Chapter 3laber/03graphs-Dijkstra.pdf · Chapter 3 Graphs Slides by ... hbo.com sorpranos.com . 9...

1

Chapter 3 Graphs

Slides by Kevin Wayne. Copyright © 2005 Pearson-Addison Wesley. All rights reserved.

3.1 Basic Definitions and Applications

3

Undirected Graphs

Undirected graph. G = (V, E)

V = nodes (non-empty)

E = edges between pairs of nodes.

Captures pairwise relationship between objects.

Graph size parameters: n = |V|, m = |E|.

V = { 1, 2, 3, 4, 5, 6, 7, 8 }

E = { 1-2, 1-3, 2-3, 2-4, 2-5, 3-5, 3-7, 3-8, 4-5, 5-6,7-8}

n = 8

m = 11

4

Undirected Graphs

Undirected graph. G = (V, E)

u and v are adjacent (neighbors) in G iff there is an edge between u and

v in G

The degree d(u) of a vertex u is the number of neighbors of u

1 and 3 are adjacent

2 and 8 are not adjacent

d(3)=5

d(4)=2

5

Undirected Graphs

Important Property: For every graph G, the sum of degrees of G equals

twice the number of edges.

m=11

Sum of degrees =22

6

Undirected Graphs

Loops

Edge whose two endpoints are the same

Parallel edges

Two Edges with the same endpoints

Simple Graph

A simple graph is a graph with neither loops nor parallel edges

Most of the time we’l’l be considering simple graphs

m <= n(n-1)/2 for simple graphs

– Bound is tight for complete graphs

7

Some Graph Applications

transportation

Graph

street intersections

Nodes Edges

highways

communication computers fiber optic cables

World Wide Web web pages hyperlinks

social people relationships

food web species predator-prey

software systems functions function calls

scheduling tasks precedence constraints

circuits gates wires

8

World Wide Web

Web graph.

Node: web page.

Edge: hyperlink from one page to another.

cnn.com

cnnsi.com novell.com netscape.com timewarner.com

hbo.com

sorpranos.com

9

9-11 Terrorist Network

Social network graph.

Node: people.

Edge: relationship between two

people.

Reference: Valdis Krebs, http://www.firstmonday.org/issues/issue7_4/krebs

10

Ecological Food Web

Food web graph.

Node = species.

Edge = from prey to predator.

Reference: http://www.twingroves.district96.k12.il.us/Wetlands/Salamander/SalGraphics/salfoodweb.giff

11

Graph Representation: Adjacency Matrix

Adjacency matrix. n-by-n matrix with Auv = 1 if (u, v) is an edge.

Two representations of each edge.

Space proportional to n2.

Checking if (u, v) is an edge takes (1) time.

Identifying all edges takes (n2) time.

1 2 3 4 5 6 7 8

1 0 1 1 0 0 0 0 0

2 1 0 1 1 1 0 0 0

3 1 1 0 0 1 0 1 1

4 0 1 0 0 1 0 0 0

5 0 1 1 1 0 1 0 0

6 0 0 0 0 1 0 0 0

7 0 0 1 0 0 0 0 1

8 0 0 1 0 0 0 1 0

12

Graph Representation: Adjacency Matrix

Line Graph: n vertices and n-1 edges.

• Adjacency matrix is full of 0’s

Facebook

• 750M vertices

• Assumption: each person has 130 friends in average

550 Petabytes to store approximately 50 Billion edges;

1 2 3 4 5 6 7 8

1 0 1 1 0 0 0 0 0

2 1 0 1 1 1 0 0 0

3 1 1 0 0 1 0 1 1

4 0 1 0 0 1 0 0 0

5 0 1 1 1 0 1 0 0

6 0 0 0 0 1 0 0 0

7 0 0 1 0 0 0 0 1

8 0 0 1 0 0 0 1 0

13

Graph Representation: Adjacency List

Adjacency list. Node indexed array of lists.

Two representations of each edge.

Space proportional to m + n.

Checking if (u, v) is an edge takes O(deg(u)) time.

Identifying all edges takes (m + n) time.

1 2 3

2

3

4 2 5

5

6

7 3 8

8

1 3 4 5

1 2 5 8 7

2 3 4 6

5

degree = number of neighbors of u

3 7

14

Graph Representation: Adjacency List

Line Graph: n vertices and n-1 edges.

• One vector of size n and one list of size n-1

Facebook

• 750M vertices

• Assumption: each person has 130 friends in average

100 Gigabytes to store approximately 50 Billion edges;

14

1 2 3

2

3

4 2 5

5

6

7 3 8

8

1 3 4 5

1 2 5 8 7

2 3 4 6

5

3 7

15

Paths and Connectivity

Def. A path in an undirected graph G = (V, E) is a sequence P of nodes

v1, v2, …, vk-1, vk with the property that each consecutive pair vi, vi+1 is

joined by an edge in E.

Def. A path is simple if all nodes are distinct.

Def. An undirected graph is connected if for every pair of nodes u and

v, there is a path between u and v.

16

Cycles

Def. A cycle is a path v1, v2, …, vk-1, vk in which v1 = vk, k >3, and the

first k-1 nodes are all distinct.

cycle C = 1-2-4-5-3-1

17

Distance

Def. The distance between vertexes s and t in a graph G is the number

of edges of the shortest path connecting s to t in G.

Distance(1,4) =2 Distance(6,3)= 2 Distance(7,8) =1

18

Trees

Def. An undirected graph is a tree if it is connected and does not

contain a cycle.

