Graph Algorithms

p2.

圖形 G 是由兩個集合 V 和 E 所構成的，可以寫成G=(V,E) 。其中 V 是非空的由有限個頂點(Vertex) 所構成的集合， E 則是由頂點對所構成的集合，這些頂點對叫做邊 (edge) 。 V(G) 和 E(G)各表示組成 G 的頂點集合和邊集合。依據 E 中邊之型態，所組成之圖形又可分成下列兩種： E 中之邊沒有方向性，亦即（Ｖ 1, Ｖ 2 ）和（Ｖ 2, Ｖ

1 ）表示相同的邊，如此構成之圖形稱作無向圖形（ undirected graph ）。

E 之邊沒有方向性，頂點對（邊）以＜Ｖ 1, Ｖ 2 ＞表示，其中Ｖ２表頭（ head ），Ｖ 1 表尾（ tail ）；很自然地，＜Ｖ 2, Ｖ 1 ＞和，＜Ｖ 1, Ｖ 2 ＞是完全不同的兩個邊。以此種邊集合構成之圖形稱作有向圖形（ directed graph ，又稱 digraph ）。

p3.

complete graph ：各種頂點的組合均存在之圖形稱作complete graph 。無向圖形若有 n 個頂點，則最大可能之邊組合有（）＝　　　　種，而有向圖形將有兩倍於此為 n(n-1) 種組合；含有 n 個頂點之 complete graph 將有如上數目之邊。

subgraph ：若 Ĝ 為Ｇ之 subgraph ，則 Ĝ 為一 graph ，且Ｖ (Ĝ) ＜ V(V) ， E(G) ＜ E(G) 。

path ：由圖形中頂點所構成的序列Ｖ p, Ｖ 1, Ｖ 2,… ＶN,Ｖ q ，若其中 ( Ｖ p, Ｖ 1), ( Ｖ 1, Ｖ 2),… ， ( ＶN, Ｖ q)均為圖形中之邊，則此序列稱做一 path ；一 path 中 edge之數目稱做該 path 之長度；在有向圖形之情況下，則要求＜Ｖp, Ｖ 1 ＞ , ＜Ｖ 1, Ｖ 2 ＞ ,… ，＜ＶN, Ｖ q ＞均為有向edge ，而此 path 又稱做 directed path 。

Simple path ：在上面之定義中，其除了Ｖ p 和Ｖ q 之外，其他vertex 均不重複出現的 path 稱之。在 simple path 中，若Ｖ p=Ｖ q ，則稱之為 cycle 或 circuit 。一 graph 若有 cycle 則稱cyclic ，反之則可稱做 acyclic 。

一 path 若不為 simple ，則其必有相交之情況。

p4.

頂點和邊之關係 , 以“ adjacent” 和“ incident” 描述之：

Ｖ 1 Ｖ 2 Ｖ 1 Ｖ 2

Ｅ 1 Ｅ 2

( 有向 )

( 無向 )

Ｖ 1 和Ｖ 2 為 adjacent Ｖ 1 adjacent fromＶ 2Ｅ 1 incident on Ｖ 1( Ｖ 2) Ｖ 2 adjacent to Ｖ 1

Ｅ 2 incident to Ｖ 1( Ｖ 2)

p5.

相連（ Connected ） connected of two vertices ：若一圖形中之兩頂點間存在

任何 path ，則稱他們為 connected 的。在有向圖形中，若兩點頂間存在自其中某頂點到另一頂點和回來之有向 path ，則稱此兩頂點為 strongly connected 。

connected of a graph ：若一圖形其所有頂點對均為 6 connected ，則此 graph 為 connected ：相似地，若一有向圖形所有頂點對間均為 strongly connected ，該圖形亦為 strongly connected 。

degree (of a vertex) ：表示一個 vertex 所adjacent 之其他 vertices 之個數。在有向圖形中，degree 又分成 in-degree 和 out-degree ，其中前者表示該 vertex 所 adjacent from 之vertex 個數，而後者表示所 adjacent to 之vertex 個數。

p6.

connected component ：對無向圖形而言，其connected component （或 component ）表示其孤立的 connected 的子圖，這可能有好幾個。在有向圖形中， strongly connected component 表示一盡量伸展而仍然保持 strongly connected 的子圖（不一定要孤立）。以下四個圖形G1 到G4 前三個均只有一個 component （G3沒有 strongly connected component ），而G4 有兩個 component 。

1

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p7.

