1. CSCE 411 Design and Analysis of Algorithms
Set 8: Graph Algorithms
Prof. Jennifer Welch
Spring 2012
CSCE 411, Spring 2012: Set 8

2. Adjacency List Representationb b c a e b a d e c a d c d d b c e e b d
Space-efficient for sparse graphs
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3. Adjacency Matrix Representation
a b c d e a c d b e a 1 1 b 1 1 1 c 1 1 d 1 1 1 e 1 1
Check for an edge in constant time
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4. Graph Traversals Ways to traverse/search a graph Breadth-First Search
Visit every vertex exactly once
Breadth-First Search
Depth-First Search
CSCE 411, Spring 2012: Set 8

5. Breadth First Search (BFS)
Input: G = (V,E) and source s in V
for each vertex v in V do
mark v as unvisited
mark s as visited
enq(Q,s) // FIFO queue Q
while Q is not empty do
u := deq(Q)
for each unvisited neighbor v of u do
mark v as visited
enq(Q,v)
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6. BFS Example
<board work> a c d b s
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7. BFS Tree
We can make a spanning tree rooted at s by remembering the "parent" of each vertex
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8. Breadth First Search #2 Input: G = (V,E) and source s in V
for each vertex v in V do
mark v as unvisited
parent[v] := nil
mark s as visited
parent[s] := s
enq(Q,s) // FIFO queue Q
CSCE 411, Spring 2012: Set 8

9. Breadth First Search #2 while Q is not empty do u := deq(Q)
for each unvisited neighbor v of u do
mark v as visited
parent[v] := u
enq(Q,v)
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10. BFS Tree Example a c d b s
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11. BFS Trees BFS tree is not necessarily unique for a given graph
Depends on the order in which neighboring vertices are processed
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12. BFS Numbering
During the breadth-first search, assign an integer to each vertex
Indicate the distance of each vertex from the source s
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13. Breadth First Search #3 Input: G = (V,E) and source s in V
for each vertex v in V do
mark v as unvisited
parent[v] := nil
d[v] := infinity
mark s as visited
parent[s] := s
d[s] := 0
enq(Q,s) // FIFO queue Q
CSCE 411, Spring 2012: Set 8

14. Breadth First Search #3 while Q is not empty do u := deq(Q)
for each unvisited neighbor v of u do
mark v as visited
parent[v] := u
d[v] := d[u] + 1
enq(Q,v)
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15. BFS Numbering Example d = 1 a c d b s d = 2 d = 0 d = 2 d = 1
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16. Shortest Path Tree Theorem: BFS algorithm
visits all and only vertices reachable from s
sets d[v] equal to the shortest path distance from s to v, for all vertices v, and
sets parent variables to form a shortest path tree
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17. Proof Ideas
Use induction on distance from s to show that the d-values are set properly.
Basis: distance 0. d[s] is set to 0.
Induction: Assume true for all vertices at distance x-1 and show for every vertex v at distance x.
Since v is at distance x, it has at least one neighbor at distance x-1. Let u be the first of these neighbors that is enqueued.
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18. Proof Ideas c dist=x+1 dist=x-1 s u v dist=x d dist=x-1
Key property of shortest path distances: if v has distance x,
it must have a neighbor with distance x-1,
no neighbor has distance less than x-1, and
no neighbor has distance more than x+1
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19. Proof Ideas Fact: When u is dequeued, v is still unvisited.
because of how queue operates and since d never underestimates the distance
By induction, d[u] = x-1.
When v is enqueued, d[v] is set to
d[u] + 1= x
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20. BFS Running Time Initialization of each vertex takes O(V) time
Every vertex is enqueued once and dequeued once, taking O(V) time
When a vertex is dequeued, all its neighbors are checked to see if they are unvisited, taking time proportional to number of neighbors of the vertex, and summing to O(E) over all iterations
Total time is O(V+E)
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21. Depth-First Search Input: G = (V,E) for each vertex u in V do
mark u as unvisited
for each unvisited vertex u in V do
call recursiveDFS(u)
recursiveDFS(u):
mark u as visited
for each unvisited neighbor v of u do
call recursiveDFS(v)
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22. DFS Example <board work> a c d b e f g h
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23. Disconnected Graphs What if graph is disconnected or is directed?
call DFS on several vertices to visit all vertices
purpose of second for-loop in non-recursive wrapper a c b d e
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24. DFS Tree Actually might be a DFS forest (collection of trees)
Keep track of parents
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25. Depth-First Search #2 recursiveDFS(u):
Input: G = (V,E)
for each vertex u in V do
mark u as unvisited
parent[u] := nil
for each unvisited vertex u in V do
parent[u] := u // a root
call recursive DFS(u)
recursiveDFS(u):
mark u as visited
for each unvisited neighbor v of u do
parent[v] := u
call recursiveDFS(v)
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26. More Properties of DFS
Need to keep track of more information for each vertex:
discovery time: when its recursive call starts
finish time: when its recursive call ends
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27. Depth-First Search #3 recursiveDFS(u): time := 0
Input: G = (V,E)
for each vertex u in V do
mark u as unvisited
parent[u] := nil
time := 0
for each unvisited vertex u in V do
parent[u] := u // a root
call recursive DFS(u)
recursiveDFS(u):
mark u as visited
time++
disc[u] := time
for each unvisited neighbor v of u do
parent[v] := u
call recursiveDFS(v)
fin[u] := time
CSCE 411, Spring 2012: Set 8

