From 08be3e96a32d95c277905ca7fa015d88195b257e Mon Sep 17 00:00:00 2001
From: Camil Staps
Date: Tue, 27 Oct 2015 18:08:50 +0100
Subject: Start report practical 1

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+\section{Algorithm description}
+\label{sec:algorithm}
+
+\subsection{Basic structure}
+\label{sec:algorithm:basic-structure}
+The basic structure of the algorithm is fairly simple. For any graph $G=(V,E)$, we pick a vertex $v\in V$. Either that vertex is in an m.i.s. (and none of its neighbours), or it is not. This gives the recurrence: \refeq[basic-structure]{\ms(G) = \max(1 + \ms(G\ex \iN(v)), \ms(G\ex v)).} The rest of the algorithm will be optimising this basic structure by cleverly choosing $v$ and pruning the search tree.
+
+\begin{lemmaproof}[one-neighbour]{For a graph $G=(V,E)$ and vertex $v\in V$, if $v$ is not in an m.i.s. there is a neighbour $w\in N(v)$ which is in an m.i.s.}
+    By contradiction. Suppose there were an m.i.s. $ms_1$ that does not contain $v$ or one of its neighbours. Then we may create a new m.i.s. $ms_2=ms_1\with v$. We then have $|ms_2| = |ms_1|+1 > |ms_1|$. Therefore, $ms_1$ was not maximal.
+\end{lemmaproof}
+
+\begin{lemmaproof}[two-neighbours]{For a graph $G=(V,E)$ and vertex $v\in V$, there is an m.i.s. with $v$, or there is an m.i.s. with two distinct elements $w_1,w_2\in N(v)$.}
+    By contradiction. If $v$ is in an m.i.s., we're done. If $v$ is not in an m.i.s., we know by \autoref{lem:one-neighbour} that there is a $w_1\in N(v)$ that is, say in m.i.s. $ms$. Suppose there were no other neighbour $w_2\in N(v), w_2\neq w_1$ with $w_2\in ms$. Then $ms\ex w_1\with v$ is also an m.i.s., and it does contain $v$.
+\end{lemmaproof}
+
+At this point we define, as Robson \cite{robson}, the function $\ms^n(G,S)$ which is the $\ms(G)$ given that the m.i.s. should contain at least $n$ elements of $S$.
+
+Based on \autoref{lem:one-neighbour} and \ref{lem:two-neighbours} we may then rewrite \autoref{eq:basic-structure} to \refeq[with-two-neighbours]{\ms(G) = \max(1+\ms(G\ex\iN(v)), \ms^2(G\ex v, N(v))).}
+
+\subsection{Picking a good $v$}
+\label{sec:algorithm:good-v}
+Until now, we have assumed that we have some vertex $v$. Let us now discuss how to pick a `good' $v$, i.e. one that allows for the most efficient use of \autoref{eq:basic-structure} and \ref{eq:with-two-neighbours}.
+
+\begin{lemmaproof}[basic-structure-high-degree]{It is most efficient to use \autoref{eq:basic-structure} with a $v$ with large $d(v)$.}
+    The left-hand side, $\ms(G\ex\iN(v))$, will run faster with large $d(v)$, because it will run on a smaller graph. For the right-hand side it doesn't matter.
+\end{lemmaproof}
+
+\begin{lemmaproof}[two-neighbours-low-degree]{It is most efficient to use \autoref{eq:with-two-neighbours} with a $v$ with small $d(v)$.}
+    A straightforward implementation of $\ms^2(G\ex v,N(v))$ will consider all pairs of vertices in $N(v)$, which is quadratic. In the left-hand side, $\ms(G\ex\iN(v))$ is only linearly faster with large $d(v)$. Therefore, we should prefer small $d(v)$.
+\end{lemmaproof}
+
+In general, we'd like to use \autoref{eq:with-two-neighbours}, because the right-hand side of the $\max$ is more restricted than in \autoref{eq:basic-structure}. However, if we have a $v$ with large $d(v)$, this may not be the case. 
+
+So, we should try to pick a $v$ with small $d(v)$. We apply \autoref{eq:with-two-neighbours} if $d(v)$ is small enough. If $d(v)$ is too large to ensure efficient usage of \autoref{eq:with-two-neighbours}, we apply \autoref{eq:basic-structure}.
