ALgorithm Full First Ctvcxvcxv Syllabus

8/12/2019 ALgorithm Full First Ctvcxvcxv Syllabus

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GRAPHALGORITHMS

Dr. Tanzima Hashem

Assistant Professor

CSE, BUET


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GRAPHS

A graph G = (V, E)

V = set of vertices

E = set of edges = subset of V V

Thus |E| = O(|V|2)


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GRAPHVARIATIONS

Variations:

A connected graphhas a path from every vertex to

every other

In an undirected graph:Edge (u,v) = edge (v,u)

No self-loops

In a directedgraph:

Edge (u,v) goes from vertex u to vertex v,notated uv

Self loops are allowed.


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GRAPHS

We will typically express running times in terms of |E|

and |V| (often dropping the |s)

If |E| |V|2the graph is dense

If |E| |V| the graph is sparse If you know you are dealing with dense or sparse

graphs, different data structures may make sense


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REPRESENTINGGRAPHS

Assume V = {1, 2, , n}

An adjacency matrixrepresents the graph as a n x n

matrix A:

A[i,j] = 1 if edge (i,j) E (or weight of edge)= 0 if edge (i,j) E


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GRAPHS: ADJACENCYMATRIX

Example:

1

2 4

3

a

d

b c

A 1 2 3 4

1

2

3??

4


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Example:

1

2 4

3

a

d

b c

A 1 2 3 4

1 0 1 1 0

2 0 0 1 0

3 0 0 0 0

4 0 0 1 0


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Space:(V2).

Not memory efficient for large graphs.

Time:to list all vertices adjacent to u: (V).

Time:to determine if (u, v)E: (1).


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The adjacency matrix is a dense representation

Usually too much storage for large graphs

But can be very efficient for small graphs

Most large interesting graphs are sparse E.g., planar graphs, in which no edges cross, have

|E| = O(|V|) by Eulers formula

For this reason the adjacency listis often a more

appropriate respresentation


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GRAPHS: ADJACENCYLIST

Adjacency list: for each vertex v V, store a list of

vertices adjacent to v

Example:

Adj[1] = {2,3} Adj[2] = {3}

Adj[3] = {}

Adj[4] = {3}

Variation: can also keepa list of edges coming into vertex

1

2 4

3


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GRAPHS: ADJACENCYLIST

For directed graphs:

Sum of lengths of all adj. lists is

out-degree(v) = |E|

v

V

Total storage:(V+E)

For undirected graphs:

Sum of lengths of all adj. lists is

degree(v) = 2|E|vV

Total storage:(V+E)

No. of edges leaving v

No. of edges incident on v. Edge (u,v) is

incident on vertices uand v.


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GRAPHDEFINITIONS

Path

Sequence of nodes n1, n2, nk

Edge exists between each pair of nodes ni ,ni+1

ExampleA, B, Cis a path


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GRAPHDEFINITIONS

Path

Sequence of nodes n1, n2, nk

Edge exists between each pair of nodes ni ,ni+1

ExampleA, B, Cis a path

A, E, Dis not a path


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GRAPHDEFINITIONS

Cycle

Path that ends back at starting node

Example

A, E, A


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GRAPHDEFINITIONS

Cycle

Path that ends back at starting node

Example

A, E, AA, B, C, D, E, A

Simple path

No cycles in path

Acyclic graph

No cycles in graph


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GRAPHSEARCHING

Given: a graph G = (V, E), directed or undirected

Goal: methodically explore every vertex and every

edge

Ultimately: build a tree on the graph Pick a vertex as the root

Choose certain edges to produce a tree

Note: might also build a forestif graph is not

connected


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BREADTH-FIRSTSEARCH

Explore a graph, turning it into a tree

One vertex at a time

Expand frontier of explored vertices across the

breadthof the frontier Builds a tree over the graph

Pick a source vertexto be the root

Find (discover) its children, then their children,

etc.


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BREADTH-FIRSTSEARCH

Input:Graph G = (V, E), either directed orundirected, and source vertex s V.

Output:

d[v]=

distance (smallest # of edges, or shortestpath) from s to v, for all v V. d[v] = if vis notreachable from s.

[v] = u such that (u, v)is last edge on shortestpath s v.

uis vspredecessor.