1 4

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路徑長度為 n 之鄰接矩陣

p8.

在找出路徑矩陣 p 時，首先要找到 a2 ，然後 a3 ，一直到 an ，最後再將 a1 ， a2 …， ， an 加起來得到 Sn ，再由 Sn 得到 p 。

Warshall’s algorithm:1.      p←a （把鄰接矩陣 a 拷貝至 p ）2.      for （ k=1 ； k<=n ； k++ ） for （ i=1 ； i<=n ； i++ ） for （ j=1 ； j<=n ； j++ ） p[i][j]=p[i][j] | p[i][k]&p[k][j] ；

Warshall’s algorithm

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

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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

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The operation of BFS on an undirected graph(a)

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p13.

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

][u

Algorithm ：

Complexity ：

BFS(G,s)

for each vertex u in V[G] – {s}

do color[u] <- white1.2.

d[u] <- infinity3.pi[u] <- NIL4.

color[s] <- gray5.

d[s] <- 06.

pi[s] <- NIL7.

Q <- {s}8.While Q .ne. {}9.

do u <- head[Q]10.

for each v in Adj[u]11.

do if color[v] = white12.

then color[v] <- gray13.

d[v] <- d[u]+114.Pi[v] <- u15.

Enqueue(Q,v)16.

Dequeue(Q)17.

color[u] <- black18.

)( VEO

p15.

Properties of Breadth-first search ： After execution of BFS, all nodes reachable from the

source s are colored black. After execution of BFS, d[v] 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 ( ) at the end.

After execution of BFS, the edges( ) for v reachable from s from a breadth-first tree.

vu],[

vu],[

p16.

Lemma1 ： G=(V,E) ： a directed or undirected graph. s ： an arbitrary vertex ： the shortest-path distance from s to v.

),( vs

Then for any 1 ),(),(,),( usvsEvu Proof ：

s

u

v

1),( us

),( vs

Lemma2 ：

G=(V,E) ： a directed or undirected graph. s ： an arbitrary vertex ： the distance from s to u computed by the

algorithm )(ud

Suppose that BFS is run on G from s.

Proof ：

Then on termination, for each vertex , the value d[v] computed byBFS satisfies

Vv).,(][ vsvd

By induction on the number of times a vertex is placed in the queue Q.

Basis ： when s is placed in Q. ).,(][ sssd 0

),(][ vsvd for all }.{sVv

Induction Step ： Consider a white vertex v discovered during the search from a vertex u.

Inductive hypothesis implies ).,(][ usud

By lemma1, ),(),(][][ ususudvd 11

From then on, d[v] wont be changed again.

Lemma3 ：

Q ： <v1,v2,…,vr>~ the queue when BFS executes on a graph G=(V,E).

Then Proof ：

11 ][][ vdvd r

By induction on the number of queue operations.

Basis ： when Q has only s. .][][ 1 sdsd

Induction Step ：

11 21 ][][][ vdvdvd r

by inductive hypothesis.

head tail

and ][][ 1 ii vdvd for i=1,2,…,r-1.

1> ： after dequeue ：v2 becomes the new head in Q.

121 rivdvd ii ,...,for ][][2> ： after enqueue a new vertex v into the Q.

Let vr+1 be v.

Note that vi’s adjacency list is being scanned.

neighbors

Thus, 111 ][][ vdvd r

And ][][][ 11 1 rr vdvdvd

The rest , for I=1,…,r-1, remain unaffected. ][][ 1 ii vdvd

p19.

Thm ： 1. During the execution of BFS on G=(V,E), BFS

discovers every vertex that is reachable from s, and on termination 2. For any vertex reachable form s, one of the

shortest paths from s to v is the shortest path form s to

followed by the edge

Vv).,(][ vsvd

sv][v

).],[( vv

Proof ：If v unreachable from s,

By induction on k, we want to prove for each there is exactly on point duringThe execution of BFS at which 1. V is grayed

2. D[v]=k 3. if then 4. v is inserted into Q.

.),(][ vssd By BFS, v is never discovered.

