Groups, Designs and Linear Algebra:
Orbit incidence matrices
Donald L. Kreher
Department of Mathematical Sciences
Michigan Technological University
CIMPA School
July-August 2013
t-wise balanced designs
Given a set X = {x1 , x2 , . . . , xv } of v -points and a positive
integer t < v we ask:
Can a non-trivial collection
B = {B1 , B2 , . . . , Bb }
of subsets of X be found so that every t-element subset of X is
contained in exactly one of them?
t-wise balanced designs
Given a set X = {x1 , x2 , . . . , xv } of v -points and a positive
integer t < v we ask:
Can a non-trivial collection
B = {B1 , B2 , . . . , Bb }
of subsets of X be found so that every t-element subset of X is
contained in exactly one of them?
Example:
t
= 2
= {0, 1, 2, 3, 4, 5, 6}
{0, 1, 2}, {0, 3, 4}, {0, 5, 6}, {1, 3, 5},
B =
{1, 4, 6}, {2, 3, 6}, {2, 4, 5}
X
Every pair is in exactly one of the chosen subsets.
Checking the blocks.
012
034
056
135
146
236
245
Checking the blocks.
012
01
02
03
04
05
06
12
13
14
15
16
23
24
25
26
34
35
36
45
46
56
034
056
135
146
236
245
Checking the blocks.
0
01
02
03
04
05
06
12
13
14
15
16
23
24
25
26
34
35
36
45
46
56
012
√
√
···
···
···
·√
··
···
···
···
···
···
···
···
···
···
···
···
···
···
···
034
056
135
146
236
245
1
2
Checking the blocks.
0
01
02
03
04
05
06
12
13
14
15
16
23
24
25
26
34
35
36
45
46
56
012
√
√
···
···
···
·√
··
···
···
···
···
···
···
···
···
···
···
···
···
···
···
034
···
·√
··
056
135
146
236
245
1
4
√
···
···
···
···
···
···
···
···
···
···
·√
··
···
···
···
···
···
2
3
Checking the blocks.
0
01
02
03
04
05
06
12
13
14
15
16
23
24
25
26
34
35
36
45
46
56
012
√
√
···
···
···
·√
··
···
···
···
···
···
···
···
···
···
···
···
···
···
···
034
···
·√
··
√
···
···
···
···
···
···
···
···
···
···
·√
··
···
···
···
···
···
056
···
···
···
·√
··
√
···
···
···
···
···
···
···
···
···
···
···
···
···
·√
··
135
146
236
245
1
4
5
2
6
3
Checking the blocks.
0
01
02
03
04
05
06
12
13
14
15
16
23
24
25
26
34
35
36
45
46
56
012
√
√
···
···
···
·√
··
···
···
···
···
···
···
···
···
···
···
···
···
···
···
034
···
·√
··
√
···
···
···
···
···
···
···
···
···
···
·√
··
···
···
···
···
···
056
···
···
···
·√
··
√
···
···
···
···
···
···
···
···
···
···
···
···
···
·√
··
135
···
···
···
···
···
···
·√
··
·√
··
···
···
···
···
···
·√
··
···
···
···
···
146
236
245
1
4
5
2
6
3
Checking the blocks.
0
01
02
03
04
05
06
12
13
14
15
16
23
24
25
26
34
35
36
45
46
56
012
√
√
···
···
···
·√
··
···
···
···
···
···
···
···
···
···
···
···
···
···
···
034
···
·√
··
√
···
···
···
···
···
···
···
···
···
···
·√
··
···
···
···
···
···
056
···
···
···
·√
··
√
···
···
···
···
···
···
···
···
···
···
···
···
···
·√
··
135
···
···
···
···
···
···
·√
··
·√
··
···
···
···
···
···
·√
··
···
···
···
···
146
···
···
···
···
···
···
···
·√
··
·√
··
···
···
···
···
···
···
···
·√
··
···
236
245
1
4
5
2
6
3
Checking the blocks.
0
01
02
03
04
05
06
12
13
14
15
16
23
24
25
26
34
35
36
45
46
56
012
√
√
···
···
···
·√
··
···
···
···
···
···
···
···
···
···
···
···
···
···
···
034
···
·√
··
√
···
···
···
···
···
···
···
···
···
···
·√
··
···
···
···
···
···
056
···
···
···
·√
··
√
···
···
···
···
···
···
···
···
···
···
···
···
···
·√
··
135
···
···
···
···
···
···
·√
··
·√
··
···
···
···
···
···
·√
··
···
···
···
···
146
···
···
···
···
···
···
···
·√
··
·√
··
···
···
···
···
···
···
···
·√
··
···
236
···
···
···
···
···
···
···
···
···
···
·√
··
···
·√
··
···
·√
··
···
···
···
245
1
4
5
2
6
3
Checking the blocks.
0
01
02
03
04
05
06
12
13
14
15
16
23
24
25
26
34
35
36
45
46
56
012
√
√
···
···
···
·√
··
···
···
···
···
···
···
···
···
···
···
···
···
···
···
034
···
·√
··
√
···
···
···
···
···
···
···
···
···
···
·√
··
···
···
···
···
···
056
···
···
···
·√
··
√
···
···
···
···
···
···
···
···
···
···
···
···
···
·√
··
135
···
···
···
···
···
···
·√
··
·√
··
···
···
···
···
···
·√
··
···
···
···
···
146
···
···
···
···
···
···
···
·√
··
·√
··
···
···
···
···
···
···
···
·√
··
···
236
···
···
···
···
···
···
···
···
···
···
·√
··
···
·√
··
···
·√
··
···
···
···
245
···
···
···
···
···
···
···
···
···
···
···
·√
··
√
···
···
···
·√
··
···
···
1
4
5
2
6
3
Checking the blocks.
0
01
02
03
04
05
06
12
13
14
15
16
23
24
25
26
34
35
36
45
46
56
012
√
√
···
···
···
·√
··
···
···
···
···
···
···
···
···
···
···
···
···
···
···
034
···
·√
··
√
···
···
···
···
···
···
···
···
···
···
·√
··
···
···
···
···
···
056
···
···
···
·√
··
√
···
···
···
···
···
···
···
···
···
···
···
···
···
·√
··
135
···
···
···
···
···
···
·√
··
·√
··
···
···
···
···
···
·√
··
···
···
···
···
146
···
···
···
···
···
···
···
·√
··
·√
··
···
···
···
···
···
···
···
·√
··
···
236
···
···
···
···
···
···
···
···
···
···
·√
··
···
·√
··
···
·√
··
···
···
···
245
···
···
···
···
···
···
···
···
···
···
···
·√
··
√
···
···
···
·√
··
···
···
1
4
5
2
6
3
G = h(0, 3, 2)(1, 4, 6), (1, 4)(2, 3)i


(0)(1)(2)(3)(4)(5)(6) 







(1, 4)(2, 3)






(0, 2, 3)(1, 6, 4)
=
(0, 2)(4, 6)








(0,
3, 2)(1, 4, 6)






(0, 3)(1, 6)
is an obvious automorphism group.
Orbits
Let G be a possible automorphism group.
B1
B2
T
B3
∆=
g
=Γ
g
B10
T0
Orbit of
t-element
subsets.
B20
B30
Orbit of
k -element
subsets.
Orbit incidence matrix



I, (0, 3, 2)(1, 4, 6),

Let G=h(0, 3, 2)(1, 4, 6), (0, 2, 3)(1, 6, 4)i = (0, 2, 3)(1, 6, 4), (1, 4)(2, 3), .


(0, 2)(4, 6), (0, 3)(1, 6)
Orbit incidence matrix

I, (0, 3, 2)(1, 4, 6),

Let G=h(0, 3, 2)(1, 4, 6), (0, 2, 3)(1, 6, 4)i = (0, 2, 3)(1, 6, 4), (1, 4)(2, 3), .


(0, 2)(4, 6), (0, 3)(1, 6)
{46, 16, 14}
36, 34, 26
12, 04, 01
{24, 13, 06}
{23, 03, 02}
{56, 45, 15}
{35, 25, 05}


Orbit incidence matrix

I, (0, 3, 2)(1, 4, 6),

Let G=h(0, 3, 2)(1, 4, 6), (0, 2, 3)(1, 6, 4)i = (0, 2, 3)(1, 6, 4), (1, 4)(2, 3), .


(0, 2)(4, 6), (0, 3)(1, 6)
046
016
134
014 136
346 246
146 126 124
{46, 16, 14}
36, 34, 26
12, 04, 01
{24, 13, 06}
{23, 03, 02}
{56, 45, 15}
{35, 25, 05}


036
045
026
356
123
125
013 034
126 015 056 035
024 236
145 345 135 235
234 012 023 456 256 245 025
Orbit incidence matrix

I, (0, 3, 2)(1, 4, 6),

Let G=h(0, 3, 2)(1, 4, 6), (0, 2, 3)(1, 6, 4)i = (0, 2, 3)(1, 6, 4), (1, 4)(2, 3), .


(0, 2)(4, 6), (0, 3)(1, 6)
046
016
134
014 136
346 246
146 126 124
{46, 16, 14}
36, 34, 26
12, 04, 01
{24, 13, 06}
{23, 03, 02}
{56, 45, 15}
{35, 25, 05}


036
045
026
356
123
125
013 034
126 015 056 035
024 236
145 345 135 235
234 012 023 456 256 245 025
Orbit incidence matrix

I, (0, 3, 2)(1, 4, 6),

Let G=h(0, 3, 2)(1, 4, 6), (0, 2, 3)(1, 6, 4)i = (0, 2, 3)(1, 6, 4), (1, 4)(2, 3), .


(0, 2)(4, 6), (0, 3)(1, 6)
046
016
134
014 136
346 246
146 126 124
{46, 16, 14}
36, 34, 26
12, 04, 01
{24, 13, 06}
{23, 03, 02}
{56, 45, 15}
{35, 25, 05}
2


036
045
026
356
123
125
013 034
126 015 056 035
024 236
145 345 135 235
234 012 023 456 256 245 025
Orbit incidence matrix

I, (0, 3, 2)(1, 4, 6),

Let G=h(0, 3, 2)(1, 4, 6), (0, 2, 3)(1, 6, 4)i = (0, 2, 3)(1, 6, 4), (1, 4)(2, 3), .


(0, 2)(4, 6), (0, 3)(1, 6)
046
016
134
014 136
346 246
146 126 124
{46, 16, 14}
36, 34, 26
12, 04, 01
2
{24, 13, 06}
2
{23, 03, 02}
0
{56, 45, 15}
0
{35, 25, 05}
0
1


036
045
026
356
123
125
013 034
126 015 056 035
024 236
145 345 135 235
234 012 023 456 256 245 025
Orbit incidence matrix

I, (0, 3, 2)(1, 4, 6),

Let G=h(0, 3, 2)(1, 4, 6), (0, 2, 3)(1, 6, 4)i = (0, 2, 3)(1, 6, 4), (1, 4)(2, 3), .


(0, 2)(4, 6), (0, 3)(1, 6)
046
016
134
014 136
346 246
146 126 124


036
045
026
356
123
125
013 034
126 015 056 035
024 236
145 345 135 235
234 012 023 456 256 245 025
{46, 16, 14}
36, 34, 26
12, 04, 01
1
1
2
0
0
0
1
0
0
0
0
1
1
1
1
0
0
1
0
0
{24, 13, 06}
0
0
2
2
0
0
0
0
1
0
{23, 03, 02}
0
0
0
2
1
1
0
0
0
1
{56, 45, 15}
0
0
0
0
0
0
2
2
1
0
{35, 25, 05}
0
0
0
0
0
0
0
2
1
2
Orbit incidence matrix

I, (0, 3, 2)(1, 4, 6),

Let G=h(0, 3, 2)(1, 4, 6), (0, 2, 3)(1, 6, 4)i = (0, 2, 3)(1, 6, 4), (1, 4)(2, 3), .


(0, 2)(4, 6), (0, 3)(1, 6)
046
016
134
014 136
346 246
146 126 124


036
045
026
356
123
125
013 034
126 015 056 035
024 236
145 345 135 235
234 012 023 456 256 245 025
{46, 16, 14}
36, 34, 26
12, 04, 01
1
1
2
0
0
0
1
0
0
0
0
1
1
1
1
0
0
1
0
0
{24, 13, 06}
0
0
2
2
0
0
0
0
1
0
{23, 03, 02}
0
0
0
2
1
1
0
0
0
1
{56, 45, 15}
0
0
0
0
0
0
2
2
1
0
{35, 25, 05}
0
0
0
0
0
0
0
2
1
2
Orbit incidence matrix

I, (0, 3, 2)(1, 4, 6),

Let G=h(0, 3, 2)(1, 4, 6), (0, 2, 3)(1, 6, 4)i = (0, 2, 3)(1, 6, 4), (1, 4)(2, 3), .


(0, 2)(4, 6), (0, 3)(1, 6)
046
016
134
014 136
346 246
146 126 124


036
045
026
356
123
125
013 034
126 015 056 035
024 236
145 345 135 235
234 012 023 456 256 245 025
{46, 16, 14}
36, 34, 26
12, 04, 01
1
1
2
0
0
0
1
0
0
0
0
1
1
1
1
0
0
1
0
0
{24, 13, 06}
0
0
2
2
0
0
0
0
1
0
{23, 03, 02}
0
0
0
2
1
1
0
0
0
1
{56, 45, 15}
0
0
0
0
0
0
2
2
1
0
{35, 25, 05}
0
0
0
0
0
0
0
2
1
2
Formal definitions
A t-wise balanced design (tBD) with parameters t-(v , K, λ) is a pair
(X , B) where:
I
X is a v -element set of points;
I
B is a family subsets of X called blocks;
I
K is a set of possible block sizes;
I
If B ∈ B, then |B| ∈ K.
I
Every t-element subset of X is in exactly λ blocks.
Formal definitions
A t-wise balanced design (tBD) with parameters t-(v , K, λ) is a pair
(X , B) where:
I
X is a v -element set of points;
I
B is a family subsets of X called blocks;
I
K is a set of possible block sizes;
I
If B ∈ B, then |B| ∈ K.
I
Every t-element subset of X is in exactly λ blocks.
If K = {k }, we obtain a t-(v , k , λ) design (t-design).
Formal definitions
A t-wise balanced design (tBD) with parameters t-(v , K, λ) is a pair
(X , B) where:
I
X is a v -element set of points;
I
B is a family subsets of X called blocks;
I
K is a set of possible block sizes;
I
If B ∈ B, then |B| ∈ K.
I
Every t-element subset of X is in exactly λ blocks.
If K = {k }, we obtain a t-(v , k , λ) design (t-design).
A tBD is non-trivial, if t, v 6∈ K and
B does not contain all the k -subsets for any 0 < k < v .
In particular X is not a block.
Formal definitions
A t-wise balanced design (tBD) with parameters t-(v , K, λ) is a pair
(X , B) where:
I
X is a v -element set of points;
I
B is a family subsets of X called blocks;
I
K is a set of possible block sizes;
I
If B ∈ B, then |B| ∈ K.
I
Every t-element subset of X is in exactly λ blocks.
If K = {k }, we obtain a t-(v , k , λ) design (t-design).
A tBD is non-trivial, if t, v 6∈ K and
B does not contain all the k -subsets for any 0 < k < v .
In particular X is not a block.
A t-(v , k, 1) design is called a Steiner system.
Formal definitions
A t-wise balanced design (tBD) with parameters t-(v , K, λ) is a pair
(X , B) where:
I
X is a v -element set of points;
I
B is a family subsets of X called blocks;
I
K is a set of possible block sizes;
I
If B ∈ B, then |B| ∈ K.
I
Every t-element subset of X is in exactly λ blocks.
If K = {k }, we obtain a t-(v , k , λ) design (t-design).
A tBD is non-trivial, if t, v 6∈ K and
B does not contain all the k -subsets for any 0 < k < v .
In particular X is not a block.
A t-(v , k, 1) design is called a Steiner system.
They are also called S(t, k , v ) systems.
Formal definitions
A t-wise balanced design (tBD) with parameters t-(v , K, λ) is a pair
(X , B) where:
I
X is a v -element set of points;
I
B is a family subsets of X called blocks;
I
K is a set of possible block sizes;
I
If B ∈ B, then |B| ∈ K.
I
Every t-element subset of X is in exactly λ blocks.
If K = {k }, we obtain a t-(v , k , λ) design (t-design).
A tBD is non-trivial, if t, v 6∈ K and
B does not contain all the k -subsets for any 0 < k < v .
In particular X is not a block.
A t-(v , k, 1) design is called a Steiner system.
They are also called S(t, k , v ) systems.
A S(2, 3, v ) design is called a Steiner triple system.
Formal definitions
A t-wise balanced design (tBD) with parameters t-(v , K, λ) is a pair
(X , B) where:
I
X is a v -element set of points;
I
B is a family subsets of X called blocks;
I
K is a set of possible block sizes;
I
If B ∈ B, then |B| ∈ K.
I
Every t-element subset of X is in exactly λ blocks.
If K = {k }, we obtain a t-(v , k , λ) design (t-design).
A tBD is non-trivial, if t, v 6∈ K and
B does not contain all the k -subsets for any 0 < k < v .
In particular X is not a block.
A t-(v , k, 1) design is called a Steiner system.
They are also called S(t, k , v ) systems.
A S(2, 3, v ) design is called a Steiner triple system.
A S(3, 4, v ) design is called a Steiner quadruple system.
Formal definitions: Orbit incidence Matrices
Given group G ≤ S YM(X ) and 0 ≤ t ≤ k ≤ v = |X | let
R = {∆
1 , ∆2 , . . . , ∆Nt } be the orbits of t-element subsets;
C =
Γ1 , Γ2 , . . . , ΓNk
be the orbits of k-element subsets.
The (t, k)-orbit incidence matrix is the matrix
Atk : R × C → Z
given by
Atk [∆, Γ] = |{K ∈ Γ : K ⊇ T }|,
where T ∈ ∆ is a fixed representative.
Kramer-Mesner 1973 Observation
A t-(v , k , λ) design exists with G ≤ S YM(X ) as an
automorphism group if and only if there is a
(0,1)-solution U to the matrix equation
Atk U = λJ,
where: J = [1, 1, 1, . . . , 1]T .
Kramer-Mesner 1973 Observation
A t-(v , k , λ) design exists with G ≤ S YM(X ) as an
automorphism group if and only if there is a
(0,1)-solution U to the matrix equation
Atk U = λJ,
where: J = [1, 1, 1, . . . , 1]T .
The orbit incidence matrix Atk is also called the Kramer-Mesner matrix.
Our method for constructing t-designs
I
Choose parameters t, k , v , and λ;
I
Find a candidate for an automorphism group G;
I
Generate the incidence matrix Atk ;
I
Solve the system of equations Atk U = λJ for one, some or all
(0,1)-vectors U;
I
Check for any special properties you may require of the found
solutions;
I
Apply any known recursive methods to the solutions found to
construct more designs.
Our method for constructing t-designs
I
Choose parameters t, k , v , and λ;
I
Find a candidate for an automorphism group G;
I
Generate the incidence matrix Atk ;
I
Solve the system of equations Atk U = λJ for one, some or all
(0,1)-vectors U;
I
Check for any special properties you may require of the found
solutions;
I
Apply any known recursive methods to the solutions found to
construct more designs.
Almost every known t-design with t ∈ {4, 5} was found this way.
Every known t-design with t ≥ 6 was found this way.
Available ingredients
t = 2, 3 Latin squares, transversal designs, orthogonal
arrays of strength 2 and 3, rich source of 2 and 3
homogeneous groups, recursive constructions,
geometry, coding theory..
t = 4, 5 a few 4 and 5 homogeneous groups, union of
orbits under other groups, coding theory.
t ≥ 6 union of group orbits,
What do they want?
Classification ⇒ There are no t-homogeneous groups with t > 5.
What do they want?
Classification ⇒ There are no t-homogeneous groups with t > 5.
Unknown group theorist
Der no be any 6-designs!
What do they want?
Classification ⇒ There are no t-homogeneous groups with t > 5.
Unknown group theorist
Der no be any 6-designs!
“The existence of non-trivial t–designs with t > 5 is the
most important unsolved problem in the area.”
Cameron & van Lint, Graphs, Codes and Designs, LMS
Lecture Note Series 43, 1980, page 1.
What do they want?
Classification ⇒ There are no t-homogeneous groups with t > 5.
Unknown group theorist
Der no be any 6-designs!
“The existence of non-trivial t–designs with t > 5 is the
most important unsolved problem in the area.”
Cameron & van Lint, Graphs, Codes and Designs, LMS
Lecture Note Series 43, 1980, page 1.
O H BOY OH BOY ....
6-(33, 8, 36) Leavit, Magliveras,1982
6-(20, 9, 112) Kramer, Leavit, Magliveras,1984
6-(14, 7, 4) Kreher, Radziszowski, 1986
t-designs exist for all t Tierlinck, 1987
...oooohhh
Still not satisfied?
Still not satisfied?
Unknown group theorist
interesting
Der no be any 6-designs!
Still not satisfied?
Unknown group theorist
interesting
Der no be any 6-designs!
“The existence of Steiner systems with large t is possibly
the most important open problem in design theory”
Cameron & van Lint, Graphs Codes and Designs, LMS
Student Texts 22, 1991, page 2.
Still not satisfied?
Unknown group theorist
interesting
Der no be any 6-designs!
“The existence of Steiner systems with large t is possibly
the most important open problem in design theory”
Cameron & van Lint, Graphs Codes and Designs, LMS
Student Texts 22, 1991, page 2.
Aw, phooey!
Research Problems
Research Problems
Research problem 1. Construct a t-(v , K, 1) design with t ≥ 6.
Research Problems
Research problem 1. Construct a t-(v , K, 1) design with t ≥ 6.
Research problem 2. Find or classify all t-wise balanced designs.
Research Problems
Research problem 1. Construct a t-(v , K, 1) design with t ≥ 6.
Research problem 2. Find or classify all t-wise balanced designs.
(Well, at least the interesting ones.)
Groups, Designs and Linear Algebra:
Graphical designs
Donald L. Kreher
Department of Mathematical Sciences
Michigan Technological University
CIMPA School
July-August 2013
‘
Graphical Designs
A graphical design
is aproper t-wise balanced designs (X , B) with
parameters t- n2 , K, λ that has the symmetric group Sn as an
automorphism group.
I X is the set of v = n labelled edges of Kn .
2
I
Blocks are subgraphs.
I
If B ∈ B, then all subgraphs isomorphic to B are also in B.
6
K
X
=
=
5
6
1
2
3
 4

12, 13, 14, 15, 16,
23, 24, 25, 26, 34,


35, 36, 45, 46, 56
Points are edges!
B
=
5
2
4
=
1
3
15, 16, 56, 24
Blocks are subgraphs!
Orbits of Sn acting on E(Kn ).
The graphical design condition:
if B ∈ B, then all subgraphs isomorphic to B are also in B
means:
I
The blocks set B is a union of isomorphism classes,
I
These are orbits under the action of Sn on E(Kn ).
I
They have pictures!
1
1
2
3
6
2
5
3
4
1
2
5
3
4
2
3
2
5
3
4
5
3
6
5
1
6
2
5
3
4
6
2
5
3
4
6
5
4
1
1
3
2
4
1
6
4
2
1
6
4
1
=
6
1
6
2
5
3
6
5
4
Orbits of Sn acting on E(Kn ), cont..
1
1
2
3
6
2
5
3
4
1
2
5
3
4
2
3
2
5
3
2
5
3
2
5
3
It is easy to see that:
h
i
I the
,
-entry of A2,6 is 1 .
h
i
I the
-entry of A2,6 is 2 .
,
3
5
4
6
5
4
1
6
4
5
6
1
6
4
1
3
2
4
1
6
4
2
1
6
4
1
=
6
1
6
2
5
3
6
5
4
A graphical 2-(6, {3, 4}, 2) design.
Pictures ⇒ orbit incidence matries are easy to construct!
Here is the A2∗ = [A2,3 , A2,4 , A2,5 ] matrix for S4 acting on E(K4 ).
0
0
4
2
4
4
1
1
2
1
5
4
The subgroups isomorphic to
form a 2-(6, {3, 4}, 2) design.
and
together
Steiner graphical designs.
We will now focus on the construction of Steiner graphical
tBDs. If we were to proceed systematicaly we would first find
the orbits that contain each orbit of t-element subsets at most
once. Then among these orbits we would seek a collection that
contains all the t-sets each exactly once.
Example 1:The unique graphical Steiner 1-wise balanced designs.
1
So,
4
6
2
gives a 1-(6, 2, 1) design.
2
2
8
5
Example 2:
The two graphical Steiner 2-wise balanced designs.
1
3
1
0
1
···
2
1
0
1
0
···
gives a 2-(15, 3, 1) design,
gives a 2-(15, {3, 5}, 1) design.
Example 3:
So,
The unique graphical Steiner 3-wise balanced design.
1
0
0
0
1
0
0
0
1
0
1
0
gives a 3-(10, 4, 1) design.
Example 4:The unique graphical Steiner 4-wise balanced design.
So,
0
0
1
0
0
0
1
0
0
1
1
0
0
0
0
0
1
0
0
1
0
0
0
0
1
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
1
gives a 4-(15, {5, 7}, 1) design.
Main Theorem(Chouinard,Kramer,Kreher, 1983.)
The only graphical Steiner t-wise balanced designs are:
1-(6, 2, 1)
n=4
2-(15, 3, 1)
n=6
2-(15, {3, 5}, 1)
n=6
3-(10, 4, 1)
n=5
4-(15, {5, 7}, 1)
n=6
Proof.
Suppose (X , B) is a non-trivial graphical t-(
Then:
n
2
, K, 1) design.
I
No block has size t.
I
No block is X = E(Kn ), i.e., no block is complete.
I
B does not contain all the k-subsets for any 0 < k < v .
I
If g ∈ Sn and B ∈ B, then g(B) ∈ B.
Let V (Kn ) = {1, 2, . . . , n}.
The proof is given in a series of Lemmas.
Key lemma
Let g ∈ S YM(V ) and B ∈ B. Then |B ∩ g(B)| ≥ t ⇒ B = g(B).
g
Proof.
If |B ∩ g(B)| ≥ t, then there is a
B:
t-element subset T ⊆ E(Kn ) that is
contained in the block B and also in
the block g(B). But (X , B) is a Steiner g(B) :
tBD and so B = g(B).
t
I
We use this Key Lemma to prove the next few lemmas by taking
an appropriate choice for a t-edge subgraph T .
I
The general strategy is when given T to consider the unique
block that contains it.
I
I
Because (X , B) is non-trivial B 6= T . Therefore there is a
edge e ∈ B \ T .
Now if g ∈ S YM(V ) is such that |g(B) ∩ B| ≥ t, it follows
from Key Lemma that g(e) ∈ B.
Lemma 1: n 6= t + 1
Lemma 1: n 6= t + 1
Proof. Suppose n = t + 1.
Lemma 1: n 6= t + 1
Proof. Suppose n = t + 1.
n
Take T =
..
.
Lemma 1: n 6= t + 1
Proof. Suppose n = t + 1.
e
n
Take T =
..
.
B 6= T ⇒ B contains an edge e 6∈ T .
Lemma 1: n 6= t + 1
Proof. Suppose n = t + 1.
e
g
g(e)
Take T =
n
..
.
B 6= T ⇒ B contains an edge e 6∈ T .
Use Key to apply permutaions g that fix the edges of T
to find edges g(e) in B.
Lemma 1: n 6= t + 1
Proof. Suppose n = t + 1.
Kn−1
n
Take T =
..
.
B 6= T ⇒ B contains an edge e 6∈ T .
Use Key to apply permutaions g that fix the edges of T
to find edges g(e) in B.
Eventualy force B to be complete.
Lemma 1: n 6= t + 1
Proof. Suppose n = t + 1.
Kn−1
n
Take T =
..
.
B 6= T ⇒ B contains an edge e 6∈ T .
Use Key to apply permutaions g that fix the edges of T
to find edges g(e) in B.
Eventualy force B to be complete.
This is a contradiction.
Therefor n 6= t + 1.
Lemma 2: n ≤ 2t + 2
Lemma 2: n ≤ 2t + 2
Proof. Suppose n ≥ 2t + 3.
Lemma 2: n ≤ 2t + 2
Proof. Suppose n ≥ 2t + 3.
1
t
3
2
···
Take T =
t+1
t+2
t+3
2t+1 2t+2
···
n − 2t ≥ 3
2t
n
Lemma 2: n ≤ 2t + 2
Proof. Suppose n ≥ 2t + 3.
1
t
3
2
···
Take T =
t+1
t+2
t+3
2t+1 2t+2
···
n − 2t ≥ 3
Let B be the unique block containing T .
2t
n
Lemma 2: n ≤ 2t + 2
Proof. Suppose n ≥ 2t + 3.
1
t
3
2
···
Take T =
t+1
t+2
t+3
2t+1 2t+2
···
n − 2t ≥ 3
Let B be the unique block containing T .
Let e be an edge in B \ T .
2t
n
Lemma 2: n ≤ 2t + 2
Proof. Suppose n ≥ 2t + 3.
1
t
3
2
···
t+1
t+2
t+3
2t+1 2t+2
···
2t
n
n − 2t ≥ 3
Let B be the unique block containing T .
Let e be an edge in B \ T .
Up to isomorphism there are 3 possibilities for e.
Lemma 2: n ≤ 2t + 2
Proof. Suppose n ≥ 2t + 3.
1
t
3
2
···
B ⊇ T +e =
t+1
t+2
t+3
2t+1 2t+2
···
2t
n
n − 2t ≥ 3
Let B be the unique block containing T .
Let e be an edge in B \ T .
Up to isomorphism there are 3 possibilities for e.
e = {1, 2}
Lemma 2: n ≤ 2t + 2
Proof. Suppose n ≥ 2t + 3.
1
t
3
2
···
B ⊇ T +e =
t+1
t+2
t+3
2t+1 2t+2
···
2t
n
n − 2t ≥ 3
Let B be the unique block containing T .
Let e be an edge in B \ T .
Up to isomorphism there are 3 possibilities for e.
e = {1, 2}
e = {1, 2t + 1}
Lemma 2: n ≤ 2t + 2
Proof. Suppose n ≥ 2t + 3.
1
t
3
2
···
B ⊇ T +e =
t+1
t+2
t+3
2t+1 2t+2
2t
···
n
n − 2t ≥ 3
Let B be the unique block containing T .
Let e be an edge in B \ T .
Up to isomorphism there are 3 possibilities for e.
e = {1, 2}
e = {1, 2t + 1}
e = {2t + 1, 2t + 2}
Lemma 2: n ≤ 2t + 2, the case e = {1, 2}
1
t
3
2
···
B ⊇ T +e =
t+1
t+2
t+3
2t+1 2t+2
2t
···
n − 2t ≥ 3
n
Lemma 2: n ≤ 2t + 2, the case e = {1, 2}
1
t
3
2
···
B ⊇ T +e =
t+1
t+2
t+3
2t+1 2t+2
2t
···
n
n − 2t ≥ 3
The automorphisms of T are 2-transitive on the vertices of valency 1.
Lemma 2: n ≤ 2t + 2, the case e = {1, 2}
1
t
3
2
K2t
···
B ⊇ T +e =
t+1
t+2
t+3
2t+1 2t+2
2t
···
n
n − 2t ≥ 3
The automorphisms of T are 2-transitive on the vertices of valency 1.
Key Lemma ⇒ B contains K2t on the vertices {1, 2, .., 2t}.
Lemma 2: n ≤ 2t + 2, the case e = {1, 2}
1
t+1
K2t
···
g
B ⊇ T +e =
t
3
2
t+2
t+3
2t+1 2t+2
2t
···
n
n − 2t ≥ 3
The automorphisms of T are 2-transitive on the vertices of valency 1.
Key Lemma ⇒ B contains K2t on the vertices {1, 2, .., 2t}.
The permutation g = (1, 2t + 1) fixes 2t−1
= (t − 1)(2t − 1) ≥ t
2
edges of the K2t−1 on {2, 3, . . . , 2t}.
Lemma 2: n ≤ 2t + 2, the case e = {1, 2}
1
t
3
2
K2t+1
···
B ⊇ T +e =
t+1
t+2
t+3
2t+1 2t+2
2t
···
n
n − 2t ≥ 3
The automorphisms of T are 2-transitive on the vertices of valency 1.
Key Lemma ⇒ B contains K2t on the vertices {1, 2, .., 2t}.
The permutation g = (1, 2t + 1) fixes 2t−1
= (t − 1)(2t − 1) ≥ t
2
edges of the K2t−1 on {2, 3, . . . , 2t}.
Key Lemma ⇒ B ⊃ K2t+1 on the vertices {1, 2, .., 2t, 2t + 1}.
Lemma 2: n ≤ 2t + 2, the case e = {1, 2}
1
t+1
t+2
K2t+1
···
g
B ⊇ T +e =
t
3
2
t+3
2t+1 2t+2
2t
···
n
n − 2t ≥ 3
The automorphisms of T are 2-transitive on the vertices of valency 1.
Key Lemma ⇒ B contains K2t on the vertices {1, 2, .., 2t}.
The permutation g = (1, 2t + 1) fixes 2t−1
= (t − 1)(2t − 1) ≥ t
2
edges of the K2t−1 on {2, 3, . . . , 2t}.
Key Lemma ⇒ B ⊃ K2t+1 on the vertices {1, 2, .., 2t, 2t + 1}.
The permutation h = (2, 2t + 2)(t + 2, 2t + 3) forces the edge
{2t + 1, 2t + 2} into B.
Lemma 2: n ≤ 2t + 2, the case e = {1, 2}
1
t
3
2
Kn − Kn−2t
···
B ⊇ T +e =
t+1
t+2
t+3
2t+1 2t+2
2t
···
n
Kn−2t
n − 2t ≥ 3
The automorphisms of T are 2-transitive on the vertices of valency 1.
Key Lemma ⇒ B contains K2t on the vertices {1, 2, .., 2t}.
The permutation g = (1, 2t + 1) fixes 2t−1
= (t − 1)(2t − 1) ≥ t
2
edges of the K2t−1 on {2, 3, . . . , 2t}.
Key Lemma ⇒ B ⊃ K2t+1 on the vertices {1, 2, .., 2t, 2t + 1}.
The permutation h = (2, 2t + 2)(t + 2, 2t + 3) forces the edge
{2t + 1, 2t + 2} into B.
Permuting the isolated vertices of T forces B to be complete.
Lemma 2: n ≤ 2t + 2, the case e = {1, 2}
1
t
3
2
Kn
···
B ⊇ T +e =
t+1
t+2
t+3
2t+1 2t+2
2t
···
n
n − 2t ≥ 3
The automorphisms of T are 2-transitive on the vertices of valency 1.
Key Lemma ⇒ B contains K2t on the vertices {1, 2, .., 2t}.
The permutation g = (1, 2t + 1) fixes 2t−1
= (t − 1)(2t − 1) ≥ t
2
edges of the K2t−1 on {2, 3, . . . , 2t}.
Key Lemma ⇒ B ⊃ K2t+1 on the vertices {1, 2, .., 2t, 2t + 1}.
The permutation h = (2, 2t + 2)(t + 2, 2t + 3) forces the edge
{2t + 1, 2t + 2} into B.
Permuting the isolated vertices of T forces B to be complete.
A contradiction.
Lemma 2: n ≤ 2t + 2, the case e = {1, 2t + 1}
1
t
3
2
···
B ⊇ T +e =
t+1
t+2
t+3
2t+1 2t+2
2t
···
n − 2t ≥ 3
n
Lemma 2: n ≤ 2t + 2, the case e = {1, 2t + 1}
1
t
3
2
···
B ⊇ T +e =
t+1
t+2
t+3
2t+1 2t+2
2t
···
n
n − 2t ≥ 3
Use permutations (1, t + 1) and (2t + 1, 2t + 2, . . . , n) to force
K1,n−2t+1 + {t + 1, 2t + 1} on {1, t + 1, 2t + 1, 2t + 2, . . . , n}.
Lemma 2: n ≤ 2t + 2, the case e = {1, 2t + 1}
1
t
3
2
···
B ⊇ T +e =
t+1
t+2
t+3
2t+1 2t+2
2t
···
n
n − 2t ≥ 3
Use permutations (1, t + 1) and (2t + 1, 2t + 2, . . . , n) to force
K1,n−2t+1 + {t + 1, 2t + 1} on {1, t + 1, 2t + 1, 2t + 2, . . . , n}.
Lemma 2: n ≤ 2t + 2, the case e = {1, 2t + 1}
1
t
3
2
···
B ⊇ T +e =
t+1
t+2
t+3
2t+1 2t+2
2t
···
n
n − 2t ≥ 3
Use permutations (1, t + 1) and (2t + 1, 2t + 2, . . . , n) to force
K1,n−2t+1 + {t + 1, 2t + 1} on {1, t + 1, 2t + 1, 2t + 2, . . . , n}.
The image of B under any permutation on {t + 1, 2t + 1, 2t + 2, . . . , n}
intersects B in at least t edges.
Lemma 2: n ≤ 2t + 2, the case e = {1, 2t + 1}
1
t
3
2
···
B ⊇ T +e =
t+1
t+2
t+3
2t+1 2t+2
2t
···
n
Complete
n − 2t ≥ 3
Use permutations (1, t + 1) and (2t + 1, 2t + 2, . . . , n) to force
K1,n−2t+1 + {t + 1, 2t + 1} on {1, t + 1, 2t + 1, 2t + 2, . . . , n}.
The image of B under any permutation on {t + 1, 2t + 1, 2t + 2, . . . , n}
intersects B in at least t edges.
Thus B contains the complete graph on {t + 1, 2t + 1, 2t + 2, . . . , n}.
Lemma 2: n ≤ 2t + 2, the case e = {1, 2t + 1}
1
B ⊇ T +e =
t+1
t
3
2
g
···
t+2
t+3
2t+1 2t+2
2t
···
n
Complete
n − 2t ≥ 3
Use permutations (1, t + 1) and (2t + 1, 2t + 2, . . . , n) to force
K1,n−2t+1 + {t + 1, 2t + 1} on {1, t + 1, 2t + 1, 2t + 2, . . . , n}.
The image of B under any permutation on {t + 1, 2t + 1, 2t + 2, . . . , n}
intersects B in at least t edges.
Thus B contains the complete graph on {t + 1, 2t + 1, 2t + 2, . . . , n}.
Now g = (2, t + 1) is such that g(B) ∩ B ≥ t, because there are at
least 2 edges on B among the vertices 2t + 1, 2t + 2, . . . , n.
Lemma 2: n ≤ 2t + 2, the case e = {1, 2t + 1}
1
B ⊇ T +e =
t+1
t
3
2
g
···
t+2
t+3
2t+1 2t+2
2t
···
n
Complete
n − 2t ≥ 3
Use permutations (1, t + 1) and (2t + 1, 2t + 2, . . . , n) to force
K1,n−2t+1 + {t + 1, 2t + 1} on {1, t + 1, 2t + 1, 2t + 2, . . . , n}.
The image of B under any permutation on {t + 1, 2t + 1, 2t + 2, . . . , n}
intersects B in at least t edges.
Thus B contains the complete graph on {t + 1, 2t + 1, 2t + 2, . . . , n}.
Now g = (2, t + 1) is such that g(B) ∩ B ≥ t, because there are at
least 2 edges on B among the vertices 2t + 1, 2t + 2, . . . , n.
Therefore B contains the edge {1, 2} and we reach a contradiction
via the previous case.
Lemma 2: n ≤ 2t + 2, the case e = {2t + 1, 2t + 2}
1
t
3
2
···
B ⊇ T +e =
t+1
t+2
t+3
2t+1 2t+2
2t
···
n − 2t ≥ 3
n
Lemma 2: n ≤ 2t + 2, the case e = {2t + 1, 2t + 2}
1
t
3
2
···
B ⊇ T +e =
t+1
t+2
t+3
2t+1 2t+2
2t
···
n
n − 2t ≥ 3
Permute the isolated points of T to force B complete on the vertices
{2t + 1, 2t + 2, . . . , n}
Lemma 2: n ≤ 2t + 2, the case e = {2t + 1, 2t + 2}
1
t
3
2
···
B ⊇ T +e =
t+1
t+2
t+3
2t+1 2t+2
2t
···
n
Kn−2t
n − 2t ≥ 3
Permute the isolated points of T to force B complete on the vertices
{2t + 1, 2t + 2, . . . , n}
Lemma 2: n ≤ 2t + 2, the case e = {2t + 1, 2t + 2}
1
B ⊇ T +e =
t+1
t
3
2
g
···
t+2
t+3
2t+1 2t+2
2t
···
n
Kn−2t
n − 2t ≥ 3
Permute the isolated points of T to force B complete on the vertices
{2t + 1, 2t + 2, . . . , n}
The permutation g = (2, t + 1) fixes at least t edges.
Lemma 2: n ≤ 2t + 2, the case e = {2t + 1, 2t + 2}
1
B ⊇ T +e =
t+1
t
3
2
g
···
t+2
t+3
2t+1 2t+2
2t
···
n
Kn−2t
n − 2t ≥ 3
Permute the isolated points of T to force B complete on the vertices
{2t + 1, 2t + 2, . . . , n}
The permutation g = (2, t + 1) fixes at least t edges.
So {1, 2} = g({1, t + 1}) ∈ B.
Lemma 2: n ≤ 2t + 2, the case e = {2t + 1, 2t + 2}
1
B ⊇ T +e =
t+1
t
3
2
g
···
t+2
t+3
2t+1 2t+2
2t
···
n
Kn−2t
n − 2t ≥ 3
Permute the isolated points of T to force B complete on the vertices
{2t + 1, 2t + 2, . . . , n}
The permutation g = (2, t + 1) fixes at least t edges.
So {1, 2} = g({1, t + 1}) ∈ B.
We reach a contradiction via the first case.
Lemma 3: If t ≥ 3, then n ≤ t + 2.
Proof.
Suppose n ≥ t + 3. we again use the general strategy.
I t = 3, n ≥ 6: Lemma 2
⇒ n ∈ {6, 7, 8}. Take T to be a
3-matching with n − 6 isolated T =
···
points. Use Key Lemma to
n−6≥t −3
force the unique block B
. . . . .containing
. . . . . . . . . . .T. . to
. . .be
. . .complete.
.............................................
t = 4, n ≥ 7: Take T to be a
triangle with a pendant edge
and n − 4 isolated points. Use T =
···
Key Lemma to force the unique
n−4≥3
block B containing T to be
complete.
.....................................................................
I
t ≥ 5, n ≥ t + 3: Take T to be
the cycle Ct with n−t isolated
···
T =
points. Use Key Lemma to
n−t ≥3
force the unique block B
containing
T
to
be
complete.
.....................................................................
I
Lemma 4:
The only graphical t-(v , K, 1) designs with t ≤ 4 are:
1-(6, 2, 1)
n=4
2-(15, 3, 1)
n=6
2-(15, {3, 5}, 1)
n=6
3-(10, 4, 1)
n=5
4-(15, {5, 7}, 1)
n=6
Proof. Define Nt ⊆ N by n ∈ Nt if there exist a graphical t-BD on Kn .
Lemmas 1,2,3 ⇒ N1 ⊆ {3, 4}, N2 ⊆ {4, 5, 6}, N3 ⊆ {5}, N4 ⊆ {6}.
Checking for (0, 1)-solutions U to
[At,t+1 , At,t+2 , . . . , At,n ]U = J, for n ∈ Nt
Only the designs on the above list are found.
Lemma 5:
If t ≥ 5, then n 6= t + 2.
Proof.
Suppose t ≥ 5, then Ct−2 is a cycle with t − 2 distinct edges.
Attemps to cover the t-edge subgraph
˙ 2 ∪K
˙ 2=
Ct−2 ∪K
Easily force B to be complete.
This is of course the usual contradiction.
Derived designs
Suppose that (X , B) is any t-(v , K, λ) design and consider a
subset S ⊂ X , where s = |S| < t. Define (X 0 , B 0 ) by
X0 = X \ S
B 0 = {B \ S : S ⊆ B ∈ B}
If T 0 ⊆ X 0 , has size |T 0 | = n − s, then T = T 0 ∪ S has size t.
Therefore T is contained in λ blocks B1 , B2 , . . . , Bλ ∈ B and
hence T 0 is contained in the the λ blocks
B1 \ S, B2 , \S . . . , Bλ \ S ∈ B 0 . Thus (X 0 , B 0 ) is a
(t − s)-(v − s, K0 , λ) design called the derived design with
respect to S. The next lemma provides a situation when the
derived with respect to a subgraph of a non-trivial graphical
design is again a non-trivial graphical design.
Lemma 6:
If n ≤ t and (X , B) is a non-trivial graphical t-BD, then the
derived design with respect to a K1,n−1 is a non-trivial graphical (t − n + 1)-BD.
Proof.
Let (X 0 , B 0 ) be the derived
design with respect to S a
fixed (labeled) K1,n .
n+1
1
2
3
n
Set X = X \ S = E(Kn−1 ) and B = {B \ S : S ⊆ B ∈ B}.
0
0
Then it is easy to see that (X 0 , B) is graphical.
X0
If (X 0 , B 0 ) is trivial, then k −n+1
⊆ B 0 for some block size k of (x, B).
and count the t − n + 1-subsets of X 0 contained in
Let w = n−1
2
these k − n + 1-element subsets in two ways to obtain:
w
k −n+1
w
w −t +n−1
=
⇒
= 1.
k −n+1
t −n+1
t −n+1
k −t
Thus k − t = 0 ⇒ k = t or k − t = w − t + n − 1 ⇒ k = n2 = |X |.
Contrary to (X , B) non-trivial.
Lemma 7:
If t ≥ 5, then graphical t-wise balanced designs do not exist.
Proof. If t ≥ 5, then n ≤ t by Lemmas 1,3, and 5.
Lemma 7:
If t ≥ 5, then graphical t-wise balanced designs do not exist.
Proof. If t ≥ 5, then n ≤ t by Lemmas 1,3, and 5.
So Lemma 6 applies.
Lemma 7:
If t ≥ 5, then graphical t-wise balanced designs do not exist.
Proof. If t ≥ 5, then n ≤ t by Lemmas 1,3, and 5.
So Lemma 6 applies.
Thus if (X , B) is a graphical t-BD with t ≥ 5, by infinite descent we
may assume the derived design with respect to a K1,n is on the list:
1-(6, 2, 1)
n=4
2-(15, 3, 1)
n=6
2-(15, {3, 5}, 1)
n=6
3-(10, 4, 1)
n=5
4-(15, {5, 7}, 1)
n=6
Lemma 7:
If t ≥ 5, then graphical t-wise balanced designs do not exist.
Proof. If t ≥ 5, then n ≤ t by Lemmas 1,3, and 5.
So Lemma 6 applies.
Thus if (X , B) is a graphical t-BD with t ≥ 5, by infinite descent we
may assume the derived design with respect to a K1,n is on the list:
1-(6, 2, 1)
n=4
2-(15, 3, 1)
n=6
2-(15, {3, 5}, 1)
n=6
3-(10, 4, 1)
n=5
4-(15, {5, 7}, 1)
n=6
Therefore we need only consider parameters:
(t, n) ∈ {(5, 5), (8, 7), (8, 6), (10, 7)}
t = 5, n = 5: Take T to be C5 =
.
Use Key Lemma to force the unique block B containing T to be
complete.
...............................................................
t = 8, n = 7: Take T =
. Use Key Lemma to force the
unique block B containing T to contain S = K1,6 =
Then B \ S contains
.
a triangle with a pendant edge.
But there is no such block among the two designs with
t = 8 − 6 = 2.
...............................................................
t = 8, n = 6: Take T =
. Use Key Lemma to force the
unique block B containing T to contain S = K1,5 =
Then B \ S contains
.
a triangle with a pendant edge.
But there is no such block among the unique designs with
t = 8 − 5 = 3.
t = 10, n = 7: Take T =
Deriving with respect to S =
leaves
The only block in the 4-BD that contains this is
Therefore the block B containing T contains
Use Key Lemma to force B to be complete.
This completes the proof of Main Theorem.
Bigraphical designs
A t–wise balanced design (X , B) of type t–(m · n, K, λ) is
bigraphical if X is the set of edges of the complete bipartite graph
Km,n and whenever B is a block and α is an automorphism of Km,n
(that fixes the independent sets), then α(B) is also a block.
Theorem (Hoffman-Kreher 1994) The only bigraphical t-designs of index 1 are:
D1 :
1−(mn, n, 1)
K1,n
(2 ≤ m ≤ n)
D2 :
1−(mn, m, 1)
Km,1
(2 ≤ m ≤ n)
D3 :
1−(4, 2, 1)
D4 :
2−(9, 3, 1)
D5 :
3−(8, 4, 1)
D6 :
3−(8, 4, 1)
D7 :
3−(16, 4, 1)
D8 :
3−(16, {4, 6}, 1)
D9 :
5−(16, {6, 8}, 1)
Remarks on the 5–(16, {6, 8}, 1) design.
Remarks on the 5–(16, {6, 8}, 1) design.
11
11
11
11
11
11
11
11
10
10
10
10
10
10
10
10
01
01
01
01
01
01
01
01
00
00
00
00
00
00
00
00
11
11
11
11
11
11
11
11
10
10
10
10
10
10
10
10
01
01
01
01
01
01
01
01
00
00
00
00
00
00
00
00
Remarks on the 5–(16, {6, 8}, 1) design.
11
11
11
11
11
11
11
11
10
10
10
10
10
10
10
10
01
01
01
01
01
01
01
01
00
00
00
00
00
00
00
00
11
11
11
11
11
11
11
11
10
10
10
10
10
10
10
10
01
01
01
01
01
01
01
01
00
00
00
00
00
00
00
00
I
The 8-element blocks are the 3–dimensional affine subspaces.
I
A 6-element set {x~1 , . . . , x~6 } is a block ⇐⇒ x~1 + · · · + x~6 = ~0.
Remarks on the 5–(16, {6, 8}, 1) design.
11
11
11
11
11
11
11
11
10
10
10
10
10
10
10
10
01
01
01
01
01
01
01
01
00
00
00
00
00
00
00
00
11
11
11
11
11
11
11
11
10
10
10
10
10
10
10
10
01
01
01
01
01
01
01
01
00
00
00
00
00
00
00
00
I
The 8-element blocks are the 3–dimensional affine subspaces.
I
A 6-element set {x~1 , . . . , x~6 } is a block ⇐⇒ x~1 + · · · + x~6 = ~0.
This vector space construction is due to R.M. Wilson and our labeling
shows that our bigraphical design is isomorphic to Wilson’s original
construction.
Multi-graphical designs
Knr = Kn, n, n, . . . , n
|
{z
}
r
.....................................................................
When n = 1 theses are the graphical designs.
There are 4 of them, with t≥2. (Chouinard, Kramer and Kreher 1983)
.....................................................................
When r = 2 theses are bigraphical designs.
There are 7 of them, with t ≥ 2.
(Hoffman and Kreher 1994)
.....................................................................
When n > 1 and r > 2 there are 2 more.
(Olsen and Kreher 1998)
n = 2, r = 3
2−(12, {3, 4}, 1)
n = 2, r = 4
2−(24, {3, 4}, 1)
Research problems
Research problem 3. Investigate other interesting families of group
actions and the t-wise balanced that can be constructed from them.
For example consider the action of S YM(X ) on the 3-element subsets
of X .
.....................................................................
Chouinard has shown that for any pair (t,
λ) with
t > 1 or λ odd, there
cannot exist a non-trivial graphical t- n2 , K, λ design with
n2t + λ + 4. Thus, in particular, for each such pair (t, λ) there are only
a finite number of non-trivial graphical t-(v , K, λ) designs. He also
shows that if we further assume no repeated blocks, then for all cases
with t > 1 or λ 6= 2, there do not exist non-trivial graphical t- n2 , K, λ
designs with n ≥ 2t + λ + 4. This suggest the following problem.
Research problem 4. Consider a parameterized family of graphs
Xi , i = 1, 2, 3, . . ., Let Gi = AUT(Xi ) the automorphism of Xi acting on
E(Xi ) the edges of Xi and let C be the set of all t-wise balanced
designs (E(Xi ), B) that have Gi as an automorphism group, for some
i. Find necessary and sufficient conditions on t and λ (or just λ) for
when |C| is finite.
Groups, Designs and Linear Algebra:
The Incidence Algebra
Donald L. Kreher
Department of Mathematical Sciences
Michigan Technological University
CIMPA School
July-August 2013
Review: Orbit incidence Matrices
Given group G ≤ S YM(X ) and 0 ≤ t ≤ k ≤ v = |X | let
R = {∆
1 , ∆2 , . . . , ∆Nt } be the orbits of t-element subsets;
C =
Γ1 , Γ2 , . . . , ΓNk
be the orbits of k-element subsets.
The (t, k)-orbit incidence matrix is the matrix
Atk : R × C → Z
given by
Atk [∆, Γ] = |{K ∈ Γ : K ⊇ T }|,
where T ∈ ∆ is a fixed representative.
Review: Orbit incidence matrix


I, (0, 3, 2)(1, 4, 6),

Let G=h(0, 3, 2)(1, 4, 6), (0, 2, 3)(1, 6, 4)i = (0, 2, 3)(1, 6, 4), (1, 4)(2, 3), .


(0, 2)(4, 6), (0, 3)(1, 6)

046
016
134
014 136
346 246
146 126 124
036
045
026
356
123
125
013 034
126 015 056 035
024 236
145 345 135 235
234 012 023 456 256 245 025
{46, 16, 14}
36, 34, 26
12, 04, 01
1
1
2
0
0
0
1
0
0
0
0
1
1
1
1
0
0
1
0
0
{24, 13, 06}
0
0
2
2
0
0
0
0
1
0
{23, 03, 02}
0
0
0
2
1
1
0
0
0
1
{56, 45, 15}
0
0
0
0
0
0
2
2
1
0
{35, 25, 05}
0
0
0
0
0
0
0
2
1
2
Matrices.
I think of a matrix M with
I
rows labeled by a set R
I
columns labeled by a set C and
I
entries in a set E
as function
M :R×C →E
that is an E valued vector with coordinates R × C.
I call these R by C matrices.
Counting relations.
Recall that the adjacency matrix of a graph (V , E) is the V by V
matrix
1 if x is adjacent to y;
A[x, y ] =
0 if not.
I
A` [x, y] = the number of walks of length ` from x to y .
Let ρ ⊆ S × T and define the S by T matrix Mρ : S × T → Z by
1 if xρy;
Mρ [x, y ] =
0 if not.
For example if ρ is adjacency for the graph (V , E). Then Mρ = A.
I
If ρ ⊆ S × T and σ ⊆ T × U, then Mρ Mσ [x, y ] = |{t ∈ T : xρtσy }|
I
If ρi ⊆ Ti × Ti+1 , i = 1, 2, . . . , `, then
Mρ1 Mρ2 · · ·Mρ` [x, y]=|{(t1 , . . ., t`−1 )∈T1 ×· · ·×T`−1 :xρ1 t1 ρ2 · · ·t`−1 ρ` y }|
Group actions
Group actions
? S YM(X ) is the symmetric group on the set X .
Group actions
? S YM(X ) is the symmetric group on the set X .
? If g ∈ S YM(X ), and x ∈ X , then x g is the image of x under g.
Group actions
? S YM(X ) is the symmetric group on the set X .
? If g ∈ S YM(X ), and x ∈ X , then x g is the image of x under g.
? If G ≤ S YM(X ), then G acts on the subsets of X .
• For S ⊆ X and g ∈ G: S g = {x g : x ∈ S}.
Group actions
? S YM(X ) is the symmetric group on the set X .
? If g ∈ S YM(X ), and x ∈ X , then x g is the image of x under g.
? If G ≤ S YM(X ), then G acts on the subsets of X .
• For S ⊆ X and g ∈ G: S g = {x g : x ∈ S}.
? If G acts on a set Ω we write G|Ω.
Group actions
? S YM(X ) is the symmetric group on the set X .
? If g ∈ S YM(X ), and x ∈ X , then x g is the image of x under g.
? If G ≤ S YM(X ), then G acts on the subsets of X .
• For S ⊆ X and g ∈ G: S g = {x g : x ∈ S}.
? If G acts on a set Ω we write G|Ω.
? If G|Ω and ω ∈ Ω, then
• ω G = {ω g : g ∈ G} is the orbit of ω under (the action of) G;
• Gω = {g ∈ G : ω g = ω} is the stabilizer of ω in (the action of) G;
• Orbit counting lemma: |ω G | |Gω | = |G|.
Group actions
? S YM(X ) is the symmetric group on the set X .
? If g ∈ S YM(X ), and x ∈ X , then x g is the image of x under g.
? If G ≤ S YM(X ), then G acts on the subsets of X .
• For S ⊆ X and g ∈ G: S g = {x g : x ∈ S}.
? If G acts on a set Ω we write G|Ω.
? If G|Ω and ω ∈ Ω, then
• ω G = {ω g : g ∈ G} is the orbit of ω under (the action of) G;
• Gω = {g ∈ G : ω g = ω} is the stabilizer of ω in (the action of) G;
• Orbit counting lemma: |ω G | |Gω | = |G|.
? The set of (G|Ω)-orbits is Ω/G.
• Cauchy-Frobenius-Burnside Lemma: |Ω/G| =
1 X
χ(g),
|G|
g∈G
where χ(g) = |{ω ∈ Ω : ω g = ω}|.
Group actions
? S YM(X ) is the symmetric group on the set X .
? If g ∈ S YM(X ), and x ∈ X , then x g is the image of x under g.
? If G ≤ S YM(X ), then G acts on the subsets of X .
• For S ⊆ X and g ∈ G: S g = {x g : x ∈ S}.
? If G acts on a set Ω we write G|Ω.
? If G|Ω and ω ∈ Ω, then
• ω G = {ω g : g ∈ G} is the orbit of ω under (the action of) G;
• Gω = {g ∈ G : ω g = ω} is the stabilizer of ω in (the action of) G;
• Orbit counting lemma: |ω G | |Gω | = |G|.
? The set of (G|Ω)-orbits is Ω/G.
• Cauchy-Frobenius-Burnside Lemma: |Ω/G| =
1 X
χ(g),
|G|
g∈G
where χ(g) = |{ω ∈ Ω : ω g = ω}|.
? If G ≤ S YM(X ),
• G| Xt , the t-element subsets of X , t ≤ |X |.
• G|P(X ), the power set of X .
Algebra of G-invariant matrices.
Algebra of G-invariant matrices.
? M ATQ (Ω) is all Ω by Ω matrices with entries in the field Q.
Algebra of G-invariant matrices.
? M ATQ (Ω) is all Ω by Ω matrices with entries in the field Q.
? G ≤ S YM(X ), then G acts on M ATQ (P(X )).
• M g [S, T ] = M[S g , T g ].
Algebra of G-invariant matrices.
? M ATQ (Ω) is all Ω by Ω matrices with entries in the field Q.
? G ≤ S YM(X ), then G acts on M ATQ (P(X )).
• M g [S, T ] = M[S g , T g ].
? A LGQ (G|X ) = {M ∈ M ATQ (P(X )) : M g = M, for all g ∈ G}.
Algebra of G-invariant matrices.
? M ATQ (Ω) is all Ω by Ω matrices with entries in the field Q.
? G ≤ S YM(X ), then G acts on M ATQ (P(X )).
• M g [S, T ] = M[S g , T g ].
? A LGQ (G|X ) = {M ∈ M ATQ (P(X )) : M g = M, for all g ∈ G}.
? If M ∈ M ATQ (P(X )/G), then
Mtk is the projection of M onto the
X
t
/G ×
X
k
/G coordinates.
Algebra of G-invariant matrices.
? M ATQ (Ω) is all Ω by Ω matrices with entries in the field Q.
? G ≤ S YM(X ), then G acts on M ATQ (P(X )).
• M g [S, T ] = M[S g , T g ].
? A LGQ (G|X ) = {M ∈ M ATQ (P(X )) : M g = M, for all g ∈ G}.
? If M ∈ M ATQ (P(X )/G), then
Mtk is the projection of M onto the Xt /G × Xk /G coordinates.
? Any Xt /G by Xk /G matrix M can be embedded into M ATQ (P(X ))
by padding with zeros.
X
k /G
g
M
tk =
X
t
/G
0
0
0
0
Mtk
0
0
0
0
P(X )/G
P(X )/G
Special Matrices
Let G ≤ S YM(X ).
Define P(X )/G by P(X )/G matrices A, B and D by
Special Matrices
Let G ≤ S YM(X ).
Define P(X )/G by P(X )/G matrices A, B and D by
? A[∆, Γ] = |{K ∈ Γ : K ⊃ T0 }|, T0 is a fixed representative of orbit ∆.
Special Matrices
Let G ≤ S YM(X ).
Define P(X )/G by P(X )/G matrices A, B and D by
? A[∆, Γ] = |{K ∈ Γ : K ⊃ T0 }|, T0 is a fixed representative of orbit ∆.
Notice that Atk is our old friend the orbit incidence matrix.
Special Matrices
Let G ≤ S YM(X ).
Define P(X )/G by P(X )/G matrices A, B and D by
? A[∆, Γ] = |{K ∈ Γ : K ⊃ T0 }|, T0 is a fixed representative of orbit ∆.
Notice that Atk is our old friend the orbit incidence matrix.
? B[∆, Γ] = |{T ∈ ∆ : T ⊂ K0 }|, K0 is a fixed representative of orbit Γ.
Special Matrices
Let G ≤ S YM(X ).
Define P(X )/G by P(X )/G matrices A, B and D by
? A[∆, Γ] = |{K ∈ Γ : K ⊃ T0 }|, T0 is a fixed representative of orbit ∆.
Notice that Atk is our old friend the orbit incidence matrix.
? B[∆, Γ] = |{T ∈ ∆ : T ⊂ K0 }|, K0 is a fixed representative of orbit Γ.
|∆| if ∆ = Γ,
? D[∆, Γ] =
.
0 if not
Special Matrices
Let G ≤ S YM(X ).
Define P(X )/G by P(X )/G matrices A, B and D by
? A[∆, Γ] = |{K ∈ Γ : K ⊃ T0 }|, T0 is a fixed representative of orbit ∆.
Notice that Atk is our old friend the orbit incidence matrix.
? B[∆, Γ] = |{T ∈ ∆ : T ⊂ K0 }|, K0 is a fixed representative of orbit Γ.
|∆| if ∆ = Γ,
? D[∆, Γ] =
.
0 if not
To emphasize the dependency of A, B and D on the group action G|X
we may write A(G|X ), B(G|X ) and D(G|X ).
Special Matrices
Let G ≤ S YM(X ).
Define P(X )/G by P(X )/G matrices A, B and D by
? A[∆, Γ] = |{K ∈ Γ : K ⊃ T0 }|, T0 is a fixed representative of orbit ∆.
Notice that Atk is our old friend the orbit incidence matrix.
? B[∆, Γ] = |{T ∈ ∆ : T ⊂ K0 }|, K0 is a fixed representative of orbit Γ.
|∆| if ∆ = Γ,
? D[∆, Γ] =
.
0 if not
To emphasize the dependency of A, B and D on the group action G|X
we may write A(G|X ), B(G|X ) and D(G|X ).
Lemma 1. A(1|X ) = B(1|X ) ∈ A LGQ (G|X )
Special Matrices
Let G ≤ S YM(X ).
Define P(X )/G by P(X )/G matrices A, B and D by
? A[∆, Γ] = |{K ∈ Γ : K ⊃ T0 }|, T0 is a fixed representative of orbit ∆.
Notice that Atk is our old friend the orbit incidence matrix.
? B[∆, Γ] = |{T ∈ ∆ : T ⊂ K0 }|, K0 is a fixed representative of orbit Γ.
|∆| if ∆ = Γ,
? D[∆, Γ] =
.
0 if not
To emphasize the dependency of A, B and D on the group action G|X
we may write A(G|X ), B(G|X ) and D(G|X ).
Lemma 1. A(1|X ) = B(1|X ) ∈ A LGQ (G|X )
Proof T ⊂ K ⇒ T g ⊂ K g , for all g ∈ G.
Special Matrices
Let G ≤ S YM(X ).
Define P(X )/G by P(X )/G matrices A, B and D by
? A[∆, Γ] = |{K ∈ Γ : K ⊃ T0 }|, T0 is a fixed representative of orbit ∆.
Notice that Atk is our old friend the orbit incidence matrix.
? B[∆, Γ] = |{T ∈ ∆ : T ⊂ K0 }|, K0 is a fixed representative of orbit Γ.
|∆| if ∆ = Γ,
? D[∆, Γ] =
.
0 if not
To emphasize the dependency of A, B and D on the group action G|X
we may write A(G|X ), B(G|X ) and D(G|X ).
Lemma 1. A(1|X ) = B(1|X ) ∈ A LGQ (G|X )
Proof T ⊂ K ⇒ T g ⊂ K g , for all g ∈ G.
? W = A(1|X ) = B(1|X ), will denote this special matrix.
A and B are dual
Remark: W [T , K ] =
1
0
if T ⊆ K
if not
A and B are dual
Remark: W [T , K ] =
Lemma 2. BD = DA.
1
0
if T ⊆ K
if not
A and B are dual
Remark: W [T , K ] =
Lemma 2. BD = DA.
Proof.
1
0
if T ⊆ K
if not
A and B are dual
Remark: W [T , K ] =
1
0
if T ⊆ K
if not
Lemma 2. BD = DA.
Suppose G|X and let ∆, Γ ∈ P(X )/G. Then
X
(BD)[∆, Γ] =
B[∆, Φ]D[Φ, Γ] = B[∆, Γ]|Γ|
Proof.
Φ∈P(X )/G
A and B are dual
Remark: W [T , K ] =
1
0
if T ⊆ K
if not
Lemma 2. BD = DA.
Suppose G|X and let ∆, Γ ∈ P(X )/G. Then
X
(BD)[∆, Γ] =
B[∆, Φ]D[Φ, Γ] = B[∆, Γ]|Γ|
Proof.
Φ∈P(X )/G
=
X
K ∈Γ
|{T ∈ ∆ : T ⊆ K }|
A and B are dual
Remark: W [T , K ] =
1
0
if T ⊆ K
if not
Lemma 2. BD = DA.
Suppose G|X and let ∆, Γ ∈ P(X )/G. Then
X
(BD)[∆, Γ] =
B[∆, Φ]D[Φ, Γ] = B[∆, Γ]|Γ|
Proof.
Φ∈P(X )/G
=
X
|{T ∈ ∆ : T ⊆ K }|
K ∈Γ
=
XX
K ∈Γ T ∈∆
W [T , K ] =
XX
T ∈∆ K ∈Γ
W [T , K ]
A and B are dual
1
0
Remark: W [T , K ] =
if T ⊆ K
if not
Lemma 2. BD = DA.
Suppose G|X and let ∆, Γ ∈ P(X )/G. Then
X
(BD)[∆, Γ] =
B[∆, Φ]D[Φ, Γ] = B[∆, Γ]|Γ|
Proof.
Φ∈P(X )/G
=
X
|{T ∈ ∆ : T ⊆ K }|
K ∈Γ
=
XX
K ∈Γ T ∈∆
=
X
T ∈∆
W [T , K ] =
XX
T ∈∆ K ∈Γ
|{K ∈ Γ : K ⊇ T }|
W [T , K ]
A and B are dual
1
0
Remark: W [T , K ] =
if T ⊆ K
if not
Lemma 2. BD = DA.
Suppose G|X and let ∆, Γ ∈ P(X )/G. Then
X
(BD)[∆, Γ] =
B[∆, Φ]D[Φ, Γ] = B[∆, Γ]|Γ|
Proof.
Φ∈P(X )/G
=
X
|{T ∈ ∆ : T ⊆ K }|
K ∈Γ
=
XX
W [T , K ] =
K ∈Γ T ∈∆
=
X
XX
W [T , K ]
T ∈∆ K ∈Γ
|{K ∈ Γ : K ⊇ T }|
T ∈∆
=
|∆|A[∆, Γ] =
X
D[∆, Φ]A[Φ, Γ] = DA[∆, Γ]
Φ∈P(X )/G
Fusion.
The Fusion matrix for the group action G|X is the P(X )by P(X )/G
matrix
( 1
√
if S ∈ ∆
|∆|
F [S, ∆] =
0
if not
Fusion.
The Fusion matrix for the group action G|X is the P(X )by P(X )/G
matrix
( 1
√
if S ∈ ∆
|∆|
F [S, ∆] =
0
if not
Lemma 3. FF T is in the center of A LGQ (G|X ).
Fusion.
The Fusion matrix for the group action G|X is the P(X )by P(X )/G
matrix
( 1
√
if S ∈ ∆
|∆|
F [S, ∆] =
0
if not
Lemma 3. FF T is in the center of A LGQ (G|X ).
