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\begin{document}
\lecture{8 --- March 1, 2005}{Spring 2005}{Prof.\ Erik Demaine}{Vincent Yeung}
\section{Overview}
In this lecture we discuss two problems for dynamic directed graphs:
{\bf transitive closure} and {\bf all-pairs shortest paths}. We will
present various results and open problems for each. Then, we will
focus on one particular result due to Sankowski \cite{sankowski} for
transitive closure, namely that of $O(n^2)$ updates and $O(1)$
queries, both worst-case times.
\section{Dynamic directed graphs}
Instead of the standard insert and delete operations for edges, we
would like our data structures to support more powerful {\em bulk
update} operations. These refer to the insertion/deletion of a vertex
and an arbitrary set of incident edges.
\subsection{Transitive closure}
The dynamic transitive closure problem is essentially that of dynamic
reachability. We would like to answer queries of the form, ``Is there
a path from $v$ to $w$?'' We will now present various results for
dynamic transitive closure:
\subsubsection{Results}
\begin{itemize}
\item
bulk update: $O(n^2)$ amortized\\
query: $O(1)$ worst case \cite{demetrescu1}
\item
bulk update: $O(n^2)$ worst-case, but randomized \\
query: $O(1)$ worst case \cite{sankowski} {\bf (what we will show today)}
\item
{\bf OPEN}: Can we achieve bulk update in $o(n^2)$ worst case?
\end{itemize}
Note that, in some sense, the $O(n^2)$ result is optimal if we
approach this problem by storing the transitive closure matrix {\em
explicitly} (we don't necessarily have to do this, however, to answer
queries). This is because in the worst case, an update can modify
$\Omega(n^2)$ edges in the transitive closure.
Next are two results for which the product of the bulk update and
query times is $O(mn)$, where $m$ is the number of edges. The first
of the two results allows for some amount of trade-off between the two
operations.
\begin{itemize}
\item
bulk update: $O(m \sqrt{n}\ t)$ amortized\\
query: $O(\sqrt{n}/t)$ worst case, where $t=O(\sqrt{n})$ \cite{roditty1}
\item
bulk update: $O(m+n \lg n)$ amortized\\
query: $O(n)$ worst case \cite{roditty2}
\item
{\bf OPEN}: Can we achieve full trade-off with the product of update
and query times equal to $O(mn)$ or $O(n^2)$?
\end{itemize}
Finally, we give two results for special cases of the problem:
\begin{itemize}
\item
{\bf (acyclic graphs)}\\
update: $O(n^{1.575}\ t)$\\
query: $O(n^{0.575}/t)$, $t=O(n^{0.325})$ \cite{demetrescu1}
\item
{\bf (decremental transitive closure)}\\
update: $O(n)$ amortized\\
query: $O(1)$ worst case \cite{demetrescu1}
\end{itemize}
\subsection{All-pairs shortest paths}
Here we present various results for the all-pairs shortest paths
problem (dynamic shortest paths), which answers the question of,
``What is the weight of the shortest path from $v$ to $w$?''
\subsubsection{Results}
\begin{itemize}
\item
bulk update: $O(n^2(\lg n + \lg^2 {(1+m/n)}))$ amortized\\
query: $O(1)$ worst case \cite{thorup1}
\item
{\bf OPEN}: Is it possiblie to achieve $O(n^2)$ or $o(n^2)$ update, the latter
even for just undirected graphs?
\item
update: $O(n^{2.75})$ worst case\\
query: $O(1)$ worst case \cite{thorup2}
\item
{\bf (unweighted graph)}\\
update: $O(m\sqrt{n}\ \polylg n)$ amortized\\
query: $O(n^{3/4})$ worst case \cite{roditty3}
\item
{\bf (unweighted, undirected, and (1+$\epsilon$)-approximate)}\\
update: $O(\sqrt{m}\ nt)$ amortized\\
query: $O(\sqrt{m}/t)$ worst case, $t=O(\sqrt{m})$ \cite{roditty4}
\end{itemize}
\section{$O(n^2)$ worst case dynamic transitive closure}
As promised, we now discuss the result in \cite{sankowski}. The key
idea is to transform the problem at hand (transitive closure) into one
({\bf disjoint cycle cover}) that is solvable via linear algebra, with
the use of various matrix properties.
