tours in graph

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Graph Algorithms

Tours in Graphs

Graph Algorithms


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Special Paths and Cycles in Graphs

Euler Path: A path that traversesallthe edges of the graph

exactly once.

Euler Cycle: A cycle that traversesallthe edges of the graph

exactly once.

Hamilton Path: A simple path that visits all the vertices of

the graph exactly once.

Hamilton Cycle: A simple cycle that visitsallthe vertices of

the graph exactly once

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An Euler Path

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An Euler Cycle

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A Hamilton Path

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A Hamilton Cycle

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Complexity

An Euler Path (or an Euler Cycle), if exists, can be found

inanygraph with m edges in O(m)-time.

Finding a Hamilton Path or a Hamilton Cycle:

Ahard to solve problem. It is strongly believedthat nopolynomial time algorithm exists to solve these problems.

A simple special case of the Traveller Salesperson

Problem (TSP).

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Euler Paths and Cycles in Undirected Graphs

Edge representation path: P = (e0, e1, . . . , em1).

P is an Euler Path in a graph with m edges if

ei=ej for all 0i=j < m.

ei= (x, y) and ei+1= (y, z)

for 0i < m 1 and vertices x, y, z.

An Euler Cycle C is an Euler Path P for which em1= (x, y) and e0= (y, z) for vertices x, y, z.

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Euler Paths and Cycles in Directed Graphs

Edge representation path: P = (e0, e1, . . . , em1)

Pis an Euler Path in a directed graph with m edges if

ei=ej for all 0i=j < m.

ei= (xy) and ei+1= (yz)

for 0i < m 1 and vertices x, y, z.

An Euler Cycle C is an Euler Path P for which em1= (xy) ande0= (yz) for vertices x, y, z.

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The bridges of Konigsberg

B

A

DC

A

B

C D

NoEuler Cycle or Euler Path exist!!!

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A Toy Example

Theleftgraph has no Euler Path.

Themiddlegraph has an Euler Path but not an Euler Cycle.

Therightgraph has an Euler Cycle.

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Graphs with Euler Cycles

Theorem: An undirected andconnectedgraph has an Euler

Cycleiffall the vertices have aneven degree.

Remark: Aself-loopadds 2 to the degree of the vertex.

The only-if direction:

Let C= (e0, e1, . . . , em1) be an Euler Cycle.

Let y be a vertex.

Ifei= (x, y) then ei+1= (y, z) ( (m 1) + 1 = 0 ).

Therefore the degree ofy must be even.

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Proof: The If Direction

Assume all the degrees are even.

Construct an O(m)-time algorithm producing an Euler Cycle

represented by vertices.

Each edge is examined constant number of times withan appropriate data structure.

Main idea: Exploreunusededges as long as they exist.

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Data Structure and Variables

Edges are marked eitherusedorunused.

Initially all the edges are markedunused.

At the end all the edges are markedused. An arbitrary starting vertex x.

A main cycle C.

Initially C is empty. At the end Ccontains all the edges.

An exploring path P = (y , . . .).

Initially P = (x).

At the end P is empty.

A secondary cycle C.

Initially and at the end C is empty.

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Finding a Secondary Cycle

Let P = (y , . . . , z) be the exploring path.

While z (the last vertex in P) hasunusededges:

Let (z, w) be anunusededge.

Mark (z, w) asused.

Append w at the end ofP: P = (y , . . . , z , w).

Let the secondary cycle C

=P = (y , . . . , y). Need to prove: this process terminates onlyat y.

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Combining the Main and the Secondary Cycles

Let C= (x , . . . , a , y , b , . . . , x) be the main cycle.

Let C = (y , c , . . . , d , y) be the secondary cycle.

Then C= (x , . . . , a , y , c , . . . , d , y , b , . . . , x).

x

y

b

ca

y d

a

d

x y

b

c

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The Algorithm

Findthe first secondary cycle C:

Startthe exploring path with x.

The first main cycle C is C.

Whilethere exists anunusededge: Find y in Cwith anunusededges.

Finda secondary cycle C starting with y.

Combinethe cycles C and C into C.

Returnthe cycle C.

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Correctness: Key Observations

Since all the edges have an even degree it follows that thefinding a secondary cycle procedure can be stuck only at y

which is the first vertex of the exploring path.

If the main cycle C does not contain all the edges in the

graph, then it must contain a vertex with an unused edge

due to connectivity.

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Complexity

Each edge is explored only once when it isunusedand then

becomesusedforever.

Can be done in O(m)-time with adjacency lists.

Each edge is traversed only once while looking for a vertex

with anunusededge in the main cycle.