+
+But this latter recurrence is clearly more efficient for \emph{large} $d(v)$. Therefore, if $d(v)$ is too large to use $v$ in \autoref{eq:with-two-neighbours}, we find one of its neighbours, say $w\in N(v)$, with the largest $d(w)$, and apply \autoref{eq:basic-structure} with that vertex. In the next recurrence, $d(v)$ will be at least one smaller, because $w$ has been removed, but in the case of a graph with many three-cycles, $d(v)$ may be much smaller. So, after at most a few applications of \autoref{eq:basic-structure} we may use the more efficient \autoref{eq:with-two-neighbours} again.
+
+\subsubsection{Dominance}
+\label{sec:algorithm:good-v:dominance}
+\begin{lemmaproof}[dominance]{If in graph $G$ with vertices $v,w$ we have $v>w$, we know that $\ms(G)=\ms(G\ex w)$.}
+    If there is an m.i.s. $ms$ with $w\in ms$, $ms\ex w\with v$ is an m.i.s. as well.
+\end{lemmaproof}
+
+If we then take $w$ instead of $v$ and use \autoref{eq:basic-structure}, it may be the case that $v>w$. In this case, we may bypass \autoref{eq:basic-structure} by applying the recurrence $\ms(G)=\ms(G\ex w)$ of \autoref{lem:dominance}. After this, it may be the case that we can apply \autoref{eq:with-two-neighbours} instead, since $d(v)$ will have been decreased by one.
+
+This gives us \autoref{alg:with-best-vertex}, which combines \autoref{eq:basic-structure} and \ref{eq:with-two-neighbours}.
+
+\begin{algorithm}[h]
+    \caption{Finding the size of the m.i.s. of a graph}
+    \label{alg:with-best-vertex}
+    \begin{algorithmic}[1]
+        \REQUIRE $G=(V,E)$
+        \IF{$|G|=1$}
+            \RETURN $|G|$ \COMMENT{Recursive base case}
+        \ELSE
+            \STATE $v\gets v\in V$ with minimal $d(v)$
+            \IF{$d(v)$ is small enough}
+                \RETURN $\max(1+\ms(G\ex\iN(v)), \ms^2(G\ex v,N(v)))$ \COMMENT{\autoref{eq:with-two-neighbours}}
+            \ELSIF{$v$ dominates $w$}
+                \RETURN $\ms(G\ex w)$
+            \ELSE
+                \STATE $w\gets w\in N(v)$ with maximal $d(w)$
+                \RETURN $\max(1+\ms(G\ex\iN(w)), \ms(G\ex w))$ \COMMENT{\autoref{eq:basic-structure}}
+            \ENDIF
+        \ENDIF
+    \end{algorithmic}
+\end{algorithm}
+
+\subsection{Strongly connected components}
+\label{sec:algorithm:components}
+\begin{lemmaproof}[components]{The size of the m.i.s. of a graph is equal to the sum of the sizes of the m.i.s. of all its strongly connected components.}
+    The union of the m.i.s. of the strongly connected components is an m.i.s. of the graph.
+
+    If there were a larger m.i.s. in the graph, there has to be a $v$ in a strongly connected component $C$ in that m.i.s. which is not in the m.i.s. of $C$. If we can add $v$ to the m.i.s. of $C$, that m.i.s. wasn't maximal. If we cannot, then the larger m.i.s. of the whole graph cannot be an independent set.
+\end{lemmaproof}
+
+Based on \autoref{lem:components} we may write \refeq[with-components]{\ms(G)=\sum_{c\in C} \ms(c),} where $C$ is the set of strongly connected components of $G$. It is not difficult to see that in case a graph has multiple strongly connected components, this recurrence is more efficient than \autoref{alg:with-best-vertex}.
+
+\subsection{Further optimisations}
+\label{sec:algorithm:further-optimisations}
+Although I have referred to Robson \cite{robson} and used his notation throughout this report, this is how far I got without his help. Robson then introduces a number of further optimisations. In this section, we use $G$ for the graph at hand and $v,w$ for the vertex with the lowest degree, and its neighbour with the highest degree, respectively.
