advanced dfs, bfs, graph modeling

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Advanced DFS, BFS, Graph Modeling

19/2/2005

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Introduction

• Depth-first search (DFS)

• Breadth-first search (BFS)

• Graph Modeling– Model a graph from a problem, ie.

transform a problem into a graph problem

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What is a graph?• A set of vertices and edges

vertex

edge

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Trees and related terms

root

siblings

descendents children

ancestors

parent

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Graph Traversal• Given: a graph• Goal: visit all (or some) vertices and edges of

the graph using some strategy (the order of visit is systematic)

• DFS, BFS are examples of graph traversal algorithms

• Some shortest path algorithms and spanning tree algorithms have specific visit order

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Intuition of DFS of BFS• This is a brief idea of DFS, BFS

• DFS: continue visiting next vertex whenever there is a road, go back if no road (ie. visit to the depth of current path)– Example: a human want to visit a place, but do not know the

path

• BFS: go through all the adjacent vertices before going further (ie. spread among next vertices)– Example: set a house on fire, the fire will spread through the

house

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DFS (pseudo code)

DFS (vertex u) {mark u as visitedfor each vertex v directly reachable from u

if v is unvisitedDFS (v)

}

• Initially all vertices are marked as unvisited

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F

A

BC

D

E

DFS (Demonstration)

unvisited

visited

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“Advanced” DFS

• Apart from just visiting the vertices, DFS can also provide us with valuable information

• DFS can be enhanced by introducing:– birth time and death time of a vertex

• birth time: when the vertex is first visited• death time: when we retreat from the vertex

– DFS tree– parent of a vertex

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DFS spanning tree / forest• A rooted tree• The root is the start vertex• If v is first visited from u, then u is the parent of v in

the DFS tree• Edges are those in forward direction of DFS, ie. when

visiting vertices that are not visited before

• If some vertices are not reachable from the start vertex, those vertices will form other spanning trees (1 or more)

• The collection of the trees are called forest

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DFS (pseudo code)

DFS (vertex u) {

mark u as visited

time time+1

for each vertex v directly reachable from u

if v is unvisited

DFS (v)

time time+1

}

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A

F

B

C

D

E

GH

DFS forest (Demonstration)

unvisited

visited

visited (dead)

A B C D E F G H

birth

death

parent

A

B

C

F

E

D

G

1 2 3 13 10 4 14

12 9 8 16 11 5 15

H

6

7

- A B - A C D C

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Classification of edges

• Tree edge• Forward edge• Back edge• Cross edge

• Question: which type of edges is always absent in an undirected graph?

A

B

C

F

E

D

G

H

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Determination of edge types• How to determine the type of an arbitrary

edge (u, v) after DFS?• Tree edge

– parent [v] = u

• Forward edge– not a tree edge; and– birth [v] > birth [u]; and– death [v] < death [u]

• How about back edge and cross edge?

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Determination of edge types

Tree edge

Forward Edge Back Edge Cross Edge

parent [v] = u

not a tree edge

birth[v] > birth[u]

death[v] < death[u]

birth[v] < birth[u]

death[v] > death[u]

birth[v] > birth[u]

death[v] >death[u]

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Applications of DFS Forests

• Topological sorting (Tsort)

• Strongly-connected components (SCC)

• Some more “advanced” algorithms

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Example: Tsort• Topological order: A numbering of the

vertices of a directed acyclic graph such that every edge from a vertex numbered i to a vertex numbered j satisfies i<j

• Tsort: Number the vertices in topological order 1

2

3

4 5

6

7

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Tsort Algorithm• If the graph has more then one vertex that

has indegree 0, add a vertice to connect to all indegree-0 vertices

• Let the indegree 0 vertice be s• Use s as start vertice, and compute the DFS

forest• The death time of the vertices represent the

reverse of topological order

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S

D

B

E F

C

G

A

S A B C D E F G

birth 1 2 3 4 12 13 8 5

death 16 11 10 7 15 14 9 6

G C F B A E D D E A B F C G

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Example: SCC• A graph is strongly-connected if

– for any pair of vertices u and v, one can go from u to v and from v to u.

