advances in algorithmic graph theorysummerschool2013.ccaba.upc.edu/documents/lecture6.1... · 2013....

Advances in algorithmic graph theory

George B. Mertzios

School of Engineering and Computing Sciences,Durham University, UK

Graph and routing dynamics: models and algorithms

Part II

EULER Summer School 2013

George Mertzios (Durham) Advances in algorithmic graph theory EULER, July 2013 1 / 56

Overview

Graph searching

Tolerance graphs

an intersection model

algorithmic applications

Graph Searching

How to traverse through a graph?Graph exploring / graph searching is a fundamental graph operation.historically: labyrinth/maze traversals

modern applications: shortest paths / optimum routing

Graph Searching

How to traverse through a graph?Graph exploring / graph searching is a fundamental graph operation.historically: labyrinth/maze traversals

modern applications: shortest paths / optimum routing

Graph Searching

In a graph search:

we start at a specific initial vertex x of the graph G

we systematically visit new (i.e. undiscovered) vertices and edges of G

every new vertex is adjacent to at least one previously visited vertex(unless we visited one whole connected component of G ,so we continue with the next component)

Graph Searching

In a graph search:

we start at a specific initial vertex x of the graph G

we systematically visit new (i.e. undiscovered) vertices and edges of G

every new vertex is adjacent to at least one previously visited vertex(unless we visited one whole connected component of G ,so we continue with the next component)

every graph search outputs an ordering of the vertices of G(according to the order in which the vertices are discovered)

Graph Searching

Two most well-known graph searching techniques:1 Breadth First Search (BFS):

visit all unvisited neighbors of the currently visited vertex,before you visit any non-neighbor

visit the vertices in “layers”

always prefer to visit a vertex that is “closer” to the start vertex x

Graph Searching

⇒ computes shortest paths from x

Graph Searching

Implementation of BFS:

using a queue: First-In-First-Out (FIFO) policyadd new vertices at the end of the queueremove vertices from the beginning of the queue

Graph Searching

{1}BFS ordering:

{6 1, 2, 3}{6 1, 6 2, 3, 4, 5}{6 1, 6 2, 6 3, 4, 5, 6}{6 1, 6 2, 6 3, 6 4, 5, 6}{6 1, 6 2, 6 3, 6 4, 6 5, 6}{6 1, 6 2, 6 3, 6 4, 6 5, 6 6}

Q =Q =Q =Q =Q =

2, 3, 4, 5, 6)

Graph Searching

{1}BFS ordering:

{6 1, 2, 3}{6 1, 6 2, 3, 4, 5}{6 1, 6 2, 6 3, 4, 5, 6}{6 1, 6 2, 6 3, 6 4, 5, 6}{6 1, 6 2, 6 3, 6 4, 6 5, 6}{6 1, 6 2, 6 3, 6 4, 6 5, 6 6}

Q =Q =Q =Q =Q =

2, 3, 4, 5, 6)

Graph Searching

{1}BFS ordering:

{6 1, 2, 3}{6 1, 6 2, 3, 4, 5}{6 1, 6 2, 6 3, 4, 5, 6}{6 1, 6 2, 6 3, 6 4, 5, 6}{6 1, 6 2, 6 3, 6 4, 6 5, 6}{6 1, 6 2, 6 3, 6 4, 6 5, 6 6}

Q =Q =Q =Q =Q =

2, 3, 4, 5, 6)

Graph Searching

{1}BFS ordering:

{6 1, 2, 3}{6 1, 6 2, 3, 4, 5}{6 1, 6 2, 6 3, 4, 5, 6}{6 1, 6 2, 6 3, 6 4, 5, 6}{6 1, 6 2, 6 3, 6 4, 6 5, 6}{6 1, 6 2, 6 3, 6 4, 6 5, 6 6}

Q =Q =Q =Q =Q =

2, 3, 4, 5, 6)

Graph Searching

{1}BFS ordering:

{6 1, 2, 3}{6 1, 6 2, 3, 4, 5}{6 1, 6 2, 6 3, 4, 5, 6}{6 1, 6 2, 6 3, 6 4, 5, 6}{6 1, 6 2, 6 3, 6 4, 6 5, 6}{6 1, 6 2, 6 3, 6 4, 6 5, 6 6}

Q =Q =Q =Q =Q =

2, 3, 4, 5, 6)

Graph Searching

{1}BFS ordering:

{6 1, 2, 3}{6 1, 6 2, 3, 4, 5}{6 1, 6 2, 6 3, 4, 5, 6}{6 1, 6 2, 6 3, 6 4, 5, 6}{6 1, 6 2, 6 3, 6 4, 6 5, 6}{6 1, 6 2, 6 3, 6 4, 6 5, 6 6}

Q =Q =Q =Q =Q =

2, 3, 4, 5, 6)

Graph Searching

{1}BFS ordering:

