Graph Traversals

CSC 1300 – Discrete StructuresVillanova University

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Major ThemesGraph traversals:• Eulerian circuit/trail

– Every edge exactly once

• Hamiltonian cycle– Every vertex exactly once– Shortest cycle:

• Traveling Salesman Problem – Shortest Path:• Dijkstra’s algorithm

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The Bridges of Königsberg Puzzle18th century map of the city of Königsberg with 7 bridges over the Pregel river

Find a walk through the city that would cross each bridge once and only once before returning to the starting point

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Bridges of Königsberg: Euler’s Solution

A

C

B

D

Leonhard Euler (1707-1783) rephrased the question in terms of a multigraph

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Is there a circuit that traverses all the edges exactly once?

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Eulerian traversalsEulerian trail: Open trail that traverses all the edges Eulerian circuit: A circuit that traverses all the edges

Examples:

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Next: Determining whether a graph or multigraph has an Eulerian circuit

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Eulerian circuits: Necessary and sufficient conditions

Theorem: A connected multigraph G has an Eulerian circuit if and only if every vertex of G has even degree.

( è ) A necessary condition: G has an Eulerian circuit only if every vertex has even degree

Proof idea: Assume G has an Eulerian circuit. Observe that every time the circuit passes through a vertex, it contributes 2 to that vertex’s degree:• the circuit must enter via an edge incident with this vertex, and• leave via a different incident edge.

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(ç ) A sufficient condition:G has an Eulerian circuit if every vertex has even degreeProof idea: Assume every vertex in a multigraph has even degree. • Start at an arbitrary vertex v0, choose an arbitrary edge (v0,v1)

• Choose an arbitrary unused edge from v1 and so on. Since the vertices have even degree, there is always an edge to choose.

• … After a finite number of steps the process will arrive at the starting vertex v0, yielding a circuit with distinct edges.

• (But we may not be done! The circuit may not yet include all edges of G)

• Start again at an arbitrary vertex with untraversed edges; repeat above process. This newer traversal can now be spliced into the original one.

• Continue in this manner until all the edges have been traversed, splicing any new traversals to create a Eulerian circuit Villanova CSC 1300 - Dr Papalaskari

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Eulerian circuits: Necessary and sufficient conditions

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1 3 5 7

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Example

Gcircuit1231

Edges remaining1 3 5 7

2 4 6 8

1 3 5 7

2 4 6 81 3 5 7

2 4 6 8

circuit34653

splicing: 1 3 5 7

2 4 6 8

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12346531

splicing6786

1 3 5 7

2 4 6 8

1 3 5 7

2 4 6 8

1 3 5 7

2 4 6 8

Splicing:12346786531 Eulerian circuit

1 3 5 7

2 4 6 8

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circuit12346531

circuit6786

Example (continued)

G Edges remaining

Eulerian trails: Necessary and sufficient conditions

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Theorem A connected graph G has an Eulerian trail if and only if it has exactly two vertices of odd degree.

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Icosian GameIcosian Game This puzzle has been invented by the renowned Irish mathematician Sir William Hamilton (1805-1865) and presented to the world under the name of the Icosian Game. The game's board was a wooden board with holes representing major world cities with grooves representing connections between them. The object of the game was to find a circular route that would pass through all the cities exactly once before returning to the starting point.

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Traveling Salesman Problemfind the lowest cost Hamilton cycle

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Optimal tour of the 13,509 cities in the USA with populations greater than 500.

More about TSP: http://www.math.uwaterloo.ca/tsp/index.htmlImage:http://www.wired.com/images_blogs/wiredscience/2013/01/tsp_map.jpg

Hamiltonian Cycle?

K6C3 K4

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Hamilton Cycle: Sufficient condition(but NOT necessary)

Let G be a graph with n vertices where n ³ 3. If the degree of every vertex is at least n/2 , then G has a Hamilton cycle.

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Intuition: Graphs with lots of edges must have a Hamilton cycle

Hamilton Cycle: Necessary conditions(but NOT sufficient)

If a graph G has a Hamiltonian cycle…• k(G) = 1 (connected)• No vertices of degree 1• No cut vertices/edges• For every proper subset S of V(G),

k(G - S) ≤|S|

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Intuition: In hamiltonian graphs, you can remove some vertices without disconnecting them into too many pieces (i.e., more than the number of vertices removed).

Theorem 12.57

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• No simple necessary and sufficient conditions are known to verify existence of a Hamilton cycle for general graphs.

• This means that the problem has no known efficient solution.

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Q0 Q1 Q2 Q3 Q4

Hypercube graphs

Hypercube graphs

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Images by NerdBoy1392 - Own work, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=5514315Watchduck (a.k.a. Tilman Piesk) - Own work, Public Domain, https://commons.wikimedia.org/w/index.php?curid=2674082https://people.eecs.berkeley.edu/~jfc/cs174lecs/lec11/lec11.html

See also: https://en.wikipedia.org/wiki/Hypercube

Q0 Q1 Q2 Q3 Q4

Q4

Q3

Q4

https://commons.wikimedia.org/w/index.php%3Fcurid=5514315https://commons.wikimedia.org/w/index.php%3Fcurid=2674082https://people.eecs.berkeley.edu/~jfc/cs174lecs/lec11/lec11.htmlhttps://en.wikipedia.org/wiki/Hypercube

Find a Hamilton cycle

List the bitstrings, in order:1. __________2. __________3. __________4. __________5. __________6. __________7. __________8. __________9. __________

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0 0 0

0 0 0

Find a Hamilton cycleList the bitstrings, in order:1. __________2. __________3. __________4. __________5. __________6. __________7. __________8. __________9. __________10. __________11. __________12. __________13. __________14. __________15. __________16. __________17. __________ (back to 1.)

Villanova CSC 1300 - Dr Papalaskari 24Image by Watchduck (a.k.a. Tilman Piesk) - Own work. This W3C-unspecified vector image was created with Inkscape., Public Domain, https://commons.wikimedia.org/w/index.php?curid=11402515

https://commons.wikimedia.org/w/index.php%3Fcurid=11402515

Gray codes

Devices indicate position by closing and opening switches or sensing and digitizing rotational angle. Decimal Binary... ...2 0103 0114 100... ...

Physical switches are not ideal and if they don’t switch state in synchrony, the reading may appear as 110.

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Problem here: all three bits are changing!

Gray codes

If bits only change one at a time, we can not get large errors.Gray codes (also called “reflected binary code”) solve this problem by changing only one bit at a timeSee also: https://en.wikipedia.org/wiki/Gray_code

Villanova CSC 1300 - Dr Papalaskari 26Graphic by “Perlygatekeeper - Own work, CC BY-SA 4.0,” https://commons.wikimedia.org/w/index.php?curid=59774462

https://en.wikipedia.org/wiki/Gray_code