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# Slides credited from Hsueh-I Lu, Hsu-Chun Hsiao, & Michael Tsai

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Slides credited from Hsueh-I Lu, Hsu-Chun Hsiao, & Michael Tsai

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Mini-HW

NTU COOL

TA hours

Course recordings

Instant feedback

Classroom (crowded, sleepy, etc.)

Homework due time

Pseudo code

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Mini-HW 7 released

Due on 11/29 (Thur) 14:20

Homework 3 released soon

Due on 12/13 (Thur) 14:20 (three weeks)

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Frequently check the website for the updated information!

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### A graph G is defined as

V: a finite, nonempty set of vertices

E: a set of edges / pairs of vertices

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### Graph type

Undirected: edge 𝑢, 𝑣 = 𝑣, 𝑢

Directed: edge 𝑢, 𝑣 goes from vertex 𝑢 to vertex 𝑣; 𝑢, 𝑣 ≠ 𝑣, 𝑢

Weighted: edges associate with weights

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### 2

How many edges at most can a undirected (or directed) graph have?

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𝑢, 𝑣

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## Degree

𝑢

### is the number of edges incident on

𝑢

In-degree of 𝑢: #edges 𝑥, 𝑢 in a directed graph

Out-degree of 𝑢: #edges 𝑢, 𝑥 in a directed graph

Degree = in-degree + out-degree

Isolated vertex: degree = 0

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𝑖

𝑖

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## Path

1

1

2

𝑘−1

𝑘

𝑘

### , 𝑣)

Reachable: 𝑣 is reachable from 𝑢 if there exists a path from 𝑢 to 𝑣

## Simple Path

All vertices except for 𝑢 and 𝑣 are all distinct

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## Forest

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### Euler path

Can you traverse each edge in a connected graph exactly once without lifting the pen from the paper?

### Euler tour

Can you finish where you started?

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### BD

Euler path Euler tour

Euler path Euler tour

Euler path Euler tour

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### 𝐺 has an Euler tour iff all vertices must be even vertices

Is it possible to determine whether a graph has an Euler path or an Euler tour, without necessarily having to find one explicitly?

Even vertices = vertices with even degrees Odd vertices = vertices with odd degrees

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### Modeling applications using graph theory

What do the vertices represent?

What do the edges represent?

Undirected or directed?

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## Matrix

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### Adjacency matrix = 𝑉 × 𝑉 matrix 𝐴 with 𝐴[𝑢][𝑣] = 1 if (𝑢, 𝑣) is an edge

1 2 3 4 5 6

1 1 1

2 1 1 1

3 1 1 1

4 1 1 1

5 1

6 1 1

1

2

3

5 4

6

• For undirected graphs, 𝐴 is symmetric; i.e., 𝐴 = 𝐴𝑇

• If weighted, store weights instead of bits in 𝐴

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## List

### One list per vertex, where for 𝑢 ∈ 𝑉, 𝐴[𝑢] consists of all vertices adjacent to 𝑢

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1 2 3 4 5 6

1 4

3

2 3

2

3 2

5

1 4 6

4

6 1

2

3

5 4

6

If weighted, store weights also in adjacency lists

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## List

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### Besides graph density, you may also choose a data structure based on the performance of other operations

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Space Query an edge

Insert an edge

Delete an edge

List a vertex’s neighbors

Identify all edges

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Textbook Chapter 22 – Elementary Graph Algorithms

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### Standard graph-searching algorithms

Depth-First Search (DFS, 深度優先搜尋)

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Textbook Chapter 22.2 – Breadth-first search

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Source 𝒔

Layer 1

Layer 2

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BFS

### 𝑣. 𝑑: distance from 𝑠 to 𝑣, for all 𝑣 ∈ 𝑉

Distance is the length of a shortest path in G

𝑣. 𝑑 = ∞ if 𝑣 is not reachable from 𝑠

𝑣. 𝑑 is also the depth of 𝑣 in 𝑇BFS

### 𝑣. 𝜋 = 𝑢 if (𝑢, 𝑣) is the last edge on shortest path to 𝑣

𝑢 is 𝑣’s predecessor in 𝑇

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BFS

### As 𝑣 is discovered from 𝑢, 𝑣 and (𝑢, 𝑣) are added to 𝑇

BFS

𝑇BFS is not explicitly stored; can be reconstructed from 𝑣. 𝜋

### Color the vertices to keep track of progress:

GRAY: discovered (first time encountered)

BLACK: finished (all adjacent vertices discovered)

