Data Structure - Adjacency

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Data StructureChapter 6 Graphs

Ju inn -Dar Huang , Ph.D.

As sis tant Professo [email protected] c tu .edu .tw

November 2004

Rev . 2005, 2006


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raphs

Juinn-DarH

uang

jdhua

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copyright 2004-2006

Knigsberg Bridge Problem

A

Kneiphof

ab

c d gC

D

Bf

e

a b

c dg

e

f

A

C

Q: Is it possible to cross each bridge exactly once and

return to the starting points?

A: Euler said no.B

D

The first recorded problem that was solved using graph in 1736


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Graphs

Definition

a graph G consists of 2 sets, V and E

V is a finite, nonempty set ofvertices E is a set ofpairs of vertices; these pairs are called

edges

Representations a graph G(V, E)

V(G) represents the set of vertices of G

E(G) represents the set of edges of G


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Directed or Undirected Graphs

In an undirected graph

the edge is undirected

(u, v) and (v, u) represent the same edge

In a directed graph

the edge is directed

and represent different edges

for an edge , u is the tail and v is the head of the

edge


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Graph Examples

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1 2

0

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G1 G2 G3

V(G1) = {0, 1, 2, 3}

E(G1) = {(0, 1), (0, 2), (0,3), (1, 2), (1, 3), (2, 3)}

V(G2) = {0, 1, 2, 3, 4, 5, 6}

E(G2) = {(0, 1), (0, 2), (1, 3), (1, 4),(2, 5), (2, 6)}

V(G3) = {0, 1, 2}

E(G3) = {, ,

}

DirectedUndirectedUndirected

0

1

2

Tail

Head


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Restrictions on Graphs

A graph cannot have an edge from a vertex v

back to itself

(v, v) and are both illegal these edges are named

self edges orself loops

A graph cannot have multiple occurrences of thesame edge

or it is a multigraph

These restrictions are not necessarily essential

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Complete Graphs

An n-vertex undirected graph

has n*(n-1)/2 distinct edges at most

An n-vertex undirected graph with exactlyn(n-1)/2 edges is a complete graph

e.g., G1

An n-vertex directed graph

has n*(n-1) distinct edges at most An n-vertex directed graph with exactly n(n-1)

edges is a complete graph

e.g., G3 is not complete

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1 2

G1

0

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2G3


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Adjacency and Incidence (1/2)

If (u, v) E(G)

u and v are adjacent

(u, v) is incident on u and v e.g., vertices adjacent to vertex 1 are vertex 0, 3, and 4

e.g., edges incident on vertex 2 are (0,2), (2,5), and (2,6)

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G2


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Adjacency and Incidence (2/2)

If E(G)

u is adjacent to v (directed, i.e., v is NOT adjacent to u)

v is adjacent from u is incident to u and v

e.g., edges incident to vertex 1 are

, , and 0

1

2

G3


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Subgraphs

A subgraph of G is a graph G such that V(G) V(G) and E(G) E(G)

0 0

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1 2

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1 2

(i) (ii) (iii)

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G1

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2G3

0 0

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2


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Paths

A path from u to v in G

is a sequence of vertices u, i1, i2, , ik, v such that (u, i1)

(i1

, i2

), , (ik

, v) are edges in E(G) (undirected graph)

is a sequence of vertices u, i1, i2, , ik, v such that

, , are edges in E(G) (directed graph)

Length of a path number of the edges on it

Simple path

is a path in which all vertices except possibly for the firstand last are distinct

A cycle is a simple path in which the first and the

last is the same vertex


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Path Examples

A path 0, 1, 3, 2 in G1 also a simple path; of length 3

A path 0, 1, 3, 2, 0 in G1

also a cycle; of length 4

A path 0, 1, 3, 1 in G1

not a simple path; of length 3 A path 0, 1, 2 in G3

also a simple path; of length 2

A path 0, 1, 0 in G3 also a cycle; of length 2

0, 1, 2, 1, is NOT a path in G3 E(G3)

