Commit 5ce685aa authored by Ondřej Mička's avatar Ondřej Mička
Browse files

Dynamization: typos

parent e62e5242
......@@ -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
......
Supports Markdown
0% or .
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment