articulation point, scc

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Applications

Depth-first searchBiconnectivity

Euler circuitsStrongly-connected components

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Depth-First Search

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Recursively visit everyvertex in the graphConsiders every edge inthe graph

Assumes undirected edge(u,v) is in us and vsadjacency list

Visited flag preventsinfinite loopsRunning time O(|V|+|E|)

DFS () ;; graph G=(V,E)foreach v in V

if (! v.visited)then Visit (v)

Visit (vertex v)v.visited = trueforeach w adjacent to v

if (! w.visited)then Visit (w)

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DFS Applications

Undirected graphTest if graph is connectedRun DFS from any vertex andthen check if any vertices notvisited

Depth-first spanning tree Add edge (v,w) to spanningtree if w not yet visited(minimum spanning tree?)If graph not connected, thendepth-first spanning forest

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DFS Applications

Remembering the DFS traversal order is importantfor many applicationsLet the edges (v,w) added to the DF spanning treebe directed

Add a directed back edge (dashed) if w is already visited when considering edge (v,w), andv is already visited when considering reverse edge (w,v)

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Biconnectivity

A connected, undirectedgraph is biconnected if thegraph is still connected afterremoving any one vertex

I.e., when a node fails, thereis always an alternative route

If a graph is not biconnected,

the disconnecting vertices arecalled articulation points

Critical points of interest in

many applications6

Biconnected? Articulation points?


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DFS Applications: Finding Articulation Points

From any vertex v, perform DFS and number verticesas they are visitedNum(v) is the visit number

Let Low(v) = lowest-numbered vertex reachable fromv using 0 or more spanning tree edges and then atmost one back edge

Low(v) = minimum of Num(v)Lowest Num(w) among all back edges (v,w)Lowest Low(w) among all tree edges (v,w)

Can compute Num(v) and Low(v) in O(|E|+|V|) time7


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DFS Applications: Finding Articulation Points (Example)

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Original Graph

Depth-first treestarting at A withNum/Low values:


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Root is articulation point iff it has morethan one child

Any other vertex v is an articulationpoint iff v has some child w such thatLow(w) Num(v)

I.e., is there a child w of v that cannotreach a vertex visited before v?

If yes, then removing v will disconnect w(and v is an articulation point)9


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DFS Applications: Finding Articulation Points (Example)

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Original Graph

Depth-first treestarting at C withNum/Low values:


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High-level algorithmPerform pre-order traversal to compute NumPerform post-order traversal to compute Low

Perform another post-order traversal to detectarticulation points

Last two post-order traversals can becombinedIn fact, all three traversals can be combined

in one recursive algorithm11


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Implementation

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Check for rootomitted.


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Check for rootomitted.


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Euler Circuits

Puzzle challengeCan you draw a figure using a pen,drawing each line exactly once, withoutlifting the pen from the paper?

And, can you finish where you started?

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Euler Circuits

Solved by Leonhard Eulerin 1736 using a graphapproach (DFS)

Also called an Euler pathor Euler tour Marked the beginning of

graph theory

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Euler Circuit Problem

Assign a vertex to each intersection in thedrawing Add an undirected edge for each linesegment in the drawingFind a path in the graph that traverses eachedge exactly once, and stops where it started

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Necessary and sufficient conditionsGraph must be connectedEach vertex must have an even degree

Graph with two odd-degree vertices can havean Euler tour (not circuit)

Any other graph has no Euler tour or circuit

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AlgorithmPerform DFS from some vertex v until youreturn to v along path pIf some part of graph not included,perform DFS from first vertex v on p that

has an un-traversed edge (path p)Splice p into pContinue until all edges traversed

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Euler Circuit Example

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Start at vertex 5.

Suppose DFS visits 5, 4, 10, 5.


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Euler Circuit Example (cont.)

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Graph remaining after 5, 4, 10, 5:

Start at vertex 4.Suppose DFS visits 4, 1, 3, 7, 4, 11, 10, 7, 9, 3, 4.Splicing into previous path: 5, 4, 1, 3, 7, 4, 11, 10, 7, 9, 3, 4, 10, 5.


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Graph remaining after 5, 4, 1, 3, 7, 4, 11, 10, 7, 9, 3, 4, 10, 5:

Start at vertex 3.Suppose DFS visits 3, 2, 8, 9, 6, 3.Splicing into previous path: 5, 4, 1, 3, 2, 8, 9, 6, 3, 7, 4, 11, 10, 7, 9, 3, 4, 10, 5.


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Graph remaining after 5, 4, 1, 3, 2, 8, 9, 6, 3, 7, 4, 11, 10, 7, 9, 3, 4, 10, 5:

Start at vertex 9.Suppose DFS visits 9, 12, 10, 9.Splicing into previous path: 5, 4, 1, 3, 2, 8, 9, 12, 10, 9, 6, 3, 7, 4, 11, 10, 7, 9, 3, 4, 10, 5.No more un-traversed edges, so above path is an Euler circuit.


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Euler Circuit Algorithm

Implementation detailsMaintain circuit as a linked list to supportO(1) splicingMaintain index on adjacency lists to avoidrepeated searches for un-traversed edges

AnalysisEach edge considered only onceRunning time is O(|E|+|V|)

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DFS on Directed Graphs

Same algorithmGraph may beconnected, but notstrongly connectedStill want the DFspanning forest toretain information

about the search25

DFS () ;; graph G=(V,E)

foreach v in V if (! v.visited)then Visit (v)

Visit (vertex v)v.visited = trueforeach w adjacent to v

if (! w.visited)then Visit (w)

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DF Spanning Forest

Three types of edges in DF spanning forestBack edges linking a vertex to an ancestorForward edges linking a vertex to a descendant

Cross edges linking two unrelated vertices

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Graph: DF Spanning Forest:

back

cross


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DF Spanning Forest

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Graph

DF Spanning Forest

back

back

cross

forward

(Note: DF Spanning Forests usuallydrawn with children and new treesadded from left to right.)


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DFS on Directed Graphs

ApplicationsTest if directed graph is acyclic

Has no back edges

Topological sortReverse post-order traversal of DF spanningforest

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Strongly-ConnectedComponents

A graph is strongly connected if every vertex can bereached from every other vertex

A strongly-connected component of a graph is asubgraph that is strongly connectedWould like to detect if a graph is strongly connectedWould like to identify strongly-connected componentsof a graphCan be used to identify weaknesses in a network General approach: Perform two DFSs

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AlgorithmPerform DFS on graph GNumber vertices according to a post-order traversal ofthe DF spanning forest

Construct graph G r by reversing all edges in GPerform DFS on G r

Always start a new DFS (initial call to Visit) at thehighest-numbered vertex

Each tree in resulting DF spanning forest is astrongly-connected component

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Graph G Graph G r

DF Spanning Forest of G r Strongly-connected components:

{G}, {H,I,J}, {B,A,C,F}, {D}, {E}


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Strongly-ConnectedComponents: Analysis

CorrectnessIf v and w are in a strongly-connected componentThen there is a path from v to w and a path from

w to vTherefore, there will also be a path between v andw in G and G r

Running timeTwo executions of DFSO(|E|+|V|)

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Summary

Graphs one of the most important datastructuresStudied for centuriesNumerous applicationsSome of the hardest problems to solveare graph problems

E.g., Hamiltonian (simple) cycle, Clique

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articulation point, scc

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