Can be done if the main cycle is a linked list and ifthe algorithm remembers the last starting vertex for the

exploring path.

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Complexity

There are at most n 1 cycle combinations since a new

exploring path never reaches again theconnectingvertex.

A combination can be done in O(1)-time if the cycles are

maintained asdouble linked lists.

Hence, O(m) complexity in connected graphs (nm).

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Directed Graphs

In a strongly connected graphthere exists a directed path

between any two vertices.

The in-degree of a vertex x din(x) is the number of

edges terminating at x.

Theout-degreeof a vertex x dout(x) is the number of

edges originating at x.

Theorem: A directed and strongly connected graph has an

Euler Cycleiffdin(x) =dout(x) for each vertex x.

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Euler Paths

Theorem: An undirected and connected graph has an Euler

Path if at most 2 vertices have anodddegree.

Theorem: A strongly connected directed graph has a directed

Euler Path starting with x and ending at y, x=y, if: din(z) =dout(z) for any vertex z / {x, y}.

din(x) =dout(x) 1.

din(y) =dout(y) + 1.

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Covering Paths

Lemma: In an undirected graph the number of vertices with

odd degree iseven.

Definition: k disjointpathscovera graph G if each edge of

G belongs to one of the k paths.

Theorem: A connected directed graph with 2k vertices with

an odd degree can becoveredwith k disjoint paths.

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Proof

Match the odd-degree 2k vertices with k newedges.

All the vertices in the new graph have an even degree.

Find an Euler Cycle in the new graph.

The new edges are not adjacent in the Euler Cycle since

each vertex belongs to at most one new edge.

Omit the k new edges from the Euler Cycle.

The cyclesplitto k paths that cover all the edges.

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De-Bruijn Sequences

={0, 1, . . . , 1} an alpha-bet of letters.

There exists distinct words of length over .

= 2 and = 3:000, 001, 010, 011, 100, 101, 110, 111.

= 3 and = 2:

00, 01, 02, 10, 11, 12, 20, 21, 22.

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De-Bruijn Sequences

A cyclic sequence S,=a0, a1, . . . , aL1 of length L=

is called aDe-Bruijnsequence if for any word w of length over there exists a unique index 0 i < L such that

w=ai, ai+1, . . . , ai+1 (the addition is donemod L).

= 2 and = 3 = 00011101.

= 3 and = 2 = 001122021.

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G2,3

00

01

10

11000

001 011

100

101

110

111010

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G3,2

00

01

0210

11

12

20

21

22

0

1 2

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De-Bruign Sequences Always Exist

Lemma: For all positive integers and there exists a

directed Euler Cycle in G,.

Proof: G, is strongly connected and for any vertexin-degree= out-degree = .

Lemma: An Euler Cycle inG,impliesa De-Bruijn sequence

S,.

Proof: Follow the Euler Cycle. Initially the sequence is the

first vertex on the path. Append only the last letter of the

next vertex to the current sequence.Theorem: For all positive integers and there exists a

De-Bruijn sequence S,.

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G2,3 and S2,3

00

01

10

11000

001 011

100

101

110

111010

Euler Cycle: 00 00 01 11 11 10 01 10 00

De-Bruijn sequence: 00011101

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G3,2 and S3,2

00

01

0210

11

12

20

21 22

0

1 2

Euler Cycle: 0 0 1 1 2 2 0 2 1 0

De-Bruijn sequence: 001122021

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Undirected Hamilton Paths and Cycles

A path of vertices: P = (v0, v1, . . . , vn1).

P is a Hamilton Path in a graph with n vertices if

vi=vj for all 0i=j < n.

(vi, vi+1) is an edge for 0i < n 1.

A Hamilton Cycle C is a Hamilton Path P for which

(vn1, v0) is also an edge.

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Directed Hamilton Paths and Cycles

A directed path of vertices: P = (v0v1. . .vn1)

P is a directed Hamilton Path in a graph with n vertices if

vi=vj for all 0i=j < n.

(vivi+1) is a directed edge for 0i < n 1.

A directed Hamilton Cycle C is a directed Hamilton Path

P for which (vn1v0) is also an edge.

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The Petersen Graph

A B

C

D

E

F

G

H

I

J

There is no Hamilton Cycle.

The following is a Hamilton Path:

P = (A,B,C,D,E,J,H,F,I,G)

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The Knight-Chess Graph

Definition: TheKnight-Chess graph has 64 vertices one for

each square on the88chess board. 2vertices are adjacent

iff a knight can move from one to another in one step.

*

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The Knight-Chess Problem

Problem: Is it possible to cover all the squares of the chess

boardwithknight moves?An equivalent formulation: Does the Knight-Chess graph

has aHamilton path?

