Graph Algorithms 1

Lectures onGraph Algorithms:

searching, testing and sorting

COMP 523: Advanced Algorithmic Techniques

Lecturer: Dariusz Kowalski

Graph Algorithms 2

Overview

Previous lectures:• Recursive method (searching, sorting, …)This lecture: algorithms on graphs, in particular• Representation of graphs• Breadth-First Search (BFS)• Depth-First Search (DFS)• Testing bipartiteness • Testing for (directed) cycles• Ordering nodes in acyclic directed graphs

Graph Algorithms 3

How to represent graphs?Given an undirected graph G = (V,E) , how to represent it?1. Adjacency matrix: ith node is represented as ith row

and ith column, edge between ith and jth nodes is represented as 1 in row i and column j, and vice versa (0 if there is no edge between i and j)

2. Adjacency list: nodes are arranged as array/list, each node record has the list of its neighbors attached

1

42

3

0 1 1 11234

1 0 1 01 1 0 11 0 1 0

1 2 3 41234

2 3 41 31 2 41 3

Adjacency matrix Adjacency list

Graph Algorithms 4

Adjacency matrixAdvantages:

– Checking in constant time if an edge belongs to the graph

Disadvantages:– Representation takes memory O(n2) - versus O(m+n)– Examining all neighbors of a given node requires time

O(n) - versus O(m/n) in average

Disadvantages especially for sparse graphs!

1

42

3

0 1 1 11234

1 0 1 01 1 0 11 0 1 0

1 2 3 41234

2 3 41 31 2 41 3

Adjacency matrix Adjacency list

Graph Algorithms 5

Adjacency listAdvantages:

– Representation takes memory O(m+n) - versus O(n2)– Examining all neighbors of a given node requires time

O(m/n) in average - versus O(n)

Disadvantages:– Checking if an edge belongs to the graph requires time

O(m/n) in average - versus O(1)

Advantages especially for sparse graphs!

1

42

3

0 1 1 11234

1 0 1 01 1 0 11 0 1 0

1 2 3 41234

2 3 41 31 2 41 3

Adjacency matrix Adjacency list

Graph Algorithms 6

Breadth-First Search (BFS)Given graph G = (V,E) of diameter D and a root nodeGoal: find a spanning tree such that the distances between nodes

and the root are the same as in graph GIdea of the algorithm:• For layers i = 0,1,…,D

– while there is a node in layer i + 1 not added to the partial BFS tree : add the node and an edge connecting it with a single node in layer i

layer 0

layer 1layer 2

root

Graph Algorithms 7

Implementing BFSStructures:

– Adjacency list– Lists L0,L1,…,LD

– Array Discovered[1…n]

Algorithm:• Set L0 = {root}• For layers i = 0,1,…,D

– Initialize empty list Li+1 – For each node v in Li take next edge adjacent to v and if its second end w is

not marked as Discovered then add w to Li+1 and {v,w} to partial BFS

layer 0

layer 1layer 2

root

Graph Algorithms 8

Analysis of BFSCorrectness: (observing layer property)

By induction on layer number: each node in layer i is in the list Li

Each node w in layer i has its predecessor on the distance path in layer i-1,

consider the first of such predecessors v in list Li-1: w is added to list Li by the step where it is considered as the neighbor of vComplexity:• Time: O(m+n) - each edge is considered at most twice - since it occurs

twice on the adjacency list• Memory: O(m+n) - adjacency list takes O(m+n), lists L0,L1,…,LD have at

most n elements in total, array Discovered has n elements

layer 0

layer 1layer 2

root

Graph Algorithms 9

Depth-First Search (DFS)Given graph G = (V,E) of diameter D and a root nodeGoal: find a spanning tree such that each edge in graph G corresponds to the

ancestor relation in the tree (i.e., no cross edges between different branches)Recursive idea of the algorithm:Repeat until no neighbor of the root is free• Select a free neighbor v of the root and add the edge from root to v to partial