Let }),(:{ kvsVvVk

,kVv

,sv 1 kVuuv ,][

Basis ： k=0, Vk={s} ~ clear !

Induction Step ： Q until BFS terminates.

Once u is inserted into Q, d[u] and never change. ][uBy lemma 3, rr vvQvdvdvd ,...,],[...][][ 121

grayed

Let , then by lemma 2. 1 kVv k , ,][ kvd

Thm ：The monotonicity property, with and the inductive hypothesis impliesthat v must be discovered after all vertices in Vk-1 are enqueued, if v is discovered at all.

kvd ][

Since a path of k edges from s to v, such that kvs ),( and 1 kVu.),( Evu

At some point u must be the head in Q.Then u’s neighbors are scanned and v is discovered.

Line 13 grays v, line 14 sets d[v]=d[u]+1=k.Line 15 sets .][ uv Line 15 enqueues v.

Thus, the inductive hypothesis holds.

If then ,kVv .][ 1 kVvThus, we can obtain a shortest path from s to v by taking a shortest path from s to then traversing the edge ][v ).],[( vv

p22.

Depth-first search ： Nodes are initially white Nodes become green when first discovered Nodes become black when they are done d[v] records when v is first discovered F[v] records when v is done d[u] < f[u]

p23.

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p24.

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p26.

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p27.

(u,v) Black 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.

p28.

DFS(G)for each vertex u in V[G]

do color[u] whitepi[u] NIL

time 0for each vertex u in V[G]

do if color[u] = whitethen DFS-Visit(u)

O(V)

O(E)

p29.

DFS-Visit(u)color[u] grayd[u] time time + 1for each v in Adj[u]

do if color[v] = whitethen pi[v] u

DFS-Visit(v)color[u] blackf[u] time time + 1Finishing

time

Discovery time

p30.

The running time of DFS is O(V+E)

After execution of DFS, all nodes are colored black

After execution of DFS, the edges( ) form a collection of depth-first tree, called a depth-first forest.

v π[v],

p31.

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

p32.

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p33.

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p34.

Thm6: In any DFS of a graph G=( V,E ), for any two vertices u,v, exactly one of the following 3 conditions holds:

(1). The intervals [d[u], f[u]] and [d[u], f[u]] are disjoint,

(2). The interval [d[u], f[u]] is contained entirely within the interval [d[v], f[v]], and u is a descendant of v in the depth-first tree,

(3). or as above with v as a descendant of u

p35.

Pf:1. If d[u] < d[v]

case 1: d[v] < f[u] :

So v is finished before finishing u. Thus (3) holds

case 2: f[u] < d[v]d[u] < f[u] < d[v] < f[v] (1)

holds2. If d[v] < d[u]:

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

p36.

Cor7: v is a proper descendant of vertex u in the depth-first forest for a graph G

iff d[u] < d[v] < f[v] < f[u]

Thm8: In a DF forest of G=( V,E ), vertex v is a descendant of u iff at the time d[u] that the search discovers u, v can be reached from u along a path consisting entirely of white vertices

p37.

Pf:“”

v: descendant of u

“”Assume at time d[u], v is reachable from u along a path of white vertices, but v does not become a descendant of u in DF tree

Thus, d[u] < d[v] < f[w] <= f[u]Thm6 implies [d[v], f[v]] is contained entirely in [d[u], f[u]] By Cor7, v is a descendant of u.

d[u] < d[w], by the above cor.

Thus w is white at d[u]

f[w] <= f[u]v must be discovered after u is discovered,but before w is finished.

uv w

uw v

Descendant of u

p38.

Thm 9: In a DFS of an undirected graph G, every edge of G is either a

tree edge or a back edgePf:

Let and suppose d[u] < d[v]. 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

E ) v u, (

p39.

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

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p40.

TOPOLOGICAL-SORT(G): (V+E)

1. Call DFS(G) to compute finishing times f[v] 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

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p41.

Lemma 23.10

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 d[v], 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.

p42.