WHITE: undiscovered

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BFS(G, s)

for each vertex u in G.V-{s}

u.color = WHITE u.d = ∞

u.pi = NIL s.color = GRAY s.d = 0

s.pi = NIL Q = {}

ENQUEUE(Q, s) while Q! = {}

u = DEQUEUE(Q)

if v.color == WHITE v.color = GRAY v.d = u.d + 1 v.pi = u

ENQUEUE(Q,v) u.color = BLACK

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𝑠 0

𝑤 𝑟

1 1

𝑟 𝑡 𝑥

1 2 2

𝑡 𝑥 𝑣

2 2 2

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𝑢 𝑦

3 3

𝑦 3

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Definition of 𝛿(𝑠, 𝑣): the shortest-path distance from 𝑠 to 𝑣 = the minimum number of edges in any path from 𝑠 to 𝑣

If there is no path from 𝑠 to 𝑣, then 𝛿 𝑠, 𝑣 = ∞

The BFS algorithm finds the shortest-path distance to each reachable vertex in a graph 𝐺 from a given source vertex 𝑠 ∈ 𝑉.

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### Proof

Case 1: 𝑢 is reachable from 𝑠

𝑠- 𝑢- 𝑣 is a path from 𝑠 to 𝑣 with length 𝛿 𝑠, 𝑢 + 1

Hence, 𝛿 𝑠, 𝑣 ≤ 𝛿 𝑠, 𝑢 + 1

Case 2: 𝑢 is unreachable from 𝑠

Then 𝑣 must be unreachable too.

Hence, the inequality still holds.

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Lemma 22.1

Let 𝐺 = 𝑉, 𝐸 be a directed or undirected graph, and let 𝑠 ∈ 𝑉 be an arbitrary vertex. Then, for any edge 𝑢, 𝑣 ∈ 𝐸, 𝛿 𝑠, 𝑣 ≤ 𝛿 𝑠, 𝑢 + 1.

𝑠-𝑣的最短路徑一定會小於等於𝑠-𝑢的最短路徑距離+1

s

v

𝛿 𝑠, 𝑢 u

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### Proof by induction

Holds when 𝑛 = 1: 𝑠 is in the queue and 𝑣. 𝑑 = ∞ for all 𝑣 ∈ 𝑉 𝑠

After 𝑛 + 1 ENQUEUE ops, consider a white vertex 𝑣 that is discovered during the search from a vertex 𝑢

Lemma 22.2

Let 𝐺 = 𝑉, 𝐸 be a directed or undirected graph, and suppose BFS is run on 𝐺 from a given source vertex 𝑠 ∈ 𝑉. Then upon termination, for each vertex 𝑣 ∈ 𝑉, the value 𝑣. 𝑑 computed by BFS satisfies 𝑣. 𝑑 ≥ 𝛿 𝑠, 𝑣 .

BFS算出的d值必定大於等於真正距離

Inductive hypothesis: 𝑣. 𝑑 ≥ 𝛿 𝑠, 𝑣 after 𝑛 ENQUEUE ops

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### Proof by induction

Holds when 𝑄 = 𝑠 .

Consider two operations for inductive step:

Dequeue op: when 𝑄 = 𝑣1, 𝑣2, … , 𝑣𝑟 and dequeue 𝑣1

Enqueue op: when 𝑄 = 𝑣1, 𝑣2, … , 𝑣𝑟 and enqueue 𝑣𝑟+1

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Lemma 22.3

Suppose that during the execution of BFS on a graph 𝐺 = 𝑉, 𝐸 , the queue 𝑄 contains the vertices 𝑣1, 𝑣2, … , 𝑣𝑟 , where 𝑣1 is the head of 𝑄 and 𝑣𝑟 is the tail. Then, 𝑣𝑟. 𝑑 ≤ 𝑣1. 𝑑 + 1 and 𝑣𝑖. 𝑑 ≤ 𝑣𝑖+1. 𝑑 for 1 ≤ 𝑖 < 𝑟.

• Q中最後一個點的d值 ≤ Q中第一個點的d值+1

• Q中第i個點的d值 ≤ Q中第i+1點的d值

Inductive hypothesis:𝑣𝑟. 𝑑 ≤ 𝑣1. 𝑑 + 1 and 𝑣𝑖. 𝑑 ≤ 𝑣𝑖+1. 𝑑 after 𝑛 queue ops

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### Enqueue op

Inductive hypothesis:

𝑣1 𝑣2 … 𝑣𝑟−1 𝑣𝑟

𝑣2 … 𝑣𝑟−1 𝑣𝑟 (induction hypothesis H2)

𝑣1 𝑣2 … 𝑣𝑟−1 𝑣𝑟

(induction hypothesis H2)

𝑣 𝑣 … 𝑣 𝑣 𝑣

𝑢

Let 𝑢 be 𝑣𝑟+1’s predecessor,

Since 𝑢 has been removed from 𝑄, the new head 𝑣1 satisfies

(induction hypothesis H1) H1

H2

 H1 holds

 H2 holds

𝑢

(Q中最後一個點的d值 ≤ Q中第一個點的d值+1)

(Q中第i個點的d值 ≤ Q中第i+1點的d值)

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### Proof

Lemma 22.3 proves that 𝑣𝑖. 𝑑 ≤ 𝑣𝑖+1. 𝑑 for 1 ≤ 𝑖 < 𝑟

Each vertex receives a finite 𝑑 value at most once during the course of BFS

Hence, this is proved.