• Informally speaking, an SCC of a graph is a subset of vertices that– forms a strongly-connected subgraph– does not form a strongly-connected

subgraph with the addition of any new vertex

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SCC (Illustration)

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SCC (Algorithm)• Compute the DFS forest of the graph G to

get the death time of the vertices• Reverse all edges in G to form G’• Compute a DFS forest of G’, but always

choose the vertex with the latest death time when choosing the root for a new tree

• The SCCs of G are the DFS trees in the DFS forest of G’

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A

F

B

C

D

GH

SCC (Demonstration)

A

F

B

C

D

E

GH

A B C D E F G H

birth

death

parent

1 2 3 13 10 4 14

12 9 8 16 11 5 15

6

7

- A B - A C D C

D

G

A E B

F

C

H

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SCC (Demonstration)

D

G

A E B

F

C

H

A

F

B

C

D

GH

E

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DFS Summary

• DFS spanning tree / forest

• We can use birth time and death time in DFS spanning tree to do varies things, such as Tsort, SCC

• Notice that in the previous slides, we related birth time and death time. But in the discussed applications, birth time and death time can be independent, ie. birth time and death time can use different time counter

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DFS (vertex u) {

mark u as visited

birthtime birthtime + 1

for each vertex v directly reachable from u

if v is unvisited

DFS (v)

deathtime deathtime + 1

}

time time+1

time time+1

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Breadth-first search (BFS)

• In order to “spread”, we need to makes use of a data structure, queue ,to remember just visited vertices

• Revised:• DFS: continue visiting next

vertex whenever there is a road, go back if no road (ie. visit to the depth of current path)

• BFS: go through all the adjacent vertices before going further (ie. spread among next vertices)

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BFS (Pseudo code)while queue not empty

dequeue the first vertex u from queue

for each vertex v directly reachable from u

if v is unvisited

enqueue v to queue

mark v as visited

• Initially all vertices except the start vertex are marked as unvisited and the queue contains the start vertex only

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A

B

C

D

E

F

G

H

I

J

BFS (Demonstration)

unvisited

visited

visited (dequeued)

Queue: A B C F D E H G J I

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Applications of BFS

• Shortest paths finding

• Flood-fill (can also be handled by DFS)

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Comparisons of DFS and BFS

DFS BFS

Depth-first Breadth-first

Stack Queue

Does not guarantee shortest paths

Guarantees shortest paths

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Bidirectional search (BDS)

• Searches simultaneously from both the start vertex and goal vertex

• Commonly implemented as bidirectional BFS

start goal

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BDS Example: Bomber Man (1 Bomb)

• find the shortest path from the upper-left corner to the lower-right corner in a maze using a bomb. The bomb can destroy a wall.

S

E

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Bomber Man (1 Bomb)

S

E

1 2 3

4

1234

5

4

Shortest Path length = 7

5

6 67 78 8

9

9

10

10

11

11

12

12 12

13

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What will happen if we stop once we find a path?

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S 1 2 3 4 5 6 7 8 9

10

21 11

20 12

19 13

18 14

17 17 16 15

11 12 13 14 15 16

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9 8 7 6 5 4 3 2 1 E

Example

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Iterative deepening search (IDS)

• Iteratively performs DFS with increasing depth bound

• Shortest paths are guaranteed

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IDS

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IDS (pseudo code)DFS (vertex u, depth d) {

mark u as visitedif (d>0)

for each vertex v directly reachable from uif v is unvisited

DFS (v,d-1)}

i=0Do {

DFS(start vertex,i)Increment i

}While (target is not found)

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IDS Complexity (the details can be skipped)

• ( )=bm

• ( ) =bm

1

0

1

1

nnk

k

rr

r

1

(1 )

1

nnk

k

r rr

r

1

0

1

1

ddk

dk

bt b

b

0

1

0

1

2

1

1

1 (1 )

1 (1 )

m

ii

im

i

m

t

b

b

m b b

b b

0

m

ii

t

- b is branching factor- td is the number of vertices visited for depth d

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1

m

m

bt

b

mt

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IDS

• Conclusion: the complexity of IDS is the same as DFS

• ( )=bm

• ( ) =bm0

m

ii

tmt

- b is branching factor- td is the number of vertices visited for depth d

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Summary of DFS, BFS

• We learned some variations of DFS and BFS – Bidirectional search (BDS) – Iterative deepening search (IDS)

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What is graph modeling?

