Tirgul 11

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Tirgul 11

• BFS,DFS review
• Properties
• Use

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Breadth-First-Search(BFS)

• The BFS algorithm executes a breadth search over
  the graph.
• The search starts at a source (node s).
• We find all the adjacent neighbors of that given
  node s.
• We continue recursively to ss neighbors.
• The output is a breadth tree.

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Breadth-First-Search(BFS)

• The root of the breadth tree is s.
• It contains all the nodes that are accessible
  from s.
• h0 would be s
• h1 would be all ss adjacent nodes.
• h2 would be all ss adjacent nodes adjacent
  nodes.
• And so on
• The path between the node s and a given node v in
  the tree defines the shortest path from s to v in
  graph G.

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Breadth-First-Search(BFS)

• For example BFS over that graph
• r s t u
• v w x y
• We give the breadth tree
• s
• r w
• v t x
• u y

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The BFS Algorithm

• The algorithm colors the graph nodes in the
  process of the search
• All nodes are initially white besides the node s.
  
• A node becomes gray when it is discovered.
• It becomes black when the algorithm backtracks
  from it (all its adjacent nodes were found).
• For each node u, du is the length of the
  shortest path from node s to node u.

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The BFS Run Time

• Each node Enqueues once and Dequeues once to the
  queue.
• Enqueue, Dequeue operations take O(1).
• That is why the queue ministration takes O(V).
• We pass through each edge only once.
• For every node, we go through all its adjacent
  neighbors.
• It takes O(E).
• Altogether the run time is O(VE).

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Questions

• What is the run time of the BFS algorithm when
  the input graph is represented as an
  adjacency-matrix?
• How can it be determined that an undirected graph
  is a full two sided graph?
• What is the algorithm to determine the shortest
  path between two given nodes in a graph? What is
  its run time?

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Depth-First-Search (DFS)

• The DFS algorithm executes a depth search over
  the graph.
• The search starts each time at a different node.
• The DFS algorithm terminates the discovery of one
  path before it backtracks to discover a new path.

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The DFS Algorithm

• The algorithm colors the graph nodes in a similar
  manner as BFS does.
• All nodes are initially white.
• A node becomes gray when it is discovered.
• It becomes black when the algorithm backtracks
  through it.
• The algorithm determines two timestamps for each
  node u.
• du
• and fu.

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The DFS Algorithm

• du is the time in which the node was colored
  gray.
• fu is the time in which the node was colored
  black.

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The DFS Algorithm

• DFS(G)for each node u initialize coloruwhite,
  puNULLset time1for all nodes u with white
  color do DFS-Visit(u)
• DFS-Visit(u)colorugray dutime
  timefor each neighbor v of u with white color
  do pv u DFS-Visit(v)colorublack
  futime time

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Example of DFS

• The DFS of this graph

• is

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The DFS Run Time

• DFS-Visit() is called exactly once for each node
  in V
• It is called for all white nodes and colors them
  to gray gt O(V).
• DFS-Visit goes over each edge it finds twice.
  Finding the edge and backtracking.
• For each node, the DFS-Visit eventually goes over
  all its neighbors.
• In total, DFS-Visit goes over all edges twice gt
  O(E).
• The total run-time is O(VE).

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Two properties of DFS

• The parenthesis theorem
• for any two nodes u, v, if dultdvltfu, then
  fvltfu and v is a descendant of u.
• proof sketch
• Since dvltfu, then when v was discovered, us
  color was gray, and thus v is a descendant of u.
• Moreover, because of the recursion, u will not
  be
• finished until v is finished, and thus
  fugtfv.

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Two Properties of DFS

• This theorem is called the parenthesis theorem
  because
• We can print (u when u is discovered and u)
  when u is finished.
• This theorem tells us that the expression well
  get will have a proper parenthesis structure.
• The white-path theorem
• In a DFS tree, node v is a descendant of node u
  if and only if at time du there is a white path
  from u to v.

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Two Properties of DFS

• proof sketch
• If v is a descendant of u, there is a path from u
  to v in the DFS tree.
• Every node w on that path is also a descendant of
  u, so dwgtdu, w is white at time du.