{6 1, 2, 3}{6 1, 6 2, 3, 4, 5}{6 1, 6 2, 6 3, 4, 5, 6}{6 1, 6 2, 6 3, 6 4, 5, 6}{6 1, 6 2, 6 3, 6 4, 6 5, 6}{6 1, 6 2, 6 3, 6 4, 6 5, 6 6}

Q =Q =Q =Q =Q =

2, 3, 4, 5, 6)

Graph Searching

{1}BFS ordering:

{6 1, 2, 3}{6 1, 6 2, 3, 4, 5}{6 1, 6 2, 6 3, 4, 5, 6}{6 1, 6 2, 6 3, 6 4, 5, 6}{6 1, 6 2, 6 3, 6 4, 6 5, 6}{6 1, 6 2, 6 3, 6 4, 6 5, 6 6}

Q =Q =Q =Q =Q =

2, 3, 4, 5, 6)

Graph Searching

{1}BFS ordering:

{6 1, 2, 3}{6 1, 6 2, 3, 4, 5}{6 1, 6 2, 6 3, 4, 5, 6}{6 1, 6 2, 6 3, 6 4, 5, 6}{6 1, 6 2, 6 3, 6 4, 6 5, 6}{6 1, 6 2, 6 3, 6 4, 6 5, 6 6}

Q =Q =Q =Q =Q =

2, 3, 4, 5, 6)

(1, 2, 3, 4, 5, 6) is a BFS ordering ⇒ a distance layering from vertex 1

Graph Searching

{1}BFS ordering:

{6 1, 2, 3}{6 1, 6 2, 3, 4, 5}{6 1, 6 2, 6 3, 4, 5, 6}{6 1, 6 2, 6 3, 6 4, 5, 6}{6 1, 6 2, 6 3, 6 4, 6 5, 6}{6 1, 6 2, 6 3, 6 4, 6 5, 6 6}

Q =Q =Q =Q =Q =

2, 3, 4, 5, 6)

(1, 2, 3, 4, 5, 6) is a BFS ordering ⇒ a distance layering from vertex 1

However: the ordering (1, 2, 3, 6, 5, 4) is not a BFS ordering,but still a distance layering from vertex 1 !

Graph Searching

Two most well-known graph searching techniques:2 Depth First Search (DFS):

whenever possible, visit an unvisited neighbor v of the currently visitedvertex u, then visit an unvisited neighbor w of v , etc.

searches the graph in “depth”

always prefer to visit a vertex that is “one step further” than thecurrent vertex ⇒ does not compute shortest paths from x

Graph Searching

Implementation of DFS:using a stack: Last-In-First-Out (LIFO) policyadd new vertices at the end (top) of the stackremove vertices also from the end (top) of the stack

