1. Graph Algorithms

2. Representations of graphs：undirected graph
An undirected graph G have five vertices and seven edges
An adjacency-list representation of G
The adjacency-matrix representation of G
edge 1 2 3 4 5
vertex 1 2 5 / 2 1 5 3 4 / 3 2 4 / 4 2 5 3 / 5 4 1 1 / 1 2 3 4 5 1 1 1 2 1 1 1 1 3 1 1 4 1 1 1 5 1 1

3. Representations of graphs：directed graph
An directed graph G have six vertices and eight edges
An adjacency-list representation of G
The adjacency-matrix representation of G 1 2 3 4 5 6 1 2 4 / 5 3 6 1 2 3 4 5 6 1 1 1 2 1 3 1 1 4 1 5 1 6 1

4. Time Complexity: O(|E|+|V|)
BFS(G,s)
for each vertex u∈ G.V-{s}
u.color = WHITE
u.d = ∞
u.π = NIL
s.color = GREEN
s.d = 0
s.π = NIL
Q = empty
Enqueue(Q,s)
while Q ≠ empty
u = Dequeue(Q)
for each v∈ G.Adj[u]
if v.color == WHITE
v.color = GREEN
v.d = u.d + 1
v.π = u
Enqueue(Q,v)
u.color = BLACK
Time Complexity: O(|E|+|V|)
第一個迴圈，初始化除s外所有的節點成白色，距離無限大，前一個點為空。
之後初始化s為綠色，距離0，前一個點是空。至此花去O(|V|)
接下來將s放入Q中，之後反覆執行Dequeue，直到Q空了為止。
每個Dequeue出來的點u，均將其白色的鄰居放入Q中，
並且將他們塗成綠色，距離設定成d[u]+1以及將前一個點設定為u。
處理完所有的鄰居之後，就將u塗成紅色。
因此每個點被放入Q一次，並且一個點被檢驗是不是白色的總次數為|E|，
故此部分花費O(|E|+|V|)的時間。
Elementary Graph Algorithms

5. The operation of BFS on an undirected graph1 1 v w x y v w x y Q Q s w r 1 1
(c) r s t u
(d) r s t u 1 2 1 2 1 2 2 1 2 v w x y v w x y Q r t x Q t x v 1 2 2 1 2 2
(e) r s t u
(f) r s t u 1 2 3 1 2 3 2 1 2 2 1 2 3 v w x y v w x y Q x v u Q v u y 2 2 3 2 3 3

6. (g) r s t u
(h) r s t u 1 2 3 1 2 3 2 2 1 2 3 2 1 2 3 v w x y v w x y Q u y y 3 3 3
(i) r s t u 1 2 3 2 1 2 3 v w x y Q

7. Breadth-first search：
Initially, vertices are colored white.
Discovered vertices are colored green.
When done, discovered vertices are colored black.
u.d stores the distance from s to u.
u.π is a predecessor to u on its shortest path.
Q is a first-in first-out queue.

8. Properties of Breadth-first search：
After execution of BFS, all nodes reachable from the source s are colored black.
After execution of BFS, v.d is the distance of a shortest path from the source s to v for vertices v reachable from s.
After execution of BFS, if v is reachable from s, then one of the shortest paths to v passes through the edge( v.π, v) at the end.
After execution of BFS, the edges( v.π, v ) for v reachable from s form a breadth-first tree.

9. Lemma 22.1： G=(V,E)：a directed or undirected graph.
s：an arbitrary vertex
： the shortest-path distance from s to v.
Then for any
Proof： s u v 1

10. Lemma 22.2： G=(V,E)：a directed or undirected graph.
s：an arbitrary vertex
u.d： the distance from s to u computed by the algorithm
Suppose that BFS is run on G from s.
Then on termination, for each vertex v ∈ V, the value v.d computed by
BFS satisfies v.d ≥ δ (s, v).
Proof：
By induction on the number Enqueue operations.
Basis： when s is placed in Q. s.d =0 = δ(s, s)
for all
Induction Step： Consider a white vertex v discovered during the search from a vertex u.
Inductive hypothesis implies u.d ≥ δ (s, u).
By Lemma 22.1, v.d = u.d +1 ≥ δ(s, u) + 1 ≥ δ(s, v)
From then on, v.d will not be changed again.

11. Lemma 22.3：
Q：<v1,v2,…,vr>~ the queue when BFS executes on a graph G=(V,E).
head
tail
Then
and
for i=1,2,…,r-1.
Proof：
By induction on the number of queue operations.
Basis： when Q has only s.
Induction Step：
1>： after dequeue：
v2 becomes the new head in Q.
by inductive hypothesis.
2>：after enqueue a new vertex v into the Q.
Let vr+1 be v.
neighbors
Note that u’s adjacency list is being scanned and u is just removed.
Thus,
And
The rest , for i=1,…,r-1, remain unaffected.