Proof. Let M ∈ A LGQ (G|X ) and let S, T ∈ P(X ). Then
Fusion.
The Fusion matrix for the group action G|X is the P(X )by P(X )/G
matrix
( 1
√
if S ∈ ∆
|∆|
F [S, ∆] =
0
if not
Lemma 3. FF T is in the center of A LGQ (G|X ).
Proof. Let M ∈ A LGQ (G|X ) and let S, T ∈ P(X ). Then
X
X
(MFF )[S, T ]=
M[S, U]F [U, ∆]F [T , ∆]
T
U∈P(X ) ∆∈P(X )/G
Fusion.
The Fusion matrix for the group action G|X is the P(X )by P(X )/G
matrix
( 1
√
if S ∈ ∆
|∆|
F [S, ∆] =
0
if not
Lemma 3. FF T is in the center of A LGQ (G|X ).
Proof. Let M ∈ A LGQ (G|X ) and let S, T ∈ P(X ). Then
X
X
X M[S, U]
(MFF )[S, T ]=
M[S, U]F [U, ∆]F [T , ∆] =
|T G |
G
T
U∈P(X ) ∆∈P(X )/G
U∈T
Fusion.
The Fusion matrix for the group action G|X is the P(X )by P(X )/G
matrix
( 1
√
if S ∈ ∆
|∆|
F [S, ∆] =
0
if not
Lemma 3. FF T is in the center of A LGQ (G|X ).
Proof. Let M ∈ A LGQ (G|X ) and let S, T ∈ P(X ). Then
X
X
X M[S, U]
(MFF )[S, T ]=
M[S, U]F [U, ∆]F [T , ∆] =
|T G |
U∈P(X ) ∆∈P(X )/G
U∈T G
1 X
1 X M[S, T g ]
1 X
=
M[S, T g ]
= G
M[S, U] = G
|G
|
|G|
|T |
|T
|
T
g∈G
g∈G
U∈T G
X
X M[S g , T ]
−1
1 X
1
1
=
M[S g , T ] =
M[S g , T ] = G
|G|
|G|
|GS |
|S |
g∈G
g∈G
g∈G
X M[U, T ]
1 X
= G
M[U, T ] =
|S |
|S G |
U∈S G
XU∈S G X
=
F [S, ∆]F [U, ∆]M[U, T ] = (FF T M)[S, T ]
T
∆∈P(X )/G U∈P(X )
Fusion part 2.
Lemma 4. F T F = I in M ATQ (P(X )/G).
Fusion part 2.
Lemma 4. F T F = I in M ATQ (P(X )/G).
Proof.
Let ∆, Γ ∈ P(X )/G. Then
X
FTF [∆, Γ] =
U∈P(X )
Hence F T F [∆, Γ] = 0 if ∆ 6= Γ.
F [U, ∆]F [U, Γ].
Fusion part 2.
Lemma 4. F T F = I in M ATQ (P(X )/G).
Proof.
Let ∆, Γ ∈ P(X )/G. Then
X
FTF [∆, Γ] =
F [U, ∆]F [U, Γ].
U∈P(X )
Hence F T F [∆, Γ] = 0 if ∆ 6= Γ.
Otherwise
F T F [∆, ∆] =
X
U∈P(X )
(F [U, ∆])2 =
X 1
1 2
=1
(p
) =
|∆|
|∆|
U∈∆
U∈∆
X
A homomorphism.
Let G|X and define τ : A LGQ (G|X ) → M ATQ (P(X )/G) by
τ (M) = D −1/2 F T MFD 1/2
Lemma 5. τ is an homomorphism.
Proof. For all x, y ∈ Q and M, N ∈ A LGQ (G|X ).
It is easy to see that xτ (M) + y τ (N) = τ (xM + yN), and by the
Fusion Lemmas
τ (MN)
= D −1 F T MNFD 1/2 = D −1 F T MNFF T FD 1/2
= D −1 F T MFF T NFD 1/2 = D −1 F T MFD 1/2 D −1/2 F T NFD 1/2
= τ (M)τ (N)
Therefor τ is a homomorphism of algebras.
A homomorphism.
Let G|X and define τ : A LGQ (G|X ) → M ATQ (P(X )/G) by
τ (M) = D −1/2 F T MFD 1/2
Lemma 6. τ is an homomorphism.
Proof. For all x, y ∈ Q and M, N ∈ A LGQ (G|X ).
It is easy to see that xτ (M) + y τ (N) = τ (xM + yN), and by the
Fusion Lemmas
τ (MN)
= D −1 F T MNFD 1/2 = D −1 F T MNFF T FD 1/2
= D −1 F T MFF T NFD 1/2 = D −1 F T MFD 1/2 D −1/2 F T NFD 1/2
= τ (M)τ (N)
Therefor τ is a homomorphism of algebras.
Epimorphism.
Lemma 7. τ is an epimorphism.
Epimorphism.
Lemma 7. τ is an epimorphism.
Proof. Given M ∈ M ATQ (P(X )/G) define N ∈ A LGQ (G|X ) by
N[S, T ] = M[S G , T G ]/|T G |,
for all S, T ⊆ X .
Epimorphism.
Lemma 7. τ is an epimorphism.
Proof. Given M ∈ M ATQ (P(X )/G) define N ∈ A LGQ (G|X ) by
N[S, T ] = M[S G , T G ]/|T G |,
for all S, T ⊆ X .
Then if ∆, Γ ∈ P(X )/G we have
τ (N)[∆, Γ]
(D −1/2 F T NFD 1/2 )[∆, Γ]
1/2 X
X
|Γ|
F [U, ∆]N[U, T ]F [T , Γ]
=
|∆|
=
U∈P(X ) T ∈P(X )
=
1/2 X X
|Γ|
|∆|
U∈∆ T ∈Γ
1
1
N[U, T ] 1/2
|∆|1/2
|Γ|
=
1 XX
N[U, T ]
|∆|
=
1 X X M[∆, Γ]
= M[∆, Γ]
|∆|
|Γ|
U∈∆ T ∈Γ
U∈∆ T ∈Γ
Property (i): τ (W ) = A
Proof.
Let ∆, Γ ∈ P(X )/G. Then
τ (W )[∆, Γ]
|Γ|
|∆|
1/2
|Γ|
|∆|
1/2
=
=
(F T WF )[∆, Γ]
X
X
F [T , ∆]W [T , K ]F [K , Γ]
T ∈P(X ) K ∈P(X )
=
1 XX
W [T , K ]
|∆|
=
1 X
1 X
|{K ∈ Γ : K ⊃ T }| =
A[∆, Γ] = A[∆, Γ]
|∆|
|∆|
T ∈∆ K ∈Γ
T ∈∆
g
f
Thus τ (W
tk ) = Atk .
T ∈∆
Property (ii): τ (W T ) = B T
Proof.
Let Γ, ∆ ∈ P(X )/G. Then
T
τ (W )[Γ, ∆]
|∆|
|Γ|
1/2
|∆|
|Γ|
1/2
=
=
(F T W T F )[Γ, ∆]
X
X
F [T , Γ]W T [T , K ]F [K , ∆]
T ∈P(X ) K ∈P(X )
=
1 XX T
W [T , K ]
|Γ|
=
1 X
1 X T
|{K ∈ ∆ : K ⊃ T }| =
B [Γ, ∆] = B T [Γ, ∆]
|Γ|
|Γ|
T ∈Γ K ∈∆
T ∈Γ
gT ) = B
fT .
Thus τ (W
tk
tk
T ∈Γ
Summary
The mapping τ is a A LGQ (G|X ) → M ATQ (P(X )/G)
epimorphism that carries equations in the Wtk and WtkT
T matrices.
matrices to equations in the Atk and Btk
Application 1
Let |X | = v and suppose 0 ≤ t ≤ k ≤ ` ≤ v .
X
Wtk Wk ,` [T , L] = K ∈
: T ⊆ K ⊂ L k
`−t if T ⊆ L
k −t
=
0
if not
`−t
=
Wt,` [T , L]
k −t
Application 1
Let |X | = v and suppose 0 ≤ t ≤ k ≤ ` ≤ v .
X
Wtk Wk ,` [T , L] = K ∈
: T ⊆ K ⊂ L k
`−t if T ⊆ L
k −t
=
0
if not
`−t
=
Wt,` [T , L]
k −t
`−t
Therefor
Wtk Wk,` =
Wt,` .
k −t
Application 1
Let |X | = v and suppose 0 ≤ t ≤ k ≤ ` ≤ v .
X
Wtk Wk ,` [T , L] = K ∈
: T ⊆ K ⊂ L k
`−t if T ⊆ L
k −t
=
0
if not
`−t
=
Wt,` [T , L]
k −t
`−t
Therefor
Wtk Wk,` =
Wt,` .
k −t
Consequently
Atk Ak,`
=
`−t
At,` .
k −t
Application 2
Application 2
Let (X , B) be a t-(v , k, λ) design.
Application 2
Let (X , B) be a t-(v , k, λ) design.
Let Wt,B be the Xt × B sub-matrix of W .
Application 2
Let (X , B) be a t-(v , k, λ) design.
Let Wt,B be the Xt × B sub-matrix of W .
Wilson in 1982 gave an inclusion-exclusion proof establishing.
We,B WfT,B =
min
e,f
X
i
T
be+f
−i Wi,e Wif
i=0
when e + f ≤ t and
bji
=λ
v −i−j k−i
v −t
k−t
Consequently
Ae,B BfT,B
=
min
e,f
X
i=0
when e + f ≤ t and
bji
=λ
v −i−j k−i
v −t
k−t
i
T
be+f
−i Bi,e Aif
An inequality.
At e = f = ` ≤ t/2 this equation becomes:
T
A`,B B`,B
=
`
X
i
T
b2`−i
Bi,`
Ai,`
i=0
i
T A and
this is a sum of positive semi-definite matrices b2`−i
Bi,`
i,`
hence has rank greater than the
rank
of
any
one
of
them.
In
X
`
T
/G.
particular R ANK(b` B`` A`` ) is `
An inequality.
At e = f = ` ≤ t/2 this equation becomes:
T
A`,B B`,B
=
`
X
i
T
b2`−i
Bi,`
Ai,`
i=0
i
T A and
this is a sum of positive semi-definite matrices b2`−i
Bi,`
i,`
hence has rank greater than the
rank
of
any
one
of
them.
In
X
`
T
/G.
particular R ANK(b` B`` A`` ) is `
Theorem 8. (Kreher 1986) A t-(v , k, λ) design (X , B) with
G ≤ S YM(X ) as an automorphism group, with t ≥ 2` and
v ≥ k + ` satisfies
X
|B/G| ≥ /G
`
An inequality.
At e = f = ` ≤ t/2 this equation becomes:
T
A`,B B`,B
=
`
X
i
T
b2`−i
Bi,`
Ai,`
i=0
i
T A and
this is a sum of positive semi-definite matrices b2`−i
Bi,`
i,`
hence has rank greater than the
rank
of
any
one
of
them.
In
X
`
T
/G.
particular R ANK(b` B`` A`` ) is `
Theorem 8. (Kreher 1986) A t-(v , k, λ) design (X , B) with
G ≤ S YM(X ) as an automorphism group, with t ≥ 2` and
v ≥ k + ` satisfies
X
|B/G| ≥ /G
`
”the number of orbits of blocks is at least the number of orbits of
`-element subsets.”
Simple consequences.
Theorem 9. (Kreher 1986) A t-(v , k, λ) design (X , B) with
G ≤ S YM(X ) as an automorphism group, with t ≥ 2` and v ≥ k + `
satisfies
X
|B/G| ≥ /G
`
”the number of orbits of blocks is at least the number of orbits of
`-element subsets.”
Simple consequences.
Theorem 9. (Kreher 1986) A t-(v , k, λ) design (X , B) with
G ≤ S YM(X ) as an automorphism group, with t ≥ 2` and v ≥ k + `
satisfies
X
|B/G| ≥ /G
`
”the number of orbits of blocks is at least the number of orbits of
`-element subsets.”
If t = 2 and ` = 1 and G = {I} the trivial group. we see that this
theorem implies
”the number of blocks is at least the number points.”
This is Fisher’s inequality.
Simple consequences.
Theorem 9. (Kreher 1986) A t-(v , k, λ) design (X , B) with
G ≤ S YM(X ) as an automorphism group, with t ≥ 2` and v ≥ k + `
satisfies
X
|B/G| ≥ /G
`
”the number of orbits of blocks is at least the number of orbits of
`-element subsets.”
If t = 2 and ` = 1 and G = {I} the trivial group. we see that this
theorem implies
”the number of blocks is at least the number points.”
This is Fisher’s inequality.
A t-(v , k, lambda) design is said to be block transitive if it has an
automorphims group that is transitive on blocks. Theorem 8 yields.
Simple consequences.
Theorem 9. (Kreher 1986) A t-(v , k, λ) design (X , B) with
G ≤ S YM(X ) as an automorphism group, with t ≥ 2` and v ≥ k + `
satisfies
X
|B/G| ≥ /G
`
”the number of orbits of blocks is at least the number of orbits of
`-element subsets.”
If t = 2 and ` = 1 and G = {I} the trivial group. we see that this
theorem implies
”the number of blocks is at least the number points.”
This is Fisher’s inequality.
A t-(v , k, lambda) design is said to be block transitive if it has an
automorphims group that is transitive on blocks. Theorem 8 yields.
Theorem 10. Every block transitive t-(v , k , λ) design is
`-homogeneous for ` = bt/2c.
Simple consequences.
Theorem 9. (Kreher 1986) A t-(v , k, λ) design (X , B) with
G ≤ S YM(X ) as an automorphism group, with t ≥ 2` and v ≥ k + `
satisfies
X
|B/G| ≥ /G
`
”the number of orbits of blocks is at least the number of orbits of
`-element subsets.”
If t = 2 and ` = 1 and G = {I} the trivial group. we see that this
theorem implies
”the number of blocks is at least the number points.”
This is Fisher’s inequality.
A t-(v , k, lambda) design is said to be block transitive if it has an
automorphims group that is transitive on blocks. Theorem 8 yields.
Theorem 10. Every block transitive t-(v , k , λ) design is
`-homogeneous for ` = bt/2c.
There are no 6-homogeneous groups so therefor no block transitive
12-designs.
Conjecture.
Conjecture.
A transitive group action G|X in which Gx has equal size orbits
on X \ {x} is said to be 23 -transitive.
Conjecture.
A transitive group action G|X in which Gx has equal size orbits
on X \ {x} is said to be 23 -transitive.
The Theorem shows that a block transitive 3-designs is
transitive on `-subsets for ` ≤ 32 .
Conjecture.
A transitive group action G|X in which Gx has equal size orbits
on X \ {x} is said to be 23 -transitive.
The Theorem shows that a block transitive 3-designs is
transitive on `-subsets for ` ≤ 32 .
Research problem 5. Prove the following conjecture:
Every block transitive 3-design is 32 -transitive.
Groups, Designs and Linear Algebra:
The maximum size of a block
Donald L. Kreher
Department of Mathematical Sciences
Michigan Technological University
CIMPA School
July-August 2013
tBD review
A t–wise balanced design (tBD) of type t–(v , K, λ) is a pair (X , B)
I
X is a v –element set of points
I
B is a collection of subsets of X called blocks
I
B ∈ B ⇒ |B| ∈ K
I
every t–element subset of X is in exactly λ blocks.
proper if t < k < v for each k ∈ K
Steiner if λ = 1
Example: A proper 2BD of type 2–(9, {3, 4}, 4).
X ={a, b, c, d, 1, 2, 3, 4, 5}



