basic graph search - itu.dkitu.dk/people/pagh/ads11/08-graphsearch.pdf · circuit gate, register,...

Algorithms and Data Structures, Fall 2011

Basic Graph Search

Rasmus Pagh

Based on slides by Kevin Wayne, PrincetonAlgorithms, 4th Edition · Robert Sedgewick and Kevin Wayne · Copyright © 2002–2011

Monday, October 31, 11

Mid-term evaluation

Thanks to those of you who provided feedback

- Will try to choose case studies for lecture on November 14 that allow us to rehearse algorithm analysis basics.

- Will start posing previous exam problems for exercises (from next week). Also, there will be a trial exam on November 28.

2


Graph. Set of vertices connected pairwise by edges.

Why study graph algorithms?

• Interesting and broadly useful abstraction.

• Challenging branch of computer science and discrete math.

• Hundreds of graph algorithms known.

• Thousands of practical applications.

3

Undirected graphs


4

Protein-protein interaction network

Reference: Jeong et al, Nature Review | Genetics


5

The Internet as mapped by the Opte Project

http://en.wikipedia.org/wiki/Internet


6

10 million Facebook friends

"Visualizing Friendships" by Paul Butler


7

Relationship graph

Relationship graph at "Jefferson High"


8

Graph applications

graph vertex edge

communication telephone, computer fiber optic cable

circuit gate, register, processor wire

mechanical joint rod, beam, spring

financial stock, currency transactions

transportation street intersection, airport highway, airway route

internet class C network connection

game board position legal move

social relationship person, actor friendship, movie cast

neural network neuron synapse

protein network protein protein-protein interaction

chemical compound molecule bond


9

Graph terminology

Path. Sequence of vertices connected by edges.Cycle. Path whose first and last vertices are the same.

Two vertices are connected if there is a path between them.

Anatomy of a graph

cycle oflength 5

vertex

vertex ofdegree 3

edge

path oflength 4

connectedcomponents


10

Some graph-processing problems

Path. Is there a path between s and t?Shortest path. What is the shortest path between s and t?

Cycle. Is there a cycle in the graph?Euler tour. Is there a cycle that uses each edge exactly once?Hamilton tour. Is there a cycle that uses each vertex exactly once?

Connectivity. Is there a way to connect all of the vertices?MST. What is the best way to connect all of the vertices?Biconnectivity. Is there a vertex whose removal disconnects the graph?

Planarity. Can you draw the graph in the plane with no crossing edges?Graph isomorphism. Do two adjacency lists represent the same graph?

Challenge. Which of these problems are easy? difficult? intractable?


Graph drawing. Provides intuition about the structure of the graph.Caveat. Intuition can be misleading.

11

Graph representation

Two drawings of the same graph


Vertex representation.

• This lecture: use integers between 0 and V-1.

• Applications: convert between names and integers with symbol table.

Anomalies.

A

G

E

CB

F

D

12

Graph representation

symbol table

0

6

4

21

5

3

Anomalies

paralleledgesself-loop


13

Graph API

public class Graph public class Graph

Graph(int V) create an empty graph with V vertices

Graph(In in) create a graph from input stream

void addEdge(int v, int w) add an edge v-w

Iterable<Integer> adj(int v) vertices adjacent to v

int V() number of vertices

int E() number of edges

String toString() string representation


Maintain a two-dimensional V-by-V boolean array;for each edge v-w in graph: adj[v][w] = adj[w][v] = true.

0 1 2 3 4 5 6 7 8 9 10 11 12

0

1

2

3

4

5

6

7

8

9

10

11

12

0 1 1 0 0 1 1 0 0 0 0 0 0

1 0 0 0 0 0 0 0 0 0 0 0 0

1 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 1 1 0 0 0 0 0 0 0

0 0 0 1 0 1 1 0 0 0 0 0 0

1 0 0 1 1 0 0 0 0 0 0 0 0

1 0 0 0 1 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 1 0 0 0 0

0 0 0 0 0 0 0 1 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 1 1 1

0 0 0 0 0 0 0 0 0 1 0 0 0

0 0 0 0 0 0 0 0 0 1 0 0 1

0 0 0 0 0 0 0 0 0 1 0 1 0

14

Adjacency-matrix graph representation

two entries

for each edge

109

1211

0

6

4

21

5

3

87


Maintain vertex-indexed array of lists.(use Bag abstraction)

15

Adjacency-list graph representation

109

1211

0

6

4

21

5

3

adj[]0

1

2

3

4

5

6

7

8

9

10

11

12

5 4

0 4

9 12

11 9

0

0

8

7

9

5 6 3

3 4 0

11 10 12

6 2 1 5

Adjacency-lists representation (undirected graph)

Bag objects

representationsof the same edge

87

(alt: HashSet)


In practice. Use adjacency-lists representation.

• Algorithms based on iterating over vertices adjacent to v.

• Real-world graphs tend to be “sparse.”

16

Graph representations

representation space add edgeedge between

v and w?

iterate over vertices

adjacent to v?

adjacency matrix V 2 1 * 1 V

adjacency lists E + V 1 degree(v) degree(v)

huge number of vertices,

small average vertex degree

* disallows parallel edges


In practice. Use adjacency-lists representation.

• Algorithms based on iterating over vertices adjacent to v.

• Real-world graphs tend to be “sparse.”

17

Graph representations

huge number of vertices,

small average vertex degree

sparse (E = 200) dense (E = 1000)

Two graphs (V = 50)


18

Maze exploration: A graph search problem

Maze graphs.

• Vertex = intersection.

• Edge = passage.

Goal. Explore every intersection in the maze.

intersection passage


Algorithm.

• Unroll a ball of string behind you.

• Mark each visited intersection and each visited passage.

• Retrace steps when no unvisited options.

19

Trémaux maze exploration

Tremaux exploration

Theseus and

the Minotaur


20

Maze exploration


Goal. Systematically search through a graph.Idea. Mimic maze exploration.

Typical applications. [ahead]

• Find all vertices connected to a given source vertex.

• Find a path between two vertices.

Depth-first search

Mark v as visited.Recursively visit all unmarked vertices w adjacent to v.

