d5748a460e4b9280d495c1ced002870572fe360f
—
Martin Hafskjold Thoresen
5 years ago
328350f

Fix label spacing and replace $$ with \[ \]

1 files changed,19insertions(+),21deletions(-) M book.tex

M book.tex => book.tex +19~~-21~~

@@ 74,7 74,7 @@ However, encoding a totally random function takes $O(u \log m)$ space.This is a problem since $n = U$, which is large. Therefore we settle on a family $\mathcal{H}$ of hash functions of small size which is \emph{Universal}. Universality means that the probability of two non-equal keys having the same hash value is $O(1/m)$: $$\forall x \neq y\ P_{h \in \mathcal{H}}[h(x) = h(y)] = O(1/m)$$ \[\forall x \neq y\ P_{h \in \mathcal{H}}[h(x) = h(y)] = O(1/m)\] \begin{example} $h(x) = ((a x) \mod p) \mod m$ where $0<a<p$, $p$ is prime and $p > m$ is a universal hash function.@@ 82,7 82,7 @@ $$\forall x \neq y\ P_{h \in \mathcal{H}}[h(x) = h(y)] = O(1/m)$$\begin{definition}{k-independent} A family of hash functions $\mathcal{H}$ is k-independent if $$\forall x_1, \dots x_k\ \Pr[\bigwedge\limits_{i} h(x_i) = t_i] = O(1/m^k)$$ \[\forall x_1, \dots x_k\ \Pr[\bigwedge\limits_{i} h(x_i) = t_i] = O(1/m^k)\] \end{definition} \begin{example}@@ 100,7 100,7 @@ Tabulation hashing is a hashing scheme.We view $x$ as a vector of bit blocks $x_1, \dots, x_c$, and use a totally random hash table on \emph{each} block. Then we \texttt{xor} the blocks together to get the final value: $$h(x) = T_1(x_1) \oplus T_2(x_2) \oplus \dots \oplus T_c(x_c)$$ \[h(x) = T_1(x_1) \oplus T_2(x_2) \oplus \dots \oplus T_c(x_c)\] Since a block is $u/c$ bits, the universe for one block is of size $u^{1/c}$. One table is of size $u^{1/c}$ (we assume the hash value are machine words, as in~\cite{Patrascu:2011:PST:1993636.1993638}), so the total space used is $O(cu^{1/c})$.@@ 121,7 121,7 @@ so $\E[C_t] = O(1)$, for $m = \Omega(n)$Since we need to traverse the list for all operations, the cost is quadratic in $C_t$, so we are interested in $\E[C_t^2]$: $$ \E[C_t^2] = 1/m \sum_{s \in [m]} E[C_s^2] = 1/m \sum_{i \neq j} \Pr[h(x_i) = h(x_j)] $$ \[\E[C_t^2] = 1/m \sum_{s \in [m]} E[C_s^2] = 1/m \sum_{i \neq j} \Pr[h(x_i) = h(x_j)]\] If we have universal hashing this is $\frac{1}{m} n^2 O(\frac{1}{m}) = O(n^2/m^2) = O(1)$ for $m = \Omega(n)$. With a totally random hash function $C_t = O(\frac{\log n}{\log\log n})$ with probability $1 - 1/n^c$ for any $c$. This also holds for simple tabulation hashing.@@ 171,7 171,7 @@ If the hashing functions are fully random or $\log n$-wise independent we get $O\chapter{Static Tree Queries} \chapter{Static Tree Queries}% \label{ch:statictrees} \topics{Range Minimum Query (\code{RMQ}{i, j}), Lowest Common Ancestor (\code{LCA}{x, y}), Level Ancestor (\code{LA}{x, n}), Range Queries, Range Trees.}@@ 216,7 216,7 @@ in the array. However, since we are not interested in the actual minimum, but onthe two arrays act exactly the same. A consequence of this is that we get \emph{Universe reduction}, where we have mapped the universe $\mathcal{U}$ to $[n-1]$. \section{Constant time \LCA{} (\RMQ{})} \section{Constant time \LCA{} (\RMQ{})}% \label{sec:constant-lca} \subsubsection{Step 1: Reduction}@@ 312,7 312,7 @@ Since all nodes have constant access to a leaf node (by its ladder, of which theonly leaves need to store the jump pointers. In other words, we make all queries start at leaves. \subsubsection{Step 6: Leaf Trimming} \subsubsection{Step 6: Leaf Trimming}% \label{sec:first-leaf-trim} \begin{definition}{Maximally Deep Node} A node with $\geq \frac{1}{4} \log n$ descendants.@@ 435,7 435,7 @@ Table~\ref{tab:trie-node-repr} sums up all approaches, with query time and space4 & Weight-balanced BST & $O(P + \log k)$ & $O(T)$\\ 5 & Leaf Trimming + (1) + (4) & $O(P + \log\Sigma)$ & $O(T)$ \end{tabular} \caption{Table over node representation, query time, and space requirement.} \caption{Table over node representation, query time, and space requirement.}% \label{tab:trie-node-repr} \end{table}@@ 512,7 512,7 @@ This allows for faster searching, as we know how much of the adjacent suffix mat5 & \texttt{a\$}\\ 6 & \texttt{\$} \end{tabular} \caption{The suffixes of $T$ \caption{The suffixes of $T$% \label{tab:suffixes}} \end{subfigure} \begin{subfigure}{0.45\textwidth}@@ 527,7 527,7 @@ This allows for faster searching, as we know how much of the adjacent suffix mat4 & \texttt{na\$} & 2 \\ 2 & \texttt{nana\$} & $-$ \end{tabular} \caption{The Suffix Array of $T$. \caption{The Suffix Array of $T$.