DFS tree• Recursively find a DFS* in graph G restricted to free nodes and node v as the

root* and add it to partial DFS tree

rootroot*

root**

Graph Algorithms 10

Implementing DFSData structures:

– Adjacency listwith pointer .next

– List (stack) S– Array Discovered[1…n]

Algorithm:• Set S = {root}• Repeat until S is empty

Consider the top element v in S– if v.next is not Discovered then put v.next into the stack S and set v := v.next and

v.next to be the first neighbour of v– if v.next is Discovered then

set v.next to be the next neighbour of v if such neighbour existsotherwise remove v from the stack, add edge {v,z} to partial DFS, where z is the current top element in S

Remark: after considering the neighbor of node v we remove this neighbor from adjacency list to avoid considering it many times!

root

root**

root*

Graph Algorithms 11

Analysis of DFSCorrectness: (observing ancestor property)By induction on the number of recursive calls: Consider edge {v,w} in G : suppose v joints DFS before w. Then w is free and since it is in a connected component with v it will be in DFS* with root* (which is v), so v is the ancestor of wComplexity:• Time: O(m+n) - each edge is considered at most twice - while adding or

removing from the stack• Memory: O(m+n) - adjacency list consumes O(m+n), stack S and array

Discovered have at most n elements each

root

root**

root*

Graph Algorithms 12

Textbook and Exercises

READING:• Chapter 3 “Graphs”, Sections 3.2 and 3.3

OBLIGATORY EXERCISES: • Prove that obtained DFS and BFS trees are spanning

trees.• Prove that if DFS and BFS trees in graph G are the

same, then G is also the same like they (does not contain any other edge).

• Prove that if G has n nodes, where n is even, and each node has at least n/2 neighbours then G is connected.

Graph Algorithms 13

Testing graph properties

• Testing bipartiteness

• Testing for directed cycles

Graph Algorithms 14

How to represent graphs?Given an undirected graph G = (V,E) , how to represent it?1. Adjacency matrix: ith node is represented as ith row

and ith column, edge between ith and jth nodes is represented as 1 in row i and column j, and vice versa (0 if there is no edge between i and j)

2. Adjacency list: nodes are arranged as array/list, each node record has the list of its neighbours attached

1

42

3

0 1 1 11234

1 0 1 01 1 0 11 0 1 0

1 2 3 41234

2 3 41 31 2 41 3

Adjacency matrix Adjacency list

Graph Algorithms 15

Adjacency listAdvantages:

– Representation takes memory O(m+n) - versus O(n2)– Examining all neighbours of a given node requires time

O(m/n) in average - versus O(n)

Disadvantages:– Checking if an edge belongs to the graph requires time

O(m/n) in average - versus O(1)

Advantages especially for sparse graphs!

1

42

3

0 1 1 11234

1 0 1 01 1 0 11 0 1 0

1 2 3 41234

2 3 41 31 2 41 3

Adjacency matrix Adjacency list

Graph Algorithms 16

BipartitenessGraph G = (V,E) is bipartite iff it can be partitioned into two sets of

nodes A and B such that each edge has one end in A and the other end in B

Alternatively:• Graph G = (V,E) is bipartite iff all its cycles have even length• Graph G = (V,E) is bipartite iff nodes can be coloured using two

coloursQuestion: given a graph represented as an adjacency list, how to test if

the graph is bipartite?Note: graphs without cycles (trees) are bipartite

non bipartite

bipartite:

Graph Algorithms 17

Testing bipartitenessMethod: use BFS search treeRecall: BFS is a rooted spanning tree having the same layers as the original graph G