Builds breadth-first tree with root sthat contains allreachable vertices.


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BREADTH-FIRSTSEARCH

Associate vertex colors to guide the algorithm

White vertices have not been discovered

All vertices start out white

Grey vertices are discovered but not fully exploredThey may be adjacent to white vertices

Black vertices are discovered and fully explored

They are adjacent only to black and gray

vertices Explore vertices by scanning adjacency list of grey

vertices


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BFS(G,s)

1. foreach vertex u in V[G] {s}

2 color[u] white

3 d[u]

4 [u] nil5 color[s] gray

6 d[s] 0

7 [s] nil

8 Q

9 enqueue(Q,s)10whileQ

11 u dequeue(Q)

12 foreach vin Adj[u]

13 ifcolor[v] = white

14 color[v] gray15 d[v] d[u] + 1

16 [v] u

17 enqueue(Q,v)

18 color[u] black

white: undiscoveredgray: discovered

black: finished

Q: a queue of discovered

vertices

color[v]: color of vd[v]: distance from s to v

[u]: predecessor of v

initialization

access source s


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BREADTH-FIRSTSEARCH: EXAMPLE

r s t u

v w x y


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0

r s t u

v w x y

sQ:


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1

0

1

r s t u

v w x y

wQ: r


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1

0

1

2

2

r s t u

v w x y

rQ: t x


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1

2

0

1

2

2

r s t u

v w x y

Q: t x v


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1

2

0

1

2

2

3

r s t u

v w x y

Q: x v u


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1

2

0

1

2

2

3

3

r s t u

v w x y

Q: v u y


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1

2

0

1

2

2

3

3

r s t u

v w x y

Q: u y


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1

2

0

1

2

2

3

3

r s t u

v w x y

Q:


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ANALYSISOFBFS

Initialization takes O(|V|).

Traversal Loop After initialization, each vertex is enqueued and

dequeued at most once, and each operation takesO(1). So, total time for queuing is O(|V|).

The adjacency list of each vertex is scanned at mostonce. The total time spent in scanning adjacency lists

is O(|E|).

Summing up over all vertices => total running time ofBFS isO(|V| + |E|)


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BREADTH-FIRSTTREE

For a graph G= (V, E) with source s, thepredecessor subgraphof Gis G

= (V

, E) where

V={vV : [v] nil} {s}

E={([v], v) E : v V - {s}}

The predecessor subgraph Gis a breadth-first treeif:

V consists of the vertices reachable from sand

for all v V, there is a unique simple path from s to v

in G that is also a shortest path from sto vin G.

The edges in Eare called tree edges.

|E| = |V

| - 1.


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DEPTH-FIRSTSEARCH(DFS)

Explore edges out of the most recently discovered

vertex v.

When all edges of vhave been explored, backtrack

to explore other edges leaving the vertex from which

vwas discovered (itspredecessor).

Search as deep as possible first.

Continue until all vertices reachable from the original

source are discovered.

If any undiscovered vertices remain, then one of

them is chosen as a new source and search is

repeated from that source.


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DEPTH-FIRSTSEARCH

Input:G = (V,E), directed or undirected. No

source vertex given!

Output:

2 timestampson each vertex. d[v] = discovery t ime (v turns from white to gray)

f [v] = f in ish ing t ime(vturns from gray to black)

[v] : predecessor of v = u, such that vwas

discovered during the scan of us adjacency list. Depth-first forest


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DFS(G)

1. foreach vertex u V[G]

2. docolor[u] white

3. [u] NIL

4. time0

5. foreach vertex u V[G]

6. doifcolor[u] = white

7. thenDFS-Visit(u)

Uses a global timestamp t ime.

DFS-Visit(u)

1. color [u] GRAY // White vertex u

has been discovered

2. timetime+ 1

3. d[u] time

4. foreach v Adj[u]

5. doifcolor[v] = WHITE

6. then[v] u

7. DFS-Visit(v)8. color[u] BLACK // Blacken u;

it is finished.