\paragraph{Preliminaries:}
We will show the connection between transitive closure and disjoint
cycle cover. To do so, we observe that each of the following is
equivalent:
\begin{itemize}
\item[-] There is a path from $i$ to $j$ in the graph.
\item[$\Leftrightarrow$] If after adding the edge $(j,i)$, there is a cycle
through $(j,i)$.
\item[$\Leftrightarrow$] If after removing edges $(*,i)$ (all incoming
edges to $i$) and $(j,*)$ (all outgoing edges from $j$), and adding
the edge $(j,i)$, there is a cycle through $(j,i)$ (or through $i$, or
$j$ -- these are equivalent conditions).
\item[$\Leftrightarrow$] If after adding loops $(k,k)$ for all
vertices $k$, and removing edges $(*,i)$ and $(j,*)$ (note that this
removes $(i,i)$ and $(j,j)$), there is a disjoint set of cycles
covering all vertices. This is the {\em disjoint cycle cover}
problem. (Here, all nodes other than $i$ and $j$ can be covered by
their self-loops, so again we are looking to see if there is a cycle
through $j$.)
\item[$\Leftrightarrow$] If after the same modifications as
immediately above, there is a permutation $\pi$ on the vertices such
that $(i,\pi (i))$ is an edge $\forall i$.
\end{itemize}
\subsection{Connection to linear algebra}
We use the standard definition of a directed graph's {\bf adjacency
matrix} $A$:
\begin{displaymath}
A_{ij} = \left\{ \begin{array}{ll}
1 & \textrm{if $(i,j)$ is an edge}\\
0 & \textrm{otherwise}
\end{array} \right.
\end{displaymath}
Now we need to make the transformations discussed earlier to the
graph. We begin by adding $I$ to $A$ to put in self-loops $(k,k)$ on
each vertex (we are assuming the graph is initially loop-free). Next,
we introduce the {\em zap} operation, represented as $\lightning$, and
define $Z=(A+I) \lightning (j,i)$ as follows:
\begin{enumerate}
\item
Zero out column $i$ and row $j$ in $(A+I)$; this removes the edges
$(*,i)$ and $(j,*)$.
\item
Set the element in the $(j,i)$ position to 1; this adds the edge
$(j,i)$.
\end{enumerate}
At this point, we are ready to start using linear algebra. First, we
employ the following definition of the determinant:
$$\det Z = \sum_{\pi}
{\Big[
\textrm{sign}(\pi)
\underbrace
{\prod_{i} {Z_{i \pi(i)}}}_{\substack{\textrm{0 $\Leftrightarrow$ any edge} \\ \textrm{$(i,\pi(i))$ is missing}}}
\Big]}
$$
If there is a disjoint cycle cover, then at least one of the terms
(products) in this sum must be non-zero, so the determinant should
(hopefully) be nonzero, but unfortunately, the sign rules for
computing determinants may potentially cancel out non-zero terms.
Before we resolve this issue, though, let us first extend the
determinant idea a bit further. One can find that the {\em adjugate}
(known as the {\em adjoint} in some circles) of a square matrix $B$
contains as its entries the determinants of $B$'s $n^2$ ``zapped''
matrices (one for each possible $(j,i)$). Applying this on our matrix
$(A+I)$, we have:
$$(\adj {(A+I)})_{ij} = \det{((A+I) \lightning (j,i))}$$
In other words, $\adj {(A+I)}$ contains all the terms we need to find
the transitive closure (i.e. to answer reachability queries for any
pair $(i,j)$) -- if we resolve the cancellation issue above.
\subsection{Avoiding cancellation}
Conceptually, we want a {\bf symbolic matrix}, which we define as:
\begin{displaymath}
(\symb Z)_{ij} = \left\{ \begin{array}{ll}
a_{ij} & \textrm{if $A_{i,j} \neq 0$} \\
0 & \textrm{otherwise}
\end{array} \right.
\end{displaymath}
In other words, every nonzero entry is represented by a unique
variable. Clearly, the determinant of $\symb Z$ is a polynomial of
degree $\leq n$ in at most $n^2$ variables ($a_{ij}$, etc.). The
symbolic determinant of this matrix, stored in $(\adj \symb
(A+I))_{ij}$ is equal to a zero polynomial iff there is no edge from
$i$ to $j$. This is because all variable are distinct, so permutation
terms cannot cancel out.
