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Directed Acyclic Graph (DAG): topological ordering

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• Directed Acyclic Graph
  • Directed graphs with and without cycles:

    Directed graph with cycles: Directed graph without cycles:

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  • Directed graphs without cycles are called Directed Accylic Graphs (DAGs)

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• Inheritance hierarchy and DAGs
  • We find DAGs in the inheritance/interface hierarchy of programming languages such as Java

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  • Example:

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• Task scheduling and DAGs
  • Task scheduling:

    A job consists of a number of tasks Some tasks can only be performed after some other task(s) have been completed

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  • Example:

    Job: building a house Some of the tasks: Pour the foundation        Build the frame You cannot start the task "build the frame" before completing the task "pour the foundation"

    Notice:

    Job schedules must be acyclic Because if you have a cycle in the job dependency, then you can never start any job in that cycle....

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• Definition: topological ordering
  • Definition:

    Topological ordering of nodes = an ordering (label) of the nodes/vertices such that: For every edge (vi , vj) ∈ G:    i < j

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  • Example:

    Every edge in the graph goes from a vertex with a smaller index to one with a larger index

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  • Note:

    There may be more than 1 possible topological ordering for the vertices

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• DAG and Topological Ordering
  • Proposition 13.21 (Goodrich)

    A directed graph G has a topological ordering if and only if G   is acyclic

  • Proof:

    If a directed graph G has a topological ordering then:   G   is acyclic Proof: by contradiction Assume that G  is not acyclic I.e., there is a cycle C in G Let the edges in the cycle C be: (vi0 , vi1 )   (vi1 , vi2 )   ... (vik , vi0 )   According to the given fact that G   has a topological ordering, we have from the edges in the cycle C that: vi0   < vi1   <   vi2   ...   < vik   <   vi0        This inequality is impossible --- contradiction Therefore, the assumption that "G  is not acyclic" is false: G  is acyclic          If directed graph G   is acyclic then:   G has a topological ordering Proof: Since G  is acyclic, there is some vertex that does not have any incoming edges Let x be a vertex in G   that does not have any incoming edges We label the vertex x as v1 We now remove the vertex v1 from graph G: G1 = G - {v1}         The graph G1  is also acyclic (we cannot add cycles to a graph by removing edges), and there is some vertex that does not have any incoming edges Let y be a vertex in G1   that does not have any incoming edges We label the vertex y as v2 We now remove the vertex v2 from graph G1: G2 = G - {v1,v2}         And so on Eventually, every vertex in G2 will be removed (And we will stop) The resulting numbering of nodes is a topological order because: An edge (vi , vj ) must be deleted before the deletion of vertex vj Therefore, the vertex vi must be deleted before the vertex vj Therefore: i   <   j   holds for every edge (vi , vj )         This is the definition of topological ordering.

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• An efficient algorithm to find a topological ordering in a DAG
  • The proof above provides an algorithm for finding a topological ordering on the vertices of a DAG

  • Psuedo code:

    i = 1; while ( G ≠ &empty ) { Let x = vertex in G with indeg(x) = 0; Label x with "i"; i++; Delete all edges (x, z) from G; }

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  • Refined psuedo code:

    Stack S = contains vertices with indeg = 0 (these vertices are ready to be labeled) ============================================================== S = ∅ ; // Initialize the stack /* --------------------------------------- Compute the "indegree" for each vertex --------------------------------------- */ for ( each vertex x ∈ G ) { incounter[x] = incoming degree of x; } /* --------------------------------------- Find vertices with indeg = 0 --------------------------------------- */ for ( each vertex x ∈ G ) { if ( incounter[x] == 0 ) push(x); } i = 1; // Initial label /* --------------------------------------- Label the next vertex in S --------------------------------------- */ while ( S ≠ &empty ) { u = S.pop(); Label u with "i"; i++; /* ------------------------------------ Update indegree of affected nodes ------------------------------------ */ for ( each edge (u, z) ) do { incounter[z]--; if ( incounter[z] == 0 ) S.push(z); // z is ready to be numbered } }

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• Worked out example
  • Directed graph

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  • Initialization:

    incounter[a] = 0 incounter[b] = 1 incounter[c] = 1 incounter[d] = 1 incounter[e] = 2 incounter[f] = 2            stack = a

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  • Process top element from stack (a):

    incounter[a] = 0 incounter[b] = 0 incounter[c] = 0 incounter[d] = 1 incounter[e] = 2 incounter[f] = 2            stack = b c

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  • Process top element from stack (c):

    incounter[a] = 0 incounter[b] = 0 incounter[c] = 0 incounter[d] = 0 incounter[e] = 1 incounter[f] = 2            stack = b d

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  • Process top element from stack (d):

    incounter[a] = 0 incounter[b] = 0 incounter[c] = 0 incounter[d] = 0 incounter[e] = 1 incounter[f] = 1            stack = b

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  • Process top element from stack (b):

    incounter[a] = 0 incounter[b] = 0 incounter[c] = 0 incounter[d] = 0 incounter[e] = 0 incounter[f] = 1            stack = e

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  • Process top element from stack (e):

    incounter[a] = 0 incounter[b] = 0 incounter[c] = 0 incounter[d] = 0 incounter[e] = 0 incounter[f] = 0            stack = f

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  • Process top element from stack (f):

    incounter[a] = 0 incounter[b] = 0 incounter[c] = 0 incounter[d] = 0 incounter[e] = 0 incounter[f] = 0            stack = empty

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  • Done