abcd,a12,a13,a14,a15,a23,a24,a25,a34,a35,a45,



abcd,b12,b13,b14,b15,b23,b24,b25,b34,b35,b45,
B=
abcd,c12,c13,c14,c15,c23,c24,c25,c34,c35,c45,





abcd,d12,d13,d14,d15,d23,d24,d25,d34,d35, d45
B={xij : x ∈ {a, b, c, d}, i, j ∈ {1, 2, . . . , 5}}
∪{abcd, abcd, abcd, abcd}
Example
A 2BD of type 2-(9, {3, 4}, 4)
1
a
2
b
c
3
4
d
5
Repeat 4 times.
History
1983 E.S. Kramer, Some results on t-wise balanced designs, Ars
Combin. 15 (1983), 179–192.
I If B is a block in a Steiner tBD, then
|B|
I
I
≤ (v + t − 3)/2 for t ≥ 2.
If B is a block in a Steiner tBD, then
|B|
≤
(v − 1)/2 for t = 2, 4 while
|B|
≤
v /2 for t = 3, 5.
Conjecture: If B is a block in a Steiner tBD, then
|B|
≤ (v − 1)/2 for t even while
|B|
≤ v /2 for t odd.
2000 M. Ira and E.S. Kramer, A block-size bound for Steiner 6-wise
balanced designs, J. Combin. Designs, 8 (2000), 141-145.
I If B is a block in a Steiner 6BD, then
|B|
≤ v /2.
ItBD
An incomplete t–wise balanced design (ItBD) of type t–(v , h, K, λ) is
a triple (X , H, B)
I
X is a v –element set of points
I
H is an h–element set of points called the hole
I
B is a collection of subsets of X called blocks
I
B ∈ B ⇒ |B| ∈ K
I
every t–element subset of X is either in the hole or in exactly λ
blocks, but not both.
Example: A proper I2BD of type 2–(9, 4, {3}, 4).
X ={a, b, c, d, 1, 2, 3, 4, 5}
H={a, b, c, d}