(each node has the same distance from the root in BFS tree and in graph G)Algorithm:• Run BFS search and colour all nodes in odd layers red, others blue• Go through all edges in adjacency list and check if each of them has two different

colours at its ends - if so then G is bipartite, otherwise it is not We use the following alternative definitions in the analysis:• Graph G = (V,E) is bipartite iff all its cycles have even length, or• Graph G = (V,E) is bipartite iff it has no odd cycle

non bipartitebipartite

Graph Algorithms 18

Testing bipartiteness - why it worksProperty of layers:• Every edge is either between two consecutive layers or in a single layer (two

ends are in the same layer) - it follows from BFS property1. Suppose that graph G is not bipartite. Hence there is an odd cycle. This cycle

must have an edge in one layer, and so the ends of this edge have the same colour. Thus the algorithm answers not bipartite correctly.

2. Suppose that graph G is bipartite. Hence all its cycles have even length. Suppose, to the contrary, that the algorithm answers incorrectly that G is not bipartite. Hence it found a monochromatic edge. This edge must be in one layer. Consider one of such edges which is in the smallest possible layer. Consider a cycle containing this edge and the BFS path between its two ends. The length of this cycle is odd, thus G is not bipartite. Contradiction!

not bipartitebipartite

Graph Algorithms 19

ComplexityTime: O(m+n)• BFS search: O(m+n) • Checking colours of all edges: O(m+n)

Memory: O(m+n)• Adjacency list with colour labels: O(m+n) • Array Active and lists Li for BFS algorithm: O(n)

Graph Algorithms 20

Testing for directed cyclesDirected graph G (edges have direction - one end is the beginning,

the other one is the end of the edge)

Reversed graph Gr has the same nodes but each edge is a reversed edge from graph G

Representing directed graph: adjacency list - each node has two lists: of in-coming and out-coming edges/neighbours

Problem: does the graph have a directed cycle?

Graph Algorithms 21

How to represent directed graphs?Given a directed graph G = (V,E) , how to represent it?1. Adjacency matrix: ith node is represented as ith

row and ith column, edge from ith to jth node is represented as 1 in row i and column j, (0 if there is no directed edge from i to j)

2. Adjacency list: nodes are arranged as array/list, each node record has two lists: one stores in-neighbours, the other stores out-neighbours

1

42

3

0 1 1 11234

0 0 1 00 0 0 01 0 1 0

1 2 3 41234

2 3 431

1241 3

Adjacency matrix Adjacency list

4

1

Graph Algorithms 22

Testing cyclesTechnique: use directed DFS tree for graph GAlgorithm:• Find a node with no incoming edges - if not found answer cycled• Build a directed DFS tree, with the following modification:

– consider only edges in proper direction; – if during building a DFS tree two nodes which are already in the stack are

considered then answer cycled; otherwise answer acyclic at the end

General remark: if graph G is not connected then do the same until no free node remains

Graph Algorithms 23

Testing cycles - analysisProperty of acyclic graph (to be proved later):• There is a node with no incoming edgesAnalysis of algorithm correctness:1. Suppose that graph G is acyclic. Hence there is no directed cycle and then

while building a directed DFS tree we never try to explore the already visited node. Answer acyclic is then proper.

2. Suppose that graph G is not acyclic. It follows that there is a cycle in it. Let v be the first node in a cycle visited by DFS search. By the property of DFS, all nodes from this cycle will be reached during the search rooted in v, and so the in-neighbour of v from this cycle (or even other node before) will attempt to visit v, which causes that the algorithm stops with the correct answer cycled.

Graph Algorithms 24

ComplexityTime: O(m+n)• Finding a node with no incoming edges: O(m+n)• DFS search with checking cycle condition: O(m+n)

Memory: O(m+n)• Adjacency list: O(m+n) • Array Active and stack S for DFS algorithm: O(n)

Graph Algorithms 25

Textbook and ExercisesREADING:• Chapter 3, Sections 3.4 and 3.5

OBLIGATORY EXERCISES:• Prove the following property of layers: every edge is either between two

consecutive layers or in a single layer (two ends are in the same layer).• Prove the following property of acyclic graphs: there is a node with no

incoming edges. Is this property true for out-going edges?

