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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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.

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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