Graphs

Slides from HKUST

COMP171

Fall 2006

Graphs

Extremely useful tool in modeling problems Consist of:

Vertices Edges

DE

AC

FB

Vertex

Edge

Vertices can beconsidered “sites”or locations.

Edges representconnections.

Application 1

Air flight system

• Each vertex represents a city• Each edge represents a direct flight between two cities• A query on direct flights = a query on whether an edge exists• A query on how to get to a location = does a path exist from A to B• We can even associate costs to edges (weighted graphs), then ask “what is the cheapest path from A to B”

Application 2: The Minimum Spanning Tree

Weighted graphs: the cost to connect A and B by a communication line is 7 units.

Can we build a communication network so that every two vertex are connected with the minimum costs?

Application 3

Wireless communication

Represented by a weighted complete graph (every two vertices are connected by an edge)

Each edge represents the Euclidean distance dij between two stations Each station uses a certain power i to transmit messages. Given this

power i, only a few nodes can be reached (bold edges). A station reachable by i then uses its own power to relay the message to other stations not reachable by i.

A typical wireless communication problem is: how to broadcast between all stations such that they are all connected and the power consumption is minimized.

Definition A graph G=(V, E) consists a set of vertices, V, and a set of edges, E. Each edge is a pair of (v, w), where v, w belongs to V If the pair is unordered, the graph is undirected; otherwise it is

directed Consider a simple graph where E is not a multiple set and it doesn’t

contain elements of the form {x,x}, i.e. no loop and no multiple edges.

{c,f}

{a,c}{a,b}

{b,d} {c,d}

{e,f}

{b,e}

An undirected graph

Terminology

1. If v1 and v2 are connected, they are said to be adjacent vertices v1 and v2 are endpoints of the edge {v1, v2}

2. If an edge e is connected to v, then v is said to be incident on e. Also, the edge e is said to be incident on v.

3. The number of incident edges on v is the degree of v.

4. Basic theorem: Let n be the number (size) of vertices and m be the number of edges, then

mvreen

ii 2)(deg

1

Path between Vertices

A path is a sequence of vertices (v0, v1, v2,… vk) such that:

For 0 ≤ i < k, {vi, vi+1} is an edge

Note: a path is allowed to go through the same vertex or the same edge any number of times!

The length of a path is the number of edges on the path

A closed path is a path with the same starting and ending vertex.

Types of paths

A path is simple if and only if it does not contain a vertex more than once.

A closed path is a cycle if and only if it has no repeated edges.

A graph is connected if there is a path between any two vertices.

A tree is graph that is connected and has no cycles.

Path Examples

1. {a,c,f,e}

2. {a,b,d,c,f,e}

3. {a, c, d, b, d, c, f, e}

4. {a,c,d,b,a}

5. {a,c,f,e,b,d,c,a}

Are these paths?

Any cycles?

What is the path’s length?

Directed Graph

A graph is directed if direction is assigned to each edge.

Directed edges are denoted as arcs. Arc is an ordered pair (u, v)

Recall: for an undirected graph An edge is denoted {u,v}, which actually

corresponds to two arcs (u,v) and (v,u)

Indegree and Outdegree

Since the edges are directed, we need to consider the arcs coming “in” and going “out” Thus, we define terms Indegree(v), and Outdegree(v)

Each arc(u,v) contributes count 1 to the outdegree of u and the indegree of v

mvvv

)(outdegree)(indegreevertex vvertex

Directed Acyclic Graph

A directed path is a sequence of vertices (v0, v1, . . . , vk)

Such that (vi, vi+1) is an arc

A directed cycle is a directed path such that the first and last vertices are the same.

A directed graph is acyclic if it does not contain any directed cycles

Graph Examples

Example

0

12

3

45

6

7

8

9

Is it a DAG?

