Lecture 13: Bridges, Graphs, and Greed

COSC 311 Algorithms, Fall 2022

$\def\compare{ {\mathrm{compare}} } \def\swap{ {\mathrm{swap}} } \def\sort{ {\mathrm{sort}} } \def\insert{ {\mathrm{insert}} } \def\true{ {\mathrm{true}} } \def\false{ {\mathrm{false}} } \def\BubbleSort{ {\mathrm{BubbleSort}} } \def\SelectionSort{ {\mathrm{SelectionSort}} } \def\Merge{ {\mathrm{Merge}} } \def\MergeSort{ {\mathrm{MergeSort}} } \def\QuickSort{ {\mathrm{QuickSort}} } \def\Split{ {\mathrm{Split}} } \def\Multiply{ {\mathrm{Multiply}} } \def\Add{ {\mathrm{Add}} } \def\cur{ {\mathrm{cur}} }$

Announcements

1. Midterm next Friday
• Material up to lecture 12 covered
• Makeup date following week
• Accommodations
2. Study guide posted this weekend
• Topics
• Example questions
• Solutions

Overview

1. Bridges of Königsberg
2. Graphs
3. Greedy Algorithms

Bridges of Königsberg

Question. Is it possible to walk around Königsberg, cross every bridge exactly once, and return to where you started?

A Result

Question. Is it possible to walk around Königsberg, cross every bridge exactly once, and return to where you started?

Theorem (Euler, 1736). No.

Abstraction

1. Replace each separate landmass with a single point
• all points in each landmass are reachable from each other without crossing a bridge
2. Represent each bridge with an “edge” connecting landmasses

Graphs

A graph $G = (V, E)$ consists of

1. a set $V$ of vertices or nodes
2. a set $E$ of edges, where each edge consists of a pair of nodes
• if $e = (u, v)$ an edge, we say $u$ and $v$ are adjacent
• $G$ a multigraph if multiple edges between same pair of vertices

Example. Königsberg graph

Paths and Circuits

Note. Terminology varies from source to source.

A path $P$ of length $k$ in $G$ is a sequence of the form $v_0 e_1 v_1 e_2 v_2 \cdots e_k v_k$ where

• each $v_i$ is a vertex, and
• each $e_i$ is an edge with $e_i = (v_{i-1}, v_i)$.

$G$ is connected if for every pair of vertices $u, v \in V$, there is a path from $u$ to $v$.

$P$ is a circuit if $v_0 = v_k$.

BoK as a Graph Problem

Original Question. Is it possible to walk around Königsberg, cross every bridge exactly once, and return to where you started?

Rephrasing as Graph Problem. Given a graph $G = (V, E)$, is there a circuit that contains every edge $e \in E$ exactly once?

• A graph with this property is called Eulerian.

General Questions

Question 1. Under what conditions is a graph $G$ Eulerian?

Question 2. If $G$ is Eulerian, how can we find an Eulerian circuit?

Degrees

Let $G = (V, E)$ be a graph, and $v \in V$ a vertex. The degree of $v$, $\deg(v)$ is the number of edges in $E$ incident to $v$.

• $G$ is even if all vertices have even degree

A Necessary Condition

Claim (Euler 1736). If $G$ is Eulerian, then every vertex has even degree.

Why?

• Suppose $u$ is a vertex other than starting vertex
• Eulerian circuit visits $u$ total of $k$ times
• Each visit must
• cross one bridge to enter
• cross another to exit
• Circuit crosses every bridge to $u$ exactly once
• $\implies$ degree of $\deg(u) = 2 k$

Question

If all vertices have even degrees, is $G$ necessarily Eulerian?

A Result

Theorem (Euler 1736). If every vertex $v$ in a graph $G$ has even degree and $G$ is connected, then $G$ is Eulerian.

Proof technique:

• Devise an algorithm to find an Eulerian cycle

Algorithmic technique:

• Go wild!
• wander aimlessly crossing only uncrossed bridges
• greedily collect new bridges to cross
• continue until you reach your starting point
• reassess

Finding a Circuit

Input:

• graph $G$
• set $V$ of vertices
• set $E$ of edges
• starting vertex $v \in V$
• assume all vertices have even degree ($G$ is even)

Output:

• a circuit $P = v_0 e_1 v_1 e_2 v_2 \cdots e_k v_k$ with $v_0 = v_k = v$
• every edge $e$ incident to $v$ is contained in $P$

FindCircuit Subroutine

  FindCircuit(V, E, v):
cur <- v
P <- v
while deg(cur) > 0
e <- any edge in E incident to cur
(prev, cur) <- e
append e, cur to P
remove e from E
if deg(prev) = 0 then remove prev from V
endwhile
remove cur from V
return P


Circuit Finding

Claim. If every vertex in $G = (V, E)$ has even degree, then FindCircuit(V, E, v) returns a circuit beginning and ending at $v$.

• Loop invariant. If $\cur \neq v$, then $\cur$ and $v$ have odd degress, while all other vertices have even degrees.
• Consequence. If $\deg(\cur) = 0$, then $\cur = v$.
• $\implies$ can only get “stuck” at starting point!

Finding Eulerian Circuits

Strategy.

1. Apply FindCircuit to find a circuit $P = v_0 e_1 v_1\cdots v_k$
2. Traverse $P$
• if a vertex $v_i$ with $\deg(v_i) > 0$ is encountered,
1. apply FindCircuit to $v_i$ to get a circuit $Q$
2. splice $Q$ into $P$ at $v_i$
• continue traversing $P$ (with $Q$ spliced in)

Eulerian Circuit Pseudocode

  EulerCircuit(V, E, v):
P <- FindCircuit(V, E, v)
for each edge e = (u, w) in P do
if deg(w) > 0 then
Q <- EulerianCircuit(V, E, w)
Splice(P, Q, w)
endif
endfor


Correctness

Claim. If $G$ is even and connected, then EulerCircuit returns an Eulerian circuit.

Argue by induction on $m =$ number of edges in $G$.

Base Case, $m = 0$. If $G$ is connected and has no edges, then $G$ has only one vertex, so EulerCircuit correctly outputs an Eulerian circuit (of length $0$)

Inductive Step

Suppose EulerCircuit finds an Eulerian circuit on all connected, even graphs with fewer than $m$ edges. Then:

1. After removing $P$ from $G$, $G$ has fewer than $m$ edges
2. $G$ is still even
3. $G$ may be disconnected, but all components touch $P$
4. By inductive hypothesis, EulerCircuit finds Eulerian circuit in each component
5. Splicing together circuits gives Eulerian circuit for whole graph

Conclusion

$G$ is Eulerian if and only if $G$ is even and connected.

If $G$ is Eulerian, then an Eulerian circuit can be found by greedily traversing the graph,

Next Time

More (greedy) graph algorithms!