The Spanning Trees Formulas in a Class of Double Fixed-Step Loop Networks

Talip Atajan, Naohisa Otsuka Tokyo Denki University, Japan

Xuerong YongUniversity of Puerto Rico, USA

This Talk is Composed of

A double fixed-step loop network

An oriented spanning tree

Reliability of a network

Designing electrical circuits

Matrix tree theorem

Recurrence formula

A series of formulae for special p, q and n

New techniques

DEFINATION

4

p,qnC is a digraph on n vertices

0,1,2,..., n-1 and for each

vertex i (0<i -1), there are

exactly two arcs leaving

from vertex i to vertices

i+p,i+q (mod n).

n

A double fixed-step loop network

01

2

3

4

5

6

789

10

11

12

13

14

15

DEFINATION

An oriented spanning tree

An oriented spanning tree in a digraph D is a rooted tree with the same vertex set as D, that is, there is a node specified as theroot and from it there is a path to any vertexof D.

1

23

0

G

1

23

0 1

23

0

APPLICATION

Reliability of a network

1 11

1

number of vertices

number of edges

number of spanning trees

probability of line break

(1 )

ing

network reliability

n m nn

n

P A

n

m

A

P

6

0

1

2

35

7

APPLICATION

Designing electrical circuits

Ohm’s Law

Kirchhoff’s Law

Matrix Tree Theorem1

23

0

G

, ,

,,

0 0 0 01 0 0

1 1 0 0, H(1,1) 0 2 0

1 0 2 01 1 1

0 1 1 1

, ,

( ) , .

H ,

( ) H (1,1) 2

i j i j j i

i ji i j

e if i j e edges v to vh

ind e g v e if i j

T G det

Techniques

1

23

0

1

23

0

Techniques

Recurrence formula

1 1

, 2

1 1 2 2 2 2

1 2 1 2

1,2 2

For any integers ,and , let ( )

were proved by professor Golin, Zhang, etl. (2000)

Example

, 1,

( )

q q

p qn n

n n n n

n n n

n n

p q n T C na

a c a c a c a

a a a a a

where T C na

Techniques

The formulae for 2

1

,p d m pd mC

2

1

,

1 2

3 1 1 23,2 33

A series of formulae for were obtained

by professors Golin , Yong etl. in 2006.

where , , p areabitrary parameters and is a

variable. For example:

(2 2 2 cos ) 3( ) 3

0

p d m pd m

m m mm

m

C

d d n

mm if

T Cotherwise

†m

Our Results0

12

34

5

6789

10

11

1213

1415 0 1

234

5

6789

10

11

1213

1415 0 1

2345

6789

1011

1213

1415

0 3 6 9

13

1074

12

1525

1 14 11 8

0 3 6 9

13

10

7

4

12

15

2

5

1 14 11 8

0 3 6 9

131074

121525

1 14 11 8

316C

516C

716C

116C

1,716C

3,516C

Our Results

1 21 21

1 2

1 1

1 2

11

1

! ! !

(0,1, , )(1, , )

Let ( ) ( ) , then

( 1) ,

, 1, 2, ,

uu u mk k kmm

m

m m

i

i

k k kn

nn n

j nj

u um u u u

u k u k mu mk m

a a am m

P x x x x

where

m n

Theorem 2 (Opposed side of Waring's formula )

Our Results

,

, '

1

11

2

For counting ( ), we using the fact

( ) (2

We need t

)

where

( ) ( )

o define , 1

.

is a unit roo

,2, .

t.

p qn

p qn

npj qj

j

n nn

i n

j

T C

T C P

P x x

x x

e

j n

Our Results

1

11

1

1 1 1

2

2

1

2

1 2 1

( ) ( ) .

, 1, 2, ,

In our case , 1,

( ) ( ) (

(

,

)

, ,

) ?

2p q k p q k

nn n

j nj

k k kk

pj qj

k k k k k

j

pn qn k

S S S S

P x x x x

S a a a k n

then

j n thu

k

s

Using Newton's identities

Our Results

01 ( )mod

01

1 ( ( ))mod

1

( ) , ( ) ( )

( ) , gcd( , ) 1, ( )

0 ( ) , gcd( , ) 1, ( )

k n p n nd m iq p d d

mi

n nq p v n p iq p q p

m mi

n C if q p n q p k n p

S n C if q p n q p p m v q p

if q p n q p p m v q p

†

Theorem 1

01

234

5

6789

10

11

1213

1415

0

12 3

4 5 6

7 89

10 11 12

1314 15

0 12

345

6789

1011

1213

1415

3,516C

Our Results