Graph Searching

(1, 2, 6)4, 5, 3,

DFS ordering: {1, 2}

{1, 2, 4}{1, 2, 4, 5}

S =S =S =S =S =S =S =

{1, 2, 4, 5, 3}{1, 2, 4, 5, 3, 6}{1, 2, 4, 5, 3}

{1, 2, 4, 5}{1, 2, 4}

{1, 2}

S =S =S =S =

Graph Searching

(1, 2, 6)4, 5, 3,

{1, 2, 4}{1, 2, 4, 5}

S =S =S =S =S =S =S =

{1, 2, 4, 5, 3}{1, 2, 4, 5, 3, 6}{1, 2, 4, 5, 3}

{1, 2, 4, 5}{1, 2, 4}

{1, 2}

S =S =S =S =

Graph Searching

(1, 2, 6)4, 5, 3,

{1, 2, 4}{1, 2, 4, 5}

S =S =S =S =S =S =S =

{1, 2, 4, 5, 3}{1, 2, 4, 5, 3, 6}{1, 2, 4, 5, 3}

{1, 2, 4, 5}{1, 2, 4}

{1, 2}

S =S =S =S =

Graph Searching

(1, 2, 6)4, 5, 3,

{1, 2, 4}{1, 2, 4, 5}

S =S =S =S =S =S =S =

{1, 2, 4, 5, 3}{1, 2, 4, 5, 3, 6}{1, 2, 4, 5, 3}

{1, 2, 4, 5}{1, 2, 4}

{1, 2}

S =S =S =S =

Graph Searching

(1, 2, 6)4, 5, 3,

{1, 2, 4}{1, 2, 4, 5}

S =S =S =S =S =S =S =

{1, 2, 4, 5, 3}{1, 2, 4, 5, 3, 6}{1, 2, 4, 5, 3}

{1, 2, 4, 5}{1, 2, 4}

{1, 2}

S =S =S =S =

Graph Searching

(1, 2, 6)4, 5, 3,

{1, 2, 4}{1, 2, 4, 5}

S =S =S =S =S =S =S =

{1, 2, 4, 5, 3}{1, 2, 4, 5, 3, 6}{1, 2, 4, 5, 3}

{1, 2, 4, 5}{1, 2, 4}

{1, 2}

S =S =S =S =

Graph Searching

(1, 2, 6)4, 5, 3,

{1, 2, 4}{1, 2, 4, 5}

S =S =S =S =S =S =S =

{1, 2, 4, 5, 3}{1, 2, 4, 5, 3, 6}{1, 2, 4, 5, 3}

{1, 2, 4, 5}{1, 2, 4}

{1, 2}

S =S =S =S =

Graph Searching

(1, 2, 6)4, 5, 3,

{1, 2, 4}{1, 2, 4, 5}

S =S =S =S =S =S =S =

{1, 2, 4, 5, 3}{1, 2, 4, 5, 3, 6}{1, 2, 4, 5, 3}

{1, 2, 4, 5}{1, 2, 4}

{1, 2}

S =S =S =S =

Graph Searching

(1, 2, 6)4, 5, 3,

{1, 2, 4}{1, 2, 4, 5}

S =S =S =S =S =S =S =

{1, 2, 4, 5, 3}{1, 2, 4, 5, 3, 6}{1, 2, 4, 5, 3}

{1, 2, 4, 5}{1, 2, 4}

{1, 2}

S =S =S =S =

Graph Searching

(1, 2, 6)4, 5, 3,

{1, 2, 4}{1, 2, 4, 5}

S =S =S =S =S =S =S =

{1, 2, 4, 5, 3}{1, 2, 4, 5, 3, 6}{1, 2, 4, 5, 3}

{1, 2, 4, 5}{1, 2, 4}

{1, 2}

S =S =S =S =

Graph Searching

(1, 2, 6)4, 5, 3,

{1, 2, 4}{1, 2, 4, 5}

S =S =S =S =S =S =S =

{1, 2, 4, 5, 3}{1, 2, 4, 5, 3, 6}{1, 2, 4, 5, 3}

{1, 2, 4, 5}{1, 2, 4}

{1, 2}

S =S =S =S =

Graph Searching

(1, 2, 6)4, 5, 3,

{1, 2, 4}{1, 2, 4, 5}

S =S =S =S =S =S =S =

{1, 2, 4, 5, 3}{1, 2, 4, 5, 3, 6}{1, 2, 4, 5, 3}

{1, 2, 4, 5}{1, 2, 4}

{1, 2}

S =S =S =S =

Graph Searching

(1, 2, 6)4, 5, 3,

{1, 2, 4}{1, 2, 4, 5}

S =S =S =S =S =S =S =

{1, 2, 4, 5, 3}{1, 2, 4, 5, 3, 6}{1, 2, 4, 5, 3}

{1, 2, 4, 5}{1, 2, 4}

{1, 2}

S =S =S =

S =S =

∅{1}

Generic graph search

Let’s formalize the generic search idea(regardless if the search is BFS, DFS, or another search):

we use a generic data structure S

S stores the candidate vertices that we may visit next

Generic Search

Input: a graph G = (V , E ) with |V | = n, a start vertex s ∈ VOutput: an ordering σ of V

1 S ← {s}2 For i = 1 to n:

3 Pick and remove an unnumbered vertex v from S

4 σ(i)← v // This assigns to v the number i

5 For each unnumbered vertex w adjacent to v:

6 Add w to S

A vertex ordering σ of V is called a search ordering of G :

if σ can be generated by the Generic Search algorithm

How do we select the next (unnumbered) vertex v from S?

Generic Search does not impose any specific criteria !

if we implement S as a queue ⇒ BFS

if we implement S as a stack ⇒ DFS

Is σ = (1, 2, 4, 3, 5, 6) a search ordering? YES.

Is σ = (1, 2, 6, 3, 4, 5) a search ordering? NO.

Is σ = (1, 2, 4, 3, 5, 6) a search ordering?

Is σ = (1, 2, 6, 3, 4, 5) a search ordering?

How can we check whether an ordering σ is a search ordering?

if, for every i , σ(i) and σ(i + 1) are adjacent ⇒ σ is a search ordering

if σ(i)σ(i + 1) /∈ E for some i ∈ {1, 2, . . . , n− 1}?

Assume that σ visits vertices a, b, c in this order, where ab /∈ E , ac ∈ E .

How could b have been chosen before c in σ?

Property (S)

Given an ordering σ of V , if a < b < c in σ, ab /∈ E , and ac ∈ E , then:

there exists a vertex d < b such that db ∈ E .

Property (S)

a b cd

Property (S)

a b cd

Property (S) can characterize search orderings:

Theorem (Corneil, Krueger, SIAM J. on Discr. Math., 2008)

For any graph G = (V , E ), an ordering σ of V is a search ordering of Gif and only if σ satisfies property (S).

Proof.

(⇒): Suppose that σ is a search ordering.

