Solved Exercises on BCC Long Paths - PowerPoint Presentation

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Solved Exercises on BCC

Long Paths

Question 1

: Consider the

DFS

tree so that every edge is a tree edge or a

backward

edge.

a) Let

h

be the height of the tree. Direct the edges of the tree

towards the root and the backward edges from the descendants to

the ancestors. Show that the

outdegree

of every vertex is out most

h

b) Show that there is a simple path of length

m/n

in the graph with

m

the number of edges and

n

the number of vertices.

Solution

Consider a path from a leaf to the root.

Note that all edges are backwards ones

Consider a path from a leaf to the root.

Thus the largest number of edges is when each vertex has an edge to all ancestors

Consider a path from a leaf to the root.

Thus the largest number of edges is when each vertex has an edge to all ancestors

Consider a path from a leaf to the root.

This implies that the outdegreeof

a vertex is at most h

Consider a path from a leaf to the root.

Bounding the number of edges from above

h

˖n≥m

. This implies that

h≥m

/n.

This proof is from a paper and the previous proof was from a paper.

This shows that professionals miss simple things.

Question 2

Say that

∑

v

L(v)=1

. Characterize the type of graphs for which this holds

Solution: This simply means that L(v)=1 for every v.

Now consider the leaves. The leaves can have L(v)=1 only if they have a direct edge to the root. Conversely, if every leaf has a

backward edge to the root, since the subtree of every vertex has at least one leaf, every vertex can god down to a leaf and jump to the

root. Hence the answer is: such graphs are graphs so that every leaf

has a backward edge to the root.

Question 3

Let C, C’ be two cycles. In an undirected graph. Show that

(C-C’)+(C’+C) is

a graph with

an

Euler Cycle.

Proof: What is the degree of a vertex?

Note that the degree of every vertex is at most 4 in the union. Now consider the following picture:

Question 3

How can the degree be

1

? This need to have all green edges belong to both cycles. The red edge belongs to one of the cycles only. But then this cycle has

3

edges on

v which can not be.

Question 3

How can the degree be

3

? This need the green edge to belong to both cycles and other edges belong to only one cycle.

There are

3

edges that belong to both cycle. But the green edge belongs to one cycle and does not belong to the other. So one cycle has

3 edges on the vertex which can not be

Remarks on the sequel questions

A binary relation



is a relation that is transitive symmetric and transitive.

This means:

1)

A  A

2) If

A  B then B

 A3)

If

A  B

and

B  C

then

A  C

Example

It is an equivalence relation:

a

b

for integers if

n

divides

a-b

.It is very simple to show that it is an equivalent relaton.

Example: the number 8.

There are

8

sets

{0,8,16,….},{1,9,17,……}

until

{7,15,23,…}

C

laim: the equivalence classes are a disjoint partition of the set (not hard to prove). Now later deal with equivalent relation between edges. Two edges are equivalent if they belong to

the same

simple

cycle (later).

Equivalence relation is the most important thing in math

Examples: Lines being

parallel

is an equivalence relation. What is in common to all parallel lines is the angle. If you are willing to look at an infinite set as one item, you got your definition of

angle.

How to define subtraction when we have only addition? Transform subtraction to additions. Namely we want to define

a-b.

A pair a,c (which we think of as

a-c in our mind) is equivalent to b,d

(looking to it as b-d in our mind) if a+d=

c+b

. The fact that it is an equivalent relation. Now consider the set of

(0,c).

Again, associate this set with

–c

We invented negative numbers.

Question 4

Show that if

e

and

e

’ belong to the same simple cycle and

e and

e’’ belong to the same simple cycle then e and e’’ belong to the same simple cycle.

Proof: Let C be the cycle of e,e’

and C’ be the cycle of e’ and e’’. Consider a separation into edges in (C-C’)+(C’-C) to

simple cycles by removing a cycle and continuing (its an

Euler cycle).

We may assume that

e, e’

’ does not belong to the same

connected component.

S

ay

that

e

is in

G(1)

and

e’

is in

G(2)

.

Proof by picture informal

The red cycle contains

e,e

’.

The green part with the two red parts and plus the blue part, is a cycle with

e’ and e’’.

Note that this implies a simple path for

e ,e’’.

e’’

e

e’

BCC as an equivalence relation

Question 5

: show that the equivalence relation we just saw decomposes the edges to

BCC

.

Proof:

consider a

BCC X so that both e and

e’ are in the BCC. Let the edges be ab

and cd. Both in the BCC.Join a vertex x

to both

a

and

c

and a vertex

y

to both

b

and

d

. This does not create a separating vertex since if any of those vertex are removed the graph is still connected. There are two vertex disjoint paths from

x

to

y

. So there are

two vertex disjoint paths

for example from

a

to

b

and

c

to

d. Hence a simple path with both edges.

Other direction: say that two edges belong to two different BCC.

e

e’

Since there are no cycles

There is a unique path from

e

to

e’

in the

BCC.

Such path must cross at least one vertex separator, which separates the edges.

How to recognize bridges

We need that

L(y)≤k(y).

x

y