0

1

2

G3

0

3

1 2

G1


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Connected Graphs

In an undirected graph G, 2 vertices u and v are

said to be connected iff

there is a path from u to v (or a path from v to u) An undirected graph G is said to be connected iff

u, v V(G), u v there is a path from u to v

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G2

Connected

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G4

H1 H2

Unconnected


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Connected Components

A connected component, H, of an undirected

graph G is a maximal connected subgraphs

Maximal: G contains no other subgraph that is bothconnected and fully contains H

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H1 H2

H1 and H2 are 2 connected components


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Strongly Connected Graphs

A directed graph G is said to be strongly

connected iff

u, v V(G), u v there is a path from u to v and apath from v to u (bi-directional)

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1

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G3

not strongly connected


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Strongly Connected Components

A strongly connected component of a directed

graph is a maximal strongly connected subgraphs

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1

2

G32 strongly connected components

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1 2


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Degree

Degree of a vertex

the number of edges incident to the vertex

For a directed graph (digraph) in-degree of v: the num of edges for which v is the head

out-degree of v: the num of edges for which v is the tail

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G2

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G3

degree: 3

in-degree: 1

out-degree:2

degree: 3


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ADT Graph

class Graph {

// objects: a nonempty set of vertices and a set of edges

// where each edge is a pair of vertices

public:Graph(); // ctor

void InsertVertex(Vertex v);

void InsertEdge(Vertex u, Vertex v);

void DeleteVertex(Vertex v);void DeleteEdge(Vertex u, Vertex v);

bool IsEmpty();

List Adjancent(Vertex v);

};


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Graph Representations

3 Common representations

Adjacency matrix

Adjacency list

Adjacency multilist


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Adjacency Matrix (1/2)

Let G = (V, E) with n vertices, n1

The adjacency matrix A of G is an nxn martix in which

A[i][j] = 1 iff (u,v) (or ) E(G)

000

101

010

2

1

0

210

0

1

2

no self loops

0010

0011

1101

0110

3

2

1

0

3210

0

3

1 2

no self loops

symmetric

non-symmetric


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Adjacency Matrix (2/2)

Need n2 bits if |V(G)| = n

The adjacency matrix is symmetric for an undirected graph

store only the upper or lower triangle of the matrix

Find degrees

degree of v (undirected graph): either row sum or column sum

in-degree of v (digraph): column sum

out-degree of v (digraph): row sum

How many edges in G? Is G connected?

time complexity is O(n2) at least

When the adjacency matrix is sparse

sparse matrix techniques in Chapter 2

time complexity can be reduced to o(n+e) where e


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Adjacency List

0

1

2

G3

0

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1 2

G1

2e nodes

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2

1

0

1

3

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1

2 0

0 0

0 0

2 0

[0]

[1]

[2]

[3]

HeadNodes

Number of edges can be determined in O(n+e) time

Easily to find the degree of a node

0

1 0

2 0 0

[0]

[1][2]

HeadNodese nodes

Easily to find the out-degree of a node

List Representation


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Sequential Representation

9 11 13 15 17 18 20 22 23 2 1 3 0 0 3 1 2 5 6 4 5 7 6

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 220 1

0

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1 2

4

7

5 6

G4

H1 H2

n+1 entries 2e entries

Advantage small memory usageDisadvantage dynamic edge insertion/deletion


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Inverse Adjacency List

0

1 0

2 0 0

[0]

[1]

[2]

HeadNodes e nodes

Easily to find the out-degree of a node

0

1

2

G3

Q: How to find the in-degree of a node ?

1 0

0 0

1 0

[0]

[1]

[2]

HeadNodese nodes

Inverse adjacency list

How about combining the adjacency list and the inverse adjacency list?