9. f[u] time time + 1

PSEUDOCODE


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DFS EXAMPLE


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DFS EXAMPLE

1 | | |

|||

| |

d f


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DFS EXAMPLE

1 | | |

|||

2 | |

d f


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DFS EXAMPLE

1 | | |

||3 |

2 | |

d f


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DFS EXAMPLE

1 | | |

||3 | 4

2 | |

d f


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DFS EXAMPLE

1 | | |

|5 |3 | 4

2 | |

d f


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DFS EXAMPLE

1 | | |

|5 | 63 | 4

2 | |

d f


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DFS EXAMPLE

1 | 8 | |

|5 | 63 | 4

2 | 7 |

d f


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DFS EXAMPLE

1 | 8 | |

|5 | 63 | 4

2 | 7 |

d f


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DFS EXAMPLE

1 | 8 | |

|5 | 63 | 4

2 | 7 9 |

d f


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DFS EXAMPLE

1 | 8 | |

|5 | 63 | 4

2 | 7 9 |10

d f


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DFS EXAMPLE

1 | 8 |11 |

|5 | 63 | 4

2 | 7 9 |10

d f


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DFS EXAMPLE

1 |12 8 |11 |

|5 | 63 | 4

2 | 7 9 |10

d f


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DFS EXAMPLE

1 |12 8 |11 13|

|5 | 63 | 4

2 | 7 9 |10

d f


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DFS EXAMPLE

1 |12 8 |11 13|

14|5 | 63 | 4

2 | 7 9 |10

d f


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DFS EXAMPLE

1 |12 8 |11 13|

14|155 | 63 | 4

2 | 7 9 |10

d f


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DFS EXAMPLE

1 |12 8 |11 13|16

14|155 | 63 | 4

2 | 7 9 |10

d f


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ANALYSISOFDFS

Loops on lines 1-3 & 5-7 take (V)time, excluding

time to execute DFS-Visit.

DFS-Visit is called once for each white vertex vV

when its painted gray the first time. Lines 4-7 ofDFS-Visit is executed |Adj[v]| times. The total cost of

executing DFS-Visit is vV|Adj[v]| = (E)

Total running time of DFS is(|V| + |E|).


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TIME-STAMPSTRUCTUREINDFS

There is also a nice structure to the time stamps,

which is referred to as Parenthesis Structure.

Theorem 22.7

For all u, v, exactly one of the following holds:1. d[u] < f [u] < d[v] < f [v] or d[v] < f [v] < d[u] < f [u] and

neither u nor v is a descendant of the other.

2. d[u] < d[v] < f [v] < f [u] and v is a descendant of u.

3. d[v] < d[u] < f [u] < f [v] and u is a descendant of v.


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TIME-STAMPSTRUCTUREINDFS

So d[u] < d[v] < f [u] < f [v] cannot happen.

Like parentheses:

OK: ( ) [ ] ( [ ] ) [ ( ) ]

Not OK: ( [ ) ] [ ( ] )

Corol lary

v is a proper descendant of u if and only if

d[u] < d[v]< f [v] < f [u].


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DFS: KINDSOFEDGES

Consider a directed graph G = (V, E). After a DFS of

graph G we can put each edge into one of four

classes:

A tree edgeis an edge in a DFS-tree.

A back edgeconnects a vertex to an ancestor in a

DFS-tree. Note that a self-loop is a back edge.

A forward edgeis a non-tree edge that connects

a vertex to a descendent in a DFS-tree.

A cross edgeis any other edge in graph G. It

connects vertices in two different DFS-tree or two

vertices in the same DFS-tree neither of which is

the ancestor of the other.


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EXAMPLEOFCLASSIFYINGEDGES

a

d

b e

fc

in DFS forest

not in DFS

forest

tree

tree

tree

treeforward

back

back

cross


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61

CLASSIFYING EDGESOFADIGRAPH

(u, v) is:

Tree edgeif v is white

Back edgeif v is gray

Forward or cross - if v is black (u, v) is:

Forward edgeif v is black and d[u] < d[v] (v was

discovered after u)

Cross edgeif v is black and d[u] > d[v] (udiscovered after v)


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DFS: KINDSOFEDGES

DFS-Visit(u) with edge classification. G must be adirected graph

1. color[u] GRAY2. time time + 13. d[u] time4. foreach vertex vadjacent to u

5. do ifcolor[v] BLACK6. then ifd[u] < d[v]7. thenClassify (u, v) as a forward edge8. elseClassify (u, v) as a cross edge9. ifcolor[v] GRAY10. thenClassify (u, v) as a back edge11. ifcolor[v] WHITE12. then[v] u13. Classify (u, v) as a tree edge14. DFS-Visit(v)15. color[u] BLACK16. time time + 117. f[u] time


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DFS: KINDSOFEDGES

Suppose G be an undirected graph, then we havefollowing edge classification:

Tree Edge an edge connects a vertex with its

parent.