 a12,a13,a14,a15,a23,a24,a25,a34,a35,a45,



b12,b13,b14,b15,b23,b24,b25,b34,b35,b45,
B=

 c12,c13,c14,c15,c23,c24,c25,c34,c35,c45,



d12,d13,d14,d15,d23,d24,d25,d34,d35, d45
Examples
I
A 2BD of type 2–(9, {3, 4}, 4)
1
a
2
b
c
3
d
5
4
Repeat 4 times.
I
A I2BD of type 2–(9, 4, {3}, 4)
1
a
2
b
c
d
The hole.
3
5
4
Equivalence
A ItBD of type t–(v , h, K, λ) is the same as a tBD of type t–(v , K ∪ {h}, λ)
with a block of size h repeated λ times.
Equivalence
A ItBD of type t–(v , h, K, λ) is the same as a tBD of type t–(v , K ∪ {h}, λ)
with a block of size h repeated λ times.
A Steiner tBD is a ItBD with λ = 1 in which any block of the tBD can
be considered as the hole.
Equivalence
A ItBD of type t–(v , h, K, λ) is the same as a tBD of type t–(v , K ∪ {h}, λ)
with a block of size h repeated λ times.
A Steiner tBD is a ItBD with λ = 1 in which any block of the tBD can
be considered as the hole.
Main Theorem: If H is a hole in a ItBD with any λ, then
|H|
≤
(v − 1)/2 for t even while
|H|
≤
v /2 for t odd.
Equivalence
A ItBD of type t–(v , h, K, λ) is the same as a tBD of type t–(v , K ∪ {h}, λ)
with a block of size h repeated λ times.
A Steiner tBD is a ItBD with λ = 1 in which any block of the tBD can
be considered as the hole.
Main Theorem: If H is a hole in a ItBD with any λ, then
|H|
≤
(v − 1)/2 for t even while
|H|
≤
v /2 for t odd.
Corollary: If B is a block in a Steiner tBD, then
|B|
≤
(v − 1)/2 for t even while
|B|
≤
v /2 for t odd.
Reduction to K = {t + 1}: If a proper ItBD (X , H, B) of type t–(v , h, K, λ)
exists, then a proper ItBD of type t–(v , h, {t + 1}, λ0 ) exists for some λ0 .
Proof. K = {k1 , k2 , . . . , k` } Replace each B ∈ B of size |B| = ki with
ki
its t+1
(t+1)-subsets and repeat each
Ki
`
Y
ci =
(kj − t) times.
T 1
j=1,j6=i
Let T ⊆ X , |T | = t.
I
T ⊆ H no problem.
I
T in ri blocks of size ki in B, then r1 + r2 + · · · + r` = λ.
In the new design T is contained in
r1 c1 (k1 − t) + r2 c2 (k2 − t) + · · · + r` c` (k` − t)
o
P` n Q`
=
r
(k
−
t)
(ki − t)
i
j
i=1
j=1,j6=i
=
(r1 + r2 + · · · + r` )
`
Y
(kj − t)
j=1
= λ
Y
k∈K
0
(k − t) = λ blocks, as required.
Even suffices
If the main theorem is true when t is even,
then it is also true when t is odd.
Proof.
I
Suppose there is a t–(v , h, {t + 1}, λ)
with h > v /2 and t odd, t ≥ 3.
I
Let x be in the hole and derive through x.
I
Result: a (t − 1)–(v − 1, h − 1, {t}, λ)
I
But t − 1 is even and
(v − 1) − 1 < (2h − 1) − 1 = 2(h − 1).
A contradiction.
Basis step
There is no 2–(v , h, {3}, λ) with h > (v − 1)/2.
Proof.
I
(X , H, ) a 2–(v , h, {3}, λ)
I
derive w.r.t. x ∈ H.
I
Result: A λ–regular multigraph on X \ H.
I
Do this for all x ∈ H to get a
λh–regular multigraph G on X \ H.
I
Each edge in G is repeated at most λ times.
I
Hence counting edges incident to a fixed vertex
λh ≤ λ(v − h − 1)
Thus h ≤ (v − 1)/2.
A λ–regular multigraph
λ times.
x
v −h
h
The hole H
X \H
We need only worry about v = 2h − 1 and v = 2h.
We need to show that no ItBD exists with h > v −1
2 . Thus we must
rule out:
v = 2h, 2h − 1 , 2h − 2, 2h − 3 . . . .
| {z } |
{z
}
induction
direct proof
Suppose t 0 ≥ 4 is even and we have shown
1. there are no t 0 –(2h − 1, h, {t 0 + 1}, λ)
2. there are no t 0 –(2h, h, {t 0 + 1}, λ)
3. The Main Theorem holds for t = t 0 − 2.
Then there cannot be any
t 0 −(v , h, {t 0 + 1}, λ) with v ≤ 2h − 2
For if so, derive through two points in the hole to get a
t−(v − 2, h − 2, {t + 1}, λ)
where
v − 2 ≤ (2h − 2) − 2 = 2h − 4 = 2(h − 2)
A contradiction.
Thus the main theorem would hold for all even t 0 by induction.
No t–(2h − 1, h, {t + 1}, λ) with t even.
H
h
i
X \H
j
h−1
~t
x0
x1
x2
x3
· · · xt−1
(0, t + 1) (1, t) (2, t − 1) (3, t − 2) · · · (t − 1, 2)
(0, t)
t +1
1
(1, t − 1)
(2, t − 2)
..
.
(t − 2, 2)
(t − 1, 1)
Solve Bt,t+1~x = ~t
t
3
..
.
h h−1
t−1
h h−1
2
t −1
h−1
t
2
..
t−2
.
3
t −1
2
h
t−2
h
t−1
..
.
h−1
2
(h − 1)
Bt,t+1 is invertible, whence
−1
−1 ~
~x = Bt,t+1
(λ~t) = λBt,t+1
t.
−1
Indexing the rows and columns of Bt,t+1
by 1, 2, . . . , t
−1
Bt,t+1
[1, j] = (−1)j−1
1
j
t+1
j
for j = 1, 2, . . . , t. Therefore
x0
=
λ
t
X
j=1
(−1)j−1
1
j
t+1
j
h
h−1
j −1 t −j +1
t
λ X
h
h−1
j 1
(−1) t = −
t +1
j −1 t −j +1
j−1
j=1
−λ h − 1
=
< 0.
t +1 t −1
This is a contradiction because,
x0 counts the number of blocks disjoint from H
and consequently cannot be negative.
Therefore there is no t–(2h − 1, h, {t + 1}, λ) with h ≥ t.
No t–(2h, h, {t + 1}, λ) with t even.
Same matrix equation as before
Bt,t+1~x = λ~t
except this time
h
h
ti =
i
t −i
for i = 0, 1, . . . , t − 1. So
x0
= λ
t
X
(−1)j−1
j=1
=
=
λ
t +1
t
X
j=1
1
j
h
j −1
t+1
j
(−1)j−1
h
t −j +1
1
t
j−1
h
h
j −1 t −j +1
−λt
h
< 0,
(t + 1)(2h − t) t
again a contradiction. Thus no proper ItBD of type t–(2h, h, {t + 1}, λ)
with h ≥ t can exist. We have therefore established the following.
Theorem:
There do not exist proper ItBDs of types t–(2h − 1, h, {t + 1}, λ)
or t–(2h, h, {t + 1}, λ) for any λ and any 2 ≤ t ≤ h, when t is even.
Hence we have established:
Main Theorem: If H is a hole in an ItBD with any λ, then
|H|
≤
(v − 1)/2 for t even while
|H|
≤
v /2 for t odd.
and so the validity of Kramer’s conjecture.
Theorem: If B is a block in an t-(v , K, 1) design, then
|B|
≤
(v − 1)/2 for t even while
|B|
≤
v /2 for t odd.
Bounds are sharp
1. For each odd t ≥ 3 and each h ≥ t + 1 there is a
h−1
t−(2h, h, {t + 1}, (2h − t)t!
).
t
2. For each even t ≥ 2 and each h ≥ t + 1 there is a
h
t−(2h + 1, h, {t + 1}, (2h − t + 1)(t + 1)!
).
t +1
For t odd we construct a t−(2h, h, {t + 1}, λ)
.
with λ = (2h − t)t! h−1
t
I
Let X = H ∪ Y , |H| = |Y | = h.
I
Let G = Sym(H) × Sym(Y ).
I
Orbits of t sets are
∆i = {T ⊆ X : |T | = t and |T ∩ H| = i}
For i = 0, 1, 2, . . . , t − 1. (The type (j, t − j)–sets.)
I
Orbits of t + 1 sets are
Γj = {K ⊆ X : |K | = t + 1 and |K ∩ H| = j}
For j = 0, 1, 2, . . . , t − 1. (The type (j, t + 1 − j)–sets.)
Let At,t+1 be the matrix given by
At,t+1 [i, j] = |{K ∈ Γj : K ⊇ T }|
where T ∈ ∆i is any fixed representative. At,t+1 is square,
upper-triangular and invertible. Thus there is a unique solution ~x to
the matrix equation
At,t+1~x = λJ
where ~x is nonnegative and integral if λ = (2h − t)t! h−1
.
t
Not all blocks have to be t + 1-sets
The solution ~x to the equation A~x = λJ was
h−1
xt−i = t!
1 + (−1)i−1
t
!
h−t+i
i
h−1
i
.
i = 1, 2, . . . , t − 1, t.
Notice: the only t + 1-sets that can cover type (0, t) sets have type
(0, t + 1) i.e. orbit Γ0
or
(1, t) i.e. orbit Γ1
When i = t we have
x0 = t!
h−1
h
+
.
t
t
When i = t − 1 we have
x1 = 0.
So the t-subsets of Y are covered by taking as blocks all of the
t + 1-subsets of Y each repeated x0 times. Thus we can simply take
Y as block and repeat it λ times. Thus we have a
h−1
t−(2h, h, {t + 1, h? }, (2h − t)t!
)
t
Bounds are sharp for even t
For t even we construct a t-(2h +
1, h, {t + 1}, λ)
h
with λ = (2h − t + 1)(t + 1)! t+1
).
Proof.
I
We already know that there is a
˙ , H, ), of type
I(t + 1)BD (H ∪Y
(t + 1)−(2(h + 1), h + 1, {t + 2}, λ0 )
with λ0 = (2(h + 1) − (t + 1))(t + 1)!
I
(h+1)−1
t+1
Take the derived design through a
point x ∈ H to get a
t−(2h + 1, h, {t + 1}, λ)
) = λ.
Need t + 1-sets
There do not exist proper ItBDs of types
t–(2h + 1, h, K, λ)
(t even)
or
t–(2h, h, K, λ)
(t odd)
for any λ and any 2 ≤ t ≤ h,
with min{k : k ∈ K} ≥ t + 2.
General bound
Theorem 11. If (X , H, B) is a proper ItBD of type
t–(v , h, K, λ) with h ≥ t ≥ 2 and min K = k ≥ t + 1,
then
v + (k − t)(t − 2) − 1
h≤
.
k −t +1
Research problem 6. Can necessary and sufficient conditions be
found so that this Bound is sharp?
Groups, Designs and Linear Algebra:
Hamilton decomposition of
vertex-transitive graphs of order p2
Donald L. Kreher
Department of Mathematical Sciences
Michigan Technological University
CIMPA School
July-August 2013
The Adrian frame.
The Adrian frame.
Don, could you first remind our students
what a graph is and what a Hamilton cycle is.
The Adrian frame.
Don, could you first remind our students
what a graph is and what a Hamilton cycle is.
A graph X = (V , E) is a pair
(V , E) where
I
V is a finite set of points
I
E is a collection of 2
element subsets called
edges
1
2
7 6
8 9
5
3
4
0
The Adrian frame.
Don, could you first remind our students
what a graph is and what a Hamilton cycle is.
A graph X = (V , E) is a pair
(V , E) where
I
V is a finite set of points
I
E is a collection of 2
element subsets called
edges
A Hamilton cycle is a cycle that
includes all the vertices.
1
2
7 6
8 9
5
3
4
0
Introduction
Let A be an abelian group and S ⊆ A be inverse closed.
x ∈ S ⇔ −x ∈ S
Cayley graph X on A with connection set S
I
V (X ) = A;
I
E(X ) = {xy : x − y ∈ S}.
If A = Zn , then X is a circulant graph and we write
X = C IRC (n; S).
A Hamilton cycle in X is a cycle in
X containing all the vertices of X .
A perfect matching in X when |X |
is even is a set of |X |/2 independent edges.
Example 1: C IRC (10; {2, 3, 4, 6, 7, 8})
9
0
1
8
2
7
3
6
9
0
9
1
5
0
4
9
1
0
1
8
2
8
2
8
2
7
3
7
3
7
3
6
5
4
6
5
4
6
5
4
Example 2: C IRC (10; {2, 4, 5, 6, 8})
9
0
1
8
2
7
3
6
9
0
9
1
5
0
4
9
1
0
1
8
2
8
2
8
2
7
3
7
3
7
3
6
5
4
6
5
4
6
5
4
Hamilton decomposable.
A d regular graph X is Hamilton-decomposable, just when the
edge set of X can be partitioned
I
into d/2 Hamilton cycles when d is even, or
I
into (d − 1)/2 Hamilton cycles and a perfect matching
when d is odd.
Alspach conjecture (1984): Every connected Cayley graphs
on an abelian group is Hamilton-decomposable.
Progress I.
Valency approach.
1990 Bermond, Favaron, and Maheo: settle the conjecture
for valency 4.
2007 Dean proves: odd order connected circulant graphs of
valency 6 are Hamilton-decomposable.
He also verifies that even order connected circulant
graphs of valency 6 have Hamilton decompositions,
except for small class.
2009 Westlund, Liu, and Kreher: settle the conjecture for
valency 6 and odd order.
2012 Westlund: settles the conjecture for valency 6 and most
even order abelian groups, but not all.
Progress II.
Liu’s structural approach.
1994,1996,2003 Liu in a series of three very nice papers proved that
if the connection set S is a minimal Cayley generating
set, then C AY (G; S) is Hamilton-decomposable with
one exceptional case for |G| even.
The connection set S is a minimal Cayley generating set when
hSi = G but hS \ {s, s − 1}i is a proper subgroup of G for every s ∈ S.
Progress III.
Order approach.
2013 Alspach, Caliskan, Kreher: provide sufficient conditions
on the abelian groups G and H for when
C AY (G × H; S ∪ B) is Hamilton-decomposable, given
that C AY (G; S) is Hamilton-decomposable and B is a
basis for H.
They also settle the conjecture for rank n elementary
abelian groups and valency 2(n + 1).
2014 Alspach, Bryant, Kreher: Settle the conjecture for all
abelian groups of order p2 , p a prime.
Today goal.
1983 Marǔsič proved that every vertex-transitive graph of
order p2 , p a prime, is a Cayley graph.
To within isomorphism there are two groups of order p2 :
Zp × Zp and Zp2 . Both are abelian.
Thus, every vertex-transitive graph of order p2 is a
Cayley graph on an abelian group.
Today we show for primes p:
C AY (Zp × Zp ; S) is Hamilton decomposable.
I C AY Zp2 ; S is Hamilton decomposable.
I
Because p = 2 is trivially easy, we may assume p is an odd prime.
Four-cycle switching
⊕
⊕
=
⊕
⊕
=
Joining p cycles uses p − 1 four cycle switches.
⊕
=
The diagonal pattern
Consider the Cayley graph X = C AY (Z5 × Z5 ; {±(1, 0), ±(0, 1)})
4
3
2
1
0
0 1 2 3 4
The diagonal pattern
Consider the Cayley graph X = C AY (Z5 × Z5 ; {±(1, 0), ±(0, 1)})
4
3
2
1
0
4
3
2
1
0
0 1 2 3 4
0 1 2 3 4
4
3
2
1
0
0 1 2 3 4
The diagonal pattern
Consider the Cayley graph X = C AY (Z5 × Z5 ; {±(1, 0), ±(0, 1)})
4
3
2
1
0
4
3
2
1
0
0 1 2 3 4
0 1 2 3 4
4
3
2
1
0
0 1 2 3 4
The diagonal pattern
Consider the Cayley graph X = C AY (Z5 × Z5 ; {±(1, 0), ±(0, 1)})
4
3
2
1
0
4
3
2
1
0
0 1 2 3 4
0 1 2 3 4
4
3
2
1
0
0 1 2 3 4
The diagonal pattern
Consider the Cayley graph X = C AY (Z5 × Z5 ; {±(1, 0), ±(0, 1)})
4
3
2
1
0
4
3
2
1
0
0 1 2 3 4
0 1 2 3 4
4
3
2
1
0
0 1 2 3 4
The diagonal pattern
Consider the Cayley graph X = C AY (Z5 × Z5 ; {±(1, 0), ±(0, 1)})
4
3
2
1
0
4
3
2
1
0
0 1 2 3 4
0 1 2 3 4
4
3
2
1
0
0 1 2 3 4
The diagonal pattern
Consider the Cayley graph X = C AY (Z5 × Z5 ; {±(1, 0), ±(0, 1)})
4
3
2
1
0
4
3
2
1
0
0 1 2 3 4
0 1 2 3 4
4
3
2
1
0
0 1 2 3 4
The diagonal pattern
Consider the Cayley graph X = C AY (Z5 × Z5 ; {±(1, 0), ±(0, 1)})
4
3
2
1
0
4
3
2
1
0
0 1 2 3 4
0 1 2 3 4
4
3
2
1
0
0 1 2 3 4
The diagonal pattern
Consider the Cayley graph X = C AY (Z5 × Z5 ; {±(1, 0), ±(0, 1)})
4
3
2
1
0
4
3
2
1
0
0 1 2 3 4
0 1 2 3 4
4
3
2
1
0
0 1 2 3 4
The diagonal pattern
Consider the Cayley graph X = C AY (Z5 × Z5 ; {±(1, 0), ±(0, 1)})
4
3
2
1
0
4
3
2
1
0
0 1 2 3 4
0 1 2 3 4
4
3
2
1
0
0 1 2 3 4
Hamilton decomp. of C AY (Z7 × Z7 ; {±(1, 0), ±(0, 1), ±(0, 2)})
6
5
4
3
2
1
0
0 1 2 3 4 5 6
Hamilton decomp. of C AY (Z7 × Z7 ; {±(1, 0), ±(0, 1), ±(0, 2)})
6
5
4
3
2
1
0
6
5
4
3
2
1
0
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
Hamilton decomp. of C AY (Z7 × Z7 ; {±(1, 0), ±(0, 1), ±(0, 2)})
6
5
4
3
2
1
0
6
5
4
3
2
1
0
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
Hamilton decomp. of C AY (Z7 × Z7 ; {±(1, 0), ±(0, 1), ±(0, 2)})
6
5
4
3
2
1
0
6
5
4
3
2
1
0
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
Hamilton decomp. of C AY (Z7 × Z7 ; {±(1, 0), ±(0, 1), ±(0, 2)})
6
5
4
3
2
1
0
6
5
4
3
2
1
0
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
Hamilton decomp. of C AY (Z7 × Z7 ; {±(1, 0), ±(0, 1), ±(0, 2)})
6
5
4
3
2
1
0
6
5
4
3
2
1
0
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
Hamilton decomp. of C AY (Z7 × Z7 ; {±(1, 0), ±(0, 1), ±(0, 2)})
6
5
4
3
2
1
0
6
5
4
3
2
1
0
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
Hamilton decomp. of C AY (Z7 × Z7 ; {±(1, 0), ±(0, 1), ±(0, 2)})
6
5
4
3
2
1
0
6
5
4
3
2
1
0
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
Hamilton decomp. of C AY (Z7 × Z7 ; {±(1, 0), ±(0, 1), ±(0, 2)})
6
5
4
3
2
1
0
6
5
4
3
2
1
0
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
Hamilton decomp. of C AY (Z7 × Z7 ; {±(1, 0), ±(0, 1), ±(0, 2)})
6
5
4
3
2
1
0
6
5
4
3
2
1
0
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
Hamilton decomp. of C AY (Z7 × Z7 ; {±(1, 0), ±(0, 1), ±(0, 2)})
6
5
4
3
2
1
0
6
5
4
3
2
1
0
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
Hamilton decomp. of C AY (Z7 × Z7 ; {±(1, 0), ±(0, 1), ±(0, 2)})
6
5
4
3
2
1
0
6
5
4
3
2
1
0
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
Hamilton decomp. of C AY (Z7 × Z7 ; {±(1, 0), ±(0, 1), ±(0, 2)})
6
5
4
3
2
1
0
6
5
4
3
2
1
0
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
Hamilton decomp. of C AY (Z7 × Z7 ; {±(1, 0), ±(0, 1), ±(0, 2)})
6
5
4
3
2
1
0
6
5
4
3
2
1
0
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
Hamilton decomp. of C AY (Z7 × Z7 ; {±(1, 0), ±(0, 1), ±(0, 2)})
6
5
4
3
2
1
0
6
5
4
3
2
1
0
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
Hamilton decomp. of C AY (Z7 × Z7 ; {±(1, 0), ±(0, 1), ±(0, 2)})
6
5
4
3
2
1
0
6
5
4
3
2
1
0