+
+\begin{lemmaproof}[robson-ms-d1]{If $d(v)=1$, we know $\ms(G)=1+\ms(G\ex\iN(v))$.}
+    By contradiction. By \autoref{lem:one-neighbour} we know that if $v$ is not in any m.i.s., then $w$ is in an m.i.s., say $ms$. But then $ms\ex w\with v$ is also independent, and has the same size.
+\end{lemmaproof}
+
+For the case that $d(v)=2$ we write $w,w'$ for the neighbours of $v$.
+
+\begin{lemmaproof}[robson-ms-d2-edge]{If $d(v)=2$ and $e(w,w')$, we know $\ms(G)=1+\ms(G\ex\iN(v))$.}
+    By \autoref{lem:two-neighbours} we know that an m.i.s. will either contain $v$ or both $w$ and $w'$. But since $e(w,w')$, the latter cannot happen. Therefore, it must contain $v$ and neither of its neighbours.
+\end{lemmaproof}
+
+\begin{lemmaproof}[robson-ms-d2-noedge]{If $d(v)=2$ and $\lnot e(w,w')$, we know $$\ms(G) = \max(2+\ms(G\ex\iN(w)\ex\iN(w')), 1 + \ms^2(G\ex\iN(v), N^2(v))).$$}
+    By \autoref{lem:two-neighbours}, an m.i.s. contains either $v$ or both $w,w'$. In the second case, we remove $w$ and $w'$ and their neighbourhoods from the graph (the left-hand side of $\max$). In the first case, the m.i.s. cannot contain $w$ or $w'$ but must contain two of their neighbours other than $v$. If not, and there is an m.i.s. $ms$ with at most one such neighbour, $u$, then $ms\ex v\ex u\with w\with w'$ is also independent, and has the same size. This gives the right-hand side.
+\end{lemmaproof}
+
+\autoref{alg:with-best-vertex} combined with \autoref{lem:components}, \ref{lem:robson-ms-d1}, \ref{lem:robson-ms-d2-edge} and \ref{lem:robson-ms-d2-noedge} gives us \autoref{alg:with-robsons-optimisations}.
+
+\begin{algorithm}[h]
+    \caption{Finding the size of the m.i.s. of a graph}
+    \label{alg:with-robsons-optimisations} \label{alg:ms}
+    \begin{algorithmic}[1]
+        \REQUIRE $G=(V,E)$
+        \IF{$G$ has multiple strongly connected components}
+            \STATE $C\gets$ some strongly connected component
+            \RETURN $\ms(C) + \ms(G\ex C)$ \COMMENT{\autoref{lem:components}}
+        \ELSIF{$|G|=1$}
+            \RETURN $|G|$ \COMMENT{Recursive base case}
+        \ELSE
+            \STATE $v\gets v\in V$ with minimal $d(v)$
+            \STATE $w\gets w\in N(v)$ with maximal $d(w)$
+            \IF{$d(v)=1$}
+                \RETURN $1+\ms(G\ex\iN(v))$ \COMMENT{\autoref{lem:robson-ms-d1}}
+            \ELSIF{$d(v)=2$}
+                \STATE $w'\gets$ the other neighbour of $v$
+                \IF{$e(w,w')$}
+                    \RETURN $1+\ms(G\ex\iN(v))$ \COMMENT{\autoref{lem:robson-ms-d2-edge}}
+                \ELSE
+                    \RETURN $\max(2+\ms(G\ex\iN(w)\ex\iN(w')), 1+\ms^2(G\ex\iN(v),N^2(v)))$ \COMMENT{\autoref{lem:robson-ms-d2-noedge}}
+                \ENDIF
+            \ELSIF{$d(v)=3$}
+                \RETURN $\max(\ms^2(G\ex v,N(v)), 1+\ms(G\ex\iN(v)))$ \COMMENT{\autoref{eq:with-two-neighbours}}
+            \ELSIF{$v>w$}
+                \RETURN $\ms(G\ex w)$ \COMMENT{\autoref{lem:dominance}}
+            \ELSE
+                \RETURN $\max(\ms(G\ex w), 1+\ms(G\ex\iN(w)))$ \COMMENT{\autoref{eq:basic-structure}}
+            \ENDIF
+        \ENDIF
+    \end{algorithmic}
+\end{algorithm}
+
+\subsection{The helper function $\ms^n$}
+\label{sec:algorithm:helper-function}
+So far we have been using $\ms^n(G,S)$ as the size of the maximum independent set of $G$ given that it contains at least $n$ vertices in $S$. This section is devoted to arguing an efficient implementation of this helper function. In the above, we have only used $\ms^2$. Therefore, we do not need to go into details about the case $n\neq2$.