28. Running Time of DFS initialization takes O(V) time
second for loop in non-recursive wrapper considers each vertex, so O(V) iterations
one recursive call is made for each vertex
in recursive call for vertex u, all its neighbors are checked; total time in all recursive calls is O(E)
Total time is O(V+E)
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29. Nested Intervals Let interval for vertex v be [disc[v],fin[v]].
Fact: For any two vertices, either one interval precedes the other or one is enclosed in the other.
because recursive calls are nested
Corollary: v is a descendant of u in the DFS forest if and only if v's interval is inside u's interval.
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30. Classifying Edges Consider edge (u,v) in directed graph
G = (V,E) w.r.t. DFS forest
tree edge: v is a child of u
back edge: v is an ancestor of u
forward edge: v is a descendant of u but not a child
cross edge: none of the above
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31. Example of Classifying Edges
in DFS forest
tree b a e
not in DFS forest
forward
tree
tree
cross
back f c d
back
tree
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32. DFS Application: Topological Sort
Given a directed acyclic graph (DAG), find a linear ordering of the vertices such that if (u,v) is an edge, then u precedes v.
DAG indicates precedence among events:
events are graph vertices, edge from u to v means event u has precedence over event v
Partial order because not all events have to be done in a certain order
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33. Precedence Example Tasks that have to be done to eat breakfast:
get glass, pour juice, get bowl, pour cereal, pour milk, get spoon, eat.
Certain events must happen in a certain order (ex: get bowl before pouring milk)
For other events, it doesn't matter (ex: get bowl and get spoon)
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34. Precedence Example get glass get bowl pour juice pour cereal get spoon
pour milk
eat breakfast
Order: glass, juice, bowl, cereal, milk, spoon, eat.
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35. Why Acyclic?
Why must directed graph by acyclic for the topological sort problem?
Otherwise, no way to order events linearly without violating a precedence constraint.
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36. Idea for Topological Sort Alg.1 2 3 4 5 6 7 8 9 10 11 12 13 14
eat
milk
spoon
juice
cereal
glass
bowl
consider reverse order of finishing times:
spoon, bowl, cereal, milk, glass, juice, eat
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37. Topological Sort Algorithm
input: DAG G = (V,E)
1. call DFS on G to compute finish[v] for all vertices v
2. when each vertex's recursive call finishes, insert it on the front of a linked list
3. return the linked list
Running Time: O(V+E)
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38. Correctness of T.S. Algorithm
Show that if (u,v) is an edge, then v finishes before u finishes. Thus the algorithm correctly orders u before v.
Case 1: u is discovered before v is discovered. By the way DFS works, u does not finish until v is discovered and v finishes.
Then v finishes before u finishes. u v
disc(v)
fin(v)
disc(u)
fin(u) v u
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39. Correctness of T.S. Algorithm
Show that if (u,v) is an edge, then v finishes before u finishes. Thus the algorithm correctly orders u before v.
Case 2: v is discovered before u is discovered. Suppose u finishes before v finishes (i.e., u is nested inside v).
Show this is impossible… u v
disc(u)
fin(u)
disc(v)
fin(v) v u
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40. Correctness of T.S. Algorithm
v is discovered but not yet finished when u is discovered.
Then u is a descendant of v.
But that would make (u,v) a back edge and a DAG cannot have a back edge (the back edge would form a cycle).
Thus v finishes before u finishes.
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41. DFS Application: Strongly Connected Components
Consider a directed graph.
A strongly connected component (SCC) of the graph is a maximal set of vertices with a (directed) path between every pair of vertices
Problem: Find all the SCCs of the graph.
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42. What Are SCCs Good For? Packaging software modules:
Construct directed graph of which modules call which other modules
A SCC is a set of mutually interacting modules
Pack together those in the same SCC
from
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43. SCC Example h f a e g c b d
four SCCs
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44. How Can DFS Help? Suppose we run DFS on the directed graph.
All vertices in the same SCC are in the same DFS tree.