12. Thm 22.5: (Correctness of BFS)
1. During the execution of BFS on G=(V,E), BFS discovers
every vertex v ∈ V that is reachable from s, and on
termination v.d = δ(s, v)
2. For any vertex v ≠ s reachable from s, one of the shortest
paths from s to v is the shortest path from s to v.π followed by the edge ( v.π, v).

13. from s to v.π then traversing the edge (v.π, v).
Proof：
If v unreachable from s, s.d  (s, v)=.
By BFS, v is never discovered.
By contradiction, suppose some vertex has a d value not equal to its shortest distance.
Let v be the vertex with minimum (s, v) that has an incorrect d value.
Let u be v’s immediate predecessor on the shortest path from s to v.
So (s, v) = (s, u) +1. Because (s, u) < (s, v), we have u.d=(s, u).
(*) Thus v.d > (s, v) = (s, u) + 1= u.d + 1.
Consider the time when BFS chooses to dequeue u from Q.
At this time v is either white, gray or black.
Case 1: (v is white) line 15 set v.d = u.d contradicting (*)
Case 2: (v is black) v is already not in Q, thus v.d  u.d --- contradicting (*)
Case 3: (v is gray) v became gray while dequeuing some vertex w, which
was removed earlier than u. Thus v.d= w.d +1 and w.d  u.d
and so v.d  u.d + 1 – again a contradiction!
Thus we conclude v.d = (s, v) for all v.
All vertices reachable from s must be discovered. If v.π =u, then v.d = u.d +1.
Thus, we can obtain a shortest path from s to v by taking a shortest path
from s to v.π then traversing the edge (v.π, v).

14. Breadth-first trees:
For a graph G=(V, E) with source s, define the predecessor subgraph of G as G’=(V’, E’), where V’ = { v  V: v.π ≠ NIL}  {s} and
E’ = {(v.π, v): v  V’- {s}}.
G’ is a BF tree if V’ consists of the vertices from s to v in G’ and, for all v in V’, there is a unique simple path from s to v, which is also a shortest path
from s to v in G.
Lemma 22. 6: When applied to a graph G=(V, E), procedure BFS construct π so that G’= (V’, E’) is a BF tree.
Proof: Line 16 of BFS sets v.π =u iff (u, v) ∈ E and (s, v) < .
Thus V’ consists of the vertices in V reachable from s.
Since G’ forms a tree, it contains a unique path from s to each
vertex in V’.
By Thm 22.5, inductively, we have that every such path is a shortest path.

15. Print path演算法 可在執行過BFS演算法的圖上印出s到v的最短路徑。如無路徑也會印出沒有路徑的訊息。
Print-Path(G, s, v)
if v == s
print s
elseif v.π == NIL
print “no path from” s “to” v
else Print-Path(G,s,v.π)
print v
此演算法利用BFS建立的表格π，並利用遞迴的方式，
利用之前提過的BFS性質，
(π[v],v)是某一條最短路徑上的一邊，
找出一條path並且列印出來。
Elementary Graph Algorithms

16. Depth-first search Nodes are initially white
Nodes become gray when first discovered
Nodes become black when they are done
v.d records when v is first discovered
v.f records when v is done
u.d< u.f

17. (a) (b) 1/ 1/ 2/ (c) (d) 1/ 2/ 1/ 2/ 3/ 4/ 3/ u v w u v w x y z x y z
Discovery time
(a) u v w
(b) u v w 1/ 1/ 2/ x y z x y z
(c) u v w
(d) u v w 1/ 2/ 1/ 2/ 3/ 4/ 3/ x y z x y z

18. (e) (f) 1/ 2/ 1/ 2/ 4/ 3/ 4/5 3/ (g) (h) 1/ 2/ 1/ 2/7 4/5 3/6 4/5 3/6u v w
(f) u v w 1/ 2/ 1/ 2/ B B 4/ 3/
4/5 3/ x y z x y z
(g) u v w
(h) u v w 1/ 2/ 1/
2/7 B B
4/5
3/6
4/5
3/6 x y z x y z

19. (i) (j) 1/ 2/7 1/8 2/7 4/5 3/6 4/5 3/6 (k) (l) 1/8 2/7 9/ 1/8 2/7 9/u v w
(j) u v w 1/
2/7
1/8
2/7 B F B F
4/5
3/6
4/5
3/6 x y z x y z
(k) u v w
(l) u v w
1/8
2/7 9/
1/8
2/7 9/ B F B C F
4/5
3/6
4/5
3/6 x y z x y z