Directed Graphs Usage

Directed graphs are often used to represent order-dependent tasks That is we cannot start a task before another task finishes

We can model this task dependent constraint using arcs An arc (i,j) means task j cannot start until task i is finished

Clearly, for the system not to hang, the graph must be acyclic

i j Task j cannot start until task i is finished

University Example CS departments course structure

171151 180

211 251

272

271 252M132M111

231

201

221

361

362

381303

327336

341

343

342

332

334

104

Any directed cycles?How many indeg(171)?How many outdeg(171)?

Topological Sort Topological sort is an algorithm for a directed acyclic

graph Linearly order the vertices so that the linear order

respects the ordering relations implied by the arcs

01

2

3

45

6

7

8

9

For example:

0, 1, 2, 5, 9?0, 4, 5, 9?0, 6, 3, 7 ?

Topological Sort Algorithm Observations

Starting point must have zero indegree If it doesn’t exist, the graph would not be acyclic

Algorithm1. A vertex with zero indegree is a task that can start right away. So

we can output it first in the linear order2. If a vertex i is output, then its outgoing arcs (i, j) are no longer useful,

since tasks j does not need to wait for i anymore- so remove all i’s outgoing arcs

3. With vertex i removed, the new graph is still a directed acyclic graph. So, repeat step 1-2 until no vertex is left.

Graph Representation

Two popular computer representations of a graph. Both represent the vertex set and the edge set, but in different ways.

1. Adjacency Matrix

Use a 2D matrix to represent the graph

2. Adjacency List

Use a 1D array of linked lists

Adjacency Matrix

The graph G = (V, E) can be represented by a table, or a matrix

M = (aij)n×n aij = 1 iff (vi, vj) E,∈

assuming V = {v1, …, vn}.

00101

10001

00001

00000

11100

M

Adjacency Matrix

2D array A[0..n-1, 0..n-1], where n is the number of vertices in the graph

Each row and column is indexed by the vertex id e,g a=0, b=1, c=2, d=3, e=4

A[i][j]=1 if there is an edge connecting vertices i and j; otherwise, A[i][j]=0

The storage requirement is Θ(n2). It is not efficient if the graph has few edges. An adjacency matrix is an appropriate representation if the graph is dense: |E|=Θ(|V|2)

We can detect in O(1) time whether two vertices are connected.

Adjacency List The graph G = (V, E) can be represented by a list of vertices and

a list of its adjacent vertices for each vertex.

Adjacent vertices for each vertex:

a: {c, d, e}

b: { }

c: {a, e}

d: {a ,e}

e: {a, d, c}

Adjacency List

If the graph is not dense, in other words, sparse, a better solution is an adjacency list

The adjacency list is an array A[0..n-1] of lists, where n is the number of vertices in the graph.

Each array entry is indexed by the vertex id Each list A[i] stores the ids of the vertices adjacent to vertex i

Adjacency Matrix Example

2

4

3

5

1

76

9

8

0 0 1 2 3 4 5 6 7 8 9

0 0 0 0 0 0 0 0 0 1 0

1 0 0 1 1 0 0 0 1 0 1

2 0 1 0 0 1 0 0 0 1 0

3 0 1 0 0 1 1 0 0 0 0

4 0 0 1 1 0 0 0 0 0 0

5 0 0 0 1 0 0 1 0 0 0

6 0 0 0 0 0 1 0 1 0 0

7 0 1 0 0 0 0 1 0 0 0

8 1 0 1 0 0 0 0 0 0 1

9 0 1 0 0 0 0 0 0 1 0

Adjacency List Example

2

4

3

5

1

76

9

8

0 0

1

2

3

4

5

6

7

8

9

2 3 7 9

8

1 4 8

1 4 5

2 3

3 6

5 7

1 6

0 2 9

1 8

The array takes up Θ(n) space Define degree of v, deg(v), to be the number of edges incident to

v. Then, the total space to store the graph is proportional to:

An edge e={u,v} of the graph contributes a count of 1 to deg(u) and contributes a count 1 to deg(v)

Therefore, Σvertex vdeg(v) = 2m, where m is the total number of edges

In all, the adjacency list takes up Θ(n+m) space If m = O(n2) (i.e. dense graphs), both adjacent matrix and adjacent

lists use Θ(n2) space. If m = O(n), adjacent list outperform adjacent matrix

However, one cannot tell in O(1) time whether two vertices are connected

Storage of Adjacency List

v

vvertex

)deg(

Adjacency List vs. Matrix

Adjacency List More compact than adjacency matrices if graph has few edges Requires more time to find if an edge exists

Adjacency Matrix Always require n2 space

This can waste a lot of space if the number of edges are sparse Can quickly find if an edge exists

Representations for Directed Graphs

The adjacency matrix and adjacency list can be used

Topological Sort, the algorithm

1) Choose a vertex v of indegree 0 (what about there are several such vertices?) and output v;

2) Modify the indegree of all the successor of v by subtracting 1;

3) Repeat the above process until all vertices are output.

01

2

3

45

6

7

8

9

Topological Sort

Find all starting points

Reduce indegree(w)

Place new startvertices on the Q

Time Complexity of Topological Sorting (Using Adjacency Lists)

We never visited a vertex more than one time For each vertex, we had to examine all outgoing

edges Σ outdegree(v) = m This is summed over all vertices, not per vertex

So, our running time is exactly O(n + m)

How about the complexity using adjacency matrix?

Graph Traversal Application example

Given a graph representation and a vertex s in the graph

Find all paths from s to other vertices

Two common graph traversal algorithms Breadth-First Search (BFS)

Find the shortest paths in an unweighted graph Depth-First Search (DFS)

Topological sort Find strongly connected components

BFS and Shortest Path Problem

Given any source vertex s, BFS visits the other vertices at increasing distances away from s. In doing so, BFS discovers paths from s to other vertices for unweighted graphs.

What do we mean by “distance”? The number of edges on a path from s

2

4

3

5

1

76

9

8

0Consider s=vertex 1

Nodes at distance 1? 2, 3, 7, 91

1

1

12

22

2

s

Example

Nodes at distance 2? 8, 6, 5, 4

Nodes at distance 3? 0

BFS Algorithm

Why use queue? Need FIFO // flag[ ]: visited table

Time Complexity of BFS(Using Adjacency List)

Assume adjacency list n = number of vertices m = number of edges

Each vertex will enter Q at most once.

Each iteration takes time proportional to 1 + deg(v).

O(n + m)

Running Time

Recall: Given a graph with m edges, what is the total degree?

The total running time of the while loop is:

this is summing over all the iterations in the while loop!

See ds.soj.me for practice.

O( Σvertex v (1 + deg(v)) ) = O(n+m)

Σvertex v deg(v) = 2m

Time Complexity of BFS(Using Adjacency Matrix)

Assume adjacency matrix n = number of vertices m = number of edges

Finding the adjacent vertices of v requires checking all elements in the row. This takes linear time O(n).

Summing over all the n iterations, the total running time is O(n2).

O(n2)

So, with adjacency matrix, BFS is O(n2) independent of the number of edges m. With adjacent lists, BFS is O(n+m); if m=O(n2) like in a dense graph, O(n+m)=O(n2).

Dijkstra’s Shortest Path Algorithm(A Greedy Algorithm)

Assume the weighted graph has no negative weights (Why?).

Single source shortest path problem ： find all the shorted paths from source s to other vertices.

Basic idea: list all the shortest paths from the source s

Dijkstra’s algorithm is a greedy algorithm, and the correctness of the algorithm can be proved by contradiction.

Time complexity O(|V|2).

Example

Ideas of the algorithm (Dijkstra): enumerate the shortest paths from the source to other vertices in increasing order.