Theorem. Let G be an undirected graph on n nodes. Any two of the

following statements imply the third.

G is connected.

G does not contain a cycle.

G has n-1 edges.

19

Rooted Trees

Rooted tree. Given a tree T, choose a root node r and orient each edge

away from r.

Importance. Models hierarchical structure.

a tree the same tree, rooted at 1

v

parent of v

child of v

root r

3.2 Graph Traversal

22

Connectivity

s-t connectivity problem. Given two node s and t, is there a path

between s and t?

s-t shortest path problem. Given two node s and t, what is the length

of the shortest path between s and t?

Applications.

Maze traversal.

Kevin Bacon number/Erdos number.

Fewest number of hops in a communication network.

23

Breadth First Search

BFS intuition. Explore outward from s in all possible directions, adding

nodes one "layer" at a time.

BFS algorithm.

L0 = { s }.

L1 = all neighbors of L0.

L2 = all nodes that do not belong to L0 or L1, and that have an edge

to a node in L1.

Li+1 = all nodes that do not belong to an earlier layer, and that have

an edge to a node in Li.

Theorem. For each i, Li consists of all nodes at distance exactly i

from s. There is a path from s to t iff t appears in some layer.

s L1 L2 L n-1

BFS(G)

1 for every vertex s of G not explored yet

2 do Enqueue(S,s)

3 mark vertex s as visited

4 while S is not empty do 5 u ← Dequeue(S);

6 For each v in Adj[u] then

7 if v is unexplored then 8 mark edge (v,u) as tree edge

9 mark vertex v as visited

10 Enqueue(S,v)

Busca em Largura

Notação

• Adj[u] : vertexes adjacent to u in some order

• Dequeue(S): Remove the first element from S • Enqueue(S,v) : Add v to S

1

2

3 4

5

7

6

S

Exemplo

BFS(G)


2 do Enqueue(S,s);






10 Enqueue(S,v)

Adjacency List 1->4,5,2 2-> 4,6,1 3->7 4 -> 5,2,1 5->1,4 6->2 7>-3

Exemplo

1

2

3 4

5

7

6

1 S

BFS(G)


2 do Enqueue(S,s);






10 Enqueue(S,v)

Adjacency List 1->4,5,2 2-> 4,6,1 3->7 4 -> 5,2,1 5->1,4 6->2 7>-3

1

2

3 4

5

7

6

Exemplo

S

BFS(G)


2 do Enqueue(S,s);






10 Enqueue(S,v)

Adjacency List 1->4,5,2 2-> 4,6,1 3->7 4 -> 5,2,1 5->1,4 6->2 7>-3

1

2

3 4

5

7

6

4 S

Exemplo

BFS(G)


2 do Enqueue(S,s);






10 Enqueue(S,v)

Adjacency List 1->4,5,2 2-> 4,6,1 3->7 4 -> 5,2,1 5->1,4 6->2 7>-3

1

2

3 4

5

7

6

4 5 S

Exemplo

BFS(G)


2 do Enqueue(S,s);






10 Enqueue(S,v)

Adjacency List 1->4,5,2 2-> 4,6,1 3->7 4 -> 5,2,1 5->1,4 6->2 7>-3

1

2

3 4

5

7

6

4 5 2 S

Exemplo

BFS(G)


2 do Enqueue(S,s);






10 Enqueue(S,v)

Adjacency List 1->4,5,2 2-> 4,6,1 3->7 4 -> 5,2,1 5->1,4 6->2 7>-3

1

2

3 4

5

7

6

5 2 S

Exemplo

BFS(G)


2 do Enqueue(S,s);






10 Enqueue(S,v)

Adjacency List 1->4,5,2 2-> 4,6,1 3->7 4 -> 5,2,1 5->1,4 6->2 7>-3

Exemplo

1

2

3 4

5

7

6

5 2 S

BFS(G)


2 do Enqueue(S,s);






10 Enqueue(S,v)

Adjacency List 1->4,5,2 2-> 4,6,1 3->7 4 -> 5,2,1 5->1,4 6->2 7>-3

1

2

3 4

5

7

6

5 2 S

Exemplo

BFS(G)


2 do Enqueue(S,s);






10 Enqueue(S,v)

Adjacency List 1->4,5,2 2-> 4,6,1 3->7 4 -> 5,2,1 5->1,4 6->2 7>-3

1

2

3 4

5

7

6

2 S

Exemplo

BFS(G)


2 do Enqueue(S,s);






10 Enqueue(S,v)

Adjacency List 1->4,5,2 2-> 4,6,1 3->7 4 -> 5,2,1 5->1,4 6->2 7>-3

1

2

3 4

5

7

6

Exemplo

S

BFS(G)


2 do Enqueue(S,s);






10 Enqueue(S,v)

Adjacency List 1->4,5,2 2-> 4,6,1 3->7 4 -> 5,2,1 5->1,4 6->2 7>-3

1

2

3 4

5

7

6

6 S

Exemplo

BFS(G)


2 do Enqueue(S,s);






10 Enqueue(S,v)

Adjacency List 1->4,5,2 2-> 4,6,1 3->7 4 -> 5,2,1 5->1,4 6->2 7>-3

1

2

3 4

5

7

6

S

Exemplo

BFS(G)


2 do Enqueue(S,s);






10 Enqueue(S,v)

Adjacency List 1->4,5,2 2-> 4,6,1 3->7 4 -> 5,2,1 5->1,4 6->2 7>-3

1

2

3 4

5

7

6

3

Exemplo

S

3

BFS(G)


2 do Enqueue(S,s);