Let a < b < c in σ, where ab /∈ E and ac ∈ E

At the point when b is chosen in σ, both b and c belong to S :

c ∈ S because ac ∈ Eif b /∈ S then b can not be chosen in σ, contradiction

Since b ∈ S , a neighbor d of b must have been already chosen in σ

Therefore d < b and bd ∈ E

Proof.

(⇒): Suppose that σ is a search ordering.

At the point when b is chosen in σ, both b and c belong to S :

c ∈ S because ac ∈ Eif b /∈ S then b can not be chosen in σ, contradiction

Since b ∈ S , a neighbor d of b must have been already chosen in σ

Therefore d < b and bd ∈ E

Proof.

(⇐): Suppose that σ = (v1, v2, . . . , vn) is satisfies property (S).Proof by contradiction:

Suppose that (v1, v2, . . . , vi−1) can be obtained by the Generic Searchalgorithm, but (v1, v2, . . . , vi ) can not (for some i ≤ n)

That is: vertex vi is the first vertex that can not be chosen next

Consider the next iteration after choosing vi−1If vi ∈ S ⇒ vi can be chosen next in σ, contradiction. Thus vi /∈ S

Let u be a vertex that the algorithm can choose next

Then u ∈ S (note: many choices for u ! )

Let w be the neighbor of u that caused u to be added to S

If wvi ∈ E ⇒ vi ∈ S , contradiction. Thus wvi /∈ E

Proof.

σ has property (S) ⇒ for the triple w < vi < u in σ there existsa vertex d < vi such that dvi ∈ E

Therefore vi ∈ S , contradiction.

Therefore vi can be chosen next ⇒ (v1, v2, . . . , vi ) can be obtainedby the algorithm for every i = 1, 2, . . . , n

⇒ σ = (v1, v2, . . . , vn) is a search ordering

Proof.

Breadth First search

Using the generic search algorithm, we can formalize BFS:

we use a queue S

here again: S stores the candidate vertices that we may visit next

Breadth First Search (BFS)

Input: a graph G = (V , E ) with |V | = n, a start vertex s ∈ VOutput: a BFS ordering σ of V

1 Initialize queue S to {s}2 For i = 1 to n:

3 Dequeue v (from the beginning of the queue S)

6 If w is not already in the queue then:

7 add w to S

Using the generic search algorithm, we can formalize BFS:

we use a queue S

here again: S stores the candidate vertices that we may visit next

Breadth First Search (BFS)

Input: a graph G = (V , E ) with |V | = n, a start vertex s ∈ VOutput: a BFS ordering σ of V

1 Initialize queue S to {s}2 For i = 1 to n:

3 Dequeue v (from the beginning of the queue S)

6 If w is not already in the queue then:

7 add w to S

A vertex ordering σ of V is called a BFS ordering of G :

if σ can be generated by the BFS algorithm

How can we check whether an ordering σ is a BFS ordering?

Since BFS is a graph search ⇒ σ must have property (S)i.e. there exists a vertex d < b such that bd ∈ E

Is this enough? NO.

Property (B)

there exists a vertex d < a such that db ∈ E .

Is this enough? NO.

Property (B)

a b cd

Is this enough? NO.

Property (B)

a b cd

Is this enough? NO.

Property (B)

a b cd

Is this enough? NO.

Property (B)

a b cd

Property (B) can characterize BFS orderings:

For any graph G = (V , E ), an ordering σ of V is a BFS ordering of G ifand only if σ satisfies property (B).

Proof.

(⇒): Suppose that σ is a BFS ordering.

Then b is removed from the queue before c

⇒ b has also been added to the queue before c

Then b must have a neighbor d that was numbered in σearlier than a, since:

if a < d for all neighbors d of b, then c is added to the queue before b,contradiction

Property (B) can characterize BFS orderings:

For any graph G = (V , E ), an ordering σ of V is a BFS ordering of G ifand only if σ satisfies property (B).

Proof.

(⇒): Suppose that σ is a BFS ordering.

Then b is removed from the queue before c

⇒ b has also been added to the queue before c

Then b must have a neighbor d that was numbered in σearlier than a, since:

if a < d for all neighbors d of b, then c is added to the queue before b,contradiction

Proof.

(⇐): Suppose that σ = (v1, v2, . . . , vn) is satisfies property (B).Proof by contradiction:

Suppose that (v1, v2, . . . , vi−1) can be obtained by theBFS algorithm, but (v1, v2, . . . , vi ) can not (for some i ≤ n)

That is: vertex vi is the first vertex that can not be chosen nextby any valid execution of BFS

Let u be a vertex that the algorithm can choose next by somepossible execution of BFS (note: many choices for u ! )

Since u is dequeued before vi (⇔ u is also enqueued before vi )

⇒ u has been enqueued by one of its neighbors vj , where j < i

If vjvi ∈ E , then some BFS execution would enqueue vi before u,contradiction. Thus vjvi /∈ E

Proof.