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
Hamilton decomp. of C AY (Z7 × Z7 ; {±(1, 0), ±(0, 1), ±(0, 2)})
6
5
4
3
2
1
0
6
5
4
3
2
1
0
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
0 1 2 3 4 5 6
6
5
4
3
2
1
0
0 1 2 3 4 5 6
Joining p cycles uses p − 1 four cycle switches.
You can also use p.
⊕
=
Joining p cycles uses p − 1 four cycle switches.
You can also use p.
⊕
=
Joining p cycles uses p − 1 four cycle switches.
You can also use p.
⊕
=
Joining p cycles uses p − 1 four cycle switches.
You can also use p.
⊕
A Hamilton-cycle
=
Not a Hamilton-cycle
Joining p cycles uses p − 1 four cycle switches.
You can also use p.
⊕
A Hamilton-cycle
=
Not a Hamilton-cycle
Joining p cycles uses p − 1 four cycle switches.
You can also use p.
⊕
=
Switching p times is
better than p − 1 times,
because
f (x, y ) = (x + 1, y + 1)
can be an automorphism.
A Hamilton-cycle
Not a Hamilton-cycle
Joining p cycles uses p − 1 four cycle switches.
You can also use p.
=
⊕
D
Switching p times is
better than p − 1 times,
because
f (x, y ) = (x + 1, y + 1)
can be an automorphism
of D
A Hamilton-cycle
Not a Hamilton-cycle
The Diagonal Lemma
Let H be the 2-factor C AY (Zp × Zp ; {±(0, s)}), where
0 < s ≤ (p − 1)/2. If x − y = s and
D(u, v ) = F0 (u, v ) ⊕ F1 (u, v ) ⊕ · · · ⊕ Fp−1 (u, v ),
where
Fi (u, v ) =
(i, v + i)
(i + 1, v + i)
(i, u + i)
(i + 1, u + i)
,
then H + D(x, y ) is a Hamilton cycle.
Note:
I
f : (x, y ) 7→ (x + 1, y + 1) is an automorphism of D(u, v ).
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
6
5
4
3
2
1
0
0 1 2 3 4 5 6
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
6
5
4
3
2
1
0
0 1 2 3 4 5 6
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
6
5
4
3
2
1
0
0 1 2 3 4 5 6
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
6
5
4
3
2
1
0
0 1 2 3 4 5 6
(6, 6)
(5, 5)
(4, 4)
(3, 3)
(2, 2)
(1, 1)
(0, 0)
[0]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
6
5
4
3
2
1
0
0 1 2 3 4 5 6
[0]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
6
5
4
3
2
1
0
0 1 2 3 4 5 6
[0]
(0, 6)
(6, 5)
(5, 4)
(4, 3)
(3, 2)
(2, 1)
(1, 0)
[1]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
6
5
4
3
2
1
0
0 1 2 3 4 5 6
[0]
[1]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
6
5
4
3
2
1
0
0 1 2 3 4 5 6
[0]
[1]
(1, 6)
(0, 5)
(6, 4)
(5, 3)
(4, 2)
(3, 1)
(2, 0)
[2]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
6
5
4
3
2
1
0
0 1 2 3 4 5 6
[0]
[1]
[2]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
6
5
4
3
2
1
0
0 1 2 3 4 5 6
[0]
[1]
[2]
[3]
[4]
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
f
[0]
[1]
[2]
[3]
[4]
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
f
[0]
[1]
[2]
[3]
[4]
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
f
[0]
[1]
[2]
[3]
[4]
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
f
[0]
[1]
[2]
[3]
[4]
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
f
[0]
[1]
[2]
[3]
[4]
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
f
[0]
[1]
[2]
[3]
[4]
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
f
[0]
[1]
[2]
[3]
[4]
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
f
[0]
[1]
[2]
[3]
[4]
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
f
[0]
[1]
[2]
[3]
[4]
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
f
[0]
[1]
[2]
[3]
[4]
Quotient graph H/f
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
If H/f is a cycle, there
f
[0]
[1]
[2]
[3]
[4]
Quotient graph H/f
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
If H/f is a cycle, there
are two possibilities for H.
f
[0]
[1]
[2]
[3]
[4]
Quotient graph H/f
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
If H/f is a cycle, there
are two possibilities for H.
f
I. H is a union of p p-cycles.
[0]
[1]
[2]
[3]
[4]
Quotient graph H/f
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
If H/f is a cycle, there
are two possibilities for H.
f
I. H is a union of p p-cycles.
[0]
[1]
[2]
[3]
[4]
Quotient graph H/f
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
If H/f is a cycle, there
are two possibilities for H.
f
I. H is a union of p p-cycles.
[0]
[1]
[2]
[3]
[4]
Quotient graph H/f
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
If H/f is a cycle, there
are two possibilities for H.
f
I. H is a union of p p-cycles.
[0]
[1]
[2]
[3]
[4]
Quotient graph H/f
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
If H/f is a cycle, there
are two possibilities for H.
f
I. H is a union of p p-cycles.
[0]
[1]
[2]
[3]
[4]
Quotient graph H/f
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
If H/f is a cycle, there
are two possibilities for H.
f
I. H is a union of p p-cycles.
[0]
[1]
[2]
[3]
[4]
Quotient graph H/f
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
If H/f is a cycle, there
are two possibilities for H.
f
I. H is a union of p p-cycles.
[0]
[1]
[2]
[3]
[4]
Quotient graph H/f
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
If H/f is a cycle, there
are two possibilities for H.
f
I. H is a union of p p-cycles.
[0]
[1]
[2]
[3]
[4]
Quotient graph H/f
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
If H/f is a cycle, there
are two possibilities for H.
f
I. H is a union of p p-cycles.
[0]
[1]
[2]
[3]
[4]
Quotient graph H/f
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
If H/f is a cycle, there
are two possibilities for H.
f
I. H is a union of p p-cycles.
[0]
[1]
[2]
[3]
[4]
Quotient graph H/f
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
If H/f is a cycle, there
are two possibilities for H.
f
I. H is a union of p p-cycles.
[0]
[1]
[2]
[3]
[4]
Quotient graph H/f
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
If H/f is a cycle, there
are two possibilities for H.
f
I. H is a union of p p-cycles.
[0]
[1]
[2]
[3]
[4]
Quotient graph H/f
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
If H/f is a cycle, there
are two possibilities for H.
f
I. H is a union of p p-cycles.
[0]
[1]
[2]
[3]
[4]
Quotient graph H/f
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
If H/f is a cycle, there
are two possibilities for H.
f
I. H is a union of p p-cycles.
[0]
[1]
[2]
[3]
[4]
Quotient graph H/f
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
If H/f is a cycle, there
are two possibilities for H.
f
I. H is a union of p p-cycles.
II. H is Hamilton cycle.
[0]
[1]
[2]
[3]
[4]
Quotient graph H/f
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
If H/f is a cycle, there
are two possibilities for H.
f
I. H is a union of p p-cycles.
II. H is Hamilton cycle.
[0]
[1]
[2]
[3]
[4]
Quotient graph H/f
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
If H/f is a cycle, there
are two possibilities for H.
f
I. H is a union of p p-cycles.
II. H is Hamilton cycle.
[0]
[1]
[2]
[3]
[4]
Quotient graph H/f
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
If H/f is a cycle, there
are two possibilities for H.
f
I. H is a union of p p-cycles.
II. H is Hamilton cycle.
[0]
[1]
[2]
[3]
[4]
Quotient graph H/f
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
If H/f is a cycle, there
are two possibilities for H.
f
I. H is a union of p p-cycles.
II. H is Hamilton cycle.
[0]
[1]
[2]
[3]
[4]
Quotient graph H/f
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
If H/f is a cycle, there
are two possibilities for H.
f
I. H is a union of p p-cycles.
II. H is Hamilton cycle.
[0]
[1]
[2]
[3]
[4]
Quotient graph H/f
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
If H/f is a cycle, there
are two possibilities for H.
f
I. H is a union of p p-cycles.
II. H is Hamilton cycle.
[0]
[1]
[2]
[3]
[4]
Quotient graph H/f
[5]
[6]
The automorphism f
Suppose H is a 2-factor fixed by f : (x, y) 7→ (x + 1, y + 1). Then
f k (x, y) = (x 0 , y 0 ) ⇐⇒ x − y = x 0 − y 0 . Let [a] = {(x, y ) : y − x = a}.
Edges between orbits
generate matchings!
If H/f is a cycle, there
are two possibilities for H.
f
I. H is a union of p p-cycles.
II. H is Hamilton cycle.
[0]
[1]
[2]
[3]
[4]
Quotient graph H/f
[5]
[6]
Lemma 1
If H is a 2-factor of the Cayley graph X on Zp × Zp fixed by
f : (x, y) 7→ (x + 1, y + 1) and H/f is a cycle, then either
I
H is a union of p disjoint p-cycles
or
I
H is a Hamilton cycle.
Lemma 1
If H is a 2-factor of the Cayley graph X on Zp × Zp fixed by
f : (x, y) 7→ (x + 1, y + 1) and H/f is a cycle, then either
I
H is a union of p disjoint p-cycles
or
I
H is a Hamilton cycle.
Fi (u, v ) =
(i, v + i)
(i + 1, v + i)
(i, u + i)
(i + 1, u + i)
.
Lemma 1
If H is a 2-factor of the Cayley graph X on Zp × Zp fixed by
f : (x, y) 7→ (x + 1, y + 1) and H/f is a cycle, then either
I
H is a union of p disjoint p-cycles
or
I
H is a Hamilton cycle.
Fi (u, v ) =
(i, v + i)
(i + 1, v + i)
(i, u + i)
(i + 1, u + i)
Then f : (x, y ) 7→ (x +1, y +1)
. is an automorphism of
D(u, v ) = F0 (u, v )⊕· · ·⊕Fp−1 (u, v )
Lemma 2
If the Cayley graph X on Zp × Zp has connection set
S = {±(1, 0), ±(0, s1 ), ±(0, S2 ), . . . , ±(0, s` )},
where 0 < s1 < s2 < · · · < s` < (p − 1)/2, then X is Hamilton
decomposable.
Proof.
Let:
H0 =C AY (Zp × Zp ; ±(1, 0)) and
Hi =C AY (Zp × Zp ; ±(0, s1 )), i = 1, . . . , `
Let {ui , vi } = {i, −i}, where i = 2−1 si , i = 1, .., `.
This is the edge in the canonical matching of length si .
The Diagonal Lemma shows that
Hi0 = Hi ⊕ D(ui , vi )
is a Hamilton cycle for each , i = 1, . . . , `.
Thus we need only show that
H00 = H0 ⊕ D(u1 , v1 ) ⊕ D(u2 , v2 ) ⊕ · · · ⊕ D(u` , v` )
is also a Hamilton cycle.
Lemma 2
If the Cayley graph X on Zp × Zp has connection set
S = {±(1, 0), ±(0, s1 ), ±(0, S2 ), . . . , ±(0, s` )},
where 0 < s1 < s2 < · · · < s` < (p − 1)/2, then X is Hamilton
decomposable.
Proof.
Let:
H0 =C AY (Zp × Zp ; ±(1, 0)) and
Hi =C AY (Zp × Zp ; ±(0, s1 )), i = 1, . . . , `
Let {ui , vi } = {i, −i}, where i = 2−1 si , i = 1, .., `.
This is the edge in the canonical matching of length si .
0
6
1
5
2
4
3
Lemma 2
If the Cayley graph X on Zp × Zp has connection set
S = {±(1, 0), ±(0, s1 ), ±(0, S2 ), . . . , ±(0, s` )},
where 0 < s1 < s2 < · · · < s` < (p − 1)/2, then X is Hamilton
decomposable.
Proof.
H0 =C AY (Zp × Zp ; ±(1, 0)) and
Hi =C AY (Zp × Zp ; ±(0, s1 )), i = 1, . . . , `
Let:
Let {ui , vi } = {i, −i}, where i = 2−1 si , i = 1, .., `.
This is the edge in the canonical matching of length si .
0
6
1
5
2
4
3
0
6
1
5
2
4
3
6
5
4
3
2
1
0
0 1 2 3 4 5 6
Lemma 2
If the Cayley graph X on Zp × Zp has connection set
S = {±(1, 0), ±(0, s1 ), ±(0, S2 ), . . . , ±(0, s` )},
where 0 < s1 < s2 < · · · < s` < (p − 1)/2, then X is Hamilton
decomposable.
Proof.
H0 =C AY (Zp × Zp ; ±(1, 0)) and
Hi =C AY (Zp × Zp ; ±(0, s1 )), i = 1, . . . , `
Let:
Let {ui , vi } = {i, −i}, where i = 2−1 si , i = 1, .., `.
This is the edge in the canonical matching of length si .
0
6
1
5
2
4
3
0
6
1
5
2
4
3
6
5
4
3
2
1
0
0 1 2 3 4 5 6
Lemma 2
If the Cayley graph X on Zp × Zp has connection set
S = {±(1, 0), ±(0, s1 ), ±(0, S2 ), . . . , ±(0, s` )},
where 0 < s1 < s2 < · · · < s` < (p − 1)/2, then X is Hamilton
decomposable.
Proof.
H0 =C AY (Zp × Zp ; ±(1, 0)) and
Hi =C AY (Zp × Zp ; ±(0, s1 )), i = 1, . . . , `
Let:
Let {ui , vi } = {i, −i}, where i = 2−1 si , i = 1, .., `.
This is the edge in the canonical matching of length si .
0
6
1
5
2
4
3
0
6
1
5
2
4
3
6
5
4
3
2
1
0
0 1 2 3 4 5 6
Lemma 2
If the Cayley graph X on Zp × Zp has connection set
S = {±(1, 0), ±(0, s1 ), ±(0, S2 ), . . . , ±(0, s` )},
where 0 < s1 < s2 < · · · < s` < (p − 1)/2, then X is Hamilton
decomposable.
Proof.
H0 =C AY (Zp × Zp ; ±(1, 0)) and
Hi =C AY (Zp × Zp ; ±(0, s1 )), i = 1, . . . , `
Let:
Let {ui , vi } = {i, −i}, where i = 2−1 si , i = 1, .., `.
This is the edge in the canonical matching of length si .
0
6
1
5
2
4
3
0
6
1
5
2
4
3
0
6
1
5
2
4
3
6
5
4
3
2
1
0
0 1 2 3 4 5 6
Lemma 2
If the Cayley graph X on Zp × Zp has connection set
S = {±(1, 0), ±(0, s1 ), ±(0, S2 ), . . . , ±(0, s` )},
where 0 < s1 < s2 < · · · < s` < (p − 1)/2, then X is Hamilton
decomposable.
Proof.
H0 =C AY (Zp × Zp ; ±(1, 0)) and
Hi =C AY (Zp × Zp ; ±(0, s1 )), i = 1, . . . , `
Let:
Let {ui , vi } = {i, −i}, where i = 2−1 si , i = 1, .., `.
This is the edge in the canonical matching of length si .
0
6
1
5
2
4
3
0
6
1
5
2
4
3
0
6
5
2
4
3
0
6
1
2
4
6
5
4
3
2
1
0
1
5
3
0 1 2 3 4 5 6
Lemma 2
If the Cayley graph X on Zp × Zp has connection set
S = {±(1, 0), ±(0, s1 ), ±(0, S2 ), . . . , ±(0, s` )},
where 0 < s1 < s2 < · · · < s` < (p − 1)/2, then X is Hamilton
decomposable.
Proof.
H0 =C AY (Zp × Zp ; ±(1, 0)) and
Hi =C AY (Zp × Zp ; ±(0, s1 )), i = 1, . . . , `
Let:
Let {ui , vi } = {i, −i}, where i = 2−1 si , i = 1, .., `.
This is the edge in the canonical matching of length si .
0
6
1
5
2
4
3
0
6
1
5
2
4
3
0
6
5
2
4
3
0
6
1
5
2
4
6
5
4
3
2
1
0
1
3
0
6
1
5
2
4
0 1 2 3 4 5 6
3
Lemma 2
If the Cayley graph X on Zp × Zp has connection set
S = {±(1, 0), ±(0, s1 ), ±(0, S2 ), . . . , ±(0, s` )},
where 0 < s1 < s2 < · · · < s` < (p − 1)/2, then X is Hamilton
decomposable.
Proof.
H0 =C AY (Zp × Zp ; ±(1, 0)) and
Hi =C AY (Zp × Zp ; ±(0, s1 )), i = 1, . . . , `
Let:
Let {ui , vi } = {i, −i}, where i = 2−1 si , i = 1, .., `.
This is the edge in the canonical matching of length si .
0
6
1
5
2
4
3
0
6
1
5
2
4
3
0
6
5
2
4
3
0
6
1
5
2
4
6
5
4
3
2
1
0
1
3
0
6
1
5
2
4
0 1 2 3 4 5 6
3
0
6
1
5
2
4
3
Lemma 2
If the Cayley graph X on Zp × Zp has connection set
S = {±(1, 0), ±(0, s1 ), ±(0, S2 ), . . . , ±(0, s` )},
where 0 < s1 < s2 < · · · < s` < (p − 1)/2, then X is Hamilton
decomposable.
Proof.
H0 =C AY (Zp × Zp ; ±(1, 0)) and
Hi =C AY (Zp × Zp ; ±(0, s1 )), i = 1, . . . , `
Let:
Let {ui , vi } = {i, −i}, where i = 2−1 si , i = 1, .., `.
This is the edge in the canonical matching of length si .
0
6
1
5
2
4
3
0
6
1
5
2
4
3
0
6
5
2
4
3
0
6
1
5
2
4
6
5
4
3
2
1
0
1
3
0
6
1
5
2
4
0 1 2 3 4 5 6
3
0
6
1
5
2
4
3
0
6
1
5
2
4
3
Lemma 2
If the Cayley graph X on Zp × Zp has connection set
S = {±(1, 0), ±(0, s1 ), ±(0, S2 ), . . . , ±(0, s` )},
where 0 < s1 < s2 < · · · < s` < (p − 1)/2, then X is Hamilton
decomposable.
Proof.
Let:
H0 =C AY (Zp × Zp ; ±(1, 0)) and
Hi =C AY (Zp × Zp ; ±(0, s1 )), i = 1, . . . , `
Let {ui , vi } = {i, −i}, where i = 2−1 si , i = 1, .., `.
This is the edge in the canonical matching of length si .
The Diagonal Lemma shows that
Hi0 = Hi ⊕ D(ui , vi )
is a Hamilton cycle for each , i = 1, . . . , `.
Thus we need only show that
H00 = H0 ⊕ D(u1 , v1 ) ⊕ D(u2 , v2 ) ⊕ · · · ⊕ D(u` , v` )
is also a Hamilton cycle.
H00 = H0 ⊕ D(u1 , v1 ) ⊕ D(u2 , v2 ) ⊕ · · · ⊕ D(u` , v` ) is a Hamilton
cycle.
It is easy to see that f is an automorphism of each D(ui , vi ),
i = 1, . . . , ` and of H0 .
Thus f is also an isomorphism H00 .
It suffices to show by Lemma 1 that (1) H00 /f is a cycle and (2) H00
contains a path of length > p.
H00 /f =H0 /f ⊕D(u1 , v1 )/f ⊕D(u2`, v2 )/f ⊕ · · · ⊕D(u` , v` )/f
M
= [0][1] · · · [p−1][0] ⊕
D(ui , vi )/f
i=1
`
M [vi ] [vi −1] = [0] · · · [p−1][0] ⊕
A cycle!
i=1
[ui ]
[ui −1]
H00 = H0 ⊕ D(u1 , v1 ) ⊕ D(u2 , v2 ) ⊕ · · · ⊕ D(u` , v` ) is a Hamilton
cycle.
It is easy to see that f is an automorphism of each D(ui , vi ),
i = 1, . . . , ` and of H0 .
Thus f is also an isomorphism H00 .