10 Enqueue(S,v)

Adjacency List 1->4,5,2 2-> 4,6,1 3->7 4 -> 5,2,1 5->1,4 6->2 7>-3

1

2

3 4

5

7

6

Exemplo

S

3

BFS(G)


2 do Enqueue(S,s);






10 Enqueue(S,v)

Adjacency List 1->4,5,2 2-> 4,6,1 3->7 4 -> 5,2,1 5->1,4 6->2 7>-3

1

2

3 4

5

7

6

S

Exemplo

BFS(G)


2 do Enqueue(S,s);






10 Enqueue(S,v)

7

Adjacency List 1->4,5,2 2-> 4,6,1 3->7 4 -> 5,2,1 5->1,4 6->2 7>-3

1

2

3 4

5

7

6

S

Exemplo

BFS(G)


2 do Enqueue(S,s);






10 Enqueue(S,v)

Adjacency List 1->4,5,2 2-> 4,6,1 3->7 4 -> 5,2,1 5->1,4 6->2 7>-3


Definition A BFS tree of G = (V, E), is the tree induced by a BFS

search on G.

• The root of the tree is the startpoint of the BFS

• A node u is a parent of v if v is first visited when the BFS

traverses the neighbors of u

Observation For the same graph there can be different BFS trees.

The BFS tree topology depends on the startpoint of the BFS and the

rule employed to break ties (how the data structure is traversed)

43


Property. Let T be a BFS tree of G = (V, E), and let (x, y) be an edge of

G. Then the level of x and y differ by at most 1.

L0

L1

L2

L3

44

Breadth First Search: Analysis

Análise O(n2):

(i) Cada vértice entra na fila S no máximo uma vez Loop While é

executado no máximo n vezes.

(ii) Cada vértice tem no máximo n-1 vizinhos For é executado no

máximo n vezes

(i) e (ii) implicam em O(n2)

Análise O(m + n)

(i) custo dentre do While O(n)

(ii) Custo dentro do For para vértice u é degree(u). Somando para

todos os vertices temos uV degree(u) = 2m

(i) e (ii) implicam em O(n+m )

each edge (u, v) is counted exactly twice in sum: once in deg(u) and once in deg(v)

46

Connected Component

Definition: Connected set. S is a connected set if and only if

v is reachable from u and u is reachable from v for every u,v in S

Definition: Connected Component. S is a connected component if and

only if

S is a connected set

for every u in V-S, S {u} is not connected

47

Connected Component

Connected components.

Connected component containing node 1 = { 1, 2, 3, 4, 5, 6, 7, 8 }.

48

Flood Fill

Flood fill. Given lime green pixel in an image, change color of entire

blob of neighboring lime pixels to blue.

Node: pixel.

Edge: two neighboring lime pixels.

Blob: connected component of lime pixels.

recolor lime green blob to blue

49

Flood Fill

Flood fill. Given lime green pixel in an image, change color of entire

blob of neighboring lime pixels to blue.

Node: pixel.

Edge: two neighboring lime pixels.

Blob: connected component of lime pixels.

recolor lime green blob to blue

Busca em Profundidade

1

2

3 4

5

7

6

DFS(G)

1 Para todo v em G

2 Se v não visitado então 3 DFS-Visit(G, v)

DFS-Visit(G, v)

1 Marque v como visitado

2 Para todo w em Adj(v)

3 Se w não visitado então 4 Insira aresta (v, w) na árvore

5 DFS-Visit(G, w)

Busca em Profundidade : Exemplo

1

2

3 4

5

7

6

DFS(G)

1 Para todo v em G


DFS-Visit(G, v)




5 DFS-Visit(G, w)


1

2

3 4

5

7

6

DFS(G)

1 Para todo v em G


DFS-Visit(G, v)




5 DFS-Visit(G, w)


1

2

3 4

5

7

6

DFS(G)

1 Para todo v em G


DFS-Visit(G, v)




5 DFS-Visit(G, w)


1

2

3 4

5

7

6

DFS(G)

1 Para todo v em G


DFS-Visit(G, v)




5 DFS-Visit(G, w)


1

2

3 4

5

7

6

DFS(G)

1 Para todo v em G


DFS-Visit(G, v)




5 DFS-Visit(G, w)


1

2

3 4

5

7

6

DFS(G)

1 Para todo v em G


DFS-Visit(G, v)




5 DFS-Visit(G, w)


1

2

3 4

5

7

6

DFS(G)

1 Para todo v em G


DFS-Visit(G, v)




5 DFS-Visit(G, w)


1

2

3 4

5

7

6

DFS(G)

1 Para todo v em G


DFS-Visit(G, v)




5 DFS-Visit(G, w)

60

Depth First Search: Analysis

Uma DFS um grafo G tem complexidade de pior caso O(m + n) já que

cada vértice e aresta são visitados exatamente uma vez

Depth First Search

Definition A DFS tree of G = (V, E), is the tree induced by a DFS

search on G.

• The root of the tree is the startpoint of the DFS

• A node u is a parent of v if v is first visited when the DFS

traverses the neighbors of u

Observation For a given graph there can be different DFS trees. The

DFS tree topology depends on the startpoint of the DFS and the rule

employed to break ties (how the data structure is traversed)

Propriedades da Busca de Profundidade

Teorema: Seja T a árvore produzida por uma DFS em um grafo não direciondo G

e seja vw uma aresta de G. Se v é visitado antes de w então v é ancestral de w em T.