Back Edge a non-tree edge connects a vertex

with an ancestor.

Forward Edge There is no forward edges

because they become back edges when

considered in the opposite direction.

Cross Edge There cannot be any cross edge

because every edge of G must connect an

ancestor with a descendant.


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64

SOMEAPPLICATIONSOFBFSANDDFS

BFS

To find the shortest path from a vertex s to a vertex

v in an unweighted graph

To find the length of such a path

Find the bipartiteness of a graph.

DFS

To find a path from a vertex s to a vertex v. To find the length of such a path.

To find out if a graph contains cycles


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APPLICATIONOFBFS: BIPARTITEGRAPH

non bipartite

bipartite:

Question: given a graph G, how to test if the graph is

bipartite?


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APPLICATIONOFBFS: BIPARTITEGRAPH

Foreach vertex uin V[G] {s}docolor[u] WHITE

d[u] partition[u] 0

color[s] GRAYpartition[s] 1d[s] 0Q [s]whileQueue 'Q' is non-empty

dou head [Q]foreach vin Adj[u] do

ifpartition [u] = partition [v] then

return 0elseifcolor[v] WHITE then

color[v] gray

d[v] = d[u] + 1partition[v] 3 partition[u]ENQUEUE (Q, v)

DEQUEUE (Q)Color[u] BLACK

Return 1

A DFS


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APPLICATIONOFDFS:

DETECTINGCYCLEFORDIRECTEDGRAPH

DFS_visit(u)

color(u) GRAY

d[u] time time + 1

foreach vadjacent to u do

ifcolor[v] GRAY then

return "cycle exists"

else ifcolor[v] WHITE then

predecessor[v] u

DFS_visit(v)

color[u] BLACK

f[u] time time + 1


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APPLICATIONOFDFS:

DETECTINGCYCLEFORUNDIRECTEDGRAPH

DFS_visit(u)

color(u) GRAY

d[u] time time + 1

foreach vadjacent to u do

ifcolor[v] GRAY and [u] v then

return "cycle exists"

else ifcolor[v] WHITE then

predecessor[v] u

DFS_visit(v)

color[u] BLACK

f[u] time time + 1


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TOPOLOGICALSORT

Want to sort a directed acyclic graph (DAG).B

E

D

C

A

C EDA B

Think of original DAG as a partial order.

Want a total orderthat extends this partial order.


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TOPOLOGICALSORT

Performed on a DAG. Linear ordering of the vertices of Gsuch that if (u, v) E,

then uappears somewhere before v.

Topological-Sort (G)

1. call DFS(G)to compute finishing times f [v] for all v V

2. as each vertex is finished, insert it onto the front of a linked list

3. returnthe linked list of vertices

Time:(V + E).


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EXAMPLE

Linked List:

A B D

C E

1/


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EXAMPLE

Linked List:

A B D

C E

1/

2/3

E

2/3


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EXAMPLE

Linked List:

A B D

C E

1/4

2/3

E

2/31/4

D


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EXAMPLE

Linked List:

A B D

C E

1/4

2/3

E

2/31/4

D

5/


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EXAMPLE

Linked List:

A B D

C E

1/4

2/3

E

2/31/4

D

5/

6/


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EXAMPLE

Linked List:

A B D

C E

1/4

2/3

E

2/31/4

D

5/

6/7

6/7

C


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EXAMPLE

Linked List:

A B D

C E

1/4

2/3

E

2/31/4

D

5/8

6/7

6/7

C

5/8

B

9/


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EXAMPLE

Linked List:

A B D

C E

1/4

2/3

E

2/31/4

D

5/8

6/7

6/7

C

5/8

B

9/10

9/10

A


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PRECEDENCEEXAMPLE

Tasks that have to be done to eat breakfast:

get glass, pour juice, get bowl, pour cereal, pour

milk, get spoon, eat.

