Commit 5ce685aa by Ondřej Mička

### 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 ... ...
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