% \label{tab:suffix-array} } \end{subfigure}@@ 586,7 586,7 @@ Note that the number of suffixes when recursing is $O(2/3)$ of what we started w\subsubsection{Step 5} Now we want to use the suffix array just obtained to radix sort the suffixes in $T_2$. Note that $$T_2[i..] = T[3i+2..] = \big<T[3i+2],\ T[3i+3..]\big> = \big<T[3i+2],\ T_0[i+1..]\big>.$$ Note that \[T_2[i..] = T[3i+2..] = \big<T[3i+2],\ T[3i+3..]\big> = \big<T[3i+2],\ T_0[i+1..]\big>.\] We can take off the first letter of the suffix, and get a suffix which we know the sorted order or, since it is in $T_0$, which we sorted in Step 4. This is like Radix sort, but with only two values, both of which can be compared in $O(1)$. This allows us to sort $T_2$.@@ 654,7 654,7 @@ We can use the \emph{Potential Method} for amortized analysis.Remember that \emph{Amortized cost} is actual cost + change in potential. Let $\Phi = c \cdot \Sigma$, where $\Sigma$ is the number of modifications in all nodes latest versions. Observe that when we make a change, the amortized cost is $$\leq c + c\ [- 2cp + p \cdot \text{recursions}]\text{?}$$ \[\leq c + c\ [- 2cp + p \cdot \text{recursions}]\text{?}\] The first $c$ is for computation, the second $c$ is the added potential, since we just added one modification to a node, and $[-2cp + p \cdot \text{recursions}]\text{?}$ are for the recursions, if there is one.@@ 733,7 733,7 @@ If we want to delete the $+3$, we can add a $-3$ at the end.\subsection{Full Retroactivity} What do we need for full retroactivity? \begin{definition}{Decomposable Search Problem} \begin{definition}{Decomposable Search Problem}% \label{sec:dsp} $\text{query}(x, A\cup B) = f(\text{query}(x, A), \text{query}(x, B))$ \end{definition}@@ 768,7 768,7 @@ To see why this is the case, consider a very simply computer with two registers$X=x$, $Y\mathrel{+}=\Delta$, and $Y = XY$. On query, the machine returns $Y$, and all operations are $O(1)$. The operation sequence $$\big<Y \mathrel{+}= a_n,\ Y=XY,\ Y\mathrel{+}= a_{n-1},\ \dots,\ Y\mathrel{+}= a_0\big>$$ \[\big<Y \mathrel{+}= a_n,\ Y=XY,\ Y\mathrel{+}= a_{n-1},\ \dots,\ Y\mathrel{+}= a_0\big>\] computes the polynomial $\sum^{n}_{i} a_{i}x^{i}$. We can use this to compute the polynomial for any $X$ by retroactively inserting $X=x$ at $t=0$. However, it is not possible to reeveluate such a polynomial using field operations any faster than to just evaluate it again.@@ 800,20 800,18 @@ In order to make things easier we define a \emph{Brige}.\end{definition} Now if $t'$ is the bridge preceding $t$ we see that $$\max\{k'\mid k' \text{\ deleted at time}\geq t\} = \max\{k'\notin Q_{\text{now}} \mid k' \text{\ inserted at time}\geq t'\}$$ \[\max\{k'\mid k' \text{\ deleted at time}\geq t\} = \max\{k'\notin Q_{\text{now}} \mid k' \text{\ inserted at time}\geq t'\}\] In other terms, the largest key deleted after a time $t$ is the largest key inserted after the previous bridge that is not in the final set. Now we can store $Q_{\text{now}}$ as a BBST on values, and all insertions in a BBST on time. The latter trees nodes is also augmented with the value of the largest insert in its subtree that is not in $Q_{\text{now}}$. At last, we store a third BBST with \emph{all} the updates, ordered on time, and also augmented as follows: $$ \text{Augmentation} = \[\text{Augmentation} = \begin{cases} 0 \text{\ if \code{insert}{k}, and\ }k \in Q_\text{now}\\ +1 \text{\ if \code{insert}{k}, and\ }k \notin Q_\text{now}\\ -1 \text{\ if \algo{Delete-Min}}\\ \end{cases} $$ \end{cases}\] In addition the internal nodes store subtree sums and min/max prefix sums. We do this in order to detect brdiges, since a bridge is a prefix summing to 0. When we have to find out which element to insert into $Q_{\text{now}}$ we can walk up from the node in the insertion BST,@@ 836,7 834,7 @@ with $O(\log^2 m)$ overhead. However, it is also possible to get full retroactivBefore tackeling the real problem, we look at an easier problem. \subsection{List Labeling} We want to store integer labels in a list, such that the list, such that insert/delete here queries are constant, We want to store integer labels in a list, such that insert/delete here queries are constant, and that the list is in a strictly monotone ordering. \emph{Label Space} is the size of the labels as a function of the number of elements in the list we want to store. Table~\ref{tab:list-labeling} shows the best known updates for different sizes of the label space.@@ 874,7 872,7 @@ Dynamic partial sums, dynamic connectivity.van Emde Boas, x-fast trees, y-fast trees, fusion trees. \section{van Emde Boas Tree} \section{van Emde Boas Tree}% \label{sec:vEB} \chapter{Succinct Structures}