Basic observations:1. If (s, u,..,v, w) is a shorted path from s to w, then (s,u,

…,v) must be a shortest path form s to v.2. The shortest path from s to v should be the shortest

one among those paths from s to v which only go through known shortest paths.

Method: On every vertex v maintain a pair (known, dist), known:

whether the shortest distance from the source s to v is known, dist: the shortest distance from s through known vertices.

1. To list the next shortest path, find the vertex v with smallest dist, and mark v as known.

2. update labels for the successors of v. 3. goto 1 until all the shortest paths are found.

Example

Method: Mark every vertex v with a pair (known, dist),

known: whether the shortest distance from the source s to v is known, dist: the shortest distance from s through known vertices.

1. Starting with : mark every vertex v with (F, w(s,v)), s with (T, 0).

2. The next shortest distance from s is the one with the smallest dist among vertices (F, dist), and mark that vertex v with T.

3. Update dist of those unknown vertices which are adjacent to v ;

4. Goto 2 until all vertices are known.

Dijkstra’s Algorithm

Dijkstra(G, s):Input: s is the sourceOutput: mark every vertex with the shortest distance from s. for every vertex v { set v (F, weight(s,v)) } set s(T, 0); while ( there is a vertex with (F, _)) {

find the vertex v with the smallest distv among (F, dist)

set v(T, distv)

for every w(F, dist) adjacent to v {

if(distv + weight(v,w) < dist)

set w(F, distv + weight(v,w)) } }

Can you add more information to get

the shortest paths?

Linear, but can be improved by using

heaps.

Running Time

If we use a vector to store “dist” information for all vertices, then finding the smallest value takes O(|V|) time, and the total updating takes O(|E|) time, and the running time is O(|V|2).

If we use a binary heap to store “dist” information, the finding the smallest takes O(log|V|) time and every updating also takes O(log|V|) time, so the total running time is

O(|V|) + O(|V|log|V|) +O(|E|log|V|)

= O(|E|log|V|).

Greedy algorithms

A greedy algorithm is used in optimization problems. It makes the local optimal choice at each stage with the hope of finding the global optimum.

• Example 1. Make 87yuan using the fewest possible bills. Using greedy algorithm, one can choose

• 50, 20, 10, 5, 2, and this is the optimal solution.

• Example 2. Make 15 krons, where available bills are 1, 7 and 10.

• Using greedy algorithm, the solution is 10, 1,1,1,1,1.

• The best solution is 7, 7, and 1.

Proving Its Correctness

The greedy approach leads to: simple and intuitive algorithms efficient algorithms But , it does not always lead to an optimal

solution.

Since the greedy approach doesn’t assure the optimality of the solution we have to analyze for each particular problem if the algorithm leads to an optimal solution.

Proving the correctness of the algorithm

We can prove that the distance recorded in distance is the length of the shortest path from source to v.

Prove: when v is choosen and marked with (T, dist), dist is the shortest path from 0 to v .

distance[x]<distance[v]And x must be in S

Notice that the assumption is the weight is positive.

Red vertices are marked with T, x is the first unknown vertex on the shortest path from 0 to v

Dijkstra

Shortest path algorithm paper EWD hand writing notes http://www.cs.utexas.edu/users/EWD/

Minimum Spanning Trees

Definition of minimum spanning trees for undirected, connected, weighted graphs.

Prim’s algorithm, a greedy algorithm. Basic observation: G=(V,E), for any A V, if

an edge e has the smallest weight connection A and V-A, then e is in a MST.

A local optimal choice is also a global optimal choice.

Prim’s Algorithm

Method:

Build the MST by adding vertices and edges one by one staring with one vertex.

At some stage, let A be the set of the vertices that are already added, V-A be the set of the remaining vertices that are not added.

1. Find the smallest weighted edge e = (u,v), that connects A and V-A, that is, e has the smallest weight among those edges with uA and v V-A.