σ has property (B) ⇒ for the triple vj < vi < u in σ there existsa vertex d < vj such that dvi ∈ E

Therefore vi can be enqueued before u (⇔ vi is dequeued before u),contradiction.

Therefore vi can be chosen next by some BFS execution ⇒(v1, v2, . . . , vi ) can be obtained by the BFS algorithm for every i

⇒ σ = (v1, v2, . . . , vn) is a BFS ordering

Proof.

Recall our BFS example:1

BFS ordering:

1 2 3 4 5 6

σ = (1, 2, 3, 4, 5, 6)

ad b c

σ satisfies property (B):

triple {a, b, c}, where a = 3, b = 4, c = 5

⇒ d = 2

BFS ordering:

1 2 3 4 5 6

σ = (1, 2, 3, 4, 5, 6)

ad b c

triple {a, b, c}, where a = 3, b = 4, c = 5

⇒ d = 2

BFS ordering:

1 2 3 4 5 6

σ = (1, 2, 3, 4, 5, 6)

ad b c

triple {a, b, c}, where a = 3, b = 4, c = 5 ⇒ d = 2

Depth First search

Using the generic search algorithm, we can also formalize DFS:

we use a stack S

once again: S stores the candidate vertices that we may visit next

Depdth First Search (DFS)

Input: a graph G = (V , E ) with |V | = n, a start vertex s ∈ VOutput: a DFS ordering σ of V

1 Initialize stack S to {s}2 For i = 1 to n:

3 Pop v (from the top of the stack S)

6 If w is already in S then remove w from S

7 Push w (on the top of the stack S)

Depth First search

Using the generic search algorithm, we can also formalize DFS:

we use a stack S

once again: S stores the candidate vertices that we may visit next

Depth First search

In contrast to BFS:

we may reorder the (unnumbered) vertices of the stack S

always at the top of S : a neighbor of the last numbered vertex

Depth First search

A vertex ordering σ of V is called a DFS ordering of G :

if σ can be generated by the DFS algorithm

There can be many different executions of DFS:

many possible orders of the neighbors w of the current vertex v⇒ many possible ways to build the stack S

How can we check whether an ordering σ is a DFS ordering?

How could b have been chosen before c in σ?Since DFS is a graph search ⇒ σ must have property (S)i.e. there exists a vertex d < b such that bd ∈ E

Is this enough? NO.

Property (D)

there exists a vertex d with a < d < b such that db ∈ E .

Depth First search

Since DFS is a graph search ⇒ σ must have property (S)i.e. there exists a vertex d < b such that bd ∈ E

Is this enough? NO.

Property (D)

a b cd

Depth First search

Is this enough? NO.

Property (D)

a b cd

Depth First search

Is this enough? NO.

Property (D)

a b cd

Depth First search

Is this enough? NO.

Property (D)

a b cd

Depth First search

Property (D) can characterize DFS orderings:

For any graph G = (V , E ), an ordering σ of V is a DFS ordering of Gif and only if σ satisfies property (D).

Proof.

Depth First search

Proof.

(⇒): Suppose that σ is a DFS ordering.

Consider the iteration when b is numbered in σ

⇒ b has been pushed on the top of the stack, when one of its neighborswas previously numbered

Let d be the rightmost neighbor of b that is numbered so far in σ(d was numbered the last time that b was pushed on the top of S)

Depth First search

Proof.

(⇒): Suppose that σ is a DFS ordering.

Consider the iteration when b is numbered in σ

⇒ b has been pushed on the top of the stack, when one of its neighborswas previously numbered

Let d be the rightmost neighbor of b that is numbered so far in σ(d was numbered the last time that b was pushed on the top of S)

Depth First search

Proof.

d < b (since d was numbered before b in σ)

d 6= a (since ab /∈ E and db ∈ E )

Thus: either d < a or a < d < b. Assume that d < a

Then, at the iteration when a was numbered (i.e. after d):

c is pushed on the top of S ⇒ c is higher than b in S

no other neighbors of b (until b is numbered)⇒ c remains higher than b in S ⇒ c is numbered before b, contradiction

Therefore a < d < b ⇒ σ satisfies property (D).

Depth First search

Proof.

Depth First search

Proof.

Depth First search

Proof.

Depth First search

Proof.

c is pushed on the top of S ⇒ c is higher than b in Sno other neighbors of b (until b is numbered)

⇒ c remains higher than b in S

⇒ c is numbered before b, contradiction

Depth First search

Proof.

⇒ c remains higher than b in S ⇒ c is numbered before b, contradiction

Depth First search

Proof.

⇒ c remains higher than b in S ⇒ c is numbered before b, contradiction

Depth First search

Proof.