+
+We will write $s_1,s_2,\dots$ for the elements of $S$.
+
+\begin{lemmaproof}[helper-general]{In our algorithm, it always holds that $\ms^n(G,S)=\ms(G)$.}
+    We only call $\ms^n$ when we know for sure that $n$ elements of $S$ are in an m.i.s. of $G$.
+\end{lemmaproof}
+
+\autoref{lem:helper-general} may be used as a default case, when no clever optimisation can be found.
+
+\begin{lemmaproof}[helper-1]{If $|S|<n$, we know $\ms^n(G,S)=0$.}
+    It is impossible to pick $n$ edges from less than $n$.
+\end{lemmaproof}
+
+\begin{lemmaproof}[helper-2-edge]{If $|S|=2$ and $e(s_1,s_2)$, then $\ms^2(G,S)=0$.}
+    The only possibility to choose two vertices from $S$, choosing $s_1$ and $s_2$ does not give us an independent set.
+\end{lemmaproof}
+
+\begin{lemmaproof}[helper-2-noedge]{If $|S|=2$ and $\lnot e(s_1,s_2)$, then $\ms^2(G,S)=2+\ms(G\ex\iN(s_1)\ex\iN(S_2)$.}
+    We only have to consider the one possibility of choosing $s_1$ and $s_2$. Then we may remove their inclusive neighbourhoods from the graph.
+\end{lemmaproof}
+
+\begin{lemmaproof}[helper-3-naive]{If $|S|=3$, we have $\ms^2(G,S)=\max(\ms^2(G,S\ex s_1), \ms^2(G,S\ex s_2), \ms^2(G,S\ex s_3)).$}
+    We only have to consider three possibilities.
+\end{lemmaproof}
+
+\autoref{lem:helper-general} through \ref{lem:helper-3-naive} were the results I achieved. Robson \cite{robson} adds some optimisations to the cases $|S|=3$ and $|S|=4$. As they are fairly straightforward, I will not discuss them here. They are included in \autoref{alg:ms2}, which combines our own results into one algorithm. Robson uses slightly different notation at some points, since he assumes $d(s_1)<d(s_2)$.
+
+\begin{algorithm}[h!]
+    \caption{Finding the size of the m.i.s. of a graph, given that it should contain two nodes from a certain subgraph}
+    \label{alg:ms2}
+    \begin{algorithmic}[1]
+        \REQUIRE $G=(V,E), S\subseteq V$
+        \IF{$|S|=1$}
+            \RETURN $0$ \COMMENT{\autoref{lem:helper-1}}
+        \ELSIF{$|S|=2$}
+            \IF{$e(s_1,s_2)$}
+                \RETURN $0$ \COMMENT{\autoref{lem:helper-2-edge}}
+            \ELSE
+                \RETURN $2+\ms(G\ex\iN(s_1)\ex\iN(s_2))$ \COMMENT{\autoref{lem:helper-2-noedge}}
+            \ENDIF
+        \ELSIF{$|S|=3$}
+            \IF{$\exists_i[d(s_i)=0]$}
+                \RETURN $1+\ms^1(G\ex s_i,S\ex s_i)$ \COMMENT{Special case of \autoref{lem:components}}
+            \ELSIF{$s_1,s_2,s_3$ form a three-cycle}
+                \RETURN $0$ \COMMENT{Impossible to pick two vertices from a three-cycle}
+            \ELSIF{$\exists_{i,j,k}[i\neq k \land e(s_i,s_j) \land e(s_i,s_k)]$}
+                \RETURN $2 + \ms(G\ex\iN(s_j)\ex\iN(s_k))$ \COMMENT{Impossible to choose $s_i$ and another vertex, so we have to choose $s_j$ and $s_k$}
+            \ELSIF{$\exists_{i,j}[e(s_i,s_j)]$}
+                \STATE $s_k\gets S\ex s_1\ex s_j$
