Added a long-form example of 7.24

Author: AustinTSchaffer

OMSCS/Courses/GA/06.2 - NP - Definitions.md

@@ -6,7 +6,6 @@ tags:
 # 06.2 - NP - Definitions
 > How do we prove that a problem is computationally difficult?
 
-
 ## Outline
 - NP
 - NP-Complete

OMSCS/Courses/GA/Practice Problems/7.24 - Example Bipartite Matching.md

@@ -0,0 +1,192 @@
+---
+tags:
+  - OMSCS
+  - Algorithms
+  - Practice
+---
+# 7.24 - Example Bipartite Matching
+This document shows how to find a maximal matching in a bipartite graph, following the definitions of "Alternating Paths", "Matching Set", "Covered Vertex", and "Maximal Matching" from Practice Problem [[7.24 - Example Bipartite Matching]].
+
+Following an iterative procedure of modifying edges along an alternating path in G, we can start with any G that has a sub-maximal matching set $M$ and arrive at a maximal matching set for G.
+
+We will start with this bipartite graph G, which has an empty matching set ($M=\emptyset$), and therefore no covered vertices.
+
+```mermaid
+graph TD
+
+A((A))
+B((B))
+C((C))
+D((D))
+E((E))
+F((F))
+G((G))
+H((H))
+I((I))
+
+A <--> E
+A <--> H
+B <--> E
+B <--> F
+B <--> G
+C <--> G
+C <--> H
+C <--> I
+D <--> F
+```
+
+In this graph, every pair of connected vertices is an Alternating Path (AP) in G. We can arbitrarily select $AP=[A,E]$ as our first AP in G. If we then take all of the edges $e \in AP$, we can add them to $M$ if $e \notin M$ or remove $e$ from $M$ if $e \in M$.
+
+This results in the following graph. Edges in $M$ are bold. Covered vertices are shown as double circles.
+
+```mermaid
+graph TD
+
+A(((A)))
+B((B))
+C((C))
+D((D))
+E(((E)))
+F((F))
+G((G))
+H((H))
+I((I))
+
+A <==> E
+A <--> H
+B <--> E
+B <--> F
+B <--> G
+C <--> G
+C <--> H
+C <--> I
+D <--> F
+```
+
+We can then apply this procedure again, selecting 2 arbitrary uncovered vertices in G and finding an AP between them. For this iteration, we will pick another easy example: $AP=[H,C]$. Since $HC$ is not already in $M$, we can add it, resulting in the graph shown below.
+
+```mermaid
+graph TD
+
+A(((A)))
+B((B))
+C(((C)))
+D((D))
+E(((E)))
+F((F))
+G((G))
+H(((H)))
+I((I))
+
+A <==> E
+A <--> H
+B <--> E
+B <--> F
+B <--> G
+C <--> G
+C <==> H
+C <--> I
+D <--> F
+```
+
+There still exists APs in G, so we can apply this procedure again. Here are all of the APs that currently exist in G.
+
+- $[G,B]$
+- $[B,F]$
+- $[D,F]$
+- $[I,C,H,A,E,B]$
+- $[G, C, H, A, E, B]$
+
+This time we will pick a long AP: $AP=[I,C,H,A,E,B]$. For each of these edges, we alternate its membership within $M$
+
+- Add $IC$ to $M$
+- Remove $CH$ from $M$
+- Add $HA$ to $M$
+- Remove $AE$ from $M$
+- Add $EB$ to $M$
+
+This leaves us with the graph below.
+
+```mermaid
+graph TD
+
+A(((A)))
+B(((B)))
+C(((C)))
+D((D))
+E(((E)))
+F((F))
+G((G))
+H(((H)))
+I(((I)))
+
+A <--> E
+A <==> H
+B <==> E
+B <--> F
+B <--> G
+C <--> G
+C <--> H
+C <==> I
+D <--> F
+```
+
+Each time we apply this procedure, we first find an AP in G. An AP always has odd length, starts with an edge which is not in $M$, and ends with an edge that is not in $M$. For each subsequent edge in the AP, that edge alternates between being in $M$ and not in $M$. For a given AP of length $2k+1$, the AP has $k$ edges which are in $M$, and $k+1$ edges which are not in $M$. Therefore, we end up removing $k$ edges from $M$, and adding $k+1$ edges to $M$. This results in $|M|$ increasing by 1 each iteration. This also always results both of the endpoints of the AP becoming covered in G after applying this procedure.
+
+Every step of the process increases the amount of G which is covered by $M$, without violating $M$ being defined as a "matching set."
+
+After running the previous iteration, $[F,D]$ is the only AP which exists in G. Applying the procedure to this AP, we end up with this graph.
+
+```mermaid
+graph TD
+
+A(((A)))
+B(((B)))
+C(((C)))
+D(((D)))
+E(((E)))
+F(((F)))
+G((G))
+H(((H)))
+I(((I)))
+
+A <--> E
+A <==> H
+B <==> E
+B <--> F
+B <--> G
+C <--> G
+C <--> H
+C <==> I
+D <==> F
+```
+
+This leaves us with $M=[AG, EB, FD, CI]$. All **covered** vertices only appear once in $M$, meaning that $M$ is still a valid matching for $G$. We also have just vertex $G$ as the only **uncovered** vertex in G. There cannot exist any other APs in G, because an AP must have odd length. Since G is bipartite, we can't have an AP which starts and ends with a single vertex.
+
+Therefore, $M=[AG, EB, FD, CI]$ is a "maximal matching" for G.
+
+Other valid maximal matchings exist for $G$, including $M=[AE, BG, DF, CI]$, leaving $H$ as the only uncovered vertex in G.
+
+```mermaid
+graph TD
+
+A(((A)))
+B(((B)))
+C(((C)))
+D(((D)))
+E(((E)))
+F(((F)))
+G(((G)))
+H((H))
+I(((I)))
+
+A <==> E
+A <--> H
+B <--> E
+B <--> F
+B <==> G
+C <--> G
+C <--> H
+C <==> I
+D <==> F
+```