paths and circuits

Paths and Circuits

Lecture 52

Section 11.2

Wed, Apr 26, 2006

The Seven Bridges of Königsberg

In the city of Königsberg, two branches of the Pregel River came together, with an island at their junction.


There were seven bridges crossing the river at various places.


The challenge was to start at one point, cross each bridge exactly once, and return to the starting point.

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Euler’s Solution

Euler abstracted the bridges as a graph with four vertices and seven edges.

Euler’s Solution

Each vertex represents a land mass and each edge represents a bridge.

North Shore

South Shore

Island Peninsula

Walks and Paths

A walk from vertex v to vertex w is a finite alternating sequence of adjacent vertices and edges from v to w:

v0 e1 v1 e2 … en – 1 vn – 1 en vn,

where v0 = v and vn = w. A path from v to w is a walk that does not

repeat any edge.

Walks and Paths

A simple path is a path that does not repeat any vertices.

A closed walk is a walk that starts and ends at the same vertex.

A circuit is a closed path. A simple circuit is a circuit that does not

repeat any vertex.

Synopsis

walk = from A to B, no restrictions. path = walk, no repeated edge. closed = from A to A. circuit = closed walk. simple = no repeated vertex.

Euler Circuits

An Euler circuit is a circuit that contains every vertex and every edge of the graph.

The problem of the Seven Bridges of Königsberg is to find an Euler circuit.

Connected Graphs

A graph is connected if, for every pair of vertices v and w, there is a walk from v to w.

A connected component of a graph is a maximal connected subgraph.

Euler’s Solution

Theorem: A graph has an Euler circuit if and only if it is connected and every vertex has even degree.

Thus, an Euler circuit over the Seven Bridges of Königsberg does not exist.

The Two Bridges of Ashland

At Randolph-Macon College, they have been trying to solve the Two Bridges of Ashland problem for decades.

King’s Dominion

RMC

I-95

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Proof

Proof ():Suppose a graph G has an Euler circuit.Let v V(G).Then as we travel the circuit, each time we

pass through v, we “use up” two of the edges incident to v.

When we finish the circuit, we have used all the edges incident to v.

Proof

Thus, v had an even number of edges.Obviously, G must be connected.

Proof

Proof ():Now suppose that G is connected and that

every vertex of G has even degree.Choose a vertex v at which to begin.deg(v) > 0 since G is connected, so follow

one of the edges incident to v.Let w be the next vertex.We used one of w’s edges to get there.

Proof

w has even degree, so there is at least one more edge available that we can follow.

This happens at every vertex that we visit.Thus, the circuit is forced to terminate only

when we return to the starting vertex v.This procedure alone does not necessarily

produce an Euler circuit.

Proof

Suppose there are edges that were not used.

Follow the original circuit until a vertex is reached that is incident to one of the unused edges.

Apply the original procedure to produce a circuit that starts and ends at this vertex.

“Splice” it into the original circuit.

Proof

Continue in this way, splicing circuits into the existing circuit, until there are no unused edges remaining.

The result is an Euler circuit.

Example

Euler Paths

Theorem: A graph G has an Euler path from v to w if G is connected, v and w have odd degree, and all other vertices have even degree.

Proof:Add an edge from v to w.Then the graph has an Euler circuit.Remove the new edge from the circuit.

Hamiltonian Circuits

A Hamiltonian circuit is a simple circuit that includes every vertex of the graph.

The Traveling Salesman Problem seeks a Hamiltonian circuit of minimal length.


Theorem: If a graph G has a nontrivial Hamiltonian circuit, then G has a subgraph H such thatV(H) = V(G).H is connected.|E(H)| = |V(G)|.deg(v) = 2 for all v V(H).

These conditions are necessary, but not sufficient.


The following graph does not have a Hamiltonian circuit.

paths and circuits

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