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Orthogonal Adjacency List

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1

2

G3

Tail Head Col link for head row link for tail

( Recall sparse matrix in Chapter 2 )

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1

2 0

1 0 0

0

0 1 0 0

1 2

1 2 0 0

head nodes(shown twice)

e nodes

Adj M l ili


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Adjacency Multilist

[0]

[1]

[2][3]

0 1 N1 N3

0 2 N2 N3

0 3 0 N4

1 2 N4 N5

1 3 0 N5

2 3 0 0

N0

N1

N2

N3

N4

N5

HeadNodes

edge (0, 1)

edge (0, 2

edge (0, 3)

edge (1, 2)

edge (1, 3)

edge (2, 3)

The lists are

Vertex 0: N0 -> N1 -> N2

Vertex 1: N0 -> N3 -> N4

Vertex 2: N1 -> N3 -> N5

Vertex 3: N2 -> N4 -> N50

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1 2

G1vertex1 vertex2 list1 list2

W i ht d Ed


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Weighted Edges

If there is a weight on an edge

represents distance or cost, or

matrix: A[i][j] might be used to keep the non-zero weight list: an additional field for the weight is required

A graph with weighted edges is called a network

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Traversals on Graphs

Graph is not like tree

there is no root

there may exist multiple paths from u to v

Given a graph and a vertex v

how to visit all vertices that are reachable from v?

i.e., visit all vertices connected to v

Traversals on graphs

depth-first search (DFS)

breadth-first search (BFS)

D th Fi t S h (DFS)


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Depth-First Search (DFS)

void Graph::DFS() {visited = new bool[n]; // n vertices from 0 to n-1

for(int i = 0; i < n; ++i)

visited[i] = false;

DFS(0); // start search from Vertex 0delete [] visited;

}

void Graph::DFS(int v) {visited[v] = true;

// pseudo code

for( each vertex w adjacent to v)

if(! visited[w])DFS(w); // recursive call

}

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Traversal Order:

0, 1, 3, 7, 4, 5, 2, 6

If G is in adjacency list O(e)If G is in adjacency matrix O(n2)

Stack-Based

B dth Fi t S h (BFS)


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Breadth-First Search (BFS)

void Graph::BFS(int v) { // starting from Vertex vvisited = new bool[n]; // n vertices from 0 to n-1for(int i = 0; i < n; ++i)visited[i] = false;

Queue q;visited[v] = true;q.Insert(v);

while(! q.IsEmpty()) {v = *q.Delete(v);

for( each vertex w adjacent to v)if(! visited[w]) {visited[w] = true;q.Insert(w);

}}delete [] visited;

}

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Traversal Order:

0, 1, 2, 3, 4, 5, 6, 7

If G is in adjacency list O(e)If G is in adjacency matrix O(n2)

Queue-Based

Spanning Trees (1/3)


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A spanning tree of a graph G is a tree that

includes all vertices of G

includes edges of G

spanning

trees

A spanning tree consists ofn vertices and n-1 edges



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If a graph G is connected a DFS or BFS starting at any vertex visits all vertices

E(G) are partitioned into two sets, T and N

T (tree edges): edges traversed during the search

N(non-tree edges): remaining edges

V(G) + T spanning tree

A spanning tree resulting from DFS/BFS

depth-first/breadth-first spanning tree

If a non-tree edge is inserted into a spanning tree

a cycle is formed



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3 4

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DFS (0) spanning tree BFS (0) spanning tree

A graph can have a bunch

of spanning trees

Articulation Point


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Articulation Point

A vertex v of a connected graph G is an articulation

point iff the deletion of v and all edges incident to v

will leave behind a graph that has at least 2connected components

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Vertices 1, 3, 5, and 7 are articulation points

Biconnected Graph


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Biconnected Graph

A biconnected graph is a connected graph that

has no articulation points

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Not biconnectedBiconnected

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Biconnected graph is desirable while representing a communication network

Why?

Biconnected Components


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Biconnected Components

A biconnected component of a connected graph Gis a maximal biconnected subgraph H of G

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A connected graph Its 6 biconnected components

A biconnected graph has just one biconnected component the whole graph

Depth-First Number


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Depth-First Number

Depth-first number: the visit order during the

depth-first search

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non-tree

edge

starting

vertex

depth-first number, dfn

Back Edge and Cross Edge


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Back Edge and Cross Edge

No cross edges in a depth-first spanning tree

No back edges in a breadth-first spanning tree

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back edge

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cross edge

Find Articulation Points by DFS (1/2)


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Articulation point

the root of the depth-first spanning tree has at least 2 children

any other vertex u that has at least one child w such that its

impossible to reach ancestor of u using a path composed solely ofw, descendants of w and a back edge