Thm 23.11TOPOLOGICAL-SORT(G) produces a topological

sort of Gpf:

Suppose DFS is run to determinate 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 f[v]<f[u]. 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

f[v]<f[u]2. If v is black, then f[v]<f[u]

p43.

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, vU, 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.

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p45.

StronglyConnectedComponents(G)1. Call DFS(G) to compute finishing times f[u] for each vertex u2. 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

Time: (V+E)

p46.

Lemma 12:If 2 vertices are in the same strongly connected

component,then no path between them ever leaves the

stronglyconnecter component.

pf:Assume that: u, v in the same component w is a vertex on the path from u to v

uw wvu u, w are in the same component

u wv

p47.

Thm 13In any depth-first search, all vertices in the

same strongly connected component are placed in the same depth-first treepf:

By lemma 12 and thm 8, every vertex in the strongly

connected component becomes a descendant of r in the

depth-first tree

r: the first discovered vertex in the component

p48.

(u): forefather of u the vertex w such that u w and f[w] is

maximized f[u] f[(u)] (*) ((u) )= (u), for any u,vV, u v implies f[(v)] f[(u)]

{w: v w}{w: u w}(1) u v w(2) u w’

Since u (u) f[((u))] f[(u)]By (*), we have f[(u)] f[((u))]Thus, f[(u)]= f[((u))] and (u)= ((u)), since only one vertex can be finished at a time.

w’ w

p49.

Thm 14

In a directed graph G=(V,E), the forefather (u) of any vertex

uV in any depth-first search of G is an ancestor of u.pf:

(u)=u : trivial, u is reachable from itself. (u) u : at time d[u], (u) can be (i) black (ii) gray (iii) white

(i) If (u) is black, then f[(u)]<f[u]. impossible!!(ii) If (u) is gray, then (u) is an ancestor of u. Done!!(iii) Claim (u) can’t be white: 1. If every intermediate vertex is white, then (u) becomes a descendant of u. But then f[(u)] < f[u]. 2. If some intermediate vertex is nonwhite,

t must be gray, since there is no blackwhite edge. Then there is a white path from t to (u). So (u) is a descendant of t. Thus, f[t] > f[(u)]

Since (u) should have the maximum finishing time

u(u)all white

u(u)

Last nonwhite vertex on this path

t

p50.

Corollary 15In any DFS of a directed graph G, vertices u and

(u), for all uV, lie in the same strongly connected

component.pf:

u (u), by definition. (u) u, some (u) is an ancestor of u, by Thm

14

p51.

Thm 16

G=(V,E): a directed graphu,vV lie in the same strongly connected component they have the same forefather in a DFS of Gpf:“” u,v in the same component Every vertex reachable from u is reachable from v and vice versa. By the definition of forefather, (u)=(v) “” Assume (u)=(v)

By cor. 15, u,w are in the same component, and v,w are in the same component, too. Thus, u,v are in the same strongly connected component

wu

v

p52.

Thm 17

Strongly-Connected-Component(G) correctly computes the

strongly connected components of a directed graph.pf:

By induction on the number of depth-first trees found in the

DFS of GT, where each tree forms a strongly connectedcomponent.Basic: When the first tree produced, there are no

previous trees.Consider a depth-first tree T with root r produced in

the depth-first search of GT.

p53.

Let C(r)={vV: (v)=r}Now prove that u is placed in T uC(r)

”” By Thm 13, every vertex in C(r) ends up in the same DF tree.

Since rC(r) and r is the root of T, every element of C(r)

ends up in T.”” w with (1) f[(w)]>f[r] or (2) f[(w)]<f[r], is not placed in T

(1) at the time r is selected w will have been placed in T with

root (w) (2) if w is placed in T, it implies w r, thus by (*) and

r=(r), we have f[(w)] f[(r)]=f[r] Thus, T contains only u with (u)=r. i.e. T=C(r)

p54.

Exercise-5

在文字檔中以 3 tuple 資料表示二 vertex connected & their distance: <1,2,12> indicates that v1, v2 is connected and their distance is 12

Please construct the directed graph 以圖形顯示 具下列功能 : D F S, B F S, minimum spanning tree,

topological sort, single shortest path( 輸入起點及終點 ,列出最點路徑長度及中間所經過 vertex) and all pairs shortest path