20. (m) (n) 1/8 2/7 9/ 1/8 2/7 9/ 4/5 3/6 10/ 4/5 3/6 10/ (o) (o) 1/8 2/7u v w
(n) u v w
1/8
2/7 9/
1/8
2/7 9/ B C F B C F B
4/5
3/6
10/
4/5
3/6
10/ x y z x y z
(o) u v w
(o) u v w
1/8
2/7 9/
1/8
2/7
9/12 B C B C F F B B
4/5
3/6
10/11
4/5
3/6
10/11 x y z x y z

21. (u,v)
Back edges: if u is connected to an ancestor v in a depth first tree. (eg self-loop)
Forward edges: if u is connected to an descendant v in a depth-first tree.
Cross edges: if u is not connected to an ancestor v in the same depth-first tree.
OR if v is not connected to an ancestor u in the same depth-first tree.
OR if u and v belong to different depth-first trees.

22. DFS演算法 DFS(G) for each vertex u∈ G.V do u.color = WHITE u.π = NIL
time = 0
for each vertex u ∈ G.V
if u.color == WHITE
DFS-Visit(G,u)
Elementary Graph Algorithms

23. DFS-Visit演算法 DFS-Visit(G, u)
time = time //u has just been discovered
u.d = time
u.color = GRAY
for each v ∈ G.Adj[u] //explore edge (u,v)
if v.color == WHITE
v.π = u
DFS-Visit(G, v)
u.color = BLACK //DFS-Visit(G, u) is done
time = time + 1
u.f = time
Elementary Graph Algorithms

24. The running time of DFS is O(V+E)
After execution of DFS, all nodes are colored black
After execution of DFS, the edges( v.𝛑, v) form a collection of depth-first tree, called a depth-first forest.

25. Edge Classification
1. Tree edges( u, v ): u was discovered first using ( u,v )
2. Back edges( u, v ): v is an ancestor of u in the DFS tree
3. Forward edges( u, v ): v is a descendent of u, not a tree edge
4. Cross edges( u, v ): Other edges
Example
In a depth-first search of an undirected graph, every edge is either a tree edge or a back edge

26. 3/6 2/9 1/10 4/5 7/8 12/13 (a) 14/15 11/16 y z s x w v B C u t F (b) s
( s (z (y (x x) y) (w w) z) s) (t (v v) (u u) t)

27. (c) s C t B F z C u v B C y w C x

28. Thm 22.7: (Parenthesis theorem)
In any DFS of a graph G=( V,E ), for any two vertices u and v, exactly one of the following 3 conditions holds:
(1). The intervals [u.d, u.f] and [v.d, v.f] are disjoint,
(2). The interval [u.d, u.f] is contained entirely within the interval [v.d, v.f], and u is a descendant of v in the depth-first tree,
(3). or as above with v as a descendant of u

29. Pf:
1. If u.d < v.d
case 1: v.d < u.f :
So v is finished before finishing u. Thus (3) holds
case 2: u.f < v.d
u.d < u.f < v.d < v.f  (1) holds
2. If v.d < u.d:
Similarly, by switching the roles of u and v
v was discovered while
u was still gray.
v is a descendant of u, and all
v’s outgoing edges are explored

30. Cor8: v is a proper descendant of vertex u in
Cor8: v is a proper descendant of vertex u in the depth-first forest for a graph G
iff u.d < v.d < v.f < u.f
Thm9: (White-path theorem)
In a DF forest of G=( V,E ), vertex v is a descendant of u iff at the time u.d that the search discovers u, v can be reached from u along a path consisting entirely of white vertices

31. Pf: “” v: descendant of u “” u.d < w.d, by the above cor.
Assume at time u.d, v is reachable from u along a path of white vertices, but v does not become a descendant of u in DF tree
Thus, u.d < v.d < w.f <= u.f
Thm7 implies [v.d, v.f] is contained entirely in [u.d, u.f] By Cor8, v is a descendant of u.
u.d < w.d, by the above cor. u w v
Thus w is white at time u.d u w v
Descendant of u
w.f <= u.f
v must be discovered after u is discovered,
but before w is finished.

32. Thm 22.10: In a DFS of an undirected graph G, every edge of G is either a tree edge or a back edge
Pf:
Let and suppose u.d < v.d. Then v must be discovered and finished before finishing u
If (u, v) is explored first in the direction from u to v, then (u, v) becomes a tree edge
If (u, v) is explored first in the direction from v to u, then (u, v) is a back edge, since u is still gray at the time (u, v) is first explored

33. Topological sort
定義： A topological sort of a dag G=(V, E) is a linear ordering of all its vertices. (dag: Directed acyclic graph) 如 edge(u, v), u appears before v in the ordering
undershorts
socks
pants
shoes
belt
shirt
tie
jacket
watch
11/16
12/15
6/7
1/8
2/5
3/4
17/18
13/14
9/10
socks
undershorts
pants
shoes
watch
shirt
belt
tie
jacket