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back edge



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Articulation point

the root of the depth-first spanning tree has at least 2 children

a vertex u had a child w such that low(w) dfn(u)

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back edge

low(w) = min {

dfn(w),

min{ low(x) | x is a child of w },

min{ dfn(x) | (w, x) is a back edge } }

10966611115low

1087621345dfn

9876543210vertex

9

66611115low

87621345dfn

9876543210vertex

Find dfn & low Values


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Find dfn & low Values

void Graph::DfnLow(int x) { // apply DFS at Vertex xnum = 0; // num is an int data member of class Graph

dfn = new int[n]; // dfn is also a data member

low = new int[n]; // low is also a data member

for(int i = 0; i < n; ++i) dfn[i] = low[i] = 0;DfnLow(x, -1);

delete[] dfn; delete[] low;

}

void Graph::DfnLow(int u, int v) {dfn[u] = low[u] = ++num;

for( each vertex w adjacent from u )

if(dfn[w] == 0) { w is an unvisited vertex

DfnLow(w, u); // recursivelow[u] = min(low[u], low[w]);

} else if (w != v) // (w, v) is a back edge

low[u] = min(low[u], dfn[w]);

}

Minimum-Cost Spanning Trees


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Minimum Cost Spanning Trees

A weighted, undirected graph may have lots of

valid spanning trees

the cost of a spanning tree is the sum of the costs(weights) of the edges in the spanning tree

A minimum-cost spanning tree is a spanning tree

ofleast cost e.g., road construction

3 algorithms here

Kruskals, Prims, Sollins

all greedy methods

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Kruskals Algorithm (1/3)


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Kruskal s Algorithm (1/3)

Build a minimum-cost spanning tree T Sort edges in non-decreasing order of their costs

Select an edge in that order for inclusion in T ifthe resultant T does not contain a cycle

only n -1 edges are included at last

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Kruskal s Algorithm (2/3)

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g ( )

// E is the set of all edges in G

T = ;while((T contains less than n 1 edges) && (E is not empty))

{

choose an edge (v, w) from E of lowest cost;

delete (v, w) from E;

if((v, w) does not create a cycle in T)

add (v, w) to T;else

discard (v, w);

}

if( T contains few than n 1 edges )

cerr


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g ( )

Begin with a tree T containing one vertex any one in the original graph

add a minimum-cost edge (u, v) to T such thatT {(u, v)} is still a tree

at first, (u, v) should be an edge such that

u T and v T0

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Prims Algorithm (2/3)


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g ( )

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Prims Algorithm (3/3)


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g ( )

// assume that G has at least one vertex// TV: vertex set of the spanning tree

// T: edge set of the spanning tree

TV = {0}; // start with Vertex 0 and no edges

for( T = ; T contains fewer than n-1 edges; ) {let (u, v) be a least-cost edge s.t. u TV and V TV;if( there is no such edge) break; // G is not connectedadd v to TV;

add (u, v) to T;

}

if( T contains few than n 1 edges )

cerr


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g

At each stage, each tree selects a minimum-costedge to connect another tree

Repeat the process until the minimum-cost

spanning tree is obtained

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(0, 5), (1, 6), (2, 3), (3, 2)

(4, 3), (5, 0), (6, 1) are selectedat the 1st stage

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16

(5, 4), (1, 2), (2, 1)

are selected at the 2nd stage

Shortest Path


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jdhuan

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Give a digraph G and two vertices A and B in G

is there a path from A to B?

if there is more than one path from A to B, which is the shortest

one?