Arestas em preto: árvore DFS Arestas em preto e laranja: arestas do grafo


Nota: Se T é uma árvore produzida por uma DFS em G, existe um caminho entre v

e w em G e v é visitado antes de w então não necessariamente v é ancestral de w em

T.


Nota: Se T é uma árvore produzida por uma DFS em G, existe um caminho entre v

e w em G e v é visitado antes de w então não necessariamente v é ancestral de w em

T.

v w

Encontrando ciclos usando DFS

DFS(G)

1 Para todo v em G


DFS-Visit(G, v)




5 DFS-Visit(G, w)

vpai(w)

6 Senao

7 Se w<>pai(v)

8 Return Existe Ciclo

9 Fim Se

10 Fim Para

Encontrando ciclos usando DFS

Ida. Se w já foi visitado, é vizinho de v e não é pai de v

Note que v não pode ser pai de w porque w já foi visitado.

Temos dois casos:

a) w é um descendente de v. Isto não pode acontecer porque a busca

já teria sido interrompida por detectar um ciclo ao tenta visitar v a

partir de w.

b) w é um ancestral de v. Então o caminho de w até v na árvore mais a

aresta vw formam um ciclo

Volta. Existe um ciclo C no grafo. Então ...

Seja v o último vértice de C visitado pela DFS. Então os dois vizinhos de

v em C são ancestrais de v na árvore DFS e pelo menos um deles não é

pai de v. Portanto ao visitar tal vizinho detectamos um ciclo.

66

67

DFS numbering

DFS(G)

0. time 1

1 Para todo v em G


DFS-Visit(G, v)


1.5 pre(v) time; time++



5 DFS-Visit(G, w)

6 Fim Para

7 pos(v) time; time++

1,8

2,7

3,4

5,6

68

Articulation Points

• A node of the graph is an articulation point in an undirected graph if its removal disconnect the graph

• How to find the articulation points of G?

• How to find the articulation points of G quickly?

Articulation Point

69

Articulation Points

• Brute Force Algorithm

For every vertex v

If G-v is not connected Display ‘v is an articulation point’ Time complexity O(n(n+m))

70

Articulation Points

• Clever algorithm (approach) Consider a node v in the DFS tree for G Case 1) v is the root • v has more than one child v is an articulation

point • v has only one child v is not an articulation

point Case 2) v is a leaf v is not an articulation point Case 3) v is an internal node

If one of the children of v, say u, is such that DFS(u) only finds edges (x,y) with pre(y)>=pre(v) then v is an articulation point.

71

Articulation Points

• Clever algorithm (justification) Consider a node v in the DFS tree T for G Case 1) v is the root • v has more than one child v is an

articulation point It is not possible to have na edge between nodes in trees rooted different chidren of v

• v has only one child v is not an articulation

point T-{v} is a connected graph

72

Articulation Points

• Clever algorithm (justification) Case 2) v is a leaf v is not an articulation point because T-{v} is a connected graph

Case 3) v is an internal node If one of the children of v, say u, is such that DFS(u) only finds edges (x,y) with pre(y)>=pre(v) then v is an articulation point. If x lies in subtree rooted at u and (x,y) is an edge then y is a descendant of v. Thus, removing v disconnects T(u) from the parent of v

73

Articulation Points

r é ponto de articulacão. Propriedade da DFS garante que não pode haver arestas entre árvores vermelhas

rRrr

. . ,

74

Articulation Points

r não é ponto de articulacão

rRrr

75

Articulation Points

Nós de alguma subárvore de v não tem arestas para ancestrais de v v é ponto de articulacãó

vRrr

rRrr

76

Articulation Points

Nós de todas subárvore de v tem arestas para ancestrais de v v não é ponto de articulacão

vRrr

rRrr

77

Menor: vetor global com uma entrada para cada vértice

DFS-Visit(G, v)

Marque v como visitado

pre(v) time; time++

Menor(v)pre(v) % calcula o menor pre encontrado na DFS(v)

Para todo w em Adj(v)

Se w não visitado então Insira w na lista dos filhos de v.

DFS-Visit(G, w)

Menor(v)Min(Menor(v),Menor(w))

Fim Para

Se ( v é raiz e tem mais de um filho) ou (v não é raíz e

menor(w)>=pre(v) para algum filho w de v) entao

“ v é ponto de articulaçao”

Articulation Points

Complexidade O(m+n)

78

Articulation Points

Pontos de Articulação

Executar algoritmo para esta figura

79

Conclusion

• Simple linear O(m+n) algorithm to compute the articulation points of G

• A problem for which the DFS is the suitable choice to traverse the graph

3.4 Testing Bipartiteness

81

Bipartite Graphs

Def. An undirected graph G = (V, E) is bipartite if the nodes can be

colored red or blue such that every edge has one red and one blue end.

Applications.

Stable marriage: men = red, women = blue.

Scheduling: machines = red, jobs = blue.

a bipartite graph

82

Testing Bipartiteness

Testing bipartiteness. Given a graph G, is it bipartite?

Many graph problems become:

– easier if the underlying graph is bipartite (matching)

– tractable if the underlying graph is bipartite (independent set)

Before attempting to design an algorithm, we need to understand

structure of bipartite graphs.

v1

v2 v3

v6 v5 v4

v7

v2

v4

v5

v7

v1

v3

v6

a bipartite graph G another drawing of G

83

An Obstruction to Bipartiteness

Lemma. If a graph G is bipartite, it cannot contain an odd length cycle.

Pf. Not possible to 2-color the odd cycle, let alone G.

bipartite (2-colorable)

not bipartite (not 2-colorable)

84

Bipartite Graphs

Lemma. Let G be a connected graph, and let L0, …, Lk be the layers

produced by BFS starting at node s. Exactly one of the following holds.