(⇐): Suppose that σ = (v1, v2, . . . , vn) is satisfies property (D).Proof by contradiction:

Suppose that (v1, v2, . . . , vi−1) can be obtained by theDFS algorithm, but σi−1 = (v1, v2, . . . , vi ) can not (for some i ≤ n)

That is: vertex vi is the first vertex that can not be chosen nextby any valid execution of DFS

Let u be a vertex that the algorithm can choose next by somepossible execution of DFS (many choices for u ! )

vi < u in σu is at the top of S after vi−1 is numbered (in this execution of DFS)

Let vj be the rightmost neighbor of u in σi−1(vj was numbered the last time that u was pushed on the top of S)

Depth First search

Proof.

Let u be a vertex that the algorithm can choose next by somepossible execution of DFS (many choices for u ! )

Depth First search

Proof.

Let u be a vertex that the algorithm can choose next by somepossible execution of DFS (many choices for u ! ) Then:

Depth First search

Proof.

Let u be a vertex that the algorithm can choose next by somepossible execution of DFS (many choices for u ! ) Then:

Depth First search

Proof.

Let k ∈ {j , j + 1, . . . , i − 1}. If vkvi ∈ E then:

vi can be pushed higher than u in S in some execution of DFS(since vj is the rightmost neighbor of u in σi−1)

⇒ vi can be numbered directly after vi−1 (instead of u), a contradiction

Therefore vkvi /∈ E for every k = j , j + 1, . . . , i − 1

σ has property (D) ⇒ for the triple vj < vi < u in σ there existsa vertex vj < d < vi such that dvi ∈ E , contradiction

Therefore vi can be chosen next by some DFS execution ⇒(v1, v2, . . . , vi ) can be obtained by the DFS algorithm for every i

⇒ σ = (v1, v2, . . . , vn) is a DFS ordering

Depth First search

Proof.

Depth First search

Proof.

Depth First search

Proof.

Depth First search

Proof.

Depth First search

Proof.

Lexicographic search

BFS and DFS are special cases of the Generic Graph Search⇒ we can use BFS and DFS for specific applications

still BFS and DFS can be executed in many different ways

⇒ many different BFS orderings and DFS orderings !

To “refine” BFS and DFS: we need a way to break ties:

how to choose among many “equal” candidates?

Definition

Let k , k ′, `, `′ ∈N. The ordered pair (k, `) is lexicographically smallerthan the ordered pair (k ′, `′) (denoted (k, `) < (k ′, `′)) if:

k < k ′, ork = k ′ and ` ≤ `′.

Similarly for longer sequences of integers:(5, 4, 8, 3) < (5, 4, 8, 6)

(2, 6, 4) < (2, 6, 4, 1)(3, 5, 2, 6) < (3, 6, 8)

Definition

k < k ′, ork = k ′ and ` ≤ `′.

(2, 6, 4) < (2, 6, 4, 1)(3, 5, 2, 6) < (3, 6, 8)

Definition

k < k ′, ork = k ′ and ` ≤ `′.

(2, 6, 4) < (2, 6, 4, 1)(3, 5, 2, 6) < (3, 6, 8)

Definition

k < k ′, ork = k ′ and ` ≤ `′.

(2, 6, 4) < (2, 6, 4, 1)(3, 5, 2, 6) < (3, 6, 8)

Definition

k < k ′, ork = k ′ and ` ≤ `′.

Similarly for longer sequences of integers:(5, 4, 8, 3) < (5, 4, 8, 6)(2, 6, 4) < (2, 6, 4, 1)

(3, 5, 2, 6) < (3, 6, 8)

Definition

k < k ′, ork = k ′ and ` ≤ `′.

Similarly for longer sequences of integers:(5, 4, 8, 3) < (5, 4, 8, 6)(2, 6, 4) < (2, 6, 4, 1)(3, 5, 2, 6) < (3, 6, 8)

Lexicographic Depth First search

Modifying again the generic search algorithm, we can “refine” DFS:

we do not use a stack any more !

we use labels and lexicographic order !

Lexicogr. Depdth First Search (LDFS) (Corneil, Krueger, 2008)

Input: a graph G = (V , E ) with |V | = n, a start vertex s ∈ VOutput: a Lexicographic DFS (LDFS) ordering σ of V

1 Assign the label ε to all vertices

2 label(s)← {0} // s is still unnumbered

3 For i = 1 to n:

4 Pick an unnumbered vertex v with lexicographically largest label

7 prepend i to label(w)

Modifying again the generic search algorithm, we can “refine” DFS:

we do not use a stack any more !

we use labels and lexicographic order !