The length/cost/weight of a path is defined as the sum of

the lengths/costs/weights of the edges on that path

For a path starting vertex: source

last vertex: destination 0

3 4

1 2

5

10 20

50

15

20

10

35

30

315

45

Single Source/All Destinations (1/4)


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Non-negative edge weights

Path Length

1) 0, 3

2) 0, 3, 4

3) 0, 3, 4, 1

4) 0, 2

10

25

45

45

0

3 4

1 2

5

10 20

50

15

20

10

35

30

315

45

Starting Vertex: 0

Increasing



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Let S denotes the set of vertices to which theshortest paths have already been found

(1) if the next shortest path is to vertex u, then the path

begins at v, ends at u, and goes through only vertices

that are in S

(2) the destination of the next path generated must be

the vertex u that has the minimum distance among all

vertices not in S

(3) The vertex u selected in (2) becomes a member of S

The algorithm is first given by Edsger Dijkstra

Therefore, its sometimes called Dijkstra Algorithm



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Data Structure

// nmax: maximum number of vertices

class Graph {int length[nmax][nmax]; // length-adjacency matrix

int dist[nmax]; // min-distance

bool s[nmax];

int choose(int); // used in ShortestPath

public:

void ShortestPath(int, int);

};

// length[i][j] is set to a large number

// if is not an edge



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void Graph::ShortestPath(int n, int v) {// n vertices, source vertex = v

for(int i = 0; i < n; ++i) { // initialize

s[i] = false; dist[i] = length[v][i];

}

s[v] = true; dist[v] = 0;

for(int j = 0; j < n-2; ++j) { // determine n-1 paths

int u = choose(n);// choose u s.t. dist[u] is minimum where s[u] = false

s[u] = true;

for(int w = 0; w < n; ++w) {

// update dist[w] where s[w] is false

if( (! s[w]) && (dist[u] + length[u][w] < dist[w]) )

dist[w] = dist[u] + length[u][w]; // shorter path found

}

}

Example (1/2)


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0

1 2

34

5

67

300

800

17001000

1000

1400

1200

1500

1000250

900

San

Francisco

Denver

New

Orleans

New York

Chicago

Los

Angeles

Miami

Boston

001700

100000

1400900010000

250015000

01200

08001000

0300

0

7

6

5

4

3

2

1

0

76543210

Length-adjacencymatrix

Example (2/2)


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Distance

LA SF DEN CHI BOST NY MIA NO

[0] [1] [2] [3] [4] [5] [6] [7]

Initial -- --- + + + 1500 0 250 + +

1 {4} 5 + + + 1250 0 250 1150 1650

2 {4,5} 6 + + + 1250 0 250 1150 1650

3 {4,5,6} 3 + + 2450 1250 0 250 1150 1650

4 {4,5,6,3} 7 3350 + 2450 1250 0 250 1150 1650

5 {4,5,6,3,7} 2 3350 3250 2450 1250 0 250 1150 1650

6 {4,5,6,3,7,2} 1 3350 3250 2450 1250 0 250 1150 1650

{4,5,6,3,7,2

,1}

iteration S Vertexselected



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General Weights (Negative weights are allowed)

0 1 2

5

7 -5

Directed graph with a negative-length edge

0 1 2

-2

1 1

Directed graph with a cycle of negative length

no cycle of negative length is allowed



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When there is no cycle of negative length, ashortest path between any 2 vertices of an n-

vertex graph has at most n-1 edges

Let distk[u] be the length of a shortest path from

the source vertex v to vertex u

under the constraint that the path contains at most kedges

distk[u] = min { distk-1[u],

min{ distk-1[i] + length[i][u]} }i

Bellman and Ford Algorithm



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distk[7]k

0 1 2 3 4 5 6

1 0 6 5 5

2 0 3 3 5 5 4

3 0 1 3 5 2 4 7

4 0 1 3 5 0 4 5

5 0 1 3 5 0 4 3

6 0 1 3 5 0 4 3

distk

0

1

2

3

4

5

6

6

5

5

-2

-2

-1

-1

1

3

3

from length-adjacency matrix



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void Graph::BellmanFord(int n, int v) {

// n vertices, source vertex = v

for(int i = 0; i < n; ++i)

dist[i] = length[v][i]; // initialize, dist1[u]

for(int k = 2; k (dist[i] + length[i][u]) )

dist[u] = dist[i] + length[i][u];

}

All-Pairs Shortest Paths (1/3)


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The problem can be solved as n independentsingle-source/all-destinations problems

each iteration, a different vertex as the source vertex

Better idea?