2. Now add vertex v to A and edge (u,v) to the MST.

3. Repeat step 1 and 2 above until A=V.

Prim’s Algorithm --An Example

1. Starting from a node, 0; 2. Add a vertex v and an edge (0,v) which has a

weight as small as possible

3. Add a vertex u such that an edge connecting u and {0,v} has a weight as small as possible;

4. Repeat the process until all vertices are added.

Finally

Prim’s Algorithm-the method

To be able to find the smallest weighted edge connecting A and V-A, on every vertex u V-A, maintain the information “what is the smallest weighted edge connecting u with A”.

This information is easily initialized, then the smallest weighted edge connection A and V-A is easily found, finally, the information is easily maintained.

Prim’s Algorithm --Implementation

Method: Mark every vertex v with (added, dist, neib): whether v is added in T and the current smallest edge weight connecting v with a vertex neib in T.

1. Starting with v(false, weight(s,v), s), s(true, 0, s), i.e. T has one vertex s.

2. Find the vertex v with the smallest distv among those (false, dist, u).

3. Mark v with (true, distv, u).

Updating those (false, _, _):

for every w(false, distw, k) adjacent to v,

if (weight(v,w) < distw)

set w(false, weight(v,w), v).

4. Repeat 2 until every vertex is marked with (true,_, _).

Prim’s Algorithm --Implementation

Use X to denote the set of nodes added in the tree, D[v] (vX) to denote the distance from v to X, N[v] v’s nearest neighbor in X.

1. Starting with X={ 0}; D[v]=weight[0][v];N[v]=0; 2. Repeat the action n-1 times: a) choose a vertex vX such that D[v] is the smallest

weight; b) update X: X = X+{v}. c) update Y: Y = Y+{(v,N[v])}; d) update D: for all wX, if (weight[v][w]<D[w]) then

D[w] = weight[v][w] and N[w]=v.

Prim’s Algorithm --Implementation

1. Intialization: for all v

2. X[v]=flase; X[0]=true;

3. D[v]=weight[0][v];

4. N[v]=0;

5. 2. Repeat the following action n-1 times: a) find the smallest D[v] such that X[v]=false; b) X[v]=true; c) for all w such that (v,w)E and X[w]=false, if (weight[v][w]<D[w]) { D[w] = weight[v][w]; N[w]=v; }

Complexity

Running time: O(|V|2+|E|) = O(|V|2 ). One may improve the algorithm by using a

heap to get the smallest connecting edge, and get O(|E|log|V| + |V|log|V|) = O(|E|log|V|), which is good for sparse graphs.

DFS

Like preorder traversal for trees DFS starting at some vertex v:

visit v and continue to visit neighbors of v in a recursive pattern

To avoid cycles, a vertex is marked visited when it is visited.

Running time: O(|V|+|E|).

DFS Algorithm

Finding neighbours of v: for matrix it is n, for

adjacency list it is degree(v).

RDFS is called once for every node.

Time complexity: O(|V|+|E|) if adjacency lists are used.

Example

2

4

3

5

1

76

9

8

0Adjacency List

source

0

1

2

3

4

5

6

7

8

9

Visited Table (T/F)

F

F

F

F

F

F

F

F

F

F

Initialize visitedtable (all False)

Initialize Pred to -1

-

-

-

-

-

-

-

-

-

-

Pred

DFS spanning Trees

Captures the structure of the recursive calls- when we visit a neighbor w of v, we add w as child of v- whenever DFS returns from a vertex v, we climb up in the tree from v to its parent

Summary

Graphs can be used to model real problems Two typical graph representations: matrix and

adjacency lists Graph algorithms, the methods, applications and

complexities. Topological sort BFS and DFS Dijkstra’s algorithm for single source shortest paths Prim’s algorithm for minimum spanning trees Greedy algorithms, a problem solving strategy.

Exercises: see course web page for problem set 4. The last programming assignment is posted.