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datovky
ds2-notes
Commits
5ce685aa
Commit
5ce685aa
authored
Sep 09, 2021
by
Ondřej Mička
Browse files
Dynamization: typos
parent
e62e5242
Changes
1
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Inline
Side-by-side
vk-dynamic/dynamic.tex
View file @
5ce685aa
...
...
@@ -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
$
\sqc
a
p
$
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
$
\sqc
u
p
$
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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