Topics

• DAG’s
• Topological sort
• Longest path in a weighted dag

Directed Acyclic Graphs (DAG’s)

• DPV: Ch 3.3, 4.7
• A dag is a digraph containing no cycle. Every dag can be drawn in such a way that all edges are directed down (or from left to right).
  [EXAMPLE PICTURE HERE]
• Dags model many real-world problems, e.g., combinational circuits, prerequisite digraphs, dependency digraphs, Bayes nets/belief systems, etc.
• A vertex in a dag is called a source (sink) if no edge enters (leaves) it, i.e., it has indegree (outdegree) zero.
• Lemma. Every dag has at least one source (sink).

Topological sort

• A topological numbering of a digraph \(G=(V,E)\) is an assignment \(V\to [n]\) so that every edge is directed from a lower number to a higher number.
  [EXAMPLE PICTURE HERE]
• Theorem. A digraph has a topological numbering if and only if it's acyclic.
• Proof:
  \(\Rightarrow\): Any path in a topologically numbered digraph goes from lower to higher number. Thus, every path is simple. Hence, no cycle can exist.
  \(\Leftarrow\): Assign the highest number \(n\) to some sink \(s\). Delete \(s\). Recursively number the remaining digraph.
• To topologically sort a dag means to topologically number its vertices.

Topological sort algorithm

• The proof suggests the following high-level algorithm to topologically sort a dag \(G\).

  H := G
  repeat until H has no vertices {
      "grow" a dfs path P till a sink s is found
      T[s] ← n
      n ← n - 1
      delete s from P & H
  }

• When the algorithm ends, \(T[1..n]\) gives the topological numbering.

Implementation

• Use array \(T[1..n]\) for two purposes. \[ T[v] = \left\{ \begin{array}{ll} 0, & \mbox{if $v$ has never been on any dfs path} \\ t, & \mbox{if $v$ has been deleted and assigned topological number t} \end{array} \right. \]

Implementation, continued

• Low-level algorithm:

  top(G) {  
      num ← n  
      for each vertex v do {
          T[v] ← 0
      }
      for each vertex v do {
          if T[v] = 0 then dfs(v)
      }
  }  
  dfs(v) {  
      for each edge (v,w) do {
          if T[w] = 0 then dfs(w)  
      }
      /* v is now a sink in the high level algorithm */  
      T[v] ← num  
      num ← num - 1  
  }

Running Time

• top(G) takes time \(O(m+n)\).

Notes

1. Our algorithm shows that numbering the vertices in decreasing finish time gives a topological numbering.
2. Would numbering the vertices in order of increasing discovery time gives a valid topological numbering?
3. Knuth's topological sort algorithm works by repeatedly finding a source \(s\), assigning the next lowest number to \(s\), and deleting \(s\) from the current graph. His algorithm can also be implemented in linear time.

Longest path in a weighted dag

• Let \(G=(V,E)\) be a weighted dag, where every edge \(uv\) has an arbitrary length \(\ell[u,v]\). We can find a longest path in \(G\) in \(O(m+n)\) time.
• Idea: We'll set \(d[v]\), for all \(v\in V\), to the length of a longest path starting at \(v\). We'll compute \(d[v]\) for all \(v\in V\) , using the formula \[ d[v] = \max \{\,0,\,\ell[v,w] + d[w] : (v, w)\in E \,\} \] computed in reverse topological order.
• Exercise: Explain why this formula holds for dags but not for general graphs.

Longest path algorithm

• We calculate the values specified by the formula by modifying the dfs code as follows.

  longest(G) {  
      for each vertex v do {
          d[v] ← -1
      }
      for each vertex v do {
          if d[v] = -1 then dfs(v)
      }
  }
  dfs(v) {  
      d[v] ← 0  
      for each edge (v,w) do {  
          if d[w] = -1 then dfs(w)
          d[v] ← max { d[v], l[v,w] + d[w] }
      }   
  }

Notes

1. Give the opposite recurrence & dynamic programming solution to this problem.
2. The longest paths in a dag are known as “critical paths” when doing Critical Path Method.
3. Finding a longest path in a general graph is NP-complete.
4. Our algorithm illustrates dynamic programming in dags. Similar algorithms exist for finding a longest path from s to t; a shortest path from s, etc.