(i) No edge of G joins two nodes of the same layer, and G is bipartite.

(ii) An edge of G joins two nodes of the same layer, and G contains an

odd-length cycle (and hence is not bipartite).

Case (i)

L1 L2 L3

Case (ii)

L1 L2 L3

85

Bipartite Graphs






Pf. (i)

Suppose no edge joins two nodes in the same layer.

By previous lemma, this implies all edges join nodes on consecutive

levels.

Bipartition: red = nodes on odd levels, blue = nodes on even levels.

Case (i)

L1 L2 L3

86

Bipartite Graphs






Pf. (ii)

Suppose (x, y) is an edge with x, y in same level Lj.

Let z = lca(x, y) = lowest common ancestor.

Let Li be level containing z.

Consider cycle that takes edge from x to y,

then path from y to z, then path from z to x.

Its length is 1 + (j-i) + (j-i), which is odd. ▪

z = lca(x, y)

(x, y) path from y to z

path from z to x

87

Obstruction to Bipartiteness

Corollary. A graph G is bipartite iff it contains no odd length cycle.

5-cycle C

bipartite (2-colorable)

not bipartite (not 2-colorable)

88

Exercícios de Implementação

1. Modifique o código da BFS para que esta preencha um vetor d com n

entradas aonde d(u) é a distância da origem s até o vértice u.

2. Modifique o algoritmo de busca em profundidade para que ele atribua

números inteiros aos vértices do grafo de modo que

(i) Vértices de uma mesma componente recebam o mesmo número

(ii) Vértices de componentes diferentes recebam números diferentes

3. Modifique o código da BFS para que ela identifique se um grafo é

bipartido ou não.

3.5 Connectivity in Directed Graphs

90

Directed Graphs

Directed graph. G = (V, E)

Edge (u, v) goes from node u to node v.

Ex. Web graph - hyperlink points from one web page to another.

Directedness of graph is crucial.

Modern web search engines exploit hyperlink structure to rank web

pages by importance.

The in-degree d-(u) of a vertex u is the number of edges that arrive at u

The out-degree d+(u) of a vertex u is the number of edges that leave u

Important property

sum of indegrees = sum of outdegre = m

91

Directed Graphs

d-(u)=2

d+(u)=1 u

Representation via Adjacency List

1 2 3

2

3

1 3

1 2

92 92

1

2 3

1 2 3

2

3

1 3

1

2 3

Undirected Graph

Directed Graph

93

Graph Search

Directed reachability. Given a node s, find all nodes reachable from s.

Directed s-t shortest path problem. Given two node s and t, what is

the length of the shortest path between s and t?

Graph search. BFS extends naturally to directed graphs.

Web crawler. Start from web page s. Find all web pages linked from s,

either directly or indirectly.

94

DFS numbering

Useful information pre(v): “time” when a node is visited in a DFS pos(v): “time” when DFS(v) finishes

1/16

d

b

f

e

a

c

g

h

2/15 3/14

4/13

7/12 8/11

9/10 5/6

95

DFS numbering

DFS(G)

0. time 1

1 Para todo v em G


DFS-Visit(G, v)





5 DFS-Visit(G, w)

6 Fim Para


96

DFS numbering

Importanto property • If u and v are nodes in G and u is visited before v

in a DFS then one of the conditions hols: 1. pre(u) < pre(v) < pos(v) < pos(u). In this case, u

is an ancestor of v in the DFS tree

2. pre(u) < pos(u) < pre(v) < pos(v). In this case, u is not an ancestor of v nor v is an ancestor of u in the DFS tree

Note: it is not possible to have pre(u) < pre(v) < pos(u) < pos(v).

How to find a directed cycle

1

2 3

1 2 3

2

3

3

• When 1 tries to visit 3, 3 has already been visited and 3 is not a parent of 1.

• This does not imply on the existence of a cycle

98

DFS numbering – Finding a Directed Cycle

DFS(G)

0. time 1

1 Para todo v em G


DFS-Visit(G, v)





5 DFS-Visit(G, w)

6 Senao

7 Se pos(w) indefinido

8 Existe ciclo

9 Fim Para


99

DFS numbering – Finding a Directed Cycle

• Se w já foi visitado e o pos dele é indefinido, enão DFS(w) ainda não terminou de

executar. Logo, w é ancestral de v na árvore DFS

DFS(G)

0. time 1

1 Para todo v em G


DFS-Visit(G, v)





5 DFS-Visit(G, w)

6 Senao

7 Se post(w) indefinido

8 Existe ciclo

9 Fim Para

10 Post(v) time; time++

3.6 DAGs and Topological Ordering

101

Directed Acyclic Graphs

Def. A DAG is a directed graph that contains no directed cycles.

Ex. Precedence constraints: edge (vi, vj) means vi must precede vj.

a DAG

v2 v3

v6 v5 v4

v7 v1

Not a DAG

v2 v3

v6 v5 v4

v7 v1

102

Precedence Constraints

Precedence constraints. Edge (vi, vj) means task vi must occur before vj.

Applications.

Course prerequisite graph: course vi must be taken before vj.

Compilation: module vi must be compiled before vj. Pipeline of

computing jobs: output of job vi needed to determine input of job vj.

What is a feasible order to compile the jobs?

103


Def. A topological order of a directed graph G = (V, E) is an ordering

of its nodes as v1, v2, …, vn so that for every edge (vi, vj) we have i < j.

