cal U}$ and possible replies are
+where universes $U_Q, U_X$ are both set ${\cal U}$ and possible replies are
 $U_R = \{\hbox{true}, \hbox{false}\}$. This search problem is decomposable, the
 operator $\sqcup$ is a simple binary \alg{or}.
 
@@ -255,8 +255,8 @@ $$ f(q, x) = f(q, B_0) \sqcup f(q, B_1) \sqcup \dots \sqcup f(q, B_i).$$
 \lemma{$Q_D(n) \in \O(Q_s(n) \cdot \log n)$.}
 
 \proof
-Let $|X| = n$. Then the block structure is determined and $\sqcap$ takes
-constant time, $Q_D(n) = \sum_{i: B_i \neq \emptyset} Q_S(2^i) + \O(1)$. Since $Q_S(x)
+Let $|X| = n$. Then the block structure is determined and $\sqcup$ takes
+constant time, $Q_D(n) = \sum_{i: B_i \neq \emptyset} \left(Q_S(2^i) + \O(1)\right)$. Since $Q_S(x)
 \leq Q_S(n)$ for all $x \leq n$, the inequality holds.
 \qed
 
@@ -338,7 +338,7 @@ even in the worst case.
 Our construction can be deamortized for the price that the resulting
 semidynamic data structure will be more complicated. We do this by not
 constructing the block at once, but decomposing the construction such that on
-each operation we do a small amount of work on it until eventually the whole
+each operation we do does a small amount of work on it until eventually the whole
 block is constructed.
 
 However, insertion is not the only operation, we can also ask queries even
-- 
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