Lecture8_9Graphs2

Graphs

Lê Sỹ Vinh

Computer science department

Email: [email protected]

2

Graphs• A graph is a pair (V, E), where

– V is a set of nodes, called vertices– E is a collection of pairs of vertices, called edges– Vertices and edges are positions and store elements

• Example:– A vertex represents an airport and stores the three-letter airport code– An edge represents a flight route between two airports and stores the

mileage of the route

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Edge Types• Directed edge

– ordered pair of vertices (u,v)– first vertex u is the origin– second vertex v is the destination– e.g., a flight

• Undirected edge– unordered pair of vertices (u,v)– e.g., a flight route

• Directed graph– all the edges are directed– e.g., route network

• Undirected graph– all the edges are undirected– e.g., flight network

ORD PVD flight AA 1206

ORD PVD849miles

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Breadth-First Search

• Breadth-first search (BFS) is a general technique for traversing a graph

• A BFS traversal of a graph G – Visits all the vertices and

edges of G– Determines whether G is

connected– Computes the connected

components of G– Computes a spanning forest

of G

• BFS on a graph with n vertices and m edges takes O(n + m ) time

• BFS can be further extended to solve other graph problems– Find and report a path with

the minimum number of edges between two given vertices

– Find a simple cycle, if there is one

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Depth-First Search

• Depth-first search (DFS) is a general technique for traversing a graph

• A DFS traversal of a graph G – Visits all the vertices and

edges of G– Determines whether G is

connected– Computes the connected

components of G– Computes a spanning forest

of G

• DFS on a graph with n vertices and m edges takes O(n + m ) time

• DFS can be further extended to solve other graph problems– Find and report a path

between two given vertices– Find a cycle in the graph

• Depth-first search is to graphs what Euler tour is to binary trees

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write c.s. program

play

Topological Sorting

• Number vertices, so that (u,v) in E implies u < vwake up

eat

nap

study computer sci.

more c.s.

work out

sleep

dream about graphs

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• Running time: O(n + m). How…?

Algorithm for Topological Sorting

Method TopologicalSort(G) H G // Temporary copy of G n G.numVertices() while H is not empty do

Let v be a vertex with no outgoing edgesLabel v nn n - 1Remove v from H

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Topological Sorting Algorithm using DFS

• Simulate the algorithm by using depth-first search

• O(n+m) time.

Algorithm topologicalDFS(G, v)Input graph G and a start

vertex v of G Output labeling of the vertices

of G in the connected

component of v setLabel(v, VISITED)

for all e G.incidentEdges(v)if getLabel(e) =

UNEXPLOREDw opposite(v,e)if getLabel(w) =

UNEXPLORED

setLabel(e, DISCOVERY)

topologicalDFS(G, w)Label v with topological number n n n - 1

Algorithm topologicalDFS(G)Input dag GOutput topological ordering

of G n G.numVertices()

for all u G.vertices() setLabel(u,

UNEXPLORED)for all e G.edges()

setLabel(e, UNEXPLORED)for all v G.vertices()

if getLabel(v) = UNEXPLORED

topologicalDFS(G, v)

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Topological Sorting Example

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Topological Sorting Example

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Topological Sorting Example

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Topological Sorting Example

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Topological Sorting Example

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Topological Sorting Example

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Topological Sorting Example

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Topological Sorting Example

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Topological Sorting Example

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Topological Sorting Example

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Shortest Paths

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Weighted Graphs

• In a weighted graph, each edge has an associated numerical value, called the weight of the edge

• Edge weights may represent, distances, costs, etc.• Example:

– In a flight route graph, the weight of an edge represents the distance in miles between the endpoint airports

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Shortest Paths• Given a weighted graph and two vertices u and v, we want to find a

path of minimum total weight between u and v.– Length of a path is the sum of the weights of its edges.

• Example:– Shortest path between Providence and Honolulu

• Applications– Internet packet routing – Flight reservations– Driving directions

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Dijkstra Algorithm

Given G=(V, E), where weight (u,v) > 0 is the weight of edge (u, v) E. Find ∈the shortest path from s to e.

General idea

- d[v]: the shortest distance from s to v.

- pre[v] = u: the previous vertex of v on the shortest path from s to v.

Relaxation (s, u, v):

If d [u] + weight (u, v) < d [v]

d [v] = d [u] + weight (u, v)

pre[v] = u

s → … → u → v

d(z) = 75

d(u) = 5010

zsu

d(z) = 60

d(u) = 5010

zsu

e

e

Dijkstra algorithm

Known = {the set of vertices which the shortest paths are known}

Unknown = {the set of vertices which the shortest paths are unknown}

Repeat

Find u ∈ unknown with the smallest d[u] if u = e then

done

else {

known = known {u}

unknown = unknown – {u}

for each edge (u, v):

relaxation (s, u, v)

}

Dijkstra Algorithm {Known = {s};Unknown = {E} – {s}for v ∈ unknown {

d[v] = weight [s, v];pre[v] = s;

}u = s; // at the first stepwhile (u != e ){

Select u such that d[u] = min {d[x] | x unknown}∈Known = Known {u}Unknown = Unknown – {u}for v ∈ Unknown

if d[u] + weight [u, v] < d[v] {//relax (s, u, v)d[v] = d[u] + weight[u,v];pre[v] = u;

}}path = {e}while (e != s) {

e = pre[e]path = {e} path

}}

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Example

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Example (cont.)