+                \RETURN $1 + \ms^1(G\ex\iN(s_k), S\ex s_k)$ \COMMENT{Impossible to choose $s_i$ and $s_j$, so we choose $s_k$ and one of $s_i,s_j$}
+            \ELSIF{$\exists_{i,j}[N(s_i)\cap N(s_j)\neq\emptyset]$}
+                \RETURN $\ms^2(G\ex(N(s_i)\cap N(s_j)), S)$ \COMMENT{m.i.s. contains either $s_i$ or $s_j$, so not their common neighbours}
+            \ELSIF{$\exists_i[d(s_i)=1]$}
+                \RETURN $1+\ms^1(G\ex\iN(s_i),S\ex s_i)$ \COMMENT{Special case of \autoref{lem:one-neighbour}}
+            \ELSE
+                \RETURN $\max(1+\ms^1(G\ex\iN(s_1),S\ex s_1), 2 + \ms^2(G\ex\iN(s_2)\ex\iN(s_3)\ex s_1, N(s_1)))$ \COMMENT{Either the m.i.s. does contain $s_1$, and one of $s_2,s_3$, or it doesn't. In the latter case, it must contain $s_2,s_3$, and at least two on $s_1$'s neighbours, using an argument similar to \autoref{lem:two-neighbours}. Robson misses the `$2+$' for this second case, but from the writing it is clear that it should be included.}
+            \ENDIF
+        \ELSIF{$|S|=4$}
+            \IF{$\lnot\exists_i[d(s_i)\leq3]$}
+                \RETURN $\max(1+\ms(G\ex\iN(s_1)),\ms^2(G\ex s_1,S\ex s_1))$ \COMMENT{m.i.s. either does or doesn't contain $s_1$}
+            \ELSE
+            \RETURN $\ms(G)$ \COMMENT{More efficient than the $\max$ in this case; \autoref{lem:helper-general}}
+            \ENDIF
+        \ELSE
+            \RETURN $\ms(G)$ \COMMENT{No clever optimisation found; \autoref{lem:helper-general}}
+        \ENDIF
+    \end{algorithmic}
+\end{algorithm}
+
+\bigskip
+
+As the observant reader may have noticed, we use $\ms^1$ in the definition of $\ms^2$. We therefore also need to write an efficient algorithm to apply this function. Of course, \autoref{lem:helper-general} and \ref{lem:helper-1} apply to $\ms^1$ as well. Note, that we always call $\ms^1(\cdots,S)$ with $|S|=2$. We will therefore not give a complete algorithm $\ms^1$, but one that works in the necessary cases.
+
+\begin{lemmaproof}[helper-intersection]{If $N(s_1)\cap N(s_2)\neq\emptyset$, we know $\ms^1(G,S)=\ms^1(G\ex(N(s_1)\cap N(s_2)), S)$.}
+    This was already used in \autoref{alg:ms2}, in the case of three states. The m.i.s. will contain either $s_1$ or $s_2$, therefore cannot contain the intersection of their neighbourhoods. Removing these vertices allows for early pruning.
+\end{lemmaproof}
+
+Robson adds further optimisations to this.
+
+\begin{lemmaproof}[helper-ms1-2-2]{If $d(s_1)=d(s_2)=2$ and $\lnot e(s_1,s_2)$, an m.i.s. containing one of $s_1,s_2$ will contain either $s_1$ or $N(s_1)\with\{s_2\}$.}
+    In the case that the m.i.s. contains $s_1$, we are done. If an m.i.s. $ms$ does not contain $s_1$ but $s_2$, it must contain $N(s_1)$, because otherwise $ms\ex N(s_1)\with s_1$ is an independent set of at least the same size.
+\end{lemmaproof}
+
+If the conditions of \autoref{lem:helper-ms1-2-2} are met, write $w_1,w_2$ for the two neighbours of $s_1$.