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Corollary 22.4

Suppose that vertices 𝑣𝑖 and 𝑣𝑗 are enqueued during the execution of BFS, and that 𝑣𝑖 is enqueued before 𝑣𝑗. Then 𝑣𝑖. 𝑑 ≤ 𝑣𝑗. 𝑑 at the time that 𝑣𝑗 is enqueued.

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### Proof of (1)

All vertices 𝑣 reachable from 𝑠 must be discovered; otherwise they would have 𝑣. 𝑑 = ∞ > 𝛿 𝑠, 𝑣 . (contradicting with Lemma 22.2) Theorem 22.5 – BFS Correctness

Let 𝐺 = 𝑉, 𝐸 be a directed or undirected graph, and and suppose that BFS is run on 𝐺 from a given source vertex 𝑠 ∈ 𝑉.

1) BFS discovers every vertex 𝑣 ∈ 𝑉 that is reachable from the source 𝑠 2) Upon termination, 𝑣. 𝑑 = 𝛿 𝑠, 𝑣 for all 𝑣 ∈ 𝑉

3) For any vertex 𝑣 ≠ 𝑠 that is reachable from 𝑠, one of the shortest paths from 𝑠 to 𝑣 is a shortest path from 𝑠 to 𝑣. 𝜋 followed by the edge 𝑣. 𝜋, 𝑣

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(2)

### Proof of (2) by contradiction

Assume some vertices receive 𝑑 values not equal to its shortest-path distance

Let 𝑣 be the vertex with minimum 𝛿 𝑠, 𝑣 that receives such an incorrect 𝑑 value; clearly 𝑣 ≠ 𝑠

By Lemma 22.2, 𝑣. 𝑑 ≥ 𝛿 𝑠, 𝑣 , thus 𝑣. 𝑑 > 𝛿 𝑠, 𝑣 (𝑣 must be reachable)

Let 𝑢 be the vertex immediately preceding 𝑣 on a shortest path from 𝑠 to 𝑣, so 𝛿 𝑠, 𝑣 = 𝛿 𝑠, 𝑢 + 1

Because 𝛿 𝑠, 𝑢 < 𝛿 𝑠, 𝑣 and 𝑣 is the minimum 𝛿 𝑠, 𝑣 , we have 𝑢. 𝑑 = 𝛿 𝑠, 𝑢

𝑣. 𝑑 > 𝛿 𝑠, 𝑣 = 𝛿 𝑠, 𝑢 + 1 = 𝑢. 𝑑 + 1

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(2)

### Proof of (2) by contradiction (cont.)

𝑣. 𝑑 > 𝛿 𝑠, 𝑣 = 𝛿 𝑠, 𝑢 + 1 = 𝑢. 𝑑 + 1

When dequeuing 𝑢 from 𝑄, vertex 𝑣 is either WHITE, GRAY, or BLACK

WHITE: 𝑣. 𝑑 = 𝑢. 𝑑 + 1, contradiction

BLACK: it was already removed from the queue

By Corollary 22.4, we have 𝑣. 𝑑 ≤ 𝑢. 𝑑, contradiction

GRAY: it was painted GRAY upon dequeuing some vertex 𝑤

Thus 𝑣. 𝑑 = 𝑤. 𝑑 + 1 (by construction)

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(3) For any vertex 𝑣 ≠ 𝑠 that is reachable from 𝑠, one of the shortest paths from 𝑠 to 𝑣 is a shortest path from 𝑠 to 𝑣. 𝜋 followed by the edge 𝑣. 𝜋, 𝑣

### Proof of (3)

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 and then traversing the edge 𝑣. 𝜋, 𝑣 .