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Two Properties of DFS

• In the other direction, if there is a white path
  from u to v at time du,
• suppose that v was the first node on that path
  that is not a descendant of u,
• and let w be the predecessor of v (so w is a
  descendant of u and thus fwltfu).
• v is discovered after u, but before w is
  finished (since v is a neighbor of w),
• dvgtdu and fugtfwgtdv.
• It follows from the parenthesis theorem that v
  is a descendant of u, in contradiction.

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Finding the Strongly-Connected-Components of a
Directed Graph

• A strongly connected component of a directed
  graph G (V,E) is a maximal set of vertices U
  such that for every pair of vertices u and v in
  U, v is reachable from u and u is reachable from
  v.
• Strongly-Connected-Components(G)
• 1 call DFS(G) to compute fu for all vertices
  u
• 2 compute GT (reverse edges)
• 3 call DFS(GT) consider the vertices according
  to fu
• in decreasing order.
• 4 the depth-first trees in 3 are the
  Strongly-Connected-Components of graph G.

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Why does the Algorithm of SCC Works?

• What is the relation between DFS and SCC?
• Claim 1 and Lemma 1 connects the two.
• Claim 1 If nodes u,v are in the same SCC then
  any node w in some path from u to v is also in
  the same SCC.
• Lemma 1 In any DFS, all vertices in the same SCC
  are placed in the same depth-first tree.

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Why does the Algorithm of SCC Works?

• Proof sketch of Lemma 1
• Suppose r is the first node of an SCC discovered
  during the DFS.
• Consider some other node v in this SCC.
• All the nodes in the path from r to v are in the
  same SCC (according to the above claim),
• and thus they are all white (r is the first to be
  discovered).
• From the white-path theorem, v is placed in the
  same depth-first tree as r.

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Why does the Algorithm of SCC Works?

• Definition node w is reachable from node u if
  uw or if there is a path from u to w.
• Definition the forefather of node u
  is some node w that is reachable from u and that
  has maximal fw.
• Note that fu lt f since u is
  reachable from u.

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Forefathers and DFS trees

• Lemma 2 The forefather of u is an ancestor of u.
• proof sketch
• Suppose the forefather of u is w.
• We need to show that colorwgray when u is
  discovered.
• If colorwblack, then fugtfw in
  contradiction.
• If colorwwhite, let v be the last node on the
  path from u to w with non-white color.
• If colorvblack this means that v has no white
  neighbors which is a contradiction since the next
  node from v on the path to w is white.
• If colorvgray then since there is a white path
  from v to w, w is a descendant of v and thus
  fvgtfw.
• Since there is a path from u to v, it follows
  that v should be the forefather of u instead of
  w, in contradiction.

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Forefathers and SCCs

• Lemma 3 two nodes u, v are in the same SCC if
  and only if they have the same forefather.
• Proof sketch
• If u,v are in the same SCC, then the forefather
  of u is reachable from v, and vice-versa.
• Therefore, they must have the same forefather.
• In the other direction, if u, v have the same
  forefather (say w),
• We know that u is reachable from w since w is its
  ancestor, and w is reachable from u, and thus u,
  w are in the same SCC.
• Equivalently v, w are in the same SCC.

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Explaining the Algorithm

• Notice that the node with the highest finish time
  
• (say u) must be a forefather.
• All we need to do is to find all the nodes that
  have a path to u.
• These are exactly the nodes in the DFS over the
  transpose of G, starting at u.
• This is the first SCC.
• The next forefather must be the (remaining) white
  node with the highest finish time.
• All the nodes in his SCC and the path to them
  must be still white.
• Performing a DFS from this forefather will
  discover exactly his SCC, and so on.

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Proving the Algorithm.

• Theorem 1 The SCCs the algorithm computes are
  the correct SCCs of G.
• proof sketch
• We prove it by induction on the number of the
  SCC.
• Suppose r is the root of the next DFS tree to be
  computed.
• Then r is the forefather of itself (by the
  induction assumption).
• A node v that is included in the DFS tree of r is
  in rs SCC because r is his forefather.
• A node v that is included in rs SCC is included
  in the DFS tree of r due to Lemma 1.
• rs tree is exactly rs SCC.