DFS (to visit a vertex v)


22

Design pattern. Decouple graph data type from graph processing.

Typical client program.

• Create a Graph.

• Pass the Graph to a graph-processing routine, e.g., Search.

• Query the graph-processing routine for information.

Design pattern for graph processing

Search search = new Search(G, s); for (int v = 0; v < G.V(); v++) if (search.marked(v)) StdOut.println(v);

public class Search public class Search

Search(Graph G, int s) find vertices connected to s

boolean marked(int v) is vertex v connected to s?

int count() how many vertices connected to s?

print all vertices

connected to s


23

Depth-first search (reachability)

public class DepthFirstSearch{ private boolean[] marked;

public DepthFirstSearch(Graph G, int s) { marked = new boolean[G.V()]; dfs(G, s); }

private void dfs(Graph G, int v) { marked[v] = true; for (int w : G.adj(v)) if (!marked[w]) dfs(G, w); }

public boolean marked(int v) { return marked[v]; }}

true if connected to s

constructor marks

vertices connected to s

recursive DFS does the work

client can ask whether

vertex v is connected to s


Depth-first search properties

Proposition. DFS marks all vertices connected to s in time proportional to the sum of their degrees.

Pf.

• Correctness:

- if w marked, then w connected to s (why?)

- if w connected to s, then w marked(if w unmarked, then consider last edgeon a path from s to w that goes from amarked vertex to an unmarked one)

• Running time: each vertexconnected to s is visited once.

24

set ofunmarked

vertices

no such edgecan exist

source

v

s

set of markedvertices

w

x


Depth-first search application: preparing for a date

25

http://xkcd.com/761/


26

Depth-first search application: flood fill

Change color of entire blob of neighboring red pixels to blue.

Build a grid graph.

• Vertex: pixel.

• Edge: between two adjacent red pixels.

• Blob: all pixels connected to given pixel.

recolor red blob to blue


27

Depth-first search application: flood fill

Change color of entire blob of neighboring red pixels to blue.

Build a grid graph.

• Vertex: pixel.

• Edge: between two adjacent red pixels.

• Blob: all pixels connected to given pixel.

recolor red blob to blue


Goal. Does there exist a path from s to t ? If yes, find any such path.

28

Pathfinding in graphs

† amortizedMonday, October 31, 11

29

Pathfinding in graphs

Goal. Does there exist a path from s to t ? If yes, find any such path.

Depth-first search. After linear-time preprocessing, can recover path itself in time proportional to its length.

Idea: Mark each visited vertex by keeping track of edge taken to visit it.

public class Paths public class Paths

Paths(Graph G, int s) find paths in G from source s

boolean hasPathTo(int v) is there a path from s to v?

Iterable<Integer> pathTo(int v) path from s to v; null if no such path


30

Depth-first search (pathfinding)

public class DepthFirstPaths{ private boolean[] marked; private int[] edgeTo; private final int s; public DepthFirstPaths(Graph G, int s) { marked = new boolean[G.V()]; edgeTo = new int[G.V()]; this.s = s; dfs(G, s); } private void dfs(Graph G, int v) { marked[v] = true; for (int w : G.adj(v)) if (!marked[w]) { edgeTo[w] = v; dfs(G, w); } }

public boolean hasPathTo(int v) public Iterable<Integer> pathTo(int v)}

parent-link representation

of DFS tree

set parent link

ahead


edgeTo[] is a parent-link representation of a tree rooted at s.

31

Depth-first search (pathfinding iterator)

public boolean hasPathTo(int v) { return marked[v]; }

public Iterable<Integer> pathTo(int v) { if (!hasPathTo(v)) return null; Stack<Integer> path = new Stack<Integer>(); for (int x = v; x != s; x = edgeTo[x]) path.push(x); path.push(s); return path; }

Trace of pathTo() computation

edgeTo[] 0 1 2 2 0 3 2 4 3 5 3

5 53 3 52 2 3 50 0 2 3 5

x path


Depth-first search. Put unvisited vertices on a stack.Breadth-first search. Put unvisited vertices on a queue.

Shortest path. Find path from s to t that uses fewest number of edges.

Intuition. BFS examines vertices in increasing distance from s.32

Breadth-first search

Put s onto a FIFO queue, and mark s as visited.Repeat until the queue is empty: - remove the least recently added vertex v - add each of v's unvisited neighbors to the queue, and mark them as visited.

BFS (from source vertex s)

Breadth-!rstmaze exploration


33

Breadth-first search (pathfinding)

private void bfs(Graph G, int s) { Queue<Integer> q = new Queue<Integer>(); q.enqueue(s); marked[s] = true; while (!q.isEmpty()) { int v = q.dequeue(); for (int w : G.adj(v)) if (!marked[w]) { q.enqueue(w); marked[w] = true; edgeTo[w] = v; } } }

Trace of breadth-!rst search to !nd all paths from 0

marked[]

0 T 1 2 3 4 5

0 T 1 T 2 T 3 4 5 T

0 T 1 T 2 T 3 T 4 T 5 T

q

0

1534

215

534

34

4

edgeTo[]

0 1 2 3 4 5

0 1 0 2 0 3 4 5 0

0 1 0 2 0 3 2 4 2 5 0

0 2 1 5 1 0 2 2 0 1 3 4 3 5 4 2 4 3 2 5 3 0

0 2 1 5 1 0 2 2 0 1 3 4 3 5 4 2 4 3 2 5 3 0

0 T 1 T 2 T 3 T 4 T 5 T

0 1 0 2 0 3 2 4 2 5 0

0 2 1 5 1 0 2 2 0 1 3 4 3 5 4 2 4 3 2 5 3 0

adj[]

0 2 1 5 1 0 2 2 0 1 3 4 3 5 4 2 4 3 2 5 3 0

0 T 1 T 2 T 3 T 4 T 5 T

0 1 0 2 0 3 2 4 2 5 0

0 2 1 5 1 0 2 2 0 1 3 4 3 5 4 2 4 3 2 5 3 0

0 T 1 T 2 T 3 T 4 T 5 T

0 1 0 2 0 3 2 4 2 5 0

0 2 1 5 1 0 2 2 0 1 3 4 3 5 4 2 4 3 2 5 3 0


Proposition. BFS computes shortest path (number of edges) from sin a connected graph in time proportional to E + V.