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BFS(G, s) forms a BFS tree with all reachable 𝑣 from 𝑠

We can extend the algorithm to find a BFS forest that contains every vertex in 𝐺

BFS-Visit(G, s) s.color = GRAY s.d = 0

s.π = NIL Q = empty ENQUEUE(Q, s) while Q ≠ empty

u = DEQUEUE(Q) for v in G.adj[u]

if v.color == WHITE //explore full graph and builds up

a collection of BFS trees BFS(G)

for u in G.V

u.color = WHITE u.d = ∞

u.π = NIL for s in G.V

if(s.color == WHITE)

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Textbook Chapter 22.3 – Depth-first search

### 45

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Search as deep as possible and then backtrack until finding a new path

1

2

3 4

8

9 12 13

14

5 6

7

10 11

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### Color the vertices to keep track of progress:

GRAY: discovered (first time encountered)

BLACK: finished (all adjacent vertices discovered)

WHITE: undiscovered 47

// Explore full graph and builds up a collection of DFS trees

DFS(G)

for each vertex u in G.V u.color = WHITE

u.pi = NIL

time = 0 // global timestamp for each vertex u in G.V

if u.color == WHITE DFS-VISIT(G, u)

DFS-Visit(G, u) time = time + 1

u.d = time // discover time u.color = GRAY

if v.color == WHITE v.pi = u

DFS-VISIT(G, v) u.color = BLACK

time = time + 1

u.f = time // finish time

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### Parenthesis Theorem

Parenthesis structure: represent the discovery of vertex 𝑢 with a left

parenthesis “(𝑢” and represent its finishing by a right parenthesis “𝑢)”. In DFS, the parentheses are properly nested.

### White Path Theorem

In a DFS forest of a directed or undirected graph 𝐺 = 𝑉, 𝐸 ,

vertex 𝑣 is a descendant of vertex 𝑢 in the forest  at the time 𝑢. 𝑑 that the search discovers 𝑢, there is a path from 𝑢 to 𝑣 in 𝐺 consisting entirely of WHITE vertices

Tree Edge

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### Parenthesis Theorem

Parenthesis structure: represent the discovery of vertex 𝑢 with a left parenthesis “(𝑢” and represent its finishing by a right parenthesis

“𝑢)”. In DFS, the parentheses are properly nested.

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Properly nested: (x (y y) x) Not properly nested: (x (y x) y)

Proof in textbook p. 608

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### White Path Theorem

In a DFS forest of a directed or undirected graph 𝐺 = 𝑉, 𝐸 ,

vertex 𝑣 is a descendant of vertex 𝑢 in the forest  at the time 𝑢. 𝑑 that the search discovers 𝑢, there is a path from 𝑢 to 𝑣 in 𝐺 consisting entirely of WHITE vertices

### Proof.

Since 𝑣 is a descendant of 𝑢, 𝑢. 𝑑 < 𝑣. 𝑑

Hence, 𝑣 is WHITE at time 𝑢. 𝑑

In fact, since 𝑣 can be any descendant of 𝑢, any vertex on the path from 𝑢

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### Classification of Edges in 𝐺

Tree Edge (GRAY to WHITE)

Edges in the DFS forest

Found when encountering a new vertex 𝑣 by exploring 𝑢, 𝑣

Back Edge (GRAY to GRAY)

𝑢, 𝑣 , from descendant 𝑢 to ancestor 𝑣 in a DFS tree

Forward Edge (GRAY to BLACK)

𝑢, 𝑣 , from ancestor 𝑢 to descendant 𝑣. Not a tree edge.

Cross Edge (GRAY to BLACK)

Any other edge between trees or subtrees. Can go between vertices in same DFS tree or in different DFS trees

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In an undirected graph, back edge = forward edge.

To avoid ambiguity, classify edge as the first type in the list that applies.

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### Edge classification by the color of 𝑣 when visiting 𝑢, 𝑣

WHITE: tree edge

GRAY: back edge

BLACK: forward edge or cross edge

𝑢. 𝑑 < 𝑣. 𝑑  forward edge

𝑢. 𝑑 > 𝑣. 𝑑  cross edge

Theorem 22.10

In DFS of an undirected graph, there are only tree edges and back edges

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### a maximal subset 𝑈 of 𝑉 s.t. any two nodes in 𝑈 are connected in 𝐺

Why must the connected components of a graph be disjoint? 55

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1

2

5

3 4

6

7

8 9

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57

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### 59

Important announcement will be sent to @ntu.edu.tw mailbox

& post to the course website

jobs

▪ Step 2: Run DFS on the transpose

Greedy-Choice Property : making locally optimal (greedy) choices leads to a globally optimal

 “Greedy”: always makes the choice that looks best at the moment in the hope that this choice will lead to a globally optimal solution.  When to

Miroslav Fiedler, Praha, Algebraic connectivity of graphs, Czechoslovak Mathematical Journal 23 (98) 1973,

Given a connected graph G together with a coloring f from the edge set of G to a set of colors, where adjacent edges may be colored the same, a u-v path P in G is said to be a

✓ Express the solution of the original problem in terms of optimal solutions for subproblems. Construct an optimal solution from

✓ Express the solution of the original problem in terms of optimal solutions for subproblems.. Construct an optimal solution from