G a topological ordering for G

v2 v3

v6 v5 v4

v7 v1

v1 v2 v3 v4 v5 v6 v7

104


v2 v3

v6 v5

No topological order

v2 v3

v6 v5

v4

Topological orders: v6->v2->v3->v5->v4 v6->v2->v5->v3->v4

Def. A topological order of a directed graph G = (V, E) is an ordering of

its nodes as v1, v2, …, vn so that for every edge (vi, vj) we have i < j.

105


What is the relation between DAG’s and topological orderings?

106


Lemma. If G has a topological order, then G is a DAG.

Pf. (by contradiction)

Suppose that G has a topological order v1, …, vn and that G also has a

directed cycle C. Let's see what happens.

Let vi be the lowest-indexed node in C, and let vj be the node just

before vi; thus (vj, vi) is an edge.

By our choice of i, we have i < j.

On the other hand, since (vj, vi) is an edge and v1, …, vn is a

topological order, we must have j < i, a contradiction. ▪

v1 vi vj vn

the supposed topological order: v1, …, vn

the directed cycle C

107


Lemma. If G has a topological order, then G is a DAG.

Q. Does every DAG have a topological ordering?

Q. If so, how do we compute one?

108


Lemma. If G is a DAG, then G has a node with no incoming edges.

Pf. (by contradiction)

Suppose that G is a DAG and every node has at least one incoming

edge. Let's see what happens.

Pick any node v, and begin following edges backward from v. Since v

has at least one incoming edge (u, v) we can walk backward to u.

Then, since u has at least one incoming edge (x, u), we can walk

backward to x.

Repeat until we visit a node, say w, twice.

Let C denote the sequence of nodes encountered between

successive visits to w. C is a cycle. ▪

w x u v

109


Lemma. If G is a DAG, then G has a topological ordering.

Pf. (by induction on n)

Base case: true if n = 1.

Given DAG on n > 1 nodes, find a node v with no incoming edges.

G - { v } is a DAG, since deleting v cannot create cycles.

By inductive hypothesis, G - { v } has a topological ordering.

Place v first in topological ordering; then append nodes of G - { v }

in topological order. This is valid since v has no incoming edges. ▪

DAG

v

03demo-graphs.ppt#1. Busca em Largura

110

v1

Topological Ordering Algorithm: Example

Topological order:

v2 v3

v6 v5 v4

v7 v1

111

v2


Topological order: v1

v2 v3

v6 v5 v4

v7

112

v3


Topological order: v1, v2

v3

v6 v5 v4

v7

113

v4


Topological order: v1, v2, v3

v6 v5 v4

v7

114

v5


Topological order: v1, v2, v3, v4

v6 v5

v7

115

v6


Topological order: v1, v2, v3, v4, v5

v6

v7

116

v7


Topological order: v1, v2, v3, v4, v5, v6

v7

117


Topological order: v1, v2, v3, v4, v5, v6, v7.

v2 v3

v6 v5 v4

v7 v1

v1 v2 v3 v4 v5 v6 v7

118

Topological Sorting Algorithm: Running Time

Algorithm 1

Scan the graph to fill a vector count that stores, for each node

v, the number of remaining edges that are incident in v

i0

While i< n

v node with minimum value in count

i++

If v has value larger than 0

Return G is not a DAG

End If

Add v to the topological order

Remove v from count

Update the vector count for the nodes adjacent to v

End

119


Analysis : count stored as a vector

O(n+m) to compute the count

The loop executes at most n times

O(n) to find the node v with minimum degree

O(1) to remove v

O(d(u)) to update the neighbors of v

O( n2 + m)

Analysis : count stored as a heap

O(n+m) to compute the vector count

The loop executes at most n times

O(1) to find the node v with minimum degree

O(log n) to remove v

O( d+(u) log n) to update the neighbors of v

O( n log n + m log n)

120


Theorem. Algorithm finds a topological order in O(m + n) time.

Pf.

Maintain the following information:

– count[w] = remaining number of incoming edges

– S = set of remaining nodes with no incoming edges

Initialization: O(m + n) via single scan through graph.

Update: to delete v

– remove v from S

– decrement count[w] for all edges from v to w, and add w to S if

count[w] hits 0

– this is O(1) per edge ▪

Strong Connectivity

Def. Node u and v are mutually reachable if there is a path from u to v

and also a path from v to u.

Def. A graph is strongly connected if every pair of nodes is mutually

reachable.

How to decide wether a given graph is strongly connected?

Strong Connectivity

Algorithm 1

SC true

For all u,v in V

If DFS(u) does not reach v

SC False

End If

End

Return SC

Analysis O( n2 (m+n))

Strong Connectivity

Obs. When we execute a DFS (BFS) starting at u we find all nodes

reachable from u

Algorithm 2

SC true

For all u in V

If DFS(u) does not reach |V| nodes

SC False

End If

End

Return SC

Analysis O( n (m+n) )

124

Strong Connectivity

Def. Node u and v are mutually reachable if there is a path from u to v

and also a path from v to u.

Def. A graph is strongly connected if every pair of nodes is mutually

reachable.

Lemma. Let s be any node. G is strongly connected iff every node is

reachable from s, and s is reachable from every node.

Pf. Follows from definition.

Pf. Path from u to v: concatenate u-s path with s-v path.