Define Ak[i][j] to be the length of the shortest path

from i to j going through no vertex of index > k An-1[i][j] will be the length of the shortest i-to-j path in an

n-vertex graph

Ak[i][j] = min{ Ak-1[i][j], Ak-1[i][k] + Ak-1[k][j] }, k 0,

where A-1[i][j] = length[i][j]



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void Graph::AllLengths(int n) {// length[n][n] is the length-adjacency matrix

// a[i][j] is the length of the shortest path between i and j

for( int i = 0; i < n; ++i)

for ( int j = 0; j < n; ++i)

a[i][j] = length[i][j]; // constructA-1

for(int k = 0; k < n; ++k)

for(int i = 0; i < n; ++i)

for(int j = 0; j < n; ++j)

if( a[i][j] > (a[i][k] + a[k][j]) )a[i][j] = a[i][k] + a[k][j];

}



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A-1 0 1 2

0 0 4 11

1 6 0 22 3 0

A0 0 1 2

0 0 4 11

1 6 0 22 3 7 0

A1 0 1 2

0 0 4 6

1 6 0 2

2 3 7 0

2

0 1

6

2

4

11

3

A2 0 1 2

0 0 4 6

1 5 0 2

2 3 7 0

A-1 A0

A1 A2

Transitive Closure (1/3)


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Given a digraph G with unweighted edges,determine if there is a path from vertex i to vertex j

positive length: transitive closure

non-negative length: reflexive transitive closure

Transitive closure matrix A+

A+[i][j] = 1 iff there is a path oflength > 0 from i to j

Transitive closure matrix A*

A*[i][j] = 1 iff there is a path oflength 0 from i to j



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ng

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To find A+ similar to the all-pairs shortest paths problem

first, let length matrix be the adjacency matrix, then

void Graph::TransitiveClosure(int n) {// length[n][n] is the adjacency matrixfor( int i = 0; i < n; ++i)

for ( int j = 0; j < n; ++i)a[i][j] = length[i][j];

for(int k = 0; k < n; ++k)for(int i = 0; i < n; ++i)

for(int j = 0; j < n; ++j)a[i][j] = a[i][j] || (a[i][k] && a[k][j]);

}



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11100

11100

11100

11100

11110

4

3

2

1

0

43210

00000

10000

01000

00100

00010

4

3

2

1

0

43210

0 1 2 3 4

Digraph G

11100

11100

11100

11110

11111

4

3

2

1

0

43210

Adjacency matrix

A+

A*

Activity-on-Vertex (AOV) Networks (1/3)


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Course number Course name Prerequisites

C1 Programming I None

C2 Discrete Mathematics None

C3 Data Structures C1, C2

C4 Calculus I None

C5 Calculus II C4

C6 Linear Algebra C5

C7 Analysis of Algorithms C3, C6C8 Assembly Language C3

C9 Operating Systems C7, C8

C10 Programming Languages C7

C11 Compiler Design C10

C12 Artificial Intelligence C7

C13 Computational Theory C7

C14 Parallel Algorithms C13

C15 Numerical Analysis C5



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C1

C2

C3

C5C4 C6 C15

C7

C8

C13

C12

C10

C9

C14

C11



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A digraph G in which the vertices represent activities andthe edges represent precedence relations between

activities is an activity-on-vertex (AOV) network

In an AOV network

vertex i is a predecessorof vertex j iff there is a path from i to j

i is an immediate predecessorof j iff is an edge j is a successorof i if i is a predecessor of j

j is an immediate successorof i if i is an immediate predecessor

of j

Partial Order


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jdhuang

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du.tw


A relation is transitive if a b and b c a c e.g., for integer a, b, c; a > b and b > c a > c

A relation

is irreflexive on a set Sif a a for no a S

e.g., you cannot find any integer a that a > a

A relation is a partial orderif it is transitive andirreflexive

Precedence Relation


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jdhuang

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du.tw


Is precedence relation a partial order? Precedence relation is transitive

a is a predecessor of b and b is a predecessor of c a

is a predecessor of c

Is precedence relation irreflexive?

