Dawn J' Lawrie

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Announcements

• Programming Assignment 3
• Code due tonight at midnight
• Print out and Analysis due Friday beginning of
  class
• Today's Office Hours
• 11-12 and 2-3 PM

2
Nontrivial Graph-Processing

• Lecture 17

3
Simple Path

• What is a simple path?
• Given what we have studied so far, can we answer
  the question
• Does a simple path exist between v and w?

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Find a Specific Simple Path

• Use Depth-First-Search
• Solve the recursive problem
• Assume I know a path from v to t, is there a path
  from t to w?

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Simple path search

• template ltclass GraphTypegt
• class SPath
• private
• const GraphType graph
• vector ltboolgt visited
• vector ltintgt path
• int length
• bool found

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Simple path search

• bool searchR(int v, int w, int depth)
• if (v w) // base case
• pathdepth v
• length depth1
• return true
• visitedv true
• typename GraphadjIterator adj(graph, v)
• for (int t adj.beg() !adj.end() t
  adj.nxt())
• if (!visitedt)
• if (searchR(t, w, depth1))
• pathdepth v
• return true
• return false

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Simple path search

• public
• SPath(const Graph G, int v, int w)
• graph(G), visited(G.V(), false), path(G.V(),
  -1),
• length(0)
• found searchR(v,w,0)
• bool exists() const
• return found
• void displayPath() const
• for (int i 0 i lt length i)
• if (i ! 0)
• cout ltlt " - "
• cout ltlt pathi
• cout ltlt endl

8
Hamiltonian Path

• Given two vertices, is there a simple path
  connecting them that visits every vertex in the
  graph exactly once?
• Hamiltonian Tour Is there a cycle that visits
  every vertex in the graph exactly oncegt

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Hamiltonian Path

• bool searchR(int v, int w, int depth)
• if (v w) // base case
• pathdepth v
• length depth1
• return true // remove
  line
• return (length graph.V()) // new line
• visitedv true
• typename GraphadjIterator adj(graph, v)
• for (int t adj.beg() !adj.end() t
  adj.nxt())
• if (!visitedt)
• if (searchR(t, w, depth1))
• pathdepth v
• return true
• visitedv false // new line
• return false

10
Euler path

• Classic problem studied by Leonhard Euler and
  presented to Russian Academy in 1735

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Euler path

• Is there a path connecting two given vertices
  that used each edge in the graph exactly once?
• Not necessarily a simple path
• A graph has a Euler tour if and only if it is
  connected and all its vertices are of even degree

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Proof

• Must be connected
• By definition a tour is a path connecting each
  pair of vertices
• Vertex v must be of even degree
• When we traverse the tour (starting from anywhere
  else), we enter v on one edge and leave on a
  different edge
• Number of edges incident on the vertex must be
  twice the number of visits

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Proof of converse

• Consider any connected graph that has more than
  one edge, with all vertices of even degree
• Starting at any vertex v, we follow and remove
  any edge
• Continue until arriving at a vertex that has no
  more edges
• Process terminates
• Outcomes
• We must end up back a v because v is the only
  vertex of odd degree

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Converse Proof continued

• We may not have a full tour
• All the vertices in the group that remain have
  even degree but may not be connected
• Each connected component is a Euler tour by
  inductive hypothesis
• Consider these detours

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Finding a Euler Path

• Cannot implement a direct recursive method
  because it would be as slow as the algorithm for
  finding Hamiltonian Path
• Is there a linear time solution?

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Linear-time Euler path

• template ltclass Graphgt
• int EPathltGraphgttour(int v)
• bool finished false
• while (!finished)
• typename GraphadjInterator adj(graph, v)
• int w adj.beg()
• if (adj.end())
• finished true
• else
• pStack.push(v)
• graph.remove(Edge(v, w))
• v w
• return v

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Linear-time Euler path

• template ltclass Graphgt
• void EPathltGraphgtshow()
• int curV v
• if (!found)
• return
• while(tour(curV) curV !pStack.empty())
• curV sStack.pop()
• cout ltlt "-" ltlt curV
• cout ltlt endl