Directed acyclic graphs

Transcription

1 Directed acyclic graphs Madhavan Mukund Chennai Mathematical Institute March, 2

2 Directed Graphs x v y Edges have direction. Cannot be traversed in the opposite direction. w

3 The Task Scheduling Problem A complicated project involves a list of several tasks to be fulfilled and there are dependencies among the tasks. Some tasks have to be completed before some other can be taken up. Here is an example:

4 The Task Scheduling Problem A complicated project involves a list of several tasks to be fulfilled and there are dependencies among the tasks. Some tasks have to be completed before some other can be taken up. Here is an example: Can we list the tasks in an order consistent with the dependencies? (i.e. if p depends on q then q should appear earlier in the listing.)

5 The Task Scheduling Problem A complicated project involves a list of several tasks to be fulfilled and there are dependencies among the tasks. Some tasks have to be completed before some other can be taken up. Here is an example: Can we list the tasks in an order consistent with the dependencies? (i.e. if p depends on q then q should appear earlier in the listing.) Topologically ordered listing

6 Directed Acyclic Graphs A directed graph is said to be acyclic if it has no cycles

7 Directed Acyclic Graphs A directed graph is said to be acyclic if it has no cycles A directed graph can be topologically ordered only if it is acyclic.

8 Some definitions indegree(v) is the number of incoming edges at v.

9 Some definitions indegree(v) is the number of incoming edges at v

10 Some definitions outdegree(v) is the number of outgoing edges at v

11 Some definitions outdegree(v) is the number of outgoing edges at v

12 An Observation If v,v 2,...v n is a topologically ordered listing then indegree(v ) =.

13 An Observation If v,v 2,...v n is a topologically ordered listing then indegree(v ) =. Claim: Every DAG has a vertex whose indegree is.

14 An Observation If v,v 2,...v n is a topologically ordered listing then indegree(v ) =. Claim: Every DAG has a vertex whose indegree is. Start at any vertex and keep walking backwards as long as the indegree is nonzero.

15 Every DAG has a topological ordering Clearly, if the DAG has only one vertex then it can be topologically ordered.

16 Every DAG has a topological ordering Clearly, if the DAG has only one vertex then it can be topologically ordered. Suppose the DAG has n+ vertices. Pick any vertex (v) with indegree = and delete it.

17 Every DAG has a topological ordering Clearly, if the DAG has only one vertex then it can be topologically ordered. Suppose the DAG has n+ vertices. Pick any vertex (v) with indegree = and delete it. The smaller graph with n vertices can be topologically ordered (by the induction hypothesis). Let w,w 2,...w n be a topological order of the smaller graph.

18 Every DAG has a topological ordering Clearly, if the DAG has only one vertex then it can be topologically ordered. Suppose the DAG has n+ vertices. Pick any vertex (v) with indegree = and delete it. The smaller graph with n vertices can be topologically ordered (by the induction hypothesis). Let w,w 2,...w n be a topological order of the smaller graph. v,w,...,w n is a topological ordering of the full graph.

19 An Algorithm

20 An Algorithm :

21 An Algorithm : 2 :

22 An Algorithm 3 4 : 2 : 5 :

23 An Algorithm 3 : 2 : 5 : 4 :

24 An Algorithm : 2 : 5 : 4 : 3 :

25 An Algorithm 7 8 : 2 : 5 : 4 : 3 : 6 :

26 An Algorithm 8 : 2 : 5 : 4 : 3 : 6 : 7 :

27 An Algorithm : 2 : 5 : 4 : 3 : 6 : 7 : 8

28 Complexity Direct adjacency matrix implementation takes O(n 2 ) time. Row i of adjacency matrix has outgoing edges from vertex i Column i of adjacency matrix has incoming edges to vertex i Compute indegree[i] initially by scanning all columns O(n 2 ) In each iteration,2,...,n Identify an unvisited vertex i with indegree[i] = O(n) Scan row i and update indegree[j] for each outgoing neighbour of i O(n) Loop takes O(n 2 )

29 Complexity... Improve to O(n + m) using adjacency list representation For each vertex i, scan list nbr[i] and increment indegree[j] for each j in the list O(m) Build a queue of vertices with indegree O(n) While the queue is not empty Pick up i from head of queue Update indegree[i] for all neighbours of i If indegree[j] becomes, add j to queue Loop takes O(m)

30 Longest Path in a DAG How do we compute the length of the longest path in a DAG?

31 Longest Path in a DAG How do we compute the length of the longest path in a DAG? For each vertex find the length of the longest path ending at the vertex.

32 Longest Paths For a vertex v with indegree = this length l(v) =.

33 Longest Paths For a vertex v with indegree = this length l(v) =

34 Longest Paths For a vertex v with indegree = this length l(v) = For any vertex v with incoming edges from w,w 2,...,w k l(v) = +Max(l(w ),l(w 2 ),...,l(w k ))

35 Longest Paths For a vertex v with indegree = this length l(v) = For any vertex v with incoming edges from w,w 2,...,w k l(v) = +Max(l(w ),l(w 2 ),...,l(w k )) Does this give an algorithm?