But there might be several different SCCs in the same DFS tree.
Example: start DFS from vertex h in previous graph
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45. Main Idea of SCC Algorithm
DFS tells us which vertices are reachable from the roots of the individual trees
Also need information in the "other direction": is the root reachable from its descendants?
Run DFS again on the "transpose" graph (reverse the directions of the edges)
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46. SCC Algorithm input: directed graph G = (V,E)
call DFS(G) to compute finishing times
compute GT // transpose graph
call DFS(GT), considering vertices in decreasing order of finishing times
each tree from Step 3 is a separate SCC of G
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47. SCC Algorithm Example h f a e g c b d input graph - run DFS
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48. After Step 1 fin(c) fin(d) fin(b) fin(e) fin(a) fin(h) fin(g) fin(f) 12 3 4 5 6 7 8 9 10 11 12 13 14 15 16 a b c d e g h f
Order of vertices for Step 3: f, g, h, a, e, b, d, c
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49. After Step 2 h f a e g c b d
transposed input graph - run DFS with specified order of vertices
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50. After Step 3 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 f h g a b d e c
SCCs are {f,h,g} and {a,e} and {b,c} and {d}.
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51. Running Time of SCC Algorithm
Step 1: O(V+E) to run DFS
Step 2: O(V+E) to construct transpose graph, assuming adjacency list rep.
Step 3: O(V+E) to run DFS again
Step 4: O(V) to output result
Total: O(V+E)
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52. Correctness of SCC Algorithm
Proof uses concept of component graph, GSCC, of G.
Vertices are the SCCs of G; call them C1, C2, …, Ck
Put an edge from Ci to Cj iff G has an edge from a vertex in Ci to a vertex in Cj
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53. Example of Component Graph
{a,e}
{f,h,g}
{d}
{b,c}
based on example graph from before
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54. Facts About Component Graph
Claim: GSCC is a directed acyclic graph.
Why?
Suppose there is a cycle in GSCC such that component Ci is reachable from component Cj and vice versa.
Then Ci and Cj would not be separate SCCs.
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55. Facts About Component Graph
Consider any component C during Step 1 (running DFS on G)
Let d(C) be earliest discovery time of any vertex in C
Let f(C) be latest finishing time of any vertex in C
Lemma: If there is an edge in GSCC from component C' to component C, then
f(C') > f(C).
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56. Proof of Lemma Case 1: d(C') < d(C).
Suppose x is first vertex discovered in C'.
By the way DFS works, all vertices in C' and C become descendants of x.
Then x is last vertex in C' to finish and finishes after all vertices in C.
Thus f(C') > f(C).
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57. Proof of Lemma Case 2: d(C') > d(C).
Suppose y is first vertex discovered in C.
By the way DFS works, all vertices in C become descendants of y.
Then y is last vertex in C to finish.
Since C'  C, no vertex in C' is reachable from y, so y finishes before any vertex in C' is discovered.
Thus f(C') > f(C).
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58. SCC Algorithm is Correct
Prove this theorem by induction on number of trees found in Step 3 (running DFS on GT).
Hypothesis is that the first k trees found constitute k SCCs of G.
Basis: k = 0. No work to do !
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59. SCC Algorithm is Correct
Induction: Assume the first k trees constructed in Step 3 (running DFS on GT) correspond to k SCCs; consider the (k+1)st tree.
Let u be the root of the (k+1)st tree.
u is part of some SCC, call it C.
By the inductive hypothesis, C is not one of the k SCCs already found and all so vertices in C are unvisited when u is discovered.
By the way DFS works, all vertices in C become part of u's tree
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60. SCC Algorithm is Correct
Show only vertices in C become part of u's tree. Consider an outgoing edge from C. w z u C' C G: w z u C' C
GT:
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61. SCC Algorithm is Correct
By lemma, in Step 1 (running DFS on G) the last vertex in C' finishes after the last vertex in C finishes.
Thus in Step 3 (running DFS on GT), some vertex in C' is discovered before any vertex in C is discovered.
Thus in Step 3, all of C', including w, is already visited before u's DFS tree starts w z u C' C G:
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