Certain events must happen in a certain order (ex:

get bowl before pouring milk)

For other events, it doesn't matter (ex: get bowl and

get spoon)


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PRECEDENCEEXAMPLE

get glass

pour juice

get bowl

pour cereal

pour milkget spoon

eat breakfast

Order: glass, juice, bowl, cereal, milk, spoon, eat.


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PRECEDENCEEXAMPLE

Topological Sort

eat

juice

glass

milk

cereal

bowl

spoon

consider reverse order of finishing times:

spoon, bowl, cereal, milk, glass, juice, eat

1 2 3 4 5 6 7 8 9 10 11 12 13 14


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PRECEDENCEEXAMPLE

What if we started withjuice?

eat

juice

glass milk

cereal

bowl

spoon

consider reverse order of finishing times:

spoon, bowl, cereal, milk, glass, juice, eat

1 2 3 4 5 6 7 8 9 10 11 12 13 14


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WHYACYCLIC?

Why must directed graph by acyclic for thetopological sort problem?

Otherwise, no way to order events linearly without

violating a precedence constraint.


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CORRECTNESSPROOF

Show if (u, v)E, then f [v] < f [u].

When we explore (u, v), what are their colors? Note, u is graywe are exploring it Is v gray?

No, because then v would be an ancestor of u.(u, v)is a back edge.a cycle (dag has no back edges).

Is v white?Then vbecomes descendant of u.By parenthesis theorem, d[u] < d[v] < f [v] < f [u].

Is v black?Then v is already finished.Since were exploring (u, v), we have not yet finished

u.Therefore, f [v] < f [u].


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STRONGLYCONNECTEDCOMPONENTS

Gis strongly connected if every pair (u, v) of verticesin G is reachable from one another.

A strongly connected component(SCC) of G is a

maximal set of vertices C V such that for all u, v

C, both u v and v uexist.


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COMPONENTGRAPH

GSCC= (VSCC, ESCC).

VSCChas one vertex for each SCC in G.

ESCChas an edge if theres an edge between the

corresponding SCCs in G.

GSCC for the example considered:


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GSCC ISADAG

Proof:

Suppose there is a path v v in G.

Then there are paths u u vand v v u in G.

Therefore, u and vare reachable from each other, so theyare not in separate SCCs.

Lemma 22.13

Let C and C be distinct SCCs in G, let u, vC, u, vC, and suppose there isa path u uin G. Then there cannot also be a path v v in G.


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TRANSPOSEOFADIRECTEDGRAPH

GT= transposeof directed G.

GT= (V, ET), ET= {(u, v): (v, u)E}.

GTis G with all edges reversed.

Can create GTin (V + E)time if using adjacency

lists.

G and GThave the same SCCs. (u and v are

reachable from each other in G if and only if

reachable from each other in GT.)


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SCC EXAMPLE

h f a e

g c b d

four SCCs


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HOWCANDFS HELP?

Suppose we run DFS on the directed graph.

All vertices in the same SCC are in the same DFS

tree.

But there might be several different SCCs in the

same DFS tree.

Example: start DFS from vertex h in previous

graph


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MAINIDEAOFSCC ALGORITHM

DFS tells us which vertices are reachable from theroots of the individual trees

Also need information in the "other direction": is the

root reachable from its descendants?

Run DFS again on the "transpose" graph (reverse thedirections of the edges)


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ALGORITHMTODETERMINESCCS

SCC(G)

1. call DFS(G)to compute finishing times f [u] for all u

2. compute GT

3. call DFS(GT), but in the main loop, consider vertices in

order of decreasing f [u] (as computed in first DFS)

4. output the vertices in each tree of the depth-first forest

formed in second DFS as a separate SCC

Time:(V + E).