Path from v to u: concatenate v-s path with s-u path. ▪

s

v

u

ok if paths overlap

125

Reverse Graph

Def. The reverse graph of a graph G=(V,E) is a graph GR=(V,E’) where

E’ ={ (v,u) | (u,v) belongs to E}

Observation: The reverse graph of a graph G can be constructed in

O(m+n) time

126

Example: reverse graph G R

d

b

f

e

a

c

g

h

d

b

f

e

a

c

g

h

G

G R

127

Strong Connectivity: Algorithm

Theorem. Can determine if G is strongly connected in O(m + n) time.

Pf.

Pick any node s.

Run BFS from s in G.

Run BFS from s in Grev.

Return true iff all nodes reached in both BFS executions.

Correctness follows immediately from previous lemma. ▪

reverse orientation of every edge in G

strongly connected not strongly connected

128

Strongly connected components

129


Definition: A subgraph induced by a set of vertices C is a strongly connected component C of a directed graph G = (V,E) if the following conditions hold:

(a) For any two vertices u and v in C , there is a path from u to v and from v to u and for every (b) For every C’, with C C’, the condition (a) does not hold. • Equivalence classes of the binary path(u,v) relation,

denoted by u ~ v.

130


Applications: networking, communications. Problem: compute the strongly connected components of G in linear time Θ(|V|+|E|).

131

Example: strongly connected components

d

b

f

e

a

c

g

h

132

Example: strongly connected components

d

b

f

e

a

c

g

h

133

Strongly connected components graph

Definition: the SCC graph G~ = (V~,E~) of the graph G = (V,E) is as follows:

V~ = {C1, …, Ck}. Each SCC is a vertex.

E~ = {(Ci,Cj)| i≠j and there is (x,y)E, where xCi and yCj}.

G~ is a directed acyclic graph (no directed cycles)!

Definition: the reverse graph G R = (V,ER) of the graph G = (V,E) is G with

its edge directions reversed: E R = {(u,v)| (v,u)E}.

Observation 1. The strongly connected components are the same in G and

G R

134

Example: reverse graph G R

d

b

f

e

a

c

g

h

d

b

f

e

a

c

g

h

G

G R

135

Example: SCC graph

C1

C2

C4

C3

d

b

f

e

a

c

g

h

136

SCC algorithm: Approach

Source node: node with in-degree 0 Sink node: node with out-degree 0 Recall that every DAG has both a source and a sink (see DAG slides)

137

SCC algorithm: Approach

H G While H is not empty v node that lies in a sink node of H~ (*) C SCC retuned by DFS(H,v) H H - C End If we manage to execute (*) we are done

141

DFS numbering

Useful information pre(v): “time” when a node is visited in a DFS post(v): “time” when DFS(v) finishes

1/16

d

b

f

e

a

c

g

h

2/15 3/14

4/13

7/12 8/11

9/10 5/6

142

DFS numbering

DFS(G)

0. time 1

1 Para todo v em G


DFS-Visit(G, v)





5 DFS-Visit(G, w)

6 Fim Para

7 Post(v) time; time++

143

DFS numbering

Property If C an D are strongly connected components and there is an edge from a node in C to a node in D then, for every DFS, the highest pos() in C is bigger than the highest pos() in D

Proof. Let c be the first node visited in C and d be the

first node visited in D during a DFS Case 1) c is visited before d. DFS(c) visit all nodes in D (they are reachable

from c due to the existing edge) Thus, post(c) > post(x) for every x in D

144

DFS numbering

Property If C an D are strongly connected components and there is an edge from a node in C to a node in D then, for every DFS, the highest pos() in C is bigger than the highest pos() in D

Proof. Let c be the first node visited in C and d be the

first node visited in D during a DFS Case 2) d is visited before c. DFS(d) visit all nodes in D because all of them are

reachable from d DFS(d) does not visit nodes from C since they are

not reachable from d. Thus , post(x) <post(y) for every pair of nodes x,y,

with x in D and y in C

147

DFS numbering

Observation 2. The node of G with highest post

number lies in a source node in G~

148

SCC algorithm

Idea: compute the SCC graph G~ = (V~,E~) with two DFS, one for G and one for its reverse GR, visiting the vertices in reverse order.

SCC(G) 1. DFS(G) to compute pos[v], ∀vV 2. Compute GR 3. DFS(G R) in the order of decreasing pos[v] 4. Output the vertices of each tree in the DFS forest

as a separate SCC.

150

Example: computing SCC

1/6

d

b

f

e

a

c

g

h

3/4

10/15

11/12 9/16

7/8 13/14

2/5

151


d

b

f

e

a

c

g

h 8

1/6

d

b

f

e

a

c

g

h

3/4

10/15

11/12 9/16

7/8 13/14

2/5

16

152


d

b

f

e

a

c

g

h 8

1/6

d

b

f

e

a

c

g

h

3/4

10/15

11/12 9/16

7/8 13/14

2/5

16

C1

C4

C2

C3 1

2

3 4

153

Conclusion

• Simple linear O(m+n) algorithm to compute the SCC of G

• Problem for which the DFS is the suitable choice to traverse the graph

154

Modelagem de problemas com Grafos

Problema

• Seja um grafo G=(V,E) com n vértices representando a planta de um

edifício. Inicialmente temos dois robos localizados em dois vértices

de V, a e b, que devem alcançar os vértices c e d de V.

• No passo i+1 um dos dois robos deve caminhar para um vértice

adjacente ao vértice que ele se encontra no momento i. Exiba um

algoritmo polinomial para resolver o seguinte problema:

• Entrada: Grafo G=(V,E) , quatro vértices: a,b,c e d e um inteiro r.

• Saída: SIM se é possível os robos partirem dos vértices a e b e

chegarem em c e d, respectivamente, sem que em nenhum momento

eles estejam a distância menor do que r. NÃO, caso contrário.

