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\begin{document}
\lecture{10: General {\sc graph-iso}}{V. Arvind}{V.
Arvind}{Ramprasad Saptharishi}
\section{Overview}
In the last few classes, we solved some special cases of the graph
isomorphism problem. Now we shall use those ideas to give an algorithm
for the general {\sc graph-iso} problem. This, of course, is not a
polynomial time algorithm but is interesting nevertheless.
Since any graph on $n$ vertices has degree bounded by $n$, just a
naive simulation of the bounded degree {\sc graph-iso} would only give
us a $n^{n^2}$ which is worse than the brute-force approach (which is
a $O(n!) = (cn)^n$ algorithm). We shall see a $O(n^{n^{2/3}})$
algorithm for {\sc graph-iso}
\section{Colourings of Graphs}
A {\em colouring} of a graph is just associating a colour to every
vertex of the graph. Formally, it is a map $f:V\longrightarrow
\inbrace{1,2,\cdots, |V|}$. And we can further assume that the range
of $f$ is an initial segment of $\inbrace{1,2,\cdots, |V|}$.
A colouring $f_1$ is said to be a {\em refinement} of $f$, denoted by $f\leq
f_1$, if $f_1(x) \leq f_1(y) \implies f(x)\leq f(y)$ for all
vertices $x,y.$ A refinement $f'$ of $f$ is said to be {\em proper} if
$f'\neq f.$
\subsection{Colour-Degree Refinement}
For this lecture, this is the refinement that we would be dealing
with. Let $(X,f)$ be a coloured graph. The new refinement is defined
as follows:
Define $g(x) = \inparen{f(x),k_1(x), k_2(x),\cdots, k_{|V|}(x)}$ where
$k_i(x)$ is the number of neighbours of $x$ that are coloured
$i.$ Since we need to map these tuples to $\inbrace{1,2,\cdots, |V|}$,
lexicographically sort the tuples and map them to
$\inbrace{1,2,\cdots, |V|}.$ Let us call this induced colouring as
$f'$.
The choice of the tuple, whose first coordinate is $f(x)$, it is clear
that $f'$ is a refinement of $f$. \\
One can continuously keep refining a colouring, and it spreads out
vertices with more refinements. Hence, we can properly refine at most
$n$ times. A colouring that cannot be properly refined by the
colour-degree refinement is said to be a {\em stable colouring.}
\subsection{Propagating Refinements}
The central idea is to keep propagate refinements as much as
possible. The problem is that the colour-degree refinements may not be
able to spread the vertices enough, it might get stuck in a stable
colouring much earlier.
But how does refinements help? Where are we heading?
\begin{proposition}
Let $(X_1,f_1)$ and $(X_2,f_2)$ are two coloured graphs with $f_1$
and $f_2$ their corresponding refinements. Then
$$
(X_1, f_1)\equiv (X_2, f_2) \Longleftrightarrow (X_1,f_1') \equiv (X_2,f_2')
$$
\end{proposition}
\begin{proof}
Of course!
\end{proof}
The idea is to get to a stage where we can effeciently solve the
problem. But, as remarked earlier, what do we do when we get stuck up
at a stable colouring? We {\em individualize} a vertex. \medskip
\begin{itemize}
\item Keep refining until stable refinement
\item Pick some vertex, and give it a colour that has not been
assigned to any vertex.
\item Repeat
\end{itemize}
The trick is to find a good a good vertex to individualize in order to
propagate the refinements. We shall see that there is a small sequence
of vertices, which on the 'refine until stable, individualize, repeat'
gets us where we want.
\section{Colour Valence and {\sc graph-iso}}
\begin{definition}
A coloured graph $(X,f)$ is said to have colour valence $d$ if for
all vertices $x$ and colours $i$, either $x$ is adjacent to at most
$d$ neighbours of colour $i$ (valence of $x$ is bounded by $d$) or
it is not adjacent to atmost $d$ vertices of colour $i$ (covalence
of $x$ is bounded by $d$).
\end{definition}
The following theorem shows that there exists a small sequence of
vertices that can get us to a constant colour valence.
\begin{theorem}
If $(X,f)$ has colour valence $d$, then there exists a sequence of
nodes $\inbrace{x_1, x_2, \cdots, x_k}$ with $k\leq 2n/d$ such that
$(X,f')$, (colouring obtained by stabilizing and individualizing
these vertices) has colour valence $d/2$.