1

2

3

4

5

6

7

8

9

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Tournaments

Definition: A tournament is a simple directed graph such

that for each pair of vertices u and v, either the directededge u v exists or the directed edge v u exists but

not both and not none.

Observation: There are exactlyn2

directed edges in atournament with n vertices.

Observation: The underlying graph of a tournament with n

vertices is the complete graph Kn.

Theorem: A tournament always has a Hamilton path.

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A Tournament with 6 Vertices

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A Hamilton Path in the Tournament

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A Hamilton Cycle in the Tournament

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Algorithm to find Hamilton Path in a Tournament

(1) Startwith the path P1= (v1) for an arbitrary vertex v1.

(2) Let the current path be Pi= (v1v2 vi)

for 1in.

(3) Ifi=n,terminatewith the Hamilton Path Pn.

(4) Let v be a vertex not in the path.

(5) Insert v into Pi to get the path Pi+1.

(6) GotoStep (2).

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Path Augmentation

If(vv1) is an edge,then Pi+1= (vv1 vi).

If(viv) is an edge,then Pi+1= (v1 viv).

Otherwise 1j


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The Algorithm is Correct

The path augmentation is always successful.

Therefore, eventually Pn exists which is a Hamilton Path.

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The Algorithm is Efficient

Inserting a vertex to a path can be done inO(n)time using

the adjacency matrix.

There are n iterations.

The total complexity is O(n2).

With abinary searchfor the insertion point, the algorithm

probs the adjacency matrix O(n log n) times. But theoverall complexity is still O(n2).

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A Hamilton Cycle Greedy Algorithm: Outline

As long as possible, constructa path byaddingvertices to

both end-vertices of the path.

Close this path into a cycle by either connecting both

end-vertices or byfindingaswitch vertex.

Connecta new vertex to the cycle andbreakit to be a new

longer path.

Repeat the above process until either a Hamilton Cycle isfound or an operation is impossible.

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Converting a Path to a Cycle

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Converting a Cycle to a Path

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A Hamilton Cycle Greedy Algorithm: Description

(1) Initially, let P = (x) be a path with an arbitrary vertex x.

(2) Expandthe path P from both ends until impossible. Let

P = (x0 x1 xh)

where there are no edges from x0 and xh outside P.

(3) If(x0, xh) is an edge thenconstructthe cycle

C= (x0 x1 xh x0).Gotostep 6.

(4) If for some 0 < i < h the edges (x0, xi+1) and (xi, xh)

exist, thenconstructthe cycleC= (x0 x1 xi xh xh1 xi+1 x0).

Gotostep 6.

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Greedy Algorithm to find an Undirected Hamilton Cycle

(5) Terminate Unsuccessfullywith the path P.

(6) Ifh=n 1thenTerminate Successfullywith the HamiltonCycle C.

(7) If there is no edge from C outside of C, then Terminate

Unsuccessfullywith the cycle C.

(8) Let (xi, x) be an arbitrary edge from C to outside of C,

thenconstructthe path

P = (x xi xi+1 xh x0 xi1).

(9) Gotostep 2 with a longer path.

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Sometimes Hamilton Cycles Exist

Theorem: Let G be a connected graph with n vertices. If

d(u) +d(v)n for any two vertices u =v in G, then G

has a Hamilton Cycle.

Corollary: Let G be a connected graph with n vertices. Ifd(u)n/2 for any vertex u in G, then G has a Hamilton

Cycle.

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Proof of the Theorem

Step 4, whenever executed, is always successful.

Therefore, the algorithm never reaches step 5.

The algorithm never terminates in step 7 since the graph isconnected.

The algorithmterminates sucessfulywith a Hamilton Cycle

in step 6 since the path is longer in each iteration.

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Why Step 4 is Always Successful?

Assume that step 4failsfor hn 1 with the path

P = (x0 x1 xh1 xh). Let xi1, xi2, . . . , xik be the neighbors ofx0 in P.

xi11, xi21, . . . , xik1 cannotbe neighbors ofxh.

d(xh)h kn 1 k.

d(x0) + d(xh)< n.

Acontradiction.

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Complexity

Represent the graph with an adjacency matrix.

Augmenting a path by one vertex at its end-point canbe done in O(n) time for a total of O(n2) for all the

augmentations.

Converting a path into a cycle can be done in O(n) timefor a total ofO(n2) for all such conversions.

All the conversions of cycles into paths can be done in

O(n2) by scanning the adjacency matrix only once.

The algorithm time complexity is O(n2).

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tours in graph

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