Lexicogr. Depdth First Search (LDFS) (Corneil, Krueger, 2008)

Input: a graph G = (V , E ) with |V | = n, a start vertex s ∈ VOutput: a Lexicographic DFS (LDFS) ordering σ of V

2 label(s)← {0} // s is still unnumbered

3 For i = 1 to n:

7 prepend i to label(w)

we prepend the labels of all neighbors of the current vertex vwith the position of v in σ

⇒ LDFS always chooses vertices that are adjacent to as many verticeswe have recently numbered, as possible

Recall our example graph:

The digits in each label appear in decreasing order

The ordering (1, 2, 4, 5, 6, 3) is a DFS but not an LDFS ordering

(1, 2, 5,

LDFS ordering:

3, 6,4)

labels0

1 1σ =

2 23 34 4

(1, 2, 5,

LDFS ordering:

3, 6,4)

labels0

1 1σ =

2 23 34 4

(1, 2, 5,

LDFS ordering:

3, 6,4)

labels0

1 1σ =

2 23 34 4

(1, 2, 5,

LDFS ordering:

3, 6,4)

labels0

1 1σ =

2 23 34 4

(1, 2, 5,

LDFS ordering:

3, 6,4)

labels0

1 1σ =

2 23 34 4

(1, 2, 5,

LDFS ordering:

3, 6,4)

labels0

1 1σ =

2 23 34 4

(1, 2, 5,

LDFS ordering:

3, 6,4)

labels0

1 1σ =

2 23 34 4

(1, 2, 5,

LDFS ordering:

3, 6,4)

labels0

1 1σ =

2 23 34 4

(1, 2, 5,

LDFS ordering:

3, 6,4)

labels0

1 1σ =

2 23 34 4

(1, 2, 5,

LDFS ordering:

3, 6,4)

labels0

1 1σ =

2 23 34 4

(1, 2, 5,

LDFS ordering:

3, 6,4)

labels0

1 1σ =

2 23 34 4

(1, 2, 5,

LDFS ordering:

3, 6,4)

labels0

1 1σ =

2 23 34 4

(1, 2, 5,

LDFS ordering:

3, 6,4)

labels0

1 1σ =

2 23 34 4

How can we check whether an ordering σ is an LDFS ordering?

Since LDFS is a DFS ⇒ σ must have property (D)i.e. there exists a vertex d with a < d < b such that bd ∈ E

Is this enough? NO.

Property (LD)

there exists a vertex d with a < d < b such that db ∈ E and dc /∈ E .

a b cd

Is this enough? NO.

Property (LD)

a b cd

Is this enough? NO.

Property (LD)

a b cd

Is this enough? NO.

Property (LD)

a b cd

Property (LD) can characterize LDFS orderings:

For any graph G = (V , E ), an ordering σ of V is an LDFS ordering of Gif and only if σ satisfies property (LD).

Proof.

Notation:

the ordering σ is a mapping σ : N→ V

⇒ σ(i) = v means: “the ith vertex in σ is v”

⇒ σ−1(v) = i means: “the position of v in σ is i”

Proof.

Notation:

the ordering σ is a mapping σ : N→ V

⇒ σ(i) = v means: “the ith vertex in σ is v”

⇒ σ−1(v) = i means: “the position of v in σ is i”

for two vertices u, v : labelu(v) is the label assigned to vertex vimmediately before vertex u was numbered

Proof.

(⇒): Suppose that σ is an LDFS ordering.

Since b was numbered before c ⇒ labelb(b) ≥ labelb(c)

Since σ−1(a) appears in labelb(c) but not in labelb(b):

⇒ labelb(b) includes a digit j > σ−1(a) that does not appear in labelb(c)

Proof.

(⇒): Suppose that σ is an LDFS ordering.

Since σ−1(a) appears in labelb(c) but not in labelb(b):

⇒ labelb(b) includes a digit j > σ−1(a) that does not appear in labelb(c)

Proof.

Let d = σ(j). Then:

d < b in σ, since d was numbered before b

db ∈ E , since j = σ−1(d) appears in labelb(b)

d > a in σ, since j = σ−1(d) > σ−1(a)

dc /∈ E , since j = σ−1(d) does not appear in labelb(c)

⇒ σ satisfies property (LD)

Proof.

(⇐): Suppose that σ = (v1, v2, . . . , vn) is satisfies property (LD).Proof by contradiction:

Suppose that (v1, v2, . . . , vi−1) can be obtained by the LDFSalgorithm, but σi−1 = (v1, v2, . . . , vi ) can not (for some i ≤ n)

i.e. vi is the first vertex that can not be chosen next by LDFS

Let u be a vertex that the algorithm can choose next by LDFS(many equal labels ⇒ many choices for u ! )

At the point when u is chosen next by the algorithm:

u is preferred over vi ⇒ label(u) > label(vi )

let j be the largest digit in label(u) that is not in label(vi )

⇒ vj is the rightmost vertex in σi−1 with vju ∈ E and vjvi /∈ E

Proof.

σ has property (LD) ⇒ for the triple vj < vi < u in σ there existsa vertex vj < d < vi such that dvi ∈ E and du /∈ E

⇒ the position of d in σ (i.e. σ−1(d)) is in label(vi ) but not in label(u)

vj < d ⇒ j < σ−1(d) ⇒ label(vi ) > label(u) (by the choice of j),contradiction

Therefore vi can be chosen next by LDFS ⇒ (v1, v2, . . . , vi ) can beobtained by the LDFS algorithm for every i

⇒ σ = (v1, v2, . . . , vn) is an LDFS ordering

Proof.