it depends on whether the given network contains anydirected cycles

A precedence relation on an network is irreflexive if

the network is a directed acyclic graph (DAG) i.e., if it is a DAG, the precedence relation is a partial

order on it

topological order

Topological Order


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jdhuang

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du.tw


A topological order is a linear ordering of the vertices of aDAG such that, for any 2 vertices i and j, if i is a

predecessor of j, then i precedes j in the linear ordering

There may be several feasible topological orders e.g.,

C1, C2, C4, C5, C3, C6 ,

C4, C5, C2, C1, C6, C3,

C1

C2

C3

C5C4 C6 C15

C7

C8

C13

C12

C10

C9

C14

C11

Topological Sorting (1/3)


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du.tw


// pseudo code here

Input the AOV network. Let n be the number of vertices.

for(int i = 0; i < n; ++i) {

if(every vertex has a predecessor) return;

// network has a cycle and thus is infeasible

pick a vertex v that has no predecessors;

// if there are multiple vertices, pick one arbitrarily

cout


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jdhuang

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du.tw


0

1

2

3

4

5

1

2

3

4

5

1

2 4

5

1

4

5

1

44

Initial Vertex 0 deleted Vertex 3 deleted

Vertex 2 deleted Vertex 5 deleted Vertex 1 deleted

one of feasible topological orders: 0, 3, 2, 5, 1, 4

Example

Topological Sorting (3/3)


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Internal data structure

1

4 04

5

2

5 0

4 0

[0]

[1][2]

[3]

[4]

0

11

1

3 02 0[5]

3 0

count first data link

Indicate how many immediate predecessors

Activity-on-Edge (AOE) Networks


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event interpretation

0 Start of project

1 Completion of activity a1

4 Completion of activities a4 and a5

7 Completion of activities a8 and a9

8 Completion of project

1

0 4

2

6

8

7

3 5

start finish

a1 = 6 a4 = 1

a2 = 4

a5

= 1

a7= 9

a10 = 2

a3 = 5

a6 = 2

a9 = 4a11 = 4

No activity can be launched

until its predecessors are

all completedactivityevent

Q1:

How fast can the project be done?

Q2:What are the critical paths?

Q3:

How to speedup the project?

a8= 7

Critical Path (1/2)


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Critical path a path oflongest length

e.g., 0, 1, 4, 6, 8 or 0, 1, 4, 7, 8 of length 18

The earliest time that an event i can start is the length ofthe longest path from the start vertex 0 to vertex i

The earliest time of an event determines the earliest start

time for all activities represented by edges outgoing fromthat vertex

denoted by e(i) for activity ai

The latest time of an activity ai, l(i) the latest time ai can start without increasing the final project

duration

An activity aiis a critical activity ife(i) = l(i)

Critical Path (2/2)


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Meaning of l(i) e(i) criticality of an activity ai

Speedup non-critical activities will not reduce theproject duration

Speedup a critical activity may not reduce the

project duration unless that activity is on all critical paths

Critical path analysis helps

evaluate project performance

identify project bottlenecks

Earliest Time Calculation (1/2)


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For an event j ee[j] : earliest event time

le[j] : latest event time

For an activity ai represented by

e[i] = ee[k]

l[i] = le[l] duration of activity ai

How to calculate ee[j]?

let P(j) be the set of all js immediate predecessors

in what order? topological order

ee[j] = max { ee[i] + duration of }

iP(j)

Earliest Time Calculation (2/2)


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adjacency list with in-degree count

Latest Time Calculation (1/2)


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How to calculate le[j]? let S(j) be the set of all js immediate successors

in what order? reverse topological order

le[j] = min { le[i] - duration of }

iS(j)

Latest Time Calculation (2/2)


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inverse adjacency list with out-degree count

Critical Activities


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e = l

critical activity

Critical Network


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All paths from 0 to 8 are critical

Speed up a1 and a4 can reduce the project time

Final Review


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Terminology Representations

adjacency matrix/list/multilist

Graph traversals DFS and BFS

Connected and biconnected components

Minimum-cost spanning tree

Shortest path

non-negative/general weights

single-source-all destinations and all pairs

transitive closure

Topological order/sorting

AOV and AOE networks

Documents

Data Structure - Adjacency