Dynamization: Semidynamization and worst-case version

vk-dynamic/dynamic.tex

@@ -73,12 +73,12 @@ limit}. We use it to determine if a tree is balanced enough.
 	$\O(\log_{1/\alpha} n)$.
 }
 
-\proof{
+\proof
 Choose an arbitrary path from the root to a leaf and track the node
 sizes. The root has size $n$. Each subsequent node has its size at most
 $\alpha n$. Once we reach a leaf, its size is 1. Thus the path can
 contain at most $\log_{1/\alpha} n$ edges.
-}
+\qed
 
 Therefore, we want to keep the nodes balanced between any operations. If any
 node becomes unbalanced, we take the highest such node $v$ and rebuild its
@@ -101,7 +101,7 @@ show this more in detail for insertion.
 	n)$, with constant factor dependent on $\alpha$.
 }
 
-\proof{
+\proof
 We define a potential as a sum of ``badness'' of all tree nodes. Each node will
 contribute by the difference of sizes of its left and right child. To make
 sure that perfectly balanced subtrees do not contribute, we clamp difference of
@@ -132,6 +132,249 @@ contributions stay the same. Thus, the potential decreases by at least
 $(2\alpha - 1) \cdot s(v) \in \Theta(s(v))$. By multiplying the potential by a
 suitable constant, the real cost $\Theta(s(v))$ of rebuild will be fully
 compensated by the potential decrease, yielding zero amortized cost.
+\qed
+
+\section{General semidynamization}
+
+Let us have a static data structure $S$. We do not need to know how the data
+structure is implemented internally. We would like to use $S$ as a ``black
+box'' to build a (semi)dynamic data structure $D$ which supports queries of $S$
+but also allows element insertion.
+
+This is not always possible, the data structure needs to support a specific
+type of queries answering {\I decomposable search problems}.
+
+\defn{
+A {\I search problem} is a mapping $f: U_Q \times 2^{U_X} \to U_R$ where $U_Q$
+is an universe of queries, $U_X$ is an universe of elements and $U_R$ is set of
+possible answers.
+}
+
+\defn{
+A search problem is {\I decomposable}, if there exists an operator $\sqcup: U_R
+\times U_R$ computable in time $\O(1)$\foot{
+The constant time constraint is only needed for a good time complexity of $D$.
+If it is not met, the construction will still work correctly. Most practical composable
+problems meet this condition.}
+such that $\forall A, B \subseteq U_X$, $A \cap B = \emptyset$ and $\forall q
+\in U_Q$: $$ f(q, A \cup B) = f(q, A) \sqcup f(q, B).$$
 }
 