Pf.

• Correctness: queue always consists of zero or more vertices of distance k from s, followed by zero or more vertices of distance k + 1.

• Running time: each vertex connected to s is visited once.

Breadth-first search properties

34

0

4

2

1

53

standard drawing

0

4

2

1

5

3

dist = 0 dist = 1 dist = 2


DFS and BFS space

In the worst case, need to store V vertices on the stack/queue.

Possible to do this with less space?

In 2004, Omer Reingold showed that it is possible to determine the existence of a path using just a constant number of integers (in addition to the input).

However, the running time of Reingold’s algorithm is high, between V10 and V18.

35


36

Breadth-first search application: routing

Fewest number of hops in a communication network.

ARPANET, July 1977


37

Kevin Bacon graph

• Include a vertex for each performer and for each movie.

• Connect a movie to all performers that appear in that movie.

• Compute shortest path from s = Kevin Bacon.

KevinBacon

KathleenQuinlan

MerylStreep

NicoleKidman

JohnGielgud

KateWinslet

BillPaxton

DonaldSutherland

The StepfordWives

Portraitof a Lady

Dial Mfor Murder

Apollo 13

To Catcha Thief

The EagleHas Landed

ColdMountain

Murder on theOrient Express

VernonDobtcheff

An AmericanHaunting

Jude

Enigma

Eternal Sunshineof the Spotless

Mind

TheWoodsman

WildThings

Hamlet

Titanic

AnimalHouse

GraceKellyCaligola

The RiverWild

LloydBridges

HighNoon

The DaVinci Code

Joe Versusthe Volcano

PatrickAllen

TomHanks

SerrettaWilson

GlennClose

JohnBelushi

YvesAubert Shane

Zaza

PaulHerbert

performervertex

movievertex

Symbol graph example (adjacency lists)

...Tin Men (1987)/DeBoy, David/Blumenfeld, Alan/... /Geppi, Cindy/Hershey, Barbara...Tirez sur le pianiste (1960)/Heymann, Claude/.../Berger, Nicole (I)...Titanic (1997)/Mazin, Stan/...DiCaprio, Leonardo/.../Winslet, Kate/...Titus (1999)/Weisskopf, Hermann/Rhys, Matthew/.../McEwan, GeraldineTo Be or Not to Be (1942)/Verebes, Ernö (I)/.../Lombard, Carole (I)...To Be or Not to Be (1983)/.../Brooks, Mel (I)/.../Bancroft, Anne/...To Catch a Thief (1955)/París, Manuel/.../Grant, Cary/.../Kelly, Grace/...To Die For (1995)/Smith, Kurtwood/.../Kidman, Nicole/.../ Tucci, Maria......

movies.txt

V and E not explicitly

specified

"/"delimiter

http://oracleofbacon.org


Def. Vertices v and w are connected if there is a path between them.

Goal. Preprocess graph to answer queries: is v connected to w ? in constant time.

38

Connectivity queries

public class CC public class CC

CC(Graph G) find connected components in G

boolean connected(int v, int w)connected(int v, int w) are v and w connected?

int count() number of connected components

int id(int v) component identifier for v


Def. A connected component is a maximal set of connected vertices.

39

Connected components


Def. A connected component is a maximal set of connected vertices.

39


63 connected components


Goal. Partition vertices into connected components.

40


Initialize all vertices v as unmarked.

For each unmarked vertex v, run DFS to identify all vertices discovered as part of the same component.


13130 54 30 19 126 45 40 211 129 100 67 89 115 3

tinyG.txt

Input format for Graph constructor (two examples)

2501273244 246239 240238 245235 238233 240232 248231 248229 249228 241226 231...(1261 additional lines)

mediumG.txtV

EV

E

13130 54 30 19 126 45 40 211 129 100 67 89 115 3

tinyG.txt

Input format for Graph constructor (two examples)

2501273244 246239 240238 245235 238233 240232 248231 248229 249228 241226 231...(1261 additional lines)

mediumG.txtV

EV

E


Graph-processing challenge 1

Problem. Is a graph bipartite?

How difficult?

• Any COS 126 student could do it.

• Need to be a typical diligent COS 226 student.

• Hire an expert.

• Intractable.

• No one knows.

• Impossible.

41

0-10-20-50-61-32-32-44-54-6

0

6

4

21

5

3

0-10-20-50-61-32-32-44-54-6

0

6

4

21

5

3



Problem. Find a cycle.

How difficult?



• Hire an expert.

• Intractable.

• No one knows.

• Impossible.

42

0-10-20-50-61-32-32-44-54-6

0

6

4

21

5

3

0

6

4

21

5

3



Problem. Find a cycle that visits every vertex.Assumption. Need to visit each vertex exactly once.

How difficult?



• Hire an expert.

• Intractable.

• No one knows.

• Impossible.

43

0-10-20-50-61-22-63-43-54-54-60-5-3-4-6-2-1-0

0

6

4

21

5

3



Problem. Find a cycle that visits every vertex.Assumption. Need to visit each vertex exactly once.

How difficult?



• Hire an expert.

• Intractable.

• No one knows.

• Impossible.

44

Hamiltonian cycle

(classical NP-complete problem)

0-10-20-50-61-22-63-43-54-54-60-5-3-4-6-2-1-0

0

6

4

21

5

3

✓

Andreas Björklund



Problem. Are two graphs identical except for vertex names?

How difficult?



• Hire an expert.

• Intractable.

• No one knows.

• Impossible.

45

0-10-20-50-63-43-54-54-6

0

6

4

21

5

3

3

1

5

2

4

0

6

0↔4, 1↔3, 2↔2, 3↔6, 4↔5, 5↔0, 6↔1

0-40-50-61-41-52-43-45-6



Problem. Are two graphs identical except for vertex names?

How difficult?



• Hire an expert.

• Intractable.

• No one knows.