OTHER EXERCISES:• Design and analyse time and memory taken by BFS for directed graphs.• What happens if you try to use a different method of searching to check

bipartiteness or acyclicity?

Graph Algorithms 26

Ordering nodes in acyclic graphs(topological order)

Graph Algorithms 27

Directed Acyclic Graphs (DAG)Directed graph G (edges have directions - one end is the beginning,

the other one is the end of the edge) Reversed graph Gr has the same nodes but each edge is a reversed

edge from graph GTest if a given direct graph is acyclic (DAG): done already,

directed DFS approach in time and memory O(m+n)Problem: is it possible to order all nodes topologically, which

means that if (v,w) is a directed edge in G then v is before w?Note: there may be many topological orders

graph G reversed graph Gr

Graph Algorithms 28

Opposite propertyFact: If graph G has a topological order then it is DAG.Proof:Suppose, to the contrary, that

G has a topological order and it also has a cycle. Let v be the smallest element in the cycle according to the

topological order. It follows that its predecessor in the cycle must be smaller (in the final order) than v, but this contradicts the fact that v is the smallest element on the cycle.

We get a contradiction with our assumption, which means that G must not have a cycle.

Graph Algorithms 29

Key propertyProperty:• In DAG there is a node with no incoming edges• After removing this node with the incident edges, the remaining graph is DAGProof:• Node with no incoming edges:

Suppose, to the contrary, that every node has at least one incoming edge. It follows that starting from any node we can always “leave” a node by incoming edge (in opposite direction) until we return to a visited node, which creates a cycle. This is a contradiction.

• After removing node still a DAG remains:There could not be a cycle in smaller graph, since it would be in the original graph.

Graph Algorithms 30

Algorithm finding topological orderAlgorithm:• Repeat until all nodes are in the ordered list

1. Find a node with no incoming edges and place the node next in the order2. Remove this node and all its outgoing edges from the graph

Efficient implementation:• Preprocessing:

– for each node keep info if it is active or not (additional array Active of size n needed) - at the beginning all nodes are active

– go through the list and for every node store the number of incoming edges from other active nodes (additional array Counter of size n needed)

– select from list S those nodes that have no incoming edges

• During the algorithm:– While removing the node (make inactive), go through its outgoing active neighbours and decrease

their counters by one - if some becomes zero put this active node into list S– Node with no incoming edges is taken (and then removed) from list S

1

2

1

3

2

1

3

42

1

Graph Algorithms 31

ComplexityTime: O(m+n)• Preprocessing:

– Initialization of array Active: O(n)– Filling array Counter - counting edges: O(m+n)– Initialization of list S - checking array Counter: O(n)

• During the algorithm:– While removing a node - go through its out-neighbours: total O(m+n)– Selecting from list S: total O(n)

Memory: O(m+n)• Adjacency list: O(m+n) • Arrays Active and Counter, list S: O(n)

Graph Algorithms 32

Conclusions

Graph algorithms in time O(m + n)• Searching in graphs

– BFS– DFS

• Testing graph properties – bipartiteness (based on BFS)– if a directed graph is acyclic (based on DFS)

• Topological order

Graph Algorithms 33

Textbook and ExercisesREADING:• Chapter 3, Section 3.6

EXERCISES: (reminder)• Prove that if DFS and BFS trees in graph G are the same, then G is also the

same like they (does not contain any other edge).• Prove that if G has n nodes, where n is even, and each node has at least n/2

neighbours then G is connected.• Prove the following property of layers: every edge is either between two

consecutive layers or in a single layer (two ends are in the same layer).• Prove the following property of acyclic graphs: there is a node with no

incoming edges. Is this property true for out-going edges?• Design and analyze time and memory taken by BFS for directed graphs.• What happens if you try to use a different method of searching to check

bipartiteness or acyclicity?