Lexicographic Breadth First search

Modifying once again the generic search algorithm, we can “refine” BFSusing labels and lexicographic order:

we do not use a queue any more !

Lexicogr. Breadth First Search (LBFS) (Rose, Tarjan, Lueker, 1976)

Input: a graph G = (V , E ) with |V | = n, a start vertex s ∈ VOutput: a Lexicographic BFS (LBFS) ordering σ of V

2 label(s)← {n + 1} // s is still unnumbered

3 For i = 1 to n:

7 append (n− i) to label(w)

Modifying once again the generic search algorithm, we can “refine” BFSusing labels and lexicographic order:

we do not use a queue any more !

Lexicogr. Breadth First Search (LBFS) (Rose, Tarjan, Lueker, 1976)

Input: a graph G = (V , E ) with |V | = n, a start vertex s ∈ VOutput: a Lexicographic BFS (LBFS) ordering σ of V

2 label(s)← {n + 1} // s is still unnumbered

3 For i = 1 to n:

7 append (n− i) to label(w)

we append the labels of all neighbors of the current vertex vwith the position of v in σ

⇒ LBFS always chooses vertices that are adjacent to as many verticeswe have numbered early, as possible

The digits in each label appear in decreasing order (like in LDFS)

The ordering (1, 2, 3, 4, 5, 6) is a BFS but not an LBFS ordering

(1, 2, 5,3,

labels

LBFS ordering: 74,6)

(1, 2, 5,3,

labels

(1, 2, 5,3,

labels

(1, 2, 5,3,

labels

(1, 2, 5,3,

labels

(1, 2, 5,3,

labels

(1, 2, 5,3,

labels

(1, 2, 5,3,

labels

(1, 2, 5,3,

labels

(1, 2, 5,3,

labels

(1, 2, 5,3,

labels

(1, 2, 5,3,

labels

How can we check whether an ordering σ is an LBFS ordering?

Since LBFS is a BFS ⇒ σ must have property (B)i.e. there exists a vertex d < a such that db ∈ E

Is this enough? NO.

Property (LB)

there exists a vertex d < a such that db ∈ E and dc /∈ E .

a b cd

Is this enough? NO.

Property (LB)

a b cd

Is this enough? NO.

Property (LB)

a b cd

Is this enough? NO.

Property (LB)

a b cd

Property (LB) can characterize LBFS orderings:

Theorem (Rose, Tarjan, Lueker, SIAM Journal on Computing, 1976)

For any graph G = (V , E ), if an ordering σ of V is an LBFS ordering of Gthen σ satisfies property (LB).

Proof.

Later also the other direction has been proved:

Theorem (Brandstadt, Dragan, Nicolai, Discrete Math., 1997)

For any graph G = (V , E ), if an ordering σ of V satisfies property (LB)then σ is an LBFS ordering of G.

Proof.

Later also the other direction has been proved:

Recall:

for two vertices u, v : labelu(v) is the label assigned to vertex vimmediately before vertex u was numbered

Proof.

Suppose that σ is an LBFS ordering.

Since (n− σ−1(a)) appears in labelb(c) but not in labelb(b):

⇒ labelb(b) includes a digit (n− j) > (n− σ−1(a)) thatdoes not appear in labelb(c)

Proof.

Suppose that σ is an LBFS ordering.

Since (n− σ−1(a)) appears in labelb(c) but not in labelb(b):

⇒ labelb(b) includes a digit (n− j) > (n− σ−1(a)) thatdoes not appear in labelb(c)

Proof.

db ∈ E , since (n− j) = (n− σ−1(d)) appears in labelb(b)

d < a in σ, since (n− j) = (n− σ−1(d)) > (n− σ−1(a))

dc /∈ E , since (n− j) = (n− σ−1(d)) does not appear in labelb(c)

⇒ σ satisfies property (LB)

Proof.

Suppose that σ = (v1, v2, . . . , vn) is satisfies property (LB).

Suppose that (v1, v2, . . . , vi−1) can be obtained by the LBFSalgorithm, but σi−1 = (v1, v2, . . . , vi ) can not (for some i ≤ n)

Let u be a vertex that the algorithm can choose next by LBFS(many equal labels ⇒ many choices for u ! )

let (n− j) be the largest digit in label(u) that is not in label(vi )

⇒ vj is the leftmost vertex in σi−1 with vju ∈ E and vjvi /∈ E

Proof.

σ has property (LB) ⇒ for the triple vj < vi < u in σ there existsa vertex d < vj such that dvi ∈ E and du /∈ E

⇒ the digit (n− σ−1(d)) is in label(vi ) but not in label(u)