• Impossible.

46

graph isomorphism is

longstanding open problem

0-10-20-50-63-43-54-54-6

0

6

4

21

5

3

✓3

1

5

2

4

0

6

0-40-50-61-41-52-43-45-6

0↔4, 1↔3, 2↔2, 3↔6, 4↔5, 5↔0, 6↔1


Digraph. Set of vertices connected pairwise by directed edges.

47

Directed graphs

A directed graph (digraph)

directedcycle

directed pathfrom 0 to 2

vertex of outdegree 3

and indegree 1


48

Road network

Vertex = intersection; edge = one-way street.

©2008 Google - Map data ©2008 Sanborn, NAVTEQ™ - Terms of Use


Vertex = synset; edge = hypernym relationship.

49

WordNet graph

http://wordnet.princeton.edu

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50

Digraph applications

digraph vertex directed edge

transportation street intersection one-way street

web web page hyperlink

food web species predator-prey relationship

WordNet synset hypernym

scheduling task precedence constraint

financial bank transaction

cell phone person placed call

infectious disease person infection

game board position legal move

citation journal article citation

object graph object pointer

inheritance hierarchy class inherits from

control flow code block jump


51

Some digraph problems

Path. Is there a directed path from s to t ?

Shortest path. What is the shortest directed path from s to t ?

Topological sort. Can you draw the digraph so that all edges point upwards?

Strong connectivity. Is there a directed path between all pairs of vertices?

Transitive closure. For which vertices v and w is there a path from v to w ?

PageRank. What is the importance of a web page?

NEW

NEW

NEW

NEW


Same as for undirected graphs.

In practice. Use adjacency-list representation.

• Algorithms based on iterating over vertices adjacent from v.

• Real-world digraphs tend to be sparse.

52

Digraph representations

representation spaceinsert edge

from v to w

edge from

v to w?

iterate over vertices

adjacent from v?

adjacency matrix V 2 1 † 1 V

adjacency list E + V 1 outdegree(v) outdegree(v)

† disallows parallel edges


Same method as for undirected graphs.

• Every undirected graph is a digraph (with edges in both directions).

• DFS is a digraph algorithm.

53

Depth-first search in digraphs

Mark v as visited.Recursively visit all unmarked vertices w adjacent from v.

DFS (to visit a vertex v)


Same method as for undirected graphs.

• Every undirected graph is a digraph (with edges in both directions).

• BFS is a digraph algorithm.

Proposition. BFS computes shortest paths (fewest number of edges).

54

Breadth-first search in digraphs

Is w reachable from v in this digraph?

v

w

s

Put s onto a FIFO queue, and mark s as visited.Repeat until the queue is empty: - remove the least recently added vertex v - for each unmarked vertex adjacent from v: add to queue and mark as visited..

BFS (from source vertex s)


Every data structure is a digraph.

• Vertex = object.

• Edge = reference.Roots. Objects known to be directly accessible by program (e.g., stack).

Reachable objects. Objects indirectly accessible by program(starting at a root and following a chain of pointers).

55

Reachability application: mark-sweep garbage collector

PROBLEM SESSION

Another approach to garbage collection is to

maintain a count of the number of references

to each object, and remove objects with

count 0.

- Argue that this will not always get rid of all

unused objects.

- How can we find cycles in a directed graph?


Goal. Given a set of tasks to be completed with precedence constraints,in which order should we schedule the tasks?

Graph model. vertex = task; edge = precedence constraint.

56

Precedence scheduling

tasksprecedence constraint graph

0

1

4

52

6

3

feasible schedule

0. Algorithms

1. Complexity Theory

2. Artificial Intelligence

3. Intro to CS

4. Cryptography

5. Scientific Computing

6. Advanced Programming


57

Topological sort

DAG. Directed acyclic graph.

Topological sort. Redraw DAG so all edges point up.

Solution. DFS. What else? topological order

directed edges DAG

0→5 0→2

0→1 3→6

3→5 3→4

5→4 6→4

6→0 3→2

1→4

0

1

4

52

6

3


58

Depth-first search order

public class DepthFirstOrder{ private boolean[] marked; private Stack<Integer> reversePost;

public DepthFirstOrder(Digraph G) { reversePost = new Stack<Integer>(); marked = new boolean[G.V()]; for (int v = 0; v < G.V(); v++) if (!marked[v]) dfs(G, v); }

private void dfs(Digraph G, int v) { marked[v] = true; for (int w : G.adj(v)) if (!marked[w]) dfs(G, w); reversePost.push(v); } public Iterable<Integer> reversePost() { return reversePost; }}

returns all vertices in

“reverse DFS postorder”


Reverse DFS postorder in a DAG

59

marked[] reversePost

0

1

4

52

6

3

0→5

0→2

0→1

3→6

3→5

3→4

5→4

6→4

6→0

3→2

1→4



59

dfs(0) 1 0 0 0 0 0 0 -


0

1

4

52

6

3

0→5

0→2

0→1

3→6

3→5

3→4

5→4

6→4

6→0

3→2

1→4



59

dfs(0) 1 0 0 0 0 0 0 - dfs(1) 1 1 0 0 0 0 0 -


0

1

4

52

6

3

0→5

0→2

0→1

3→6

3→5

3→4

5→4

6→4

6→0

3→2

1→4



59

dfs(0) 1 0 0 0 0 0 0 - dfs(1) 1 1 0 0 0 0 0 - dfs(4) 1 1 0 0 1 0 0 -


0

1

4

52

6

3

0→5

0→2

0→1

3→6

3→5

3→4

5→4

6→4

6→0

3→2

1→4



59

dfs(0) 1 0 0 0 0 0 0 - dfs(1) 1 1 0 0 0 0 0 - dfs(4) 1 1 0 0 1 0 0 - 4 done 1 1 0 0 1 0 0 4


0

1

4

52

6

3

0→5

0→2

0→1

3→6

3→5

3→4

5→4

6→4

6→0

3→2

1→4



59

dfs(0) 1 0 0 0 0 0 0 - dfs(1) 1 1 0 0 0 0 0 - dfs(4) 1 1 0 0 1 0 0 - 4 done 1 1 0 0 1 0 0 4 1 done 1 1 0 0 1 0 0 4 1