155

Modelando com Grafos

Solução

Seja H=(V ’,E ’ ) um grafo representando as configurações possíveis

(posições dos robos) do problema. Cada nó de H corresponde a um par

ordenado de vértices de V cuja distância é menor ou igual a r. Logo

existem no máximo |V|2 vértices em V ’.

Um par de nós u e v de H tem uma aresta se e somente em um passo é

possível alcançar a configuração v a partir da configuração u. Mais

formalmente, se uv é uma aresta de E’, com u=(u1,u2) e v=(v1,v2), então

uma das alternativas é válida

(i) u1=v1 e (u2,v2) pertence a E

(ii) u2=v2 e (u1,v1) pertence a E

O problema, portanto, consiste em decidir se existe um camìnho entre o

nó x=(a,b) e o nó y=(c,d) em H.

156


Solução

Para construir o grafo H basta realizar n BFS’s no grafo G, cada uma

delas partindo de um vértice diferente. Ao realizar uma BFS a partir de

um nó s obtemos o conjunto de todos os vértices que estão a distância

maior ou igual a r de s. A obtenção do conjunto V’ tem custo O(n(m+n))

e a do conjunto de arestas E’ tem custo O(n3) .

Decidir se existe um camìnho entre o nó x=(a,b) e o nó y=(c,d) em H tem

complexidade O(|V’|+|E’|). Como |V’| tem O(n2) vértices e |E’| tem

O(n3) arestas, o algoritmo executa em O(n3) . Note que |E’| é O(n3)

porque cada vértice de H tem no máximo 2(n-1) vizinhos

• Análise mais cuidadosa mostra que | E ‘ | tem O(mn) arestas

157


Exercício. Mostrar que o conjuto E ‘ tem O(mn) arestas e analisar o

impacto disso para o algoritmo

4.4 Shortest Paths in a Graph

shortest path from Princeton CS department to Einstein's house

159

Shortest Path Problem

Shortest path network.

Directed graph G = (V, E).

Source s, destination t.

Length ce = length of edge e. (non-negative numbers)

Shortest path problem: find shortest directed path from s to t.

Cost of path s-2-3-5-t = 9 + 23 + 2 + 16 = 50.

s

3

t

2

6

7

4

5

23

18

2

9

14

15 5

30

20

44

16

11

6

19

6

cost of path = sum of edge costs in path

160

Dijkstra's Algorithm

Approach

Find the node closest to the s, then the second closest, then the

third closest, and so on …

Key observation:

– the shortest path from s to the k-th closest node can be

decomposed as the shortest path from s to the i-th closest node

(for some i<k) and an edge from the i-th closest node to the k-th

closest node.

s

v

u

d(u)

S

ce

161

Dijkstra's Shortest Path Algorithm

s

3

t

2

6

7

4

5

24

18

2

9

14

15 5

30

20

44

16

11

6

19

6

15

9

14

0


s

3

t

2

6

7

4

5

24

18

2

9

14

15 5

30

20

44

16

11

6

19

6

15

9

14

0

32

163


s

3

t

2

6

7

4

5

18

2

9

14

15 5

30

20

44

16

11

6

19

6

15

9

14

0

34

32

164

Dijkstra's Algorithm

Dijkstra's algorithm.

Maintain a set of explored nodes S for which we have determined

the shortest path distance d(u) from s to u.

Initialize S = { s }, d(s) = 0.

Repeatedly choose unexplored node v which minimizes

add v to S, and set d(v) = (v).

Complexity (Naïve Implementation)

n loops, one for each node

(n+m) to find the the node with minimum

Thus, complexity is O(n(n+m)) time

,)(min)(:),(

eSuvue

cudv

shortest path to some u in explored part, followed by a single edge (u, v)

166

Dijkstra's Algorithm: Implementation

For each unexplored node, explicitly maintain

Next node to explore = node with minimum (v).

When exploring v, for each incident edge e = (v, w), update

Efficient implementation. Maintain a priority queue of unexplored

nodes, prioritized by (v).’

† Individual ops are amortized bounds

PQ Operation

Insert

ExtractMin

ChangeKey

Binary heap

log n

log n

log n

Array

n

n

1

IsEmpty 1 1

Priority Queue

Total m log n n2

Dijkstra

n

n

m

n

.)(min)( :),(

eSuvue

cudv

.})( ),( {min )( ecvww

04demo-dijkstra.ppt#1. Dijkstra's Shortest Path Algorithm

168

Dijkstra's Algorithm: Proof of Correctness

Invariant. For each node u S, d(u) is the length of the shortest s-u path.

Pf. (by induction on |S|)

Base case: |S| = 1 is trivial.

Inductive hypothesis: Assume true for |S| = k 1.

Let v be next node added to S, and let u-v be the chosen edge.

The shortest s-u path plus (u, v) is an s-v path of length (v).

Consider any s-v path P. We'll see that it's no shorter than (v).

Let x-y be the first edge in P that leaves S,

and let P' be the subpath to x.

P is already too long as soon as it leaves S.

c (P) c (P') + c (x,y) d(x) + c(x, y) (y) (v)

nonnegative weights

inductive hypothesis

defn of (y) Dijkstra chose v instead of y

S

s

y

v

x

P

u

P'

Date post:	09-Sep-2018
Category:	Documents
Upload:	trinhtuong
View:	221 times
Download:	0 times

Chapter 3laber/03graphs-Dijkstra.pdf · Chapter 3 Graphs Slides by ... hbo.com sorpranos.com . 9...

Documents