34. TOPOLOGICAL-SORT(G): (V+E)
1. Call DFS(G) to compute finishing times v.f for each vertex v (V+E)
2. As each vertex is finished, insert it onto the front of a linked list O(1)
3. Return the linked list of vertices
undershorts
socks
pants
shoes
belt
shirt
tie
jacket
watch
11/16
12/15
6/7
1/8
2/5
3/4
17/18
13/14
9/10
socks
undershorts
pants
shoes
watch
shirt
belt
tie
jacket
11/16
12/15
6/7
1/8
3/4
13/14
17/18
9/10
2/5

35. Lemma 22.11
A directed graph G is acyclic iff DFS(G) yields no back edges.
pf:
 Suppose there is a back edge (u,v), v is an ancestor of u.
Thus there is a path from v to u and a cycle exists.
 Suppose G has a cycle c. We show DFS(G) yields a back edge.
Let v be the first vertex to be discovered in c, and
(u,v) be the preceding edge in c.
At time v.d, there is a path of white vertices from v to u.
By the white-path thm., u becomes a descendant of v in the DF forest (u,v) is a back edge.

36. Thm 22.12
TOPOLOGICAL-SORT(G) produces a topological sort of G
pf:
Suppose DFS is run to determine finishing times for vertices.
It suffices to show that for any pair of u,v , if there is an edge from u to v, then v.f< u.f.
When (u,v) is explored by DFS(G), v cannot be gray.
Therefore v must be either white or black.
1. If v is white, it becomes a descendant of u, so v.f < u.f
2. If v is black, then v.f < u.f

37. Strongly connected components:
A strongly connected component of a directed graph G(V,E) is a maximal set of vertices U  V s.t. for every pair u, vU, u and v are reachable from each other.
Given G=(V,E), define GT=(V,ET), where ET={(u,v): (v,u)E} Given a G with adjacency-list representation, it takes O(V+E) to create GT.
G and GT have the same strongly connected components.

38. G: GT: 13/14 11/16 12/15 3/4 1/10 2/7 8/9 5/6 a b c d e f g h abe cd h

39. StronglyConnectedComponents(G)
1. Call DFS(G) to compute finishing times u.f 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 u.f
4. Output the vertices of each tree in the depth-first forest of
step 3 as a separate strongly connected component
Time: (V+E)

40. Lemma 22.13:
Let C and C’ be distinct strongly connected components in directed graph G=(V, E), let u, v in C and u’, v’ in C’, and suppose there is a path from u to u’ in in G.
Then there cannot also be a path from v’ to v in G.
Def: Let U  V, then we define
d(U) = min { u.d }
u U
f (U) = max { u.f }

41. Lemma 22.14
Let C and C’ be distinct strongly connected components in directed graph G=(V, E). Suppose that there is an edge (u, v) in E, where u in C and v in C’. Then f (C) > f ( C’).
pf:
(1) If d(C) < d(C’): let x be the 1st discovered vertex in C.
At time x.d all vertices in C and C’ are white. There is a path from x to all (white) vertices in C and to all vertices in C’:
x~↝ u  v ~↝ w. All vertices in C and C’ are descendants of x.
Thus x.f = f (C) > f(C’).
(2) If d(C) > d(C’): let y be the 1st discovered vertex in C’. At time y.d all vertices in C’ are white and there is a path from y to all vertices in C’. I.e. all vertices in C’ are descendants of y.
So y.f = f(C’). There cannot be a path from C’ to C. Thus any w in C has w.f > y.f and so f(C) > f(C’).

42. Corollary 22.15: Let C and C’ be distinct strongly connected components in directed graph G=(V, E). Suppose that there is an edge (u, v) in , where u in C and v in C’.
Then f ( C ) < f ( C’ ).
Pf: It is clear that (v, u) is in E.
By the previous lemma we have f( C’ ) > f ( C ).

43. Thm 22.16
Strongly-Connected-Component(G) correctly computes the
strongly connected components of a directed graph.
pf:
By induction on the number (k) of depth-first trees found in the DFS of GT, where each tree forms a strongly connected
component.
Basis: trivial for k=0.
Inductive step: assume each of the first k DFS trees produced in line 3 is a SCC.
Consider the (k+1)-st tree produced.
Let u be the root of this tree and let u be in SCC C.
Thus u.f = f ( C ) > f ( C’ ) for any other SCC C’ yet to be visited. Note we visit in decreasing finishing time.

44. All other vertices in C are descendant of u in its DFS.
By inductive hypothesis and the previous Corollary 15 any edges in GT that leave C must be to SCC’s already visited.
Thus, no vertex in any SCC other than C will be a descendant of u during the DFS of GT.
Thus, the vertices of the DFS tree in GT that is rooted at u form exactly one SCC. 