0

1

4

52

6

3

0→5

0→2

0→1

3→6

3→5

3→4

5→4

6→4

6→0

3→2

1→4



59

dfs(0) 1 0 0 0 0 0 0 - dfs(1) 1 1 0 0 0 0 0 - dfs(4) 1 1 0 0 1 0 0 - 4 done 1 1 0 0 1 0 0 4 1 done 1 1 0 0 1 0 0 4 1 dfs(2) 1 1 1 0 1 0 0 4 1


0

1

4

52

6

3

0→5

0→2

0→1

3→6

3→5

3→4

5→4

6→4

6→0

3→2

1→4



59

dfs(0) 1 0 0 0 0 0 0 - dfs(1) 1 1 0 0 0 0 0 - dfs(4) 1 1 0 0 1 0 0 - 4 done 1 1 0 0 1 0 0 4 1 done 1 1 0 0 1 0 0 4 1 dfs(2) 1 1 1 0 1 0 0 4 1 2 done 1 1 1 0 1 0 0 4 1 2


0

1

4

52

6

3

0→5

0→2

0→1

3→6

3→5

3→4

5→4

6→4

6→0

3→2

1→4



59

dfs(0) 1 0 0 0 0 0 0 - dfs(1) 1 1 0 0 0 0 0 - dfs(4) 1 1 0 0 1 0 0 - 4 done 1 1 0 0 1 0 0 4 1 done 1 1 0 0 1 0 0 4 1 dfs(2) 1 1 1 0 1 0 0 4 1 2 done 1 1 1 0 1 0 0 4 1 2 dfs(5) 1 1 1 0 1 1 0 4 1 2


0

1

4

52

6

3

0→5

0→2

0→1

3→6

3→5

3→4

5→4

6→4

6→0

3→2

1→4



59

dfs(0) 1 0 0 0 0 0 0 - dfs(1) 1 1 0 0 0 0 0 - dfs(4) 1 1 0 0 1 0 0 - 4 done 1 1 0 0 1 0 0 4 1 done 1 1 0 0 1 0 0 4 1 dfs(2) 1 1 1 0 1 0 0 4 1 2 done 1 1 1 0 1 0 0 4 1 2 dfs(5) 1 1 1 0 1 1 0 4 1 2 check 2 1 1 1 0 1 1 0 4 1 2


0

1

4

52

6

3

0→5

0→2

0→1

3→6

3→5

3→4

5→4

6→4

6→0

3→2

1→4



59

dfs(0) 1 0 0 0 0 0 0 - dfs(1) 1 1 0 0 0 0 0 - dfs(4) 1 1 0 0 1 0 0 - 4 done 1 1 0 0 1 0 0 4 1 done 1 1 0 0 1 0 0 4 1 dfs(2) 1 1 1 0 1 0 0 4 1 2 done 1 1 1 0 1 0 0 4 1 2 dfs(5) 1 1 1 0 1 1 0 4 1 2 check 2 1 1 1 0 1 1 0 4 1 2 5 done 1 1 1 0 1 1 0 4 1 2 5


0

1

4

52

6

3

0→5

0→2

0→1

3→6

3→5

3→4

5→4

6→4

6→0

3→2

1→4



59

dfs(0) 1 0 0 0 0 0 0 - dfs(1) 1 1 0 0 0 0 0 - dfs(4) 1 1 0 0 1 0 0 - 4 done 1 1 0 0 1 0 0 4 1 done 1 1 0 0 1 0 0 4 1 dfs(2) 1 1 1 0 1 0 0 4 1 2 done 1 1 1 0 1 0 0 4 1 2 dfs(5) 1 1 1 0 1 1 0 4 1 2 check 2 1 1 1 0 1 1 0 4 1 2 5 done 1 1 1 0 1 1 0 4 1 2 50 done 1 1 1 0 1 1 0 4 1 2 5 0


0

1

4

52

6

3

0→5

0→2

0→1

3→6

3→5

3→4

5→4

6→4

6→0

3→2

1→4



59

dfs(0) 1 0 0 0 0 0 0 - dfs(1) 1 1 0 0 0 0 0 - dfs(4) 1 1 0 0 1 0 0 - 4 done 1 1 0 0 1 0 0 4 1 done 1 1 0 0 1 0 0 4 1 dfs(2) 1 1 1 0 1 0 0 4 1 2 done 1 1 1 0 1 0 0 4 1 2 dfs(5) 1 1 1 0 1 1 0 4 1 2 check 2 1 1 1 0 1 1 0 4 1 2 5 done 1 1 1 0 1 1 0 4 1 2 50 done 1 1 1 0 1 1 0 4 1 2 5 0check 1 1 1 1 0 1 1 0 4 1 2 5 0


0

1

4

52

6

3

0→5

0→2

0→1

3→6

3→5

3→4

5→4

6→4

6→0

3→2

1→4



59

dfs(0) 1 0 0 0 0 0 0 - dfs(1) 1 1 0 0 0 0 0 - dfs(4) 1 1 0 0 1 0 0 - 4 done 1 1 0 0 1 0 0 4 1 done 1 1 0 0 1 0 0 4 1 dfs(2) 1 1 1 0 1 0 0 4 1 2 done 1 1 1 0 1 0 0 4 1 2 dfs(5) 1 1 1 0 1 1 0 4 1 2 check 2 1 1 1 0 1 1 0 4 1 2 5 done 1 1 1 0 1 1 0 4 1 2 50 done 1 1 1 0 1 1 0 4 1 2 5 0check 1 1 1 1 0 1 1 0 4 1 2 5 0check 2 1 1 1 0 1 1 0 4 1 2 5 0