But calculating the symbolic determinant takes too long. Instead, we
settle for a {\bf random matrix}, which we define as:
\begin{displaymath}
(\rand Z)_{ij} = \left\{ \begin{array}{ll}
\substack
{\textrm{independent random number chosen} \\
{\textrm{uniformly between 0 and $p-1$}}}
& \textrm{if $A_{i,j} \neq 0$} \\
0 & \textrm{otherwise}
\end{array} \right.
\end{displaymath}
So what we want to do is replace the ones in $(A+I)$ with random
numbers ($p$ is a large prime number). Then, if the determinant of
the symbolic matrix is zero, the determinant of the random matrix will
also be zero. But what about when the polynomial is not zero? The
following lemma can be used to calculate the probability that the
determinant of the random matrix will be zero in this case (which will
make our data structure answer queries incorrectly).
\begin{lemma}[Zippel-Schwartz lemma \cite{zippel,schwartz}]
$$\Pr {\{
\textrm{nonzero polynomial of degree $d$ evaluated at uniform random inputs}
\equiv 0\pmod{p} \}}\leq d/p
$$
\end{lemma}
So, in the case that $\det \symb Z \neq 0$, we have $\Pr{\{\det \rand
Z=0\}} \leq n/p \leq \frac{1}{n^{c-1}}$, if we choose a prime $p \geq
n^c$. We can extend this to all $n^2$ $Z$ matrices that our data
structure needs to handle, and we get that:
$$\Pr {\{\adj \rand {(A+I)} \textrm{\ has more zeros than\ } \adj
\symb {(A+I)}\}} \leq \frac{1}{n^{c-3}} \textrm{\ \ (by union
bound)}$$
It follows that if the data structure runs for polynomial time, we can make
the probability of error $\leq \frac{1}{n^{c'}}$ for any $c' > 0$.
\paragraph{Rebuilding:} If the data structure runs for a long time, we
need to reinitialize the matrix with other random values every
$n^{O(1)}$ operations. The bound remains $O(n^2)$ amortized, and can
be made $O(n^2)$ worst-case by deamortizing global rebuilding as in
problem 3.
\subsection{Dynamic matrix inverse}
Now comes the algorithmic part of this discussion. How do we maintain
the matrix $\adj \rand (A+I)$ in $O(n^2)$ worst case time for bulk
updates and $O(1)$ worst case time for queries? The solution is to
use the following relationship between the inverse and adjugate
matrices:
$$
B^{-1} = \frac {\adj B} {\det B} \textrm{, when it exists } (\det B \ne 0)
$$
Thus, if during bulk udpates we can maintain the inverse and
determinant of $\rand (A+I)$ in $O(n^2)$ time, we can answer queries
in constant time. This is what we will do.
\paragraph{Approach:} When a bulk update to the vertex $i$ occurs, row $i$ and column $i$
of $A$ (really $\rand (A+I)$, but we will just say $A$ for the rest of
the analysis for simplicity) may be modified. Here, we will only
consider modifications to column $i$. We represent this modification
by the $\Delta$ vector: $A'_{ji} \leftarrow A_{ji} + \Delta_j,\
\forall j$. We also define the row vector $e_i^T = \left(
\begin{array}{ccccccc} 0 & \ldots & 0& \underbrace{1}_{i^{th}\
column}&0 & \ldots & 0
\end{array} \right)
$.
Then,
$$
\Delta e_i^T =
\left( \begin{array}{c}
\Delta_1 \\
\vdots \\
\Delta_n
\end{array} \right)
\left( \begin{array}{ccccccc}
0 & \ldots &0& 1 &0 & \ldots & 0
\end{array} \right)
=
\left( \begin{array}{ccccccc}
0 & \ldots & 0&\Delta_1 &0 & \ldots & 0 \\
& & &\vdots& & & \\
0 & \ldots &0& \Delta_n &0 & \ldots & 0
\end{array} \right)
$$
Thus, the new matrix is $A'=A+\Delta e_i^T$.