• Conversion of a problem into a graph problem

• Sometimes a problem can be easily solved once its underlying graph model is recognized

• Graph modeling appears almost every year in NOI or IOI

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Basics of graph modeling

• A few steps:– identify the vertices and the edges– identify the objective of the problem– state the objective in graph terms– implementation:

• construct the graph from the input instance• run the suitable graph algorithms on the graph• convert the output to the required format

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Simple examples (1)• Given a grid maze with obstacles, find a

shortest path between two given points

start

goal

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Simple examples (2)

• A student has the phone numbers of some other students

• Suppose you know all pairs (A, B) such that A has B’s number

• Now you want to know Alan’s number, what is the minimum number of calls you need to make?

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Simple examples (2)

• Vertex: student

• Edge: whether A has B’s number

• Add an edge from A to B if A has B’s number

• Problem: find a shortest path from your vertex to Alan’s vertex

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Complex examples (1)• Same settings as simple example 1

• You know a trick – walking through an obstacle! However, it can be used for only once (Bomber Man with 1 bomb)

• What should a vertex represent?– your position only?– your position + whether you have used the

trick

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Complex examples (1)

• A vertex is in the form (position, used)

• The vertices are divided into two groups– trick used– trick not used

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Complex examples (1)

start

goal

unused

used

start goal

goal

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Complex examples (1)

• How about you can walk through obstacles for k times?

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Complex examples (2)• The famous 8-puzzle

• Given a state, find the moves that bring it to the goal state

1 2 3

4 5 6

7 8

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Complex examples (2)• What does a vertex represent?

– the position of the empty square?– the number of tiles that are in wrong

positions?– the state (the positions of the eight tiles)

• What are the edges?

• What is the equivalent graph problem?

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Complex examples (2)1 2 34 5 67 8

1 2 34 5 67 8

1 2 34 57 8 6

1 2 34 67 5 8

1 2 34 5 6

7 8

1 24 5 37 8 6

1 2 34 57 8 6

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Complex examples (3)• Theseus and Minotaur

– http://www.logicmazes.com/theseus.html– Extract:

• Theseus must escape from a maze. There is also a mechanical Minotaur in the maze. For every turn that Theseus takes, the Minotaur takes two turns. The Minotaur follows this program for each of his two turns:

• First he tests if he can move horizontally and get closer to Theseus. If he can, he will move one square horizontally. If he can’t, he will test if he could move vertically and get closer to Theseus. If he can, he will move one square vertically. If he can’t move either horizontally or vertically, then he just skips that turn.

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Complex examples (3)• What does a vertex represent?

– Theseus’ position– Minotaur’s position– Both

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Some more examples• How can the followings be modeled?

– Tilt maze (Single-goal mazes only)• http://www.clickmazes.com/newtilt/ixtilt2d.htm

– Double title maze• http://www.clickmazes.com/newtilt/ixtilt.htm

– No-left-turn maze• http://www.clickmazes.com/noleft/ixnoleft.htm

– Same as complex example 1, but you can use the trick for k times

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Competition problems• HKOI2000 S – Wormhole Labyrinth• HKOI2001 S – A Node Too Far• HKOI2004 S – Teacher’s Problem *• TFT2001 – OIMan * • TFT2002 – Bomber Man *• NOI2001 – cung1 ming4 dik7 daa2 zi6 jyun4• NOI2001 – Equation *• IOI2000 – Walls • IOI2002 – Troublesome Frog• IOI2003 – Amazing Robots

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Teacher’s Problem

• Question: A teacher wants to distribute sweets to students in an order such that, if student u tease student v, u should not get the sweet before v

• Vertex: student• Edge: directed, (v,u) is a directed edge if

student v tease u• Algorithm: Tsort

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OI Man• Question: OIMan (O) has 2 kinds of form:

H- and S-form. He can transform to S-form for m minutes by battery. He can only kill monsters (M) and virus (V) in S-form. Given the number of battery and m, find the minimum time needed to kill the virus.

• Vertex: position, form, time left for S-form, number of batteries left

• Edge: directed– Move to P in H-form (position)– Move to P/M/V in S-form (position, S-form time)– Use battery and move (position, form, S-form

time, number of batteries)

PWWPPPPPPWWPOMMV

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Equation

• Question: Find the number of solution of xi, given ki,pi. 1<=n<=6, 1<=xi<=150

• Vertex: possible values of – k1x1

p1 , k1x1p1 + k2x2

p2 , k1x1p1 + k2x2

p2 + k3x3p3 , k4x4

p4 , k4x4

p4 + k5x5p5 , k4x4

p4 + k5x5p5 + k6x6

p6