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EXAMPLEa b c d

e f g h

1/

2/3/ 4 65/7

8/11/

12/

13/ 91014

15

16

f

4

h

6

g

7

d

9

c

10

a

14

e

15

b

16

DFS on the initial graph G

DFS on GT:

start at b: visit a, e

start at c: visit d

start at g: visit fstart at h

Strongly connected components: C1= {a, b, e}, C2= {c, d}, C3= {f, g}, C4= {h}

a b c d

e f g h

COMPONENT GRAPH


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COMPONENTGRAPH

The component graphGSCC= (VSCC, ESCC):

VSCC= {v1, v2, , vk}, where vicorresponds to eachstrongly connected component Ci

There is an edge (vi, vj) ESCCif G contains a directed

edge (x, y) for some x Ciandy Cj

The component graph is a DAG

a b c d

e f g h

a b e

c d

f g h


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NOTATIONS

Extend notation for d(starting time) and f(finishing time) to sets of vertices U V:

d(U) = minuU{ d[u]} (earliest discovery time)

f(U) = maxuU{ f[u]} (latest finishing time)

a b c d

e f g h

1/

2/3/ 4 65/7

8/11/

12/

13/ 91014

15

16

C1 C2

C3 C4

d(C1)f(C1)

d(C2)f(C2)

d(C3)f(C3)

d(C4)f(C4)

=11

=16

=1

=10

=2

=7

=5

=6


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SCCSANDDFS FINISHINGTIMES

Proof:

Case 1: d(C)< d(C) Letx be the first vertex discovered in C.

At time d[x], all vertices in C and Carewhite. Thus, there exist paths of whitevertices fromx to all vertices in C andC.

By the white-path theorem, all verticesin C and Care descendants ofx indepth-first tree.

By the parenthesis theorem, f [x] = f (C)> f(C).

Lemma 22.14

Let C and Cbe distinct SCCs in G = (V, E). Suppose there is an edge (u, v)E

such that u C and v C. Then f (C)> f (C).

C C

u v

x


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Proof: Case 2: d(C)> d(C)

Let y be the first vertex discovered in C. At time d[y], all vertices in Care white

and there is a white path from y to eachvertex in Call vertices in Cbecomedescendants of y. Again, f [y] = f (C).

At time d[y], all vertices in C are alsowhite.

By earlier lemma, since there is an edge(u, v), we cannot have a path from Cto C.

So no vertex in C is reachable from y. Therefore, at time f [y], all vertices in C

are still white. Therefore, for all w C, f [w] > f [y], which

implies that f (C)> f (C).

Lemma 22.14

Let C and Cbe distinct SCCs in G = (V, E). Suppose there is an edge (u, v)E

such that u C and v C. Then f (C)> f (C).

C C

u v

yx


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Proof:

(u, v)ET(v, u)E.

Since SCCs of G and GTare the same, f(C)> f (C), by

Lemma 22.14.

Corollary 22.15

Let C and Cbe distinct SCCs in G = (V, E). Suppose there is an edge

(u, v)ET, where u C and v C. Then f(C)< f(C).


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CORRECTNESSOFSCC

When we do the second DFS, on G

T

, we start with a component Csuch that f(C) is maximum (b, in our case)

We start from band visit all vertices in C1

From corollary: f(C) > f(C) in G for all C Cthere are no edges

from C to any other SCCs in GT

DFS will visit only vertices in C1

The depth-first tree rooted at bcontains exactly the vertices of C1

C1 C2

C3 C4

a b c d

e f g h

f4

h6

g7

d9

c10

a14

e15

b16


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CORRECTNESSOFSCC

The next root chosen in the second DFS is in SCC C2such that f(C)is maximum over all SCCs other than C1

DFS visits all vertices in C2

the only edges out of C2go to C1, which weve already visited

The only tree edges will be to vertices in C2

Each time we choose a new root it can reach only: vertices in its own component

vertices in components already visited

C1 C2

C3 C4

a b c d

e f g h

f

4

h

6

g

7

d

9

c

10

a

14

e

15

b

16


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RUNNINGTIMEOFSCC ALGORITHM

Step 1: O(V+E) to run DFS Step 2: O(V+E) to construct transpose graph,

assuming adjacency list rep.

Step 3: O(V+E) to run DFS again

Step 4: O(V) to output result

Total: O(V+E)


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ACKNOWLEDGEMENTS

These slides contain material developed andcopyright by:

Roger Crawfis(Ohio State university) (Course#

CSE680, 2010)

David Luebke(Virginia University) (Course#CS332, 2009)

Jennifer Welch (Texas A&M University)

(Course#CSCE411, 2012)


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THEEND

ALgorithm Full First Ctvcxvcxv Syllabus

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