cse 780 algorithms advanced algorithms graph alg. dfs topological sort

CSE 780 Algorithms

Advanced Algorithms

Graph Alg.DFS

Topological sort

CSE 780 Algorithms

Objectives

On completion of this lecture, students should be able to:

1. Write a dfs algorithm

2. Apply dfs in topological sort and finding strongly-

connected components.

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Breadth-first Search

Works for both directed and undirected graph Starting from source node s, visits remaining

nodes of graph from small distance to large distance

Produce a BF-tree Return distance between s to any reachable node

in time O(|V| + |E|)

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Depth-first Search

Breadth-first search: Go as broad as possible at each node

Depth-first search (a different strategy): Go as deep as possible first

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More Formally

Again, a node white: unvisited gray: discovered but not finished black: finished (explored)

For every node v V d[v]: time v is discovered f[v]: time v is finished (all edges in v’s adjacency list

are explored) f[v] - d[v] = time from grey to black

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Pseudo-code

Time complexity

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Example

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Depth-first Forest

Consider all edges (p(u), u) When it happens: p(u) is grey, u is white A forest, called Depth-first forest Depends on the order of vertices Example from previous page

Property: Start (finish) time for each tree: same order as if we

visit nodes in pre-order (post-order) tree walk

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Properties of DFS

Parenthesis Theorem: Any two nodes u and v, one of the following 3 cases:

(1) u is descendant of v, and d[v] < d[u] < f[u] < f[v] (2) v is descendant of u, and d[u] < d[v] < f[v] < f[u] (3) u is not descendant of v, neither is v a descendant of u, and d[u] < f[u] < d[v] < f[v] or d[v] < f[v] < d[u] < f[u]

u is descendant of v iff d[v] < d[u] < f[u] < f[v] White-path Theorem

u is descendant of v iff at the time of d[v], there is a all white path from v to u.

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Classification of Edges

Four types of an edge (u,v) Tree edge Non-tree edges:

Back edge: u is descendant of v Forward edge: v is descendent of u Cross edge: others

Distinguish by the color of v Example

b

dc

e

f g

h

j

a

k

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Theorem

In a DFS of an undirected graph G, every edge of G is either a tree edge or a back edge.

b

dc

e

f g

h

j

a

k

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Connected Components

Undirected graph DF forests represents the set of connected components

Directed graph Does the tree rooted at u include all nodes accessible

from u ?

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Directed Acyclic Graph

DAG: directed acyclic graph Determine whether a directed graph is DAG or not

How?

A directed graph G is a dag iff a DFS of G yieldsno back edges.

How about an undirected acyclic graph?

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Topological Sort

A topological sort of a dag G = (V, E) A linear ordering A of all vertices from V If edge (u,v) E => A[u] < A[v]

undershorts

pants

beltshirt

tie

jacket

shoes

socks

watch

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Topological Sort

Using DFS

Topological-Sort(G) Call DFS(G) to compute finishing times f[v] for each v Output vertices in descreasing order of finishing time

Time complexity O (|V| + |E|)

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Exampleundershort

s

pants

beltshirt

tie

jacket

shoes

socks

watch

11, 16

12, 15

6, 7

3, 4

2, 5

1, 813, 14

17, 18

9, 10

socks , undershots , pants , shoes , watch , shirt , belt , tie , jacket

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Correctness

Theorem Topological-Sort(G) produces a topological sort of a

directed acyclic graph G

Proof: Consider any edge (u, v)

Goal: f[v] < f[u] Three possible types of edges:

Tree edge, forward edge, cross edge

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Connected Components

An undirected graph is connected if every pair of vertices is connected by a path

A graph that is not connected is naturally decomposed into several connected components.

A connected components is a maximal set of vertices, which are all connected.

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Connected Components (cont.)

The graph below has 3 connected components

{1,2,5}, {3,6} and {4}.An undirected graph is connected if it has exactly

one connected component.

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Strongly-connected components

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Strongly-connected components

The above graph has 3 strongly connected components {1,2,4,5}, {3} and {6}

A directed graph is strongly connected if it has only one strongly connected component.

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Transpose of a Graph

The transpose of a directed graph G = (V,E) is the graph GT = (V, ET)

where ET = {(u,v): (v,u) Є E}. ET consists of the edges of G with their directions

reversed. G and GT have exactly the same strongly

connected components.

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Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

CSE 780 Algorithms

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Corollary 22.15

Let C and C’ be distinct strongly connected components in directed graph G = (V,E). Suppose that there is an edge (u,v) ET, where u C and

v C’. Then f(C) < f(C’)

Note: f(C) = max u C {f[u]}

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Theorem 22.16STRONGLY-CONNECTED-COMPONENTS(G) correctly computes the strongly connected components of a directed graph G.Proof.We argue by induction that the vertices of each tree form a strongly connected component.The inductive hypothesis: The first k trees produced in line 3 are strongly connected.

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The basis for the induction, when k=0, is trivial.The inductive step: assume that each of the first trees produced is a strongly connected component.Consider the (k+1)st tree.Let the root of this tree be vertex u,and let u be in a strongly connected component C.Because of how we choose roots in line 3,f[u] = f(C) > f(C’) for any strongly connected component C’ that has yet to be visited.

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By inductive hypothesis, at the time that the searchvisits u, all other vertices of C are white.Therefore, all other vertices of C are descendants of u in its tree.Moreover, by inductive hypothesis and corollary 22.15, any edge in GT that leave C must be to strongly connected components that have already visited.Thus, no vertex in any strongly connected component other than C will be a descendant of u during the depth-first search of GT. Thus, the vertices of the depth-first tree in GT that is rooted at u form exactly one strongly connected component.

CSE 780 Algorithms

Summary

Graph search/traversal method

Breadth-first search Discover nodes in shortest distance (# links) e.g, chess

Depth-first search Parenthesis theorem e.g, topological sort