0

1

4

52

6

3

0→5

0→2

0→1

3→6

3→5

3→4

5→4

6→4

6→0

3→2

1→4



59

dfs(0) 1 0 0 0 0 0 0 - dfs(1) 1 1 0 0 0 0 0 - dfs(4) 1 1 0 0 1 0 0 - 4 done 1 1 0 0 1 0 0 4 1 done 1 1 0 0 1 0 0 4 1 dfs(2) 1 1 1 0 1 0 0 4 1 2 done 1 1 1 0 1 0 0 4 1 2 dfs(5) 1 1 1 0 1 1 0 4 1 2 check 2 1 1 1 0 1 1 0 4 1 2 5 done 1 1 1 0 1 1 0 4 1 2 50 done 1 1 1 0 1 1 0 4 1 2 5 0check 1 1 1 1 0 1 1 0 4 1 2 5 0check 2 1 1 1 0 1 1 0 4 1 2 5 0dfs(3) 1 1 1 1 1 1 0 4 1 2 5 0


0

1

4

52

6

3

0→5

0→2

0→1

3→6

3→5

3→4

5→4

6→4

6→0

3→2

1→4



59

dfs(0) 1 0 0 0 0 0 0 - dfs(1) 1 1 0 0 0 0 0 - dfs(4) 1 1 0 0 1 0 0 - 4 done 1 1 0 0 1 0 0 4 1 done 1 1 0 0 1 0 0 4 1 dfs(2) 1 1 1 0 1 0 0 4 1 2 done 1 1 1 0 1 0 0 4 1 2 dfs(5) 1 1 1 0 1 1 0 4 1 2 check 2 1 1 1 0 1 1 0 4 1 2 5 done 1 1 1 0 1 1 0 4 1 2 50 done 1 1 1 0 1 1 0 4 1 2 5 0check 1 1 1 1 0 1 1 0 4 1 2 5 0check 2 1 1 1 0 1 1 0 4 1 2 5 0dfs(3) 1 1 1 1 1 1 0 4 1 2 5 0 check 2 1 1 1 1 1 1 0 4 1 2 5 0


0

1

4

52

6

3

0→5

0→2

0→1

3→6

3→5

3→4

5→4

6→4

6→0

3→2

1→4



59

dfs(0) 1 0 0 0 0 0 0 - dfs(1) 1 1 0 0 0 0 0 - dfs(4) 1 1 0 0 1 0 0 - 4 done 1 1 0 0 1 0 0 4 1 done 1 1 0 0 1 0 0 4 1 dfs(2) 1 1 1 0 1 0 0 4 1 2 done 1 1 1 0 1 0 0 4 1 2 dfs(5) 1 1 1 0 1 1 0 4 1 2 check 2 1 1 1 0 1 1 0 4 1 2 5 done 1 1 1 0 1 1 0 4 1 2 50 done 1 1 1 0 1 1 0 4 1 2 5 0check 1 1 1 1 0 1 1 0 4 1 2 5 0check 2 1 1 1 0 1 1 0 4 1 2 5 0dfs(3) 1 1 1 1 1 1 0 4 1 2 5 0 check 2 1 1 1 1 1 1 0 4 1 2 5 0 check 4 1 1 1 1 1 1 0 4 1 2 5 0


0

1

4

52

6

3

0→5

0→2

0→1

3→6

3→5

3→4

5→4

6→4

6→0

3→2

1→4



59

dfs(0) 1 0 0 0 0 0 0 - dfs(1) 1 1 0 0 0 0 0 - dfs(4) 1 1 0 0 1 0 0 - 4 done 1 1 0 0 1 0 0 4 1 done 1 1 0 0 1 0 0 4 1 dfs(2) 1 1 1 0 1 0 0 4 1 2 done 1 1 1 0 1 0 0 4 1 2 dfs(5) 1 1 1 0 1 1 0 4 1 2 check 2 1 1 1 0 1 1 0 4 1 2 5 done 1 1 1 0 1 1 0 4 1 2 50 done 1 1 1 0 1 1 0 4 1 2 5 0check 1 1 1 1 0 1 1 0 4 1 2 5 0check 2 1 1 1 0 1 1 0 4 1 2 5 0dfs(3) 1 1 1 1 1 1 0 4 1 2 5 0 check 2 1 1 1 1 1 1 0 4 1 2 5 0 check 4 1 1 1 1 1 1 0 4 1 2 5 0 check 5 1 1 1 1 1 1 0 4 1 2 5 0


0

1

4

52

6

3

0→5

0→2

0→1

3→6

3→5

3→4

5→4

6→4

6→0

3→2

1→4



59

dfs(0) 1 0 0 0 0 0 0 - dfs(1) 1 1 0 0 0 0 0 - dfs(4) 1 1 0 0 1 0 0 - 4 done 1 1 0 0 1 0 0 4 1 done 1 1 0 0 1 0 0 4 1 dfs(2) 1 1 1 0 1 0 0 4 1 2 done 1 1 1 0 1 0 0 4 1 2 dfs(5) 1 1 1 0 1 1 0 4 1 2 check 2 1 1 1 0 1 1 0 4 1 2 5 done 1 1 1 0 1 1 0 4 1 2 50 done 1 1 1 0 1 1 0 4 1 2 5 0check 1 1 1 1 0 1 1 0 4 1 2 5 0check 2 1 1 1 0 1 1 0 4 1 2 5 0dfs(3) 1 1 1 1 1 1 0 4 1 2 5 0 check 2 1 1 1 1 1 1 0 4 1 2 5 0 check 4 1 1 1 1 1 1 0 4 1 2 5 0 check 5 1 1 1 1 1 1 0 4 1 2 5 0 dfs(6) 1 1 1 1 1 1 1 4 1 2 5 0


0

1

4

52

6

3

0→5

0→2

0→1

3→6

3→5

3→4

5→4

6→4

6→0

3→2

1→4



59

dfs(0) 1 0 0 0 0 0 0 - dfs(1) 1 1 0 0 0 0 0 - dfs(4) 1 1 0 0 1 0 0 - 4 done 1 1 0 0 1 0 0 4 1 done 1 1 0 0 1 0 0 4 1 dfs(2) 1 1 1 0 1 0 0 4 1 2 done 1 1 1 0 1 0 0 4 1 2 dfs(5) 1 1 1 0 1 1 0 4 1 2 check 2 1 1 1 0 1 1 0 4 1 2 5 done 1 1 1 0 1 1 0 4 1 2 50 done 1 1 1 0 1 1 0 4 1 2 5 0check 1 1 1 1 0 1 1 0 4 1 2 5 0check 2 1 1 1 0 1 1 0 4 1 2 5 0dfs(3) 1 1 1 1 1 1 0 4 1 2 5 0 check 2 1 1 1 1 1 1 0 4 1 2 5 0 check 4 1 1 1 1 1 1 0 4 1 2 5 0 check 5 1 1 1 1 1 1 0 4 1 2 5 0 dfs(6) 1 1 1 1 1 1 1 4 1 2 5 0 6 done 1 1 1 1 1 1 1 4 1 2 5 0 6


