Lecture 11 Graph algorithms: Order in acyclic graphs

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Lecture 11 Graph algorithmsOrder in acyclic
graphs

• COMP 523 Advanced Algorithmic Techniques
• Lecturer Dariusz Kowalski

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Overview

• Previous lectures
• Testing bipartiteness
• Testing for (directed) cycles
• This lecture
• Ordering nodes in acyclic directed graphs

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How to represent directed graphs?

• Given directed graph G (V,E) , how to represent
  it?
• Adjacency matrix ith node is represented as ith
  row and ith column, edge from ith to jth node is
  represented as 1 in row i and column j, (0 if
  there is no directed edge from i to j)
• Adjacency list nodes are arranged as array/list,
  each node record has two lists one stores
  in-neighbours other stores out-neighbours

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Adjacency matrix
Adjacency list
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Directed Acyclic Graphs (DAG)

• Directed graph G (edges have directions - one end
  is the beginning, the other one is the end of the
  edge).
• Reversed graph Gr has the same nodes but each
  edge is a reversed edge from graph G.
• Test if given direct graph is acyclic (DAG) last
  lecture, directed DFS approach in time and memory
  O(mn).
• Problem is it possible to order all nodes
  topologically, which is if (v,w) is a directed
  edge in G then v is before w?
• Note there may be many topological orders.

graph G
reversed graph Gr
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Opposite property

• Fact
• If graph G has a topological order then it is
  DAG.
• Proof
• Suppose, to the contrary, that G has topological
  order and it also has a cycle. Let v be the
  smallest element in the cycle according to the
  topological order. It follows that its successor
  in the cycle must be smaller than v, but this
  contradicts the fact that v is the smallest.
  Hence G must not have a cycle.

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Key property

• Property
• In DAG there is a node with no incoming edges
• After removing the node with the incident edges,
  remaining graph is still DAG
• Proof
• Node with no incoming edges
• Suppose, to the contrary, that every node has at
  least one incoming edge. It follows that starting
  from any node we can always leave a node by
  incoming edge (in opposite direction) until we
  return to a visited node, which makes a cycle.
  This is a contradiction.
• After removing node still a DAG remains
• There could not be a cycle in smaller graph,
  since it would be in the original one.

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Algorithm for topological order

• Algorithm
• Repeat until all nodes are in the ordered list
• Find a node with no incoming edges
• Remove this node and all its outgoing edges
• Efficient implementation
• Preprocessing
• for each node keep info if it is active or not
  (additional array Active of size n needed) - at
  the beginning all nodes are active
• go through the list and for every node store the
  number of incoming edges from other active nodes
  (additional array Counter of size n needed)
• select on list S those nodes which have no
  incoming edges.
• During the algorithm
• While removing the node (make inactive), go
  through its outgoing active neighbours and
  decrease a counter of the active out-neighbours
  by one - if it becomes zero put this active node
  to list S
• Node with no incoming edges is always taken (and
  then removed) from list S.

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Complexity

• Time O(mn)
• Preprocessing
• Initialization of array Active O(n)
• Filling array Counter - counting edges O(mn)
• Initialization of list S - checking array
  Counter O(n)
• During the algorithm
• While removing the node - go through its
  out-neigbours total O(mn)
• Selecting from list S total O(1)
• Memory O(mn)
• Adjacency list O(mn)
• Arrays Active and Counter, list S O(n)

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Conclusions

• Ordering nodes in acyclic graph in time and
  memory
• O(m n).

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Textbook and Exercises

• Section 3.6
• Prove that if a DFS and BFS trees in graph G are
  the same, than G is also the same like they (does
  not contain any other edge).
• Prove that if G has n nodes, where n is even, and
  each node has at least n/2 neighbours then G is
  connected.
• Prove the property of layers every edge is
  either between two consecutive layers or within
  one layer (two ends are in the same layer).
• Design and analyse time and memory taken by BFS
  for directed graphs.
• What happens if you try to use a different method
  of searching for checking bipartiteness or
  acyclicity?