The key idea is to write $A'=AB$, where $B=I+A^{-1}\Delta e_i^T =
I+(A^{-1}\Delta) e_i^T = I+be_i^T$. Here $b = A^{-1} \Delta$ is a
column vector $ \left( \begin{array}{c} b_1 \\ \vdots \\ b_n
\end{array} \right)
$
that is a linear combination of the columns of $A^{-1}$ and takes $O(n^2)$ time
to compute. Hence, we can write $B$ in the form:
$$
B=
\left( \begin{array}{ccccccc}
1 & & & b_1 & && \\
& \ddots & & b_2 & & & \\
& & 1 & \vdots & & & \\
& & & 1+b_i & & & \\
& & & \vdots & 1 & & \\
& & & b_{n-1}& & \ddots & \\
& & & b_n & & & 1
\end{array} \right)
$$
We know $A'^{-1} = B^{-1}A^{-1}$, so we need to find $B^{-1}$, which we
can do with $\det B$ and $\adj B$, both of which can be computed
in $O(n^2)$ with the following formulas:
\begin{eqnarray*}
&&B^{-1}=
\left( \begin{array}{ccccccc}
1 & & & -\frac{b_1}{1+b_i} & & & \\
& \ddots & & -\frac{b_2}{1+b_i} & & & \\
& & 1 & \vdots & & & \\
& & & \frac{1}{1+b_i} & & & \\
& & & \vdots & 1 & & \\
& & & -\frac{b_{n-1}}{1+b_i} & &\ddots & \\
& & & -\frac{b_n}{1+b_i} & & & 1
\end{array} \right) \\
&&\det B = 1+b_i
\end{eqnarray*}
We also see that $B^{-1}$ has $O(n)$ nonzeros. This is good because it means
we can get $A'^{-1} = B^{-1} A^{-1}$ in $O(n^2)$ time (as
opposed to the standard $O(n^3)$ time needed for matrix multiplication with the
obvious algorithm). The new determinant can also be quickly computed
since $\det{A'} = \det B\det A$.
There's a caveat, though. What if $A$ is singular? We can show that
this happens with very low probability (indeed, as low as we want to
make it) because remember in the analysis $A$ was really $\rand
(A+I)$, and from the probability analysis in the previous section, we
know that if $\det \symb (A+I) \neq 0$, then $\det \rand (A+I) \neq 0$
with high probability, and we know that $\det \symb (A+I)$ is never
zero.
%\bibliography{mybib}
\bibliographystyle{alpha}
\begin{thebibliography}{77}
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Camil Demetrescu, Giuseppe F. Italiano.
\emph{Fully Dynamic Transitive Closure: Breaking Through the O($n^2$) Barrier}.
FOCS 2000: 381-389
\bibitem{roditty2}
Liam Roditty, Uri Zwick.
\emph{A fully dynamic reachability algorithm for directed graphs with an almost linear update time}.
STOC 2004: 184-191
\bibitem{roditty1}
Liam Roditty, Uri Zwick.
\emph{Improved Dynamic Reachability Algorithms for Directed Graphs}.
FOCS 2002: 679-
\bibitem{roditty3}
Liam Roditty, Uri Zwick.
\emph{On Dynamic Shortest Paths Problems}.
ESA 2004: 580-591
\bibitem{roditty4}
Liam Roditty, Uri Zwick.
\emph{Dynamic Approximate All-Pairs Shortest Paths in Undirected Graphs}.
FOCS 2004: 499-508
\bibitem{sankowski}
Piotr Sankowski.
\emph{Dynamic Transitive Closure via Dynamic Matrix Inverse}.
FOCS 2004: 509-517.
\bibitem{schwartz}
Jacob T. Schwartz.
\emph{Fast Probabilistic Algorithms for Verification of Polynomial Identities. J}
ACM 27(4): 701-717 (1980)
\bibitem{thorup1}
Mikkel Thorup.
\emph{Fully-Dynamic All-Pairs Shortest Paths: Faster and Allowing Negative Cycles}.
SWAT 2004: 384-396
\bibitem{thorup2}
Mikkel Thorup.
\emph{Worst-case update times for fully-dynamic all-pairs shortest paths}.
STOC 2005
\bibitem{zippel}
Richard Zippel.
\emph{Probabilistic algorithms for sparse polynomials}.
EUROSAM 1979: 216-226
\end{thebibliography}
\end{document}