0

1

4

52

6

3

0→5

0→2

0→1

3→6

3→5

3→4

5→4

6→4

6→0

3→2

1→4



59

dfs(0) 1 0 0 0 0 0 0 - dfs(1) 1 1 0 0 0 0 0 - dfs(4) 1 1 0 0 1 0 0 - 4 done 1 1 0 0 1 0 0 4 1 done 1 1 0 0 1 0 0 4 1 dfs(2) 1 1 1 0 1 0 0 4 1 2 done 1 1 1 0 1 0 0 4 1 2 dfs(5) 1 1 1 0 1 1 0 4 1 2 check 2 1 1 1 0 1 1 0 4 1 2 5 done 1 1 1 0 1 1 0 4 1 2 50 done 1 1 1 0 1 1 0 4 1 2 5 0check 1 1 1 1 0 1 1 0 4 1 2 5 0check 2 1 1 1 0 1 1 0 4 1 2 5 0dfs(3) 1 1 1 1 1 1 0 4 1 2 5 0 check 2 1 1 1 1 1 1 0 4 1 2 5 0 check 4 1 1 1 1 1 1 0 4 1 2 5 0 check 5 1 1 1 1 1 1 0 4 1 2 5 0 dfs(6) 1 1 1 1 1 1 1 4 1 2 5 0 6 done 1 1 1 1 1 1 1 4 1 2 5 0 63 done 1 1 1 1 1 1 1 4 1 2 5 0 6 3


0

1

4

52

6

3

0→5

0→2

0→1

3→6

3→5

3→4

5→4

6→4

6→0

3→2

1→4



59

dfs(0) 1 0 0 0 0 0 0 - dfs(1) 1 1 0 0 0 0 0 - dfs(4) 1 1 0 0 1 0 0 - 4 done 1 1 0 0 1 0 0 4 1 done 1 1 0 0 1 0 0 4 1 dfs(2) 1 1 1 0 1 0 0 4 1 2 done 1 1 1 0 1 0 0 4 1 2 dfs(5) 1 1 1 0 1 1 0 4 1 2 check 2 1 1 1 0 1 1 0 4 1 2 5 done 1 1 1 0 1 1 0 4 1 2 50 done 1 1 1 0 1 1 0 4 1 2 5 0check 1 1 1 1 0 1 1 0 4 1 2 5 0check 2 1 1 1 0 1 1 0 4 1 2 5 0dfs(3) 1 1 1 1 1 1 0 4 1 2 5 0 check 2 1 1 1 1 1 1 0 4 1 2 5 0 check 4 1 1 1 1 1 1 0 4 1 2 5 0 check 5 1 1 1 1 1 1 0 4 1 2 5 0 dfs(6) 1 1 1 1 1 1 1 4 1 2 5 0 6 done 1 1 1 1 1 1 1 4 1 2 5 0 63 done 1 1 1 1 1 1 1 4 1 2 5 0 6 3check 4 1 1 1 1 1 1 0 4 1 2 5 0 6 3


0

1

4

52

6

3

0→5

0→2

0→1

3→6

3→5

3→4

5→4

6→4

6→0

3→2

1→4



59

dfs(0) 1 0 0 0 0 0 0 - dfs(1) 1 1 0 0 0 0 0 - dfs(4) 1 1 0 0 1 0 0 - 4 done 1 1 0 0 1 0 0 4 1 done 1 1 0 0 1 0 0 4 1 dfs(2) 1 1 1 0 1 0 0 4 1 2 done 1 1 1 0 1 0 0 4 1 2 dfs(5) 1 1 1 0 1 1 0 4 1 2 check 2 1 1 1 0 1 1 0 4 1 2 5 done 1 1 1 0 1 1 0 4 1 2 50 done 1 1 1 0 1 1 0 4 1 2 5 0check 1 1 1 1 0 1 1 0 4 1 2 5 0check 2 1 1 1 0 1 1 0 4 1 2 5 0dfs(3) 1 1 1 1 1 1 0 4 1 2 5 0 check 2 1 1 1 1 1 1 0 4 1 2 5 0 check 4 1 1 1 1 1 1 0 4 1 2 5 0 check 5 1 1 1 1 1 1 0 4 1 2 5 0 dfs(6) 1 1 1 1 1 1 1 4 1 2 5 0 6 done 1 1 1 1 1 1 1 4 1 2 5 0 63 done 1 1 1 1 1 1 1 4 1 2 5 0 6 3check 4 1 1 1 1 1 1 0 4 1 2 5 0 6 3check 5 1 1 1 1 1 1 0 4 1 2 5 0 6 3


0

1

4

52

6

3

0→5

0→2

0→1

3→6

3→5

3→4

5→4

6→4

6→0

3→2

1→4



59

dfs(0) 1 0 0 0 0 0 0 - dfs(1) 1 1 0 0 0 0 0 - dfs(4) 1 1 0 0 1 0 0 - 4 done 1 1 0 0 1 0 0 4 1 done 1 1 0 0 1 0 0 4 1 dfs(2) 1 1 1 0 1 0 0 4 1 2 done 1 1 1 0 1 0 0 4 1 2 dfs(5) 1 1 1 0 1 1 0 4 1 2 check 2 1 1 1 0 1 1 0 4 1 2 5 done 1 1 1 0 1 1 0 4 1 2 50 done 1 1 1 0 1 1 0 4 1 2 5 0check 1 1 1 1 0 1 1 0 4 1 2 5 0check 2 1 1 1 0 1 1 0 4 1 2 5 0dfs(3) 1 1 1 1 1 1 0 4 1 2 5 0 check 2 1 1 1 1 1 1 0 4 1 2 5 0 check 4 1 1 1 1 1 1 0 4 1 2 5 0 check 5 1 1 1 1 1 1 0 4 1 2 5 0 dfs(6) 1 1 1 1 1 1 1 4 1 2 5 0 6 done 1 1 1 1 1 1 1 4 1 2 5 0 63 done 1 1 1 1 1 1 1 4 1 2 5 0 6 3check 4 1 1 1 1 1 1 0 4 1 2 5 0 6 3check 5 1 1 1 1 1 1 0 4 1 2 5 0 6 3check 6 1 1 1 1 1 1 0 4 1 2 5 0 6 3


