Lecture 16: DFS, DAG, and Strongly Connected Components Shang-Hua Teng
Dec 16, 2015
Directed Acyclic Graphs
• A directed acyclic graph or DAG is a directed graph with no directed cycles:
DFS and DAGs
• Theorem: a directed graph G is acyclic iff a DFS of G yields no back edges:– => if G is acyclic, will be no back edges
• Trivial: a back edge implies a cycle
– <= if no back edges, G is acyclic• Proof by contradiction: G has a cycle a back edge
– Let v be the vertex on the cycle first discovered, and u be the predecessor of v on the cycle
– When v discovered, whole cycle is white– Must visit everything reachable from v before returning from
DFS-Visit()– So path from uv is graygray, thus (u, v) is a back edge
Topological Sort
• Topological sort of a DAG:– Linear ordering of all vertices in graph G such that
vertex u comes before vertex v if edge (u, v) G
• Real-world application: Scheduling a dependent graph, find a feasible course plan for university studies
A Topological Sort AlgorithmTopological-Sort(){
1. Call DFS to compute finish time f[v] for each vertex
2. As each vertex is finished, insert it onto the front of a linked list
3. Return the linked list of vertices}
• Time: O(V+E)• Correctness: need to prove that
(u,v) G f[u]>f[v]
Correctness of Topological Sort
• Lemma: (u,v) G f[u] > f[v]– When (u,v) is explored, u is gray, consider the
following cases:1. v is gray (u,v) is back edge. Can’t happen, if G is a
DAG.2. v if white v becomes descendent of u f[v]
< f[u] (since must finish v before backtracking and finishing u)
3. v is black v already finished f[v] < f[u]
Strongly Connected Directed graphs
• Every pair of vertices are reachable from each other
a
b
d
c
e
f
g
Strongly-ConnectedGraph G is strongly connected if, for every u
and v in V, there is some path from u to v and some path from v to u.
Strongly Connected
Not Strongly Connected
Strongly-Connected Components
A strongly connected component of a graph is a maximal subset of nodes (along with their associated edges) that is strongly connected. Nodes share a strongly connected component if they are inter-reachable.
Reduced Component Graph of Strongly Connected Components
a
b
d
c
e
f
g{ a , c , g }
{ f , d , e , b }
• Component graph GSCC=(VSCC, ESCC): one vertex for each component– (u, v) ESCC if there exists at least one directed
edge from the corresponding components
Graph of Strongly Connected Components
• Theorem: the Component graph GSCC=(VSCC, ESCC) is a DAG– Each component is maximal in the sense that no
other vertices can be added to it. If GSCC=(VSCC, ESCC) is not a DAG, then one can merge components on along a circle of GSCC
• Therefore, GSCC has a topological ordering
Finding Strongly-Connected Components
• Input: A directed graph G = (V,E)
• Output: a partition of V into disjoint sets so that each set defines a strongly connected component of G
• How should we compute the partition?
Graph of Strongly Connected Components
• Recall: Theorem: the Component graph GSCC=(VSCC, ESCC) is a DAG– Each component is maximal in the sense that no
other vertices can be added to it. If GSCC=(VSCC, ESCC) is not a DAG, then one can merge components on along a circle of GSCC
• Therefore, GSCC has a topological ordering
DFS on G Topological Sort GSCC=(VSCC, ESCC)
• Let U be a subset of V
• If we output U in VSCC in the decreasing order of f[U], then we topologically sort GSCC
• Lemma: Let U and U’ be distinct strongly connected component, suppose there is an edge (u,v) in E where u in U and v in U’. Then f[U] > f[U’]
][max)(
][min)(
ufUf
udUd
Uu
Uu
Proof of the Lemma
Lemma: Let U and U’ be distinct strongly connected component, suppose there is an edge (u,v) in E where u in U and v in U’. Then f[U] > f[U’]
Proof: Two cases1. d[U] < d[U’], say x in U is the first vertex
2. d[U’] < d[U], say y is the first, but U is not reachable from y
'Uvux
Transpose of a Digraph
Transpose of G = (V,E):
GT=(V, ET), where ET={(u, v): (v, u) E}
If G is a DAG then GT is also a DAG
If we print the topological order of G in the reverse order, then it is a topological order of GT
Strongly-Connected ComponentsStrongly-Connected-Components(G)
1. call DFS(G) to compute finishing times f[u] for each vertex u.
2. compute GT
3. call DFS(GT), but in the main loop of DFS, consider the vertices in order of decreasing f[u]
4. output the vertices of each tree in the depth-first forest of step 3 as a separate strongly connected component.
The graph GT is the transpose of G, which is visualized by reversing the arrows on the digraph.
Strong Components: example
a
d c
b a
d c
b
after step 1
a4
b3
c2
d1a4
d1 c2
b3
Graph Gr
a4
c2
b3
d1
df spanning forest for Gr
Runtime
Lines 1 and 3 are (E+V) due to DFS
Line 2 involves creating an adjacency list or matrix, and it is also O(E+V)
Line 4 is constant time
So, SCC(G) is (E+V)
Strongly-Connected Components
DFS on G, starting at c.
node ad=13f=14
node bd=11f=16
node cd=1f=10
node dd=8f=9
node ed=12f=15
node fd=3f=4
node gd=2f=7
node hd=5f=6
DFS on GT
node ad=13f=14
node bd=11f=16
node cd=1f=10
node dd=8f=9
node ed=12f=15
node fd=3f=4
node gd=2f=7
node hd=5f=6
node ad=2f=5=b
node bd=1f=6
=NIL
node cd=7f=10=NIL
node dd=8f=9=c
node ed=3f=4=a
node fd=12f=13=g
node gd=11f=14=NIL
node hd=15f=16=NIL
DFS on GT
This is GT, labeled after running DFS(GT). In order of decreasing finishing time, process the nodes in this order: b e a c d g h f.
GT:
node ad=2f=5=b
node bd=1f=6
=NIL
node cd=7f=10=NIL
node dd=8f=9=c
node ed=3f=4=a
node fd=12f=13=g
node gd=11f=14=NIL
node hd=15f=16=NIL
Strongly-Connected ComponentsThese are the 4 trees that result, yielding the strongly connected
components. Finally, merge the nodes of any given tree into a super-node, and draw
links between them, showing the resultant acyclic component graph.
a
b c
d
e
f
g h
a b c d
e f g h
abe cd
fg h
Component Graph