+
+\begin{lemmaproof}[helper-ms1-2-2-edge]{If the conditions of \autoref{lem:helper-ms1-2-2} are met and $e(w_1,w_2)$, we have $\ms^1(G,S)=1+\ms(G\ex s_1)$.}
+    It cannot be the case that an m.i.s. will contain $N(s_1)$. \autoref{lem:helper-ms1-2-2} leaves this (choosing $s_1$) as only option.
+\end{lemmaproof}
+
+\begin{lemmaproof}[helper-ms1-2-2-dominance]{If the conditions of \autoref{lem:helper-ms1-2-2} are met and $N^2(s_1)\subset N(s_2)$, we have $\ms^1(G,S)=3+\ms(G\ex\iN(s_1)\ex\iN(s_2))$.}
+    Obviously, an m.i.s. $\iN(s_1)+\iN(s_2)$ that contains $s_1$ or $s_2$ cannot be larger than three. Furthermore, $\{w_1,w_2,s_2\}$ is an independent set and dominates every other independent set of the same size.
+\end{lemmaproof}
+
+This leads us to a slightly adapted version of $\ms^2$ that we can use for $\ms^1$. Pseudocode is given in \autoref{alg:ms1}.
+
+\begin{algorithm}[h]
+    \caption{Finding the size of the m.i.s. of a graph, given that it should contain one node from a certain subgraph}
+    \label{alg:ms1}
+    \begin{algorithmic}[1]
+        \REQUIRE $G=(V,E), S\subseteq V, d(s_1)\leq d(s_2)$
+        \IF{$d(s_1)\leq1$}
+            \RETURN $\ms(G)$ \COMMENT{\autoref{lem:helper-general}} \label{alg:ms1:case-d1}
+        \ELSIF{$e(s_1,s_2)$}
+            \IF{$d(s_1)>3$}
+                \RETURN $1 + \max(\ms(G\ex\iN(s_1)), \ms(G\ex\iN(s_2)))$ \COMMENT{An m.i.s. contains either $s_1$ or $s_2$, but not both} \label{alg:ms1:case-edge-d4}
+            \ELSE
+                \RETURN $\ms(G)$ \COMMENT{More efficient than the $\max$ in this case; \autoref{lem:helper-general}}\label{alg:ms1:case-edge-d3}
+            \ENDIF
+        \ELSIF{$N(s_1)\cap N(s_2)\neq\emptyset$}
+            \RETURN $\ms^1(G\ex(N(s_1)\cap N(s_2)), S)$ \COMMENT{m.i.s. contains $s_1$ or $s_2$, so not their common neighbours}\label{alg:ms1:intersection}
+        \ELSIF{$d(s_1)=d(s_2)=2$}
+            \STATE $\{w_1,w_2\} \gets N(s_1)$
+            \IF{$e(w_1,w_2)$}
+                \RETURN $1+\ms(G\ex\iN(s_1))$ \COMMENT{\autoref{lem:helper-ms1-2-2-edge}}\label{alg:ms1:case-2-2-edge}
+            \ELSIF{$N^2(s_1)\subset N(s_2)$}
+                \RETURN $3 + \ms(G\ex\iN(s_1)\ex\iN(s_2))$ \COMMENT{\autoref{lem:helper-ms1-2-2-dominance}}\label{alg:ms1:case-2-2-dominance}
+            \ELSE
+                \RETURN $\max(1+\ms(G\ex\iN(s_1)), 3+\ms(G\ex\iN(w_1)\ex\iN(w_2)\ex\iN(s_2)))$ \COMMENT{\autoref{lem:helper-ms1-2-2}}\label{alg:ms1:case-2-2-else}
+            \ENDIF
+        \ELSE
+            \RETURN $1+\max(\ms(G\ex\iN(s_2)), \ms^2(G\ex\iN(s_1)\ex s_2, N(s_2)))$ \COMMENT{An m.i.s. contains either $s_2$ or $s_1$ and two of $s_2$'s neighbours. If an m.i.s. $ms$ would contain $s_1$ and at most one of $s_2$'s neighbours, say $w$, then there is another independent set $ms\ex w\with s_2$, which is at least as large.}\label{alg:ms1:case-general}
+        \ENDIF
+    \end{algorithmic}
+\end{algorithm}
+
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