0

1

4

52

6

3

0→5

0→2

0→1

3→6

3→5

3→4

5→4

6→4

6→0

3→2

1→4



59

dfs(0) 1 0 0 0 0 0 0 - dfs(1) 1 1 0 0 0 0 0 - dfs(4) 1 1 0 0 1 0 0 - 4 done 1 1 0 0 1 0 0 4 1 done 1 1 0 0 1 0 0 4 1 dfs(2) 1 1 1 0 1 0 0 4 1 2 done 1 1 1 0 1 0 0 4 1 2 dfs(5) 1 1 1 0 1 1 0 4 1 2 check 2 1 1 1 0 1 1 0 4 1 2 5 done 1 1 1 0 1 1 0 4 1 2 50 done 1 1 1 0 1 1 0 4 1 2 5 0check 1 1 1 1 0 1 1 0 4 1 2 5 0check 2 1 1 1 0 1 1 0 4 1 2 5 0dfs(3) 1 1 1 1 1 1 0 4 1 2 5 0 check 2 1 1 1 1 1 1 0 4 1 2 5 0 check 4 1 1 1 1 1 1 0 4 1 2 5 0 check 5 1 1 1 1 1 1 0 4 1 2 5 0 dfs(6) 1 1 1 1 1 1 1 4 1 2 5 0 6 done 1 1 1 1 1 1 1 4 1 2 5 0 63 done 1 1 1 1 1 1 1 4 1 2 5 0 6 3check 4 1 1 1 1 1 1 0 4 1 2 5 0 6 3check 5 1 1 1 1 1 1 0 4 1 2 5 0 6 3check 6 1 1 1 1 1 1 0 4 1 2 5 0 6 3done 1 1 1 1 1 1 1 4 1 2 5 0 6 3


0

1

4

52

6

3

0→5

0→2

0→1

3→6

3→5

3→4

5→4

6→4

6→0

3→2

1→4



59

dfs(0) 1 0 0 0 0 0 0 - dfs(1) 1 1 0 0 0 0 0 - dfs(4) 1 1 0 0 1 0 0 - 4 done 1 1 0 0 1 0 0 4 1 done 1 1 0 0 1 0 0 4 1 dfs(2) 1 1 1 0 1 0 0 4 1 2 done 1 1 1 0 1 0 0 4 1 2 dfs(5) 1 1 1 0 1 1 0 4 1 2 check 2 1 1 1 0 1 1 0 4 1 2 5 done 1 1 1 0 1 1 0 4 1 2 50 done 1 1 1 0 1 1 0 4 1 2 5 0check 1 1 1 1 0 1 1 0 4 1 2 5 0check 2 1 1 1 0 1 1 0 4 1 2 5 0dfs(3) 1 1 1 1 1 1 0 4 1 2 5 0 check 2 1 1 1 1 1 1 0 4 1 2 5 0 check 4 1 1 1 1 1 1 0 4 1 2 5 0 check 5 1 1 1 1 1 1 0 4 1 2 5 0 dfs(6) 1 1 1 1 1 1 1 4 1 2 5 0 6 done 1 1 1 1 1 1 1 4 1 2 5 0 63 done 1 1 1 1 1 1 1 4 1 2 5 0 6 3check 4 1 1 1 1 1 1 0 4 1 2 5 0 6 3check 5 1 1 1 1 1 1 0 4 1 2 5 0 6 3check 6 1 1 1 1 1 1 0 4 1 2 5 0 6 3done 1 1 1 1 1 1 1 4 1 2 5 0 6 3


0

1

4

52

6

3

reverse DFS

postorder is a

topological

order!

0→5

0→2

0→1

3→6

3→5

3→4

5→4

6→4

6→0

3→2

1→4


Proposition. Reverse DFS postorder of a DAG is a topological order.Pf. Consider any edge v→w. When dfs(G, v) is called:

• Case 1: dfs(G, w) has already been called and returned.Thus, w was put on the stack before v.

• Case 2: dfs(G, w) has not yet been called.It will get called directly or indirectlyby dfs(G, v) and will finish before dfs(G, v).Thus, w will be put on the stack before v.

• Case 3: dfs(G, w) has already been called, but has not returned. Can’t happen in a DAG: function call stack containspath from w to v, so v→w would complete a cycle.

dfs(0) dfs(1) dfs(4) 4 done 1 done dfs(2) 2 done dfs(5) check 2 5 done 0 done check 1 check 2 dfs(3) check 2 check 4 check 5 dfs(6) 6 done 3 done check 4 check 5 check 6 done

60

Topological sort in a DAG: correctness proof

all vertices adjacent from 3 are done before 3 is done,

so they appear after 3 in topological order

Ex:

case 1

case 2


Microsoft Excel does cycle detection (and has a circular reference toolbar!)

61

Directed cycle detection application: spreadsheet recalculation


Conclusion

Graphs can be used to model a large variety of computational problems.

Efficient graph algorithms (more to come!) can often be used as building blocks for solving given problems.

Releted course goals:• Choose among and make use of the most important algorithms and data

structures in libraries, based on knowledge of their complexity.• Design algorithms for ad hoc problems by using and combining known

algorithms and data structures.

62


basic graph search - itu.dkitu.dk/people/pagh/ads11/08-graphsearch.pdf · circuit gate, register,...

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