\end{theorem}
\begin{proof}
The proof is a greedy algorithm to pick up the vertices. Let $S_i =
\inbrace{x_1,\cdots, x_{i-1}}$. If $(X,S_i)$ has colour valence $\leq
d/2$, then we are already done, hence stop.
Else, there exists an $x\in V(X)$ and a colour $m$ such that both
valence and covalence of $x$ in $f^{-1}(m)$ is $>d/2.$ Let $N(x)$ be
the neighborhood or co-neighbourhood of $x$ (whichever violates the
valence bound). Hence we have $d/2 < |N(x)| \leq d.$ Pick $x$ as
$x_i$ and continue.
{\bf Claim:} The sets $\inbrace{N(x_i)}$ are pairwise
disjoint. \medskip
Once we prove the claim,
$$
\frac{kd}{2} \leq \sum_{i=1}^k |N(x_i)| \leq n
$$
which then forces $k\leq dn/2$.
{\em Proof of claim:} Suppose $N(x_j) \cap N(x_i) \neq \phi$ for
some $j*d$ contradicting the
colour valence of $X$ being bounded by $d$.
Hence, $\inbrace{N(x_j)}$ are pairwise disjoint, thus proving the
claim and the theorem.
\end{proof}
We can now start with a trivial colouring (every vertex given the same
colour, and with $k\leq 8n/d$ get to $(X,f')$ with colour valence
bounded by $d$.
\subsection{Enter Luks}
Let $T(n)$ be the worst case time bound for {\sc graph-iso} and
$T(n,d)$ be the worst case time bound for $d$-colourvalence-{\sc
graph-iso}.
The natural algorithm is the following:
\begin{enumerate}
\item Put trivial colouring on both graphs
\item Refine until stable.
\item Pick the set of vertices to individualize on one graph. Try all
possibilities on the other.
\item If at any try, if $x$ was the vertex picked with valence greater
than $d/2$, but valence $\leq d/2$ on the other graph, stop that
try.
\item Once you get to bounded colour valence, do the corresponding
algorithm.
\end{enumerate}
Hence, clearly $T(n) \leq n^{8n/d} T(n,d)$.
The claim is that, Luks' algorithm for bounded degree graphs work
here!
Let $(X,f)$ be the graph with a stable colouring and colour valence
$d$. Let $C_1, \cdots, C_r$ be the colour classes. Define $X(C_i)$ be
the induced subgraph on this colour class (only edges within that
class) and $X(C_i, C_j)$ as the induced bipartite graph (only edges
from one class to another).
The choice of the stable colouring then forces $X(C_i)$ to be a
regular graph (since the colour-degree-refinement would give a proper
refinement) and also $X(C_i,C_j)$ is a semi-regular\footnote{degree of
all vertices on left are the same, the same on the right} graph.
Now in $X(C_i)$, either the degree is bounded by $d$ or codegree. We
can assume that it is the degree, by complementing otherwise (and
checking if the other graph also has the same property, they
necessarily have to if they are isomorphic). And similarly for
$X(C_i,C_j)$, complement if $|E(X(C_i,C_j))| > |C_i||C_j|/2.$
$(X,f)$ now has the property that for all vertices $x$ and colours
$i$, degree of $x$ in $C_i$ is bounded above by $d.$ Hence we now have
two graphs $(X,f)$ and $(Y,g)$ with the above property. As in the Luks
algorithm, add a distinguished edge, and additionally a vertex on the
edge with a new colour. Now layer the graphs based on edges obtained
by paths from the new vertex.
Here, the kernel is again $\bigotimes \Sym(\cdot)$ since the colour
classes can't move! And each element of the product is bounded since
the degree is bounded! Hence the automorphism group is infact in
$\mathcal{B}_d$ and hence can be solved by a $O(n^{d^2})$ algorithm.
\bigskip
Now $T(n) \leq n^{8n/d+d^2}$ and an optimal choice for $d$would give us a
$O(n^{n^{2/3}})$ algorithm for {\sc graph-iso}.
\end{document}*