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