handouts/Advanced/Graph Algorithms/parts/04 applications.tex

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\section{Applications}
\problem{Maximum Cardinality Matching}
A \textit{matching} is a subset of edges in a bipartite graph. Nodes in a matching must not have more than one edge connected to them. \\
A matching is \textit{maximal} if it has more edges than any other matching.
\vspace{5mm}
\begin{minipage}[t]{0.48\textwidth}
\begin{center}
Initial Graph \\
\vspace{2mm}
\begin{tikzpicture}
% Nodes
\begin{scope}[layer = nodes]
\node[main] (A1) at (0mm, 24mm) {};
\node[main] (A2) at (0mm, 18mm) {};
\node[main] (A3) at (0mm, 12mm) {};
\node[main] (A4) at (0mm, 6mm) {};
\node[main] (A5) at (0mm, 0mm) {};
\node[main] (B1) at (20mm, 24mm) {};
\node[main] (B2) at (20mm, 18mm) {};
\node[main] (B3) at (20mm, 12mm) {};
\node[main] (B4) at (20mm, 6mm) {};
\node[main] (B5) at (20mm, 0mm) {};
\end{scope}
% Edges
\draw
(A1) edge (B2)
(A1) edge (B3)
(A2) edge (B1)
(A2) edge (B4)
(A4) edge (B3)
(A2) edge (B3)
(A5) edge (B3)
(A5) edge (B4)
;
\end{tikzpicture}
\end{center}
\end{minipage}
\hfill
\begin{minipage}[t]{0.48\textwidth}
\begin{center}
Maximal Matching \\
\vspace{2mm}
\begin{tikzpicture}
% Nodes
\begin{scope}[layer = nodes]
\node[main] (A1) at (0mm, 24mm) {};
\node[main] (A2) at (0mm, 18mm) {};
\node[main] (A3) at (0mm, 12mm) {};
\node[main] (A4) at (0mm, 6mm) {};
\node[main] (A5) at (0mm, 0mm) {};
\node[main] (B1) at (20mm, 24mm) {};
\node[main] (B2) at (20mm, 18mm) {};
\node[main] (B3) at (20mm, 12mm) {};
\node[main] (B4) at (20mm, 6mm) {};
\node[main] (B5) at (20mm, 0mm) {};
\end{scope}
% Edges
\draw[opacity = 0.4]
(A1) edge (B2)
(A1) edge (B3)
(A2) edge (B1)
(A2) edge (B4)
(A4) edge (B3)
(A4) edge (B3)
(A5) edge (B3)
(A5) edge (B4)
;
\draw
(A1) edge (B2)
(A2) edge (B1)
(A4) edge (B3)
(A5) edge (B4)
;
\end{tikzpicture}
\end{center}
\end{minipage}
\vspace{5mm}
Devise an algorithm to find a maximal matching in any bipartite graph. \\
Find an upper bound for its runtime.
\begin{solution}
Turn this into a maximum flow problem and use FF. \\
Connect a node $S$ to all nodes in the left group and a node $T$ to all nodes in the right group. All edges have capacity 1.
\vspace{2ex}
Just like FF, this algorithm will take at most $\min(\# \text{ left nodes}, \# \text{ right nodes})$ iterations.
\end{solution}
\vfill
\pagebreak
\problem{Circulations with Demand}
Say we have a network of cities and power stations. Stations produce power; cities consume it.
Each station produces a limited amount of power, and each city has limited demand.
\vspace{2ex}
We can represent this power grid as a graph, with cities and stations as nodes and transmission lines as edges.
\vspace{2ex}
A simple example is below. There are two cities ($2$ and $4$) and two stations (both $-3$).
We'll represent station capacity with a negative number, since they \textit{consume} a negative amount of energy.
\begin{center}
\begin{tikzpicture}[
node distance = 25mm,
main/.style = {
draw,
circle,
fill = white,
minimum size = 8mm
},
]
% Nodes
\begin{scope}[layer = nodes]
\node[main] (S1) {$-3$};
\node[main] (S2) [below left of = S1] {$-3$};
\node[main] (C1) [below right of = S1] {$2$};
\node[main] (C2) [below right of = S2] {$4$};
\end{scope}
% Edges
\draw[->]
(S1) edge node[label] {$3$} (S2)
(S1) edge node[label] {$3$} (C1)
(S2) edge node[label] {$2$} (C1)
(S2) edge node[label] {$2$} (C2)
(C1) edge node[label] {$2$} (C2)
;
\end{tikzpicture}
\end{center}
We'd like to know if there exists a \textit{feasible circulation} in this network---that is, can we supply our cities with the energy they need without exceeding the capacity of power plants or transmission lines?
\vspace{2ex}
\textbf{Your job:} Devise an algorithm that solve this problem.
\vspace{2ex}
\note{\textbf{Bonus:} Say certain edges have a lower bound on their capacity, meaning that we must send \textit{at least} that much flow down the edge. Modify your algorithm to account for these additional constraints.}
\begin{solution}
Create a source node $S$, and connect it to each station with an edge. Set the capacity of that edge to the capacity of the station. \\
Create a sink node $T$ and do the same for cities.
\vspace{2ex}
This is now a maximum-flow problem with one source and one sink. Apply FF.
\linehack{}
To solve the bonus problem, we'll modify the network before running the algorithm above.
\vspace{2ex}
Say an edge from $A$ to $B$ has minimum capacity $l$ and maximum capacity $u \geq l$. Apply the following transformations:
\begin{itemize}
\item Add $l$ to the capacity of $A$
\item Subtract $l$ from the capacity of $B$
\item Subtract $l$ from the total capacity of the edge.
\end{itemize}
Do this for every edge that has a lower bound then apply the algorithm above.
\end{solution}
\vfill
\pagebreak