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

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 at Mayaguez, USA

(Presented at the SIAM Workshop on Analytic Algorithmics and

Combinatorics (Jan 2009, New York)

We Will Talk About

A double fixed-step loop network

An oriented spanning tree

Reliability of a network

Designing electrical circuits

Modified matrix tree theorem

Recurrence formulas

New techniques and open problems

DEFINITION

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

DEFINITION

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

APPLICATIONS

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

APPLICATIONS

Designing electrical circuits

Ohm’s Law

Kirchhoff’s Law

Modified Matrix Tree Theorem1

23

0

G

, ,

,,

0 1 0 11 1 0

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

0 0 2 00 1 1

0 0 1 1

, ,

( ) , .

H=(h_ij) ,

( ) H (1,1) 2

i j i j i j

i ji i i

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 formulas

1 1

, 2

1 1 2 2 2 2

1 2 1 2

1,2 2

For any integers ,and , ( )

proved by Golin, Yong, Zhang (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 formula for 2

1

,p d m pd mC

2

1

,

1 2

3 1 1 23,2 33

A series of formulas for were obtained

by Golin , Yong in 2006 (but too complicated).

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

In counting ( ), we use the fact

( ) (2)

wher

We need t

e

( ) ( )

.

o

.

find , 1,2, .

p qn

p qn

npj

j

qj

j

n nn

i n

T C

T C P

P x x

x

n

e

j

x

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