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Why It Doesn’t Work for Negative-Weight Edges

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C’s true distance is 1, but it is already in the cloud with d(C)=5!

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Bellman-Ford Algorithm• Works even with negative-

weight edges• Must assume directed edges

(for otherwise we would have negative-weight cycles)

• Iteration i finds all shortest paths that use i edges.

• Running time: O(nm).• Can be extended to detect a

negative-weight cycle if it exists – How?

Algorithm BellmanFord(G, s)for all v G.vertices()

if v = ssetDistance(v, 0)

else setDistance(v, )

for i 1 to n-1 dofor each e G.edges()

{ relax edge e }u G.origin(e)z G.opposite(u,e)r getDistance(u) +

weight(e)if r < getDistance(z)

setDistance(z,r)

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Nodes are labeled with their d(v) values

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Minimum Spanning Trees

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Minimum Spanning TreesSpanning subgraph

– Subgraph of a graph G containing all the vertices of G

Spanning tree– Spanning subgraph that is itself a

(free) tree

Minimum spanning tree (MST)– Spanning tree of a weighted

graph with minimum total edge weight

• Applications– Communications networks– Transportation networks

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Cycle Property

Cycle Property:– Let T be a minimum

spanning tree of a weighted graph G

– Let e be an edge of G that is not in T and let C be the cycle formed by e with T

– For every edge f of C, weight(f) weight(e)

Proof:– By contradiction– If weight(f) > weight(e) we

can get a spanning tree of smaller weight by replacing e with f

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Partition PropertyPartition Property:

– Consider a partition of the vertices of G into subsets U and V

– Let e be an edge of minimum weight across the partition

– There is a minimum spanning tree of G containing edge e

Proof:– Let T be an MST of G– If T does not contain e, consider the cycle C

formed by e with T and let f be an edge of C across the partition

– By the cycle property,weight(f) weight(e)

– Thus, weight(f) = weight(e)– We obtain another MST by replacing f with e

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Kruskal’s AlgorithmA priority queue stores the edges outside the cloud Key: weight Element: edge

At the end of the algorithm We are left with one cloud

that encompasses the MST A tree T which is our MST

Algorithm KruskalMST(G)for each vertex V in G do

define a Cloud(v) of {v}

let Q be a priority queue.Insert all edges into Q using their weights as the keyT while T has fewer than n-1 edges

do edge e = T.removeMin()

Let u, v be the endpoints of e

if Cloud(v) Cloud(u) then

Add edge e to TMerge Cloud(v)

and Cloud(u)return T

Phạm Bảo Sơn - DSA 35

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Example

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Example

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Example

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MIA

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SFO BWI

PVD

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Example

JFK

BOS

MIA

ORD

LAXDFW

SFO BWI

PVD

8672704

187

1258

849

144740

1391

184

946

1090

1121

2342

1846 621

802

1464

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Dec 2008

Phạm Bảo Sơn - DSA 39

Example

JFK

BOS

MIA

ORD

LAXDFW

SFO BWI

PVD

8672704

187

1258

849

144740

1391

184

946

1090

1121

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1846 621

802

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Dec 2008

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Example

JFK

BOS

MIA

ORD

LAXDFW

SFO BWI

PVD

8672704

187

1258

849

144740

1391

184

946

1090

1121

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1846 621

802

1464

1235

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Dec 2008

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Example

JFK

BOS

MIA

ORD

LAXDFW

SFO BWI

PVD

8672704

187

1258

849

144740

1391

184

946

1090

1121

2342

1846 621

802

1464

1235

337

Dec 2008

Phạm Bảo Sơn - DSA 42

Example

JFK

BOS

MIA

ORD

LAXDFW

SFO BWI

PVD

8672704

187

1258

849

144740

1391

184

946

1090

1121

2342

1846 621

802

1464

1235

337

Dec 2008

Phạm Bảo Sơn - DSA 43

Example

JFK

BOS

MIA

ORD

LAXDFW

SFO BWI

PVD

8672704

187

1258

849

144740

1391

184

946

1090

1121

2342

1846 621

802

1464

1235

337

Dec 2008

Phạm Bảo Sơn - DSA 44

Example

JFK

BOS

MIA

ORD

LAXDFW

SFO BWI

PVD

8672704

187

1258

849

144740

1391

184

946

1090

1121

2342

1846 621

802

1464

1235

337

Dec 2008

Phạm Bảo Sơn - DSA 45

Example

JFK

BOS

MIA

ORD

LAXDFW

SFO BWI

PVD

8672704

187

1258

849

144740

1391

184

946

1090

1121

2342

1846 621

802

1464

1235

337

Dec 2008

Phạm Bảo Sơn - DSA 46

Example

JFK

BOS

MIA

ORD

LAXDFW

SFO BWI

PVD

8672704

187

1258

849

144740

1391

184

946

1090

1121

2342

1846 621

802

1464

1235

337

Dec 2008