+\examples
+
+\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
+$U_R = \{\hbox{true}, \hbox{false}\}$. This search problem is decomposable, the
+operator $\sqcup$ is a simple binary \alg{or}.
+
+\: Let $X$ be set of points on a plane. For a point $q$, what is the distance
+of $q$ and the point $x \in X$ closest to $q$? This is a search problem where
+$U_Q = U_X = \R^2$ and $U_R = \R^+_0$. It is also decomposable -- $\sqcup$
+returns the minimum.
+
+\: Let $X$ be set of points of a plane. Is $q$ in convex hull of $X$? This
+search problem is not decomposable -- it is enough to choose $X = \{a, b\}$ and
+$q \notin X$. If $A = \{a\}$ and $B = \{b\}$, both subqueries answer
+negatively. However, the query answer is equivalent to whether $q$ is a convex
+combination of $a$ and $b$.
+\endlist
+
+For a decomposable search problem $f$ we can thus split (decompose) any query
+into two queries on disjoint element subsets, compute results on them
+separately and then combine them in constant time to the final result. We can
+further chain the decomposition on each subset, allowing to decompose the query
+into an arbitrary amount of subsets.
+
+We can therefore use multiple data structures $S$ as blocks, and to answer a
+query we simply query all blocks, and then combine their answers using
+$\sqcup$. We will show this construction in detail.
+
+\subsection{Construction}
+
+First, let us denote a few parameters for the static and dynamic data
+structure.
+
+\nota{For a data structure $S$ containing $n$ elements and answering a
+decomposable search problem $f$ and the resulting dynamic data structure $D$:}
+
+\tightlist{o}
+\: $B_S(n)$ is time complexity of building $S$,
+\: $Q_S(n)$ is time complexity of query on $S$,
+\: $S_S(n)$ is the space complexity of $S$.
+\medskip
+\: $Q_D(n)$ is time complexity of query on $D$,
+\: $S_D(n)$ is the space complexity of $D$,
+\: $\bar I_D(n)$ is {\I amortized} time complexity of insertion to $D$.
+\endlist
+
+We assume that $Q_S(n)$, $B_S(n)/n$, $S_S(n)/n$ are all nondecreasing functions.
+
+We decompose the set $X$ into blocks $B_i$ such that $|B_i| \in \{0, 2^i\}$
+such that $\bigcup_i B_i = X$ and $B_i \cap B_j = \emptyset$ for all $i \neq
+j$. Let $|X| = n$. Since $n = \sum_i n_i 2^i$, its binary representation
+uniquely determines the block structure. Thus, the total number of blocks is at
+most $\log n$.
+
+For each nonempty block $B_i$ we build a static structure $S$ of size $2^i$.
+Since $f$ is decomposable, a query on the structure will run queries on each
+block, and then combine them using $\sqcup$:
+$$ f(q, x) = f(q, B_0) \sqcup f(q, B_1) \sqcup \dots \sqcup f(q, B_i).$$
+
+TODO image
+
+\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 $Q_S(x)
+\leq Q_S(n)$ for all $x \leq n$, the inequality holds.
+\qed
+
+Now let us calculate the space complexity of $D$.
+
+\lemma{$S_D(n) \in \O(S_S(n))$.}
+
+\proof
+For $|X| = n$ let $I = \{i \mid B_i \neq \emptyset\}$. Then for each $i \in I$
+we store a static data structure $S$ with $2^i$ elements contained in this
+block. Therefore, $Q_D(n) = \sum_{i \in I} Q_S(2^i)$. Since $S_S(n)$ is
+assumed to be nondecreasing,
+$$
+	\sum_{i \in I} Q_S(2^i)
+	\leq \sum_{i \in I} {Q_S(2^i) \over 2^i} \cdot 2^i
+	\leq {S_S(n) \over n} \cdot \sum_{i=0}^{\log n} 2^i
+	\leq {S_S(n) \over n} \cdot n.
+$$
+\qed
+
+It might be advantageous to store the elements in each block separately so that
+we do not have to inspect the static structure and extrace the elements from
+it, which may take additional time.
+
+An insertion of $x$ will act like an addition of 1 to a binary number. Let $i$
+be the smallest index such that $B_i = \emptyset$. We create a new block $B_i$
+with elements $B_0 \cup B_1 \cup \dots \cup B_{i-1} \cup \{x\}$. This new block
+has $1 + \sum_{j=0}^{i-1} 2^j = 2^i$ elements, which is the required size for
+$B_i$. At last, we remove all blocks $B_0, \dots, B_{i-1}$ and add $B_i$.
+
+TODO image
+
+\lemma{$\bar I_D(n) \in \O(B_S(n)/n \cdot \log n)$.}
+
+\proof{
+	Since the last creation of $B_i$ there had to be least $2^i$
+	insertions. Amortized over one element this cost is $B_S(2^i) / 2^i$.
+	As this function is nondecreasing, we can lower bound it by $B_S(n) /
+	n$. However, one element can participate in $\log n$ rebuilds during