36 Length via topological ordering If there is an edge from w to v then w appears before v in any topological ordering.

37 Length via topological ordering If there is an edge from w to v then w appears before v in any topological ordering. Let v,v 2,...,v N be a topological ordering of the vertices.

38 Length via topological ordering If there is an edge from w to v then w appears before v in any topological ordering. Let v,v 2,...,v N be a topological ordering of the vertices. l(v ) =.

39 Length via topological ordering If there is an edge from w to v then w appears before v in any topological ordering. Let v,v 2,...,v N be a topological ordering of the vertices. l(v ) =. Use the equation l(v) = +Max(l(w ),l(w 2 ),...,l(w k )) (where w,w 2,...w k are the vertices with edges to v) to determine l(v) for v {v 2,v 3,...v N } in that order.

40 Propagating the length We may also combine the computation of the topological ordering and the values of l(v).

41 Propagating the length We may also combine the computation of the topological ordering and the values of l(v). When the node v is processed, we have with us the value of l(v) and we propagate this information to each w with an incoming edge form v by setting l(w) = Max(l(w),l(v)+)

42 Propagating the length We may also combine the computation of the topological ordering and the values of l(v). When the node v is processed, we have with us the value of l(v) and we propagate this information to each w with an incoming edge form v by setting l(w) = Max(l(w),l(v)+)

43 Propagating the length We may also combine the computation of the topological ordering and the values of l(v). When the node v is processed, we have with us the value of l(v) and we propagate this information to each w with an incoming edge form v by setting l(w) = Max(l(w),l(v)+) 3 4 5

44 Propagating the length We may also combine the computation of the topological ordering and the values of l(v). When the node v is processed, we have with us the value of l(v) and we propagate this information to each w with an incoming edge form v by setting l(w) = Max(l(w),l(v)+) 3 4 5

45 Propagating the length We may also combine the computation of the topological ordering and the values of l(v). When the node v is processed, we have with us the value of l(v) and we propagate this information to each w with an incoming edge form v by setting l(w) = Max(l(w),l(v)+) 3 4

46 Propagating the length We may also combine the computation of the topological ordering and the values of l(v). When the node v is processed, we have with us the value of l(v) and we propagate this information to each w with an incoming edge form v by setting l(w) = Max(l(w),l(v)+) 3 4 2

47 Propagating the length We may also combine the computation of the topological ordering and the values of l(v). When the node v is processed, we have with us the value of l(v) and we propagate this information to each w with an incoming edge form v by setting l(w) = Max(l(w),l(v)+) 4 2

48 Propagating the length We may also combine the computation of the topological ordering and the values of l(v). When the node v is processed, we have with us the value of l(v) and we propagate this information to each w with an incoming edge form v by setting l(w) = Max(l(w),l(v)+) 4 2 2

49 Propagating the length We may also combine the computation of the topological ordering and the values of l(v). When the node v is processed, we have with us the value of l(v) and we propagate this information to each w with an incoming edge form v by setting l(w) = Max(l(w),l(v)+) 2 2

50 Propagating the length We may also combine the computation of the topological ordering and the values of l(v). When the node v is processed, we have with us the value of l(v) and we propagate this information to each w with an incoming edge form v by setting l(w) = Max(l(w),l(v)+) 2 2

51 Propagating the length We may also combine the computation of the topological ordering and the values of l(v). When the node v is processed, we have with us the value of l(v) and we propagate this information to each w with an incoming edge form v by setting l(w) = Max(l(w),l(v)+)

52 Propagating the length We may also combine the computation of the topological ordering and the values of l(v). When the node v is processed, we have with us the value of l(v) and we propagate this information to each w with an incoming edge form v by setting l(w) = Max(l(w),l(v)+)

53 Propagating the length We may also combine the computation of the topological ordering and the values of l(v). When the node v is processed, we have with us the value of l(v) and we propagate this information to each w with an incoming edge form v by setting l(w) = Max(l(w),l(v)+) 3 8 2

54 Propagating the length We may also combine the computation of the topological ordering and the values of l(v). When the node v is processed, we have with us the value of l(v) and we propagate this information to each w with an incoming edge form v by setting l(w) = Max(l(w),l(v)+) 3 8 4

55 Propagating the length We may also combine the computation of the topological ordering and the values of l(v). When the node v is processed, we have with us the value of l(v) and we propagate this information to each w with an incoming edge form v by setting l(w) = Max(l(w),l(v)+) 4

56 The Algorithm Initialize l(v) = for all vertices. In each round Pick a vertex v with indegree(v) =. For each w with v w, set l(w) = Max(l(w),l(v)+). Delete v from the graph.

57 What if the graph had cycles? Recall that finding shortest paths in graphs is very easy.

58 What if the graph had cycles? Recall that finding shortest paths in graphs is very easy. As shown above finding the longest path in a DAG is very easy too.

59 What if the graph had cycles? Recall that finding shortest paths in graphs is very easy. As shown above finding the longest path in a DAG is very easy too. It turns out that finding the length of the longest path in a graph (directed or undirected) is very hard. (NP-Complete) It is extremely unlikely that this problem will have an efficient solution.