+	the structure life. Therefore, each element needs to store up cost $\log n
+	\cdot B_S(n) / n$ to pay off all rebuilds.
+}
+
+\theorem{
+Let $S$ be a static data structure answering a decomposable search problem $f$.
+Then there exists a semidynamic data structure $D$ answering $f$ with parameters
+
+\tightlist{o}
+\: $Q_D(n) \in \O(Q_S(n) \cdot \log_n)$,
+\: $S_D(n) \in \O(S_S(n))$,
+\: $\bar I_D(n) \in \O(B_S(n)/n \cdot \log n)$.
+\endlist
+}
+
+\example
+
+If we use a sorted array using binary search to search elements in a static
+set, we can use this technique to create a dynamic data structure for general
+sets. It will require $\Theta(n)$ space and the query will take $\Theta(\log^2
+n)$ time as we need to binary search in each list. Since building requires
+sorting the array, building one requires $\Theta(n \log n)$ and insertion thus
+costs $\Theta(\log^2 n)$ amortized time.
+
+We can speed up insertion time. Instead of building the list anew, we can merge
+the lists in $\Theta(n)$ time, therefore speeding up insertion to $\O(\log n)$
+amortized.
+
+In general, the bound for insertion is not tight. If $B_S(n) =
+\O(n^\varepsilon)$ for $\varepsilon > 1$, the logarithmic factor is dominated
+and $\bar I_D(n) \in \O(n^\varepsilon)$.
+
+\subsection{Worst-case semidynamization}
+
+So far we have created a data structure that acts well in the long run, but one
+insertion can take long time. This may be unsuitable for applications where we
+require a low latency. In such cases, we would like that each insertion is fast
+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
+block is constructed.
+
+However, insertion is not the only operation, we can also ask queries even
+during the construction process. Thus we must keep the old structures until the
+construction finishes. As a consequence, more than one block of each size may
+exist at the same time.
+
+For each rank $i$ let $B_i^0, B_i^1, B_i^2$ be complete blocks participating in
+queries. No such block contains a duplicate element and union of all complete
+blocks contains the whole set $X$.
+
+Next let $B_i^*$ be a block in construction. Whenever two blocks $B_i^a, B_i^b$
+of same rank $i$ meet, we will immidiately start building $B_{i+1}^*$ using
+elements from $B_i^a \cup B_i^b$.
+
+This construction will require $2^{i+1}$
+steps until $B_{i+1}^*$ is finished, allocating enough time for each step. Once
+we finish $B_{i+1}^*$, we add it to the structure as one of the three full
+blocks and finally remove $B_i^a$ and $B_i^b$.
+
+We will show that, using this scheme, this amount of blocks is enough to
+bookkeep the structure.
+
+\lemma{
+At any point of the structure's life, for each rank $i$, there are at most
+three finished blocks and at most one block in construction.
+}
+
+\proof
+For an empty structure, this certainly holds.
+
+Consider a situation when two blocks $B_i^0$ and $B_i^1$ meet and $B_i^1$ has
+just been finalized. Then we start constructing $B_{i+1}^*$.  $2^{i+1}$ steps
+later $B_{i+1}$ is added and blocks $B_i^0$, $B_i^1$ are removed.
+
+There may appear a new block $B_i^2$ earlier. However, this can only happen
+$2^i$ steps later. For the fourth block $B_i^3$ to appear, another $2^i$ steps
+are required. The earliest time is then $2 \cdot 2^i = 2^{i+1}$ steps later,
+during which $B_{i+1}^*$ has been already finalized, leaving at most two blocks
+together and no block of rank $i+1$ in construction.
+\qed
+
+An insertion is now done by simply creating new block $B_0$. Next, we
+additionaly run one step of construction for each $B_j^*$. There may be up to
+$\log n$ blocks in construction.
+
+\theorem{
+Let $S$ be a static data structure answering a decomposable problem $f$. Then
+there exists semidynamic structure with parameters
+
+\tightlist{o}
+\: $Q_D(n) \in \O(Q_S(n) \cdot \log_n)$,
+\: $S_D(n) \in \O(S_S(n))$,
+\: $I_D(n) \in \O(B_S(n)/n \cdot \log n)$ worst-case.
+\endlist
+}
+
+\proof
+Since there is now a constant amount of blocks of each rank, the query time and
+space complexities have increased by a constant compared to previous
+technique.
+
+Each insertion builds a block of size 1 and then runs up to $\log n$
+construction steps, each taking $B_S(2^i)/2^i$ time. Summing this
+together, we get the required upper bound.
+\qed
+
 \endchapter