It suffices to show by Lemma 1 that (1) H00 /f is a cycle and (2) H00
contains a path of length > p.
H00 /f =H0 /f ⊕D(u1 , v1 )/f ⊕D(u2`, v2 )/f ⊕ · · · ⊕D(u` , v` )/f
M
= [0][1] · · · [p−1][0] ⊕
D(ui , vi )/f
i=1
`
M [vi ] [vi −1] = [0] · · · [p−1][0] ⊕
A cycle!
i=1
[0]
[1]
[10]
[2]
[9]
[3]
[8]
[4]
[7]
[6]
[5]
[ui ]
[ui −1]
H00 = H0 ⊕ D(u1 , v1 ) ⊕ D(u2 , v2 ) ⊕ · · · ⊕ D(u` , v` ) is a Hamilton
cycle.
It is easy to see that f is an automorphism of each D(ui , vi ),
i = 1, . . . , ` and of H0 .
Thus f is also an isomorphism H00 .
It suffices to show by Lemma 1 that (1) H00 /f is a cycle and (2) H00
contains a path of length > p.
H00 /f =H0 /f ⊕D(u1 , v1 )/f ⊕D(u2`, v2 )/f ⊕ · · · ⊕D(u` , v` )/f
M
= [0][1] · · · [p−1][0] ⊕
D(ui , vi )/f
i=1
`
M [vi ] [vi −1] = [0] · · · [p−1][0] ⊕
A cycle!
[ui ]
i=1
[0]
[1]
[10]
[2]
[9]
L
[3]
[8]
[4]
[4]
[7]
[6]
[5]
[6]
[5]
[ui −1]
H00 = H0 ⊕ D(u1 , v1 ) ⊕ D(u2 , v2 ) ⊕ · · · ⊕ D(u` , v` ) is a Hamilton
cycle.
It is easy to see that f is an automorphism of each D(ui , vi ),
i = 1, . . . , ` and of H0 .
Thus f is also an isomorphism H00 .
It suffices to show by Lemma 1 that (1) H00 /f is a cycle and (2) H00
contains a path of length > p.
H00 /f =H0 /f ⊕D(u1 , v1 )/f ⊕D(u2`, v2 )/f ⊕ · · · ⊕D(u` , v` )/f
M
= [0][1] · · · [p−1][0] ⊕
D(ui , vi )/f
i=1
`
M [vi ] [vi −1] = [0] · · · [p−1][0] ⊕
A cycle!
[ui ]
i=1
[ui −1]
[0]
[0]
[1]
[10]
[2]
[9]
=
[3]
[8]
[5]
[3]
[8]
[4]
[4]
[6]
[2]
[9]
L
[7]
[1]
[10]
[6]
[5]
[4]
[7]
[6]
[5]
H00 = H0 ⊕ D(u1 , v1 ) ⊕ D(u2 , v2 ) ⊕ · · · ⊕ D(u` , v` ) is a Hamilton
cycle.
It is easy to see that f is an automorphism of each D(ui , vi ),
i = 1, . . . , ` and of H0 .
Thus f is also an isomorphism H00 .
It suffices to show by Lemma 1 that (1) H00 /f is a cycle and (2) H00
contains a path of length > p.
H00 /f =H0 /f ⊕D(u1 , v1 )/f ⊕D(u2`, v2 )/f ⊕ · · · ⊕D(u` , v` )/f
M
= [0][1] · · · [p−1][0] ⊕
D(ui , vi )/f
i=1
`
M [vi ] [vi −1] = [0] · · · [p−1][0] ⊕
A cycle!
i=1
[0]
[1]
[10]
[2]
[9]
[3]
[8]
[4]
[7]
[6]
[5]
[ui ]
[ui −1]
H00 = H0 ⊕ D(u1 , v1 ) ⊕ D(u2 , v2 ) ⊕ · · · ⊕ D(u` , v` ) is a Hamilton
cycle.
It is easy to see that f is an automorphism of each D(ui , vi ),
i = 1, . . . , ` and of H0 .
Thus f is also an isomorphism H00 .
It suffices to show by Lemma 1 that (1) H00 /f is a cycle and (2) H00
contains a path of length > p.
H00 /f =H0 /f ⊕D(u1 , v1 )/f ⊕D(u2`, v2 )/f ⊕ · · · ⊕D(u` , v` )/f
M
= [0][1] · · · [p−1][0] ⊕
D(ui , vi )/f
i=1
`
M [vi ] [vi −1] = [0] · · · [p−1][0] ⊕
A cycle!
i=1
[0]
[0]
[1]
[10]
[10]
[2]
[9]
[9]
L
[3]
[8]
[4]
[7]
[6]
[ui ]
[5]
[1]
[ui −1]
H00 = H0 ⊕ D(u1 , v1 ) ⊕ D(u2 , v2 ) ⊕ · · · ⊕ D(u` , v` ) is a Hamilton
cycle.
It is easy to see that f is an automorphism of each D(ui , vi ),
i = 1, . . . , ` and of H0 .
Thus f is also an isomorphism H00 .
It suffices to show by Lemma 1 that (1) H00 /f is a cycle and (2) H00
contains a path of length > p.
H00 /f =H0 /f ⊕D(u1 , v1 )/f ⊕D(u2`, v2 )/f ⊕ · · · ⊕D(u` , v` )/f
M
= [0][1] · · · [p−1][0] ⊕
D(ui , vi )/f
i=1
`
M [vi ] [vi −1] = [0] · · · [p−1][0] ⊕
A cycle!
i=1
[0]
[ui −1]
[0]
[1]
[10]
[10]
[2]
[9]
[3]
[8]
[4]
[7]
[5]
[0]
[1]
[1]
[10]
[9]
L
[6]
[ui ]
[2]
[9]
=
[3]
[8]
[4]
[7]
[6]
[5]
H00 = H0 ⊕ D(u1 , v1 ) ⊕ D(u2 , v2 ) ⊕ · · · ⊕ D(u` , v` ) is a Hamilton
cycle.
It is easy to see that f is an automorphism of each D(ui , vi ),
i = 1, . . . , ` and of H0 .
Thus f is also an isomorphism H00 .
It suffices to show by Lemma 1 that (1) H00 /f is a cycle and (2) H00
contains a path of length > p.
H00 /f =H0 /f ⊕D(u1 , v1 )/f ⊕D(u2`, v2 )/f ⊕ · · · ⊕D(u` , v` )/f
M
= [0][1] · · · [p−1][0] ⊕
D(ui , vi )/f
i=1
`
M [vi ] [vi −1] = [0] · · · [p−1][0] ⊕
A cycle!
i=1
[ui −1]
[ui ]
[0]
[1]
[10]
[1]
[2]
[9]
[9]
L
[3]
[8]
[4]
[7]
[6]
[5]
[8]
[2]
H00 = H0 ⊕ D(u1 , v1 ) ⊕ D(u2 , v2 ) ⊕ · · · ⊕ D(u` , v` ) is a Hamilton
cycle.
It is easy to see that f is an automorphism of each D(ui , vi ),
i = 1, . . . , ` and of H0 .
Thus f is also an isomorphism H00 .
It suffices to show by Lemma 1 that (1) H00 /f is a cycle and (2) H00
contains a path of length > p.
H00 /f =H0 /f ⊕D(u1 , v1 )/f ⊕D(u2`, v2 )/f ⊕ · · · ⊕D(u` , v` )/f
M
= [0][1] · · · [p−1][0] ⊕
D(ui , vi )/f
i=1
`
M [vi ] [vi −1] = [0] · · · [p−1][0] ⊕
A cycle!
i=1
[ui −1]
[ui ]
[0]
[1]
[10]
[1]
[2]
[9]
[9]
L
[3]
[8]
[4]
[7]
[6]
[5]
[8]
[2]
H00 = H0 ⊕ D(u1 , v1 ) ⊕ D(u2 , v2 ) ⊕ · · · ⊕ D(u` , v` ) is a Hamilton
cycle.
It is easy to see that f is an automorphism of each D(ui , vi ),
i = 1, . . . , ` and of H0 .
Thus f is also an isomorphism H00 .
It suffices to show by Lemma 1 that (1) H00 /f is a cycle and (2) H00
contains a path of length > p.
H00 /f =H0 /f ⊕D(u1 , v1 )/f ⊕D(u2`, v2 )/f ⊕ · · · ⊕D(u` , v` )/f
M
= [0][1] · · · [p−1][0] ⊕
D(ui , vi )/f
i=1
`
M [vi ] [vi −1] = [0] · · · [p−1][0] ⊕
A cycle!
i=1
[ui −1]
[ui ]
[0]
[0]
[1]
[10]
[1]
[2]
[9]
[9]
L
[3]
[8]
[4]
[7]
[6]
[5]
[1]
[10]
[2]
[2]
[9]
=
[8]
[3]
[8]
[4]
[7]
[6]
[5]
H00 = H0 ⊕ D(u1 , v1 ) ⊕ D(u2 , v2 ) ⊕ · · · ⊕ D(u` , v` ) is a Hamilton
cycle.
[0]
[10]
[1]
[2]
[9]
[3]
[8]
[4]
[7]
[6]
H00 /f
[5]
⊕ D(6, 7) ⊕ D(10, 1) ⊕ D(9, 2)/f
H00 = H0 ⊕ D(u1 , v1 ) ⊕ D(u2 , v2 ) ⊕ · · · ⊕ D(u` , v` ) is a Hamilton
cycle.
[0]
10
9
8
[2]
[9]
7
6
5
[3]
[8]
4
3
[4]
[7]
2
1
[5]
[6]
0
H0 /f ⊕ D(6, 7) ⊕ D(10, 1) ⊕ D(9, 2)/f 0 0 1 2 3 4 5 6 7 8 9 10
[10]
[1]
H00 = H0 ⊕ D(u1 , v1 ) ⊕ D(u2 , v2 ) ⊕ · · · ⊕ D(u` , v` ) is a Hamilton
cycle.
[0]
10
9
8
[2]
[9]
7
6
5
[3]
[8]
4
3
[4]
[7]
2
1
[5]
[6]
0
H0 /f ⊕ D(6, 7) ⊕ D(10, 1) ⊕ D(9, 2)/f 0 0 1 2 3 4 5 6 7 8 9 10
[10]
[1]
H00 = H0 ⊕ D(u1 , v1 ) ⊕ D(u2 , v2 ) ⊕ · · · ⊕ D(u` , v` ) is a Hamilton
cycle.
[0]
10
9
8
[2]
[9]
7
6
5
[3]
[8]
4
3
[4]
[7]
2
1
[5]
[6]
0
H0 /f ⊕ D(6, 7) ⊕ D(10, 1) ⊕ D(9, 2)/f 0 0 1 2 3 4 5 6 7 8 9 10
[10]
[1]
H00 = H0 ⊕ D(u1 , v1 ) ⊕ D(u2 , v2 ) ⊕ · · · ⊕ D(u` , v` ) is a Hamilton
cycle.
[0]
10
9
8
[2]
[9]
7
6
5
[3]
[8]
4
3
[4]
[7]
2
1
[5]
[6]
0
H0 /f ⊕ D(6, 7) ⊕ D(10, 1) ⊕ D(9, 2)/f 0 0 1 2 3 4 5 6 7 8 9 10
[10]
[1]
H00 = H0 ⊕ D(u1 , v1 ) ⊕ D(u2 , v2 ) ⊕ · · · ⊕ D(u` , v` ) is a Hamilton
cycle.
[0]
10
9
8
[2]
[9]
7
6
5
[3]
[8]
4
3
[4]
[7]
2
1
[5]
[6]
0
H0 /f ⊕ D(6, 7) ⊕ D(10, 1) ⊕ D(9, 2)/f 0 0 1 2 3 4 5 6 7 8 9 10
[10]
[1]
H00 = H0 ⊕ D(u1 , v1 ) ⊕ D(u2 , v2 ) ⊕ · · · ⊕ D(u` , v` ) is a Hamilton
cycle.
[0]
10
9
8
[2]
[9]
7
6
5
[3]
[8]
4
3
[4]
[7]
2
1
[5]
[6]
0
H0 /f ⊕ D(6, 7) ⊕ D(10, 1) ⊕ D(9, 2)/f 0 0 1 2 3 4 5 6 7 8 9 10
[10]
[1]
H00 = H0 ⊕ D(u1 , v1 ) ⊕ D(u2 , v2 ) ⊕ · · · ⊕ D(u` , v` ) is a Hamilton
cycle.
[0]
10
9
8
[2]
[9]
7
6
5
[3]
[8]
4
3
[4]
[7]
2
1
[5]
[6]
0
H0 /f ⊕ D(6, 7) ⊕ D(10, 1) ⊕ D(9, 2)/f 0 0 1 2 3 4 5 6 7 8 9 10
[10]
[1]
H00 = H0 ⊕ D(u1 , v1 ) ⊕ D(u2 , v2 ) ⊕ · · · ⊕ D(u` , v` ) is a Hamilton
cycle.
[0]
10
9
8
[2]
[9]
7
6
5
[3]
[8]
4
3
[4]
[7]
2
1
[5]
[6]
0
H0 /f ⊕ D(6, 7) ⊕ D(10, 1) ⊕ D(9, 2)/f 0 0 1 2 3 4 5 6 7 8 9 10
[10]
[1]
H00 = H0 ⊕ D(u1 , v1 ) ⊕ D(u2 , v2 ) ⊕ · · · ⊕ D(u` , v` ) is a Hamilton
cycle.
[0]
10
9
8
[2]
[9]
7
6
5
[3]
[8]
4
3
[4]
[7]
2
1
[5]
[6]
0
H0 /f ⊕ D(6, 7) ⊕ D(10, 1) ⊕ D(9, 2)/f 0 0 1 2 3 4 5 6 7 8 9 10
[10]
[1]
H00 = H0 ⊕ D(u1 , v1 ) ⊕ D(u2 , v2 ) ⊕ · · · ⊕ D(u` , v` ) is a Hamilton
cycle.
[0]
10
9
8
[2]
[9]
7
6
5
[3]
[8]
4
3
[4]
[7]
2
1
[5]
[6]
0
H0 /f ⊕ D(6, 7) ⊕ D(10, 1) ⊕ D(9, 2)/f 0 0 1 2 3 4 5 6 7 8 9 10
[10]
[1]
H00 = H0 ⊕ D(u1 , v1 ) ⊕ D(u2 , v2 ) ⊕ · · · ⊕ D(u` , v` ) is a Hamilton
cycle.
[0]
10
9
8
[2]
[9]
7
6
5
[3]
[8]
4
3
[4]
[7]
2
1
[5]
[6]
0
H0 /f ⊕ D(6, 7) ⊕ D(10, 1) ⊕ D(9, 2)/f 0 0 1 2 3 4 5 6 7 8 9 10
[10]
[1]
H00 = H0 ⊕ D(u1 , v1 ) ⊕ D(u2 , v2 ) ⊕ · · · ⊕ D(u` , v` ) is a Hamilton
cycle.
[0]
10
9
8
[2]
[9]
7
6
5
[3]
[8]
4
3
[4]
[7]
2
1
[5]
[6]
0
H0 /f ⊕ D(6, 7) ⊕ D(10, 1) ⊕ D(9, 2)/f 0 0 1 2 3 4 5 6 7 8 9 10
[10]
[1]
Corollary
If ~u , ~v ∈ Zp × Zp are linearly independent and the Cayley graph X on
Zp × Zp has connection set
S = {±~u , ±s1~v , ±S2~v , . . . , ±s`~v },
where 0 < s1 < s2 < · · · < s` < (p − 1)/2, then X is Hamilton
decomposable.
Proof. Let M be any 2 by 2 non-singular matrix.
Set S 0 = MS = {M ~u : ~u ∈ S} and X 0 = C AY (Zp × Zp ; S 0 ) .
Then ~u 7→ M ~u is an isomorphism.
M ~x − M ~y = M(~x − ~y ).
Hence ~x − ~y ∈ S if and only if M ~x − M ~y ∈ S 0 .
Thus X and X 0 are isomorphic.
Consequently we may assume ~u = (1, 0) and ~v = (0, 1).
Then X 0 and hence X is Hamilton decomposable by Lemma 2.
.
Lemma 3
Let p be an odd prime. If ~u and ~v are linearly independent, then
C AY (Zp × Zp ; S~u ∪ S~v ) is Hamilton decomposable.
Proof. Suppose a = |S~u | ≤ b = |S~v |.
S~u
S~v
±α1~u
±β1~v
±α2~u
..
.
±αa−1~u
±β2~v
..
.
±βa−1~v
±αa~u
±βa~v
±βa+1~v
±βa+2~v
..
.
±βb ~v
Lemma 3
Let p be an odd prime. If ~u and ~v are linearly independent, then
C AY (Zp × Zp ; S~u ∪ S~v ) is Hamilton decomposable.
Proof. Suppose a = |S~u | ≤ b = |S~v |.
S~u
S~v
±α1~u
±β1~v
±α2~u
..
.
±αa−1~u
±β2~v
..
.
±βa−1~v
±αa~u
±βa~v
±βa+1~v
±βa+2~v
..
.
±βb ~v
The Corollary with ` = 1.
Lemma 3
Let p be an odd prime. If ~u and ~v are linearly independent, then
C AY (Zp × Zp ; S~u ∪ S~v ) is Hamilton decomposable.
Proof. Suppose a = |S~u | ≤ b = |S~v |.
S~u
S~v
±α1~u
±β1~v
The Corollary with ` = 1.
±α2~u
..
.
±αa−1~u
±β2~v
..
.
±βa−1~v
The Corollary with ` = 1.
±αa~u
±βa~v
±βa+1~v
±βa+2~v
..
.
±βb ~v
Lemma 3
Let p be an odd prime. If ~u and ~v are linearly independent, then
C AY (Zp × Zp ; S~u ∪ S~v ) is Hamilton decomposable.
Proof. Suppose a = |S~u | ≤ b = |S~v |.
S~u
S~v
±α1~u
±β1~v
The Corollary with ` = 1.
±α2~u
..
.
±αa−1~u
±β2~v
..
.
±βa−1~v
The Corollary with ` = 1.
±αa~u
±βa~v
±βa+1~v
±βa+2~v
..
.
±βb ~v
The Corollary with ` = 1.
Lemma 3
Let p be an odd prime. If ~u and ~v are linearly independent, then
C AY (Zp × Zp ; S~u ∪ S~v ) is Hamilton decomposable.
Proof. Suppose a = |S~u | ≤ b = |S~v |.
S~u
S~v
±α1~u
±β1~v
The Corollary with ` = 1.
±α2~u
..
.
±αa−1~u
±β2~v
..
.
±βa−1~v
The Corollary with ` = 1.
±αa~u
±βa~v
±βa+1~v
±βa+2~v
..
.
±βb ~v
The Corollary with ` = 1.
The Corollary with ` = b − a.
Lemma 4
Let p be an odd prime. If ~u and ~v are linearly independent, then
C AY (Zp × Zp ; S~u ∪ S~v ∪ Sw~ ) is Hamilton decomposable.
Lemma 4
Let p be an odd prime. If ~u and ~v are linearly independent, then
C AY (Zp × Zp ; S~u ∪ S~v ∪ Sw~ ) is Hamilton decomposable.
Proof. Several cases.
Lemma 4
Let p be an odd prime. If ~u and ~v are linearly independent, then
C AY (Zp × Zp ; S~u ∪ S~v ∪ Sw~ ) is Hamilton decomposable.
Proof. Several cases.
S~u S~v Sw~
Lemma 4
Let p be an odd prime. If ~u and ~v are linearly independent, then
C AY (Zp × Zp ; S~u ∪ S~v ∪ Sw~ ) is Hamilton decomposable.
Proof. Several cases.
S~u S~v Sw~
The Corollary
with ` = 1.
Lemma 4
Let p be an odd prime. If ~u and ~v are linearly independent, then
C AY (Zp × Zp ; S~u ∪ S~v ∪ Sw~ ) is Hamilton decomposable.
Proof. Several cases.
S~u S~v Sw~
The Corollary
with ` = 1.
S~u S~v Sw~
Lemma 4
Let p be an odd prime. If ~u and ~v are linearly independent, then
C AY (Zp × Zp ; S~u ∪ S~v ∪ Sw~ ) is Hamilton decomposable.
Proof. Several cases.
S~u S~v Sw~
The Corollary
with ` = 1.
S~u S~v Sw~
Lemma 4
Let p be an odd prime. If ~u and ~v are linearly independent, then
C AY (Zp × Zp ; S~u ∪ S~v ∪ Sw~ ) is Hamilton decomposable.
Proof. Several cases.
S~u S~v Sw~
The Corollary
with ` = 1.
S~u S~v Sw~
The Corollary
with ` > 1.
Lemma 4
Let p be an odd prime. If ~u and ~v are linearly independent, then
C AY (Zp × Zp ; S~u ∪ S~v ∪ Sw~ ) is Hamilton decomposable.
Proof. Several cases.
S~u S~v Sw~
The Corollary
with ` = 1.
S~u S~v Sw~
The Corollary
with ` > 1.
S~u S~v Sw~
Lemma 4
Let p be an odd prime. If ~u and ~v are linearly independent, then
C AY (Zp × Zp ; S~u ∪ S~v ∪ Sw~ ) is Hamilton decomposable.
Proof. Several cases.
S~u S~v Sw~
The Corollary
with ` = 1.
S~u S~v Sw~
The Corollary
with ` > 1.
S~u S~v Sw~
Lemma 4
Let p be an odd prime. If ~u and ~v are linearly independent, then
C AY (Zp × Zp ; S~u ∪ S~v ∪ Sw~ ) is Hamilton decomposable.
Proof. Several cases.
S~u S~v Sw~
The Corollary
with ` = 1.
S~u S~v Sw~
The Corollary
with ` > 1.
S~u S~v Sw~
Westlund,Liu,Kreher
Odd order 6-regular
connected Cayley graphs
are Hamilton decomposable.
Lemma 4
Let p be an odd prime. If ~u and ~v are linearly independent, then
C AY (Zp × Zp ; S~u ∪ S~v ∪ Sw~ ) is Hamilton decomposable.
Proof. Several cases.
S~u S~v Sw~
S~u S~v Sw~
S~u S~v Sw~
and so forth.
The Corollary
with ` = 1.
The Corollary
with ` > 1.
Westlund,Liu,Kreher
Odd order 6-regular
connected Cayley graphs
are Hamilton decomposable.
Theorem 1
Every connected Cayley graph X = C AY (Zp × Zp ; S), p prime, is
Hamilton decomposable.
Proof. If ~v ∈ Zp × Zp , then ~v = α~u , where
~u ∈ I = {(1, 0), (0, 1), (1, 1), (2, 1), . . . , (p − 1, 1)}
and α ∈ Zp . The vectors in I are pairwise linear independent.
Write
[
S=
S~u
~
u ∈I
where
S~u = {~v ∈ S : ~v = α~u , for some α ∈ Zp } and S~u 6= ∅.
Decompose X into connected Cayley sub-graphs of the form
C AY (Zp × Zp ; S~u ∪ S~v )
and/or
C AY (Zp × Zp ; S~u ∪ S~v ∪ Sw~ )
Apply Lemma’s 3 and 4.
Example: C AY (Z11 × Z11 ; {±(1, 0), ±(0, 1), ±(0, 2), ±(0, 4)})
10
10
9
9
8
8
7
7
6
6
5
5
4
4
3
3
2
2
1
1
0
0
0 1 2 3 4 5 6 7 8 9 10
0 1 2 3 4 5 6 7 8 9 10
10
10
9
9
8
8
7
7
6
6
5
5
4
4
3
3
2
2
1
1
0
0
0 1 2 3 4 5 6 7 8 9 10
0 1 2 3 4 5 6 7 8 9 10
Example: C AY (Z11 × Z11 ; {±(1, 0), ±(0, 1), ±(0, 2), ±(0, 4)})
10
10
9
9
8
8
7
7
6
6
5
5
4
4
3
3
2
2
1
1
0
0
0 1 2 3 4 5 6 7 8 9 10
0 1 2 3 4 5 6 7 8 9 10
10
10
9
9
8
8
7
7
6
6
5
5
4
4
3
3
2
2
1
1
0
0
0 1 2 3 4 5 6 7 8 9 10
0 1 2 3 4 5 6 7 8 9 10
2
Circulant graphs of
order p
Let X = C AY Zp2 ; S .
If a ∈ S and GCD(p, a) = 1, then the subgraph C AY Zp2 ; ±a is an
H-cycle.
Thus we may assume up to isomorphism
S = {±1, ±`1 p, ±`2 p, · · · ± `k p}
where 1 ≤ `j ≤ (p − 1)/2.
110
99
88
77
66
55
44
33
22
11
110
99
88
77
66
55
44
33
22
11
0 1 2 3 4 5 6 7 8 9 10
C AY (Z112 ; {±1})
0 1 2 3 4 5 6 7 8 9 10
C AY (Z112 ; {±2 · 11})
Hamilton decomposition of C AY Zp2 ; S
For example consider C AY (Z112 ; {±1, ±1 · 11, ±2 · 11, ±4 · 11})
0
10
1
9
2
3
8
7
4
6
5
Hamilton decomposition of C AY Zp2 ; S
For example consider C AY (Z112 ; {±1, ±1 · 11, ±2 · 11, ±4 · 11})
0
10
9
8
7
6
110
99
1
88
2 77
66
55
3
44
33
4
22
11
5
0
0 1 2 3 4 5 6 7 8 9 10
Hamilton decomposition of C AY Zp2 ; S
For example consider C AY (Z112 ; {±1, ±1 · 11, ±2 · 11, ±4 · 11})
0
10
9
8
7
6
110
99
1
88
2 77
66
55
3
44
33
4
22
11
5
0
0 1 2 3 4 5 6 7 8 9 10
Hamilton decomposition of C AY Zp2 ; S
For example consider C AY (Z112 ; {±1, ±1 · 11, ±2 · 11, ±4 · 11})
0
10
9
8
7
6
110
99
1
88
2 77
66
55
3
44
33
4
22
11
5
0
0 1 2 3 4 5 6 7 8 9 10
Hamilton decomposition of C AY Zp2 ; S
For example consider C AY (Z112 ; {±1, ±1 · 11, ±2 · 11, ±4 · 11})
0
10
9
8
7
6
110
99
1
88
2 77
66
55
3
44
33
4
22
11
5
0
0 1 2 3 4 5 6 7 8 9 10
Hamilton decomposition of C AY Zp2 ; S
For example consider C AY (Z112 ; {±1, ±1 · 11, ±2 · 11, ±4 · 11})
0
10
9
8
7
6
110
99
1
88
2 77
66
55
3
44
33
4
22
11
5
0
0 1 2 3 4 5 6 7 8 9 10
Hamilton decomposition of C AY Zp2 ; S
For example consider C AY (Z112 ; {±1, ±1 · 11, ±2 · 11, ±4 · 11})
0
10
9
8
7
6
110
99
1
88
2 77
66
55
3
44
33
4
22
11
5
0
0 1 2 3 4 5 6 7 8 9 10
Hamilton decomposition of C AY Zp2 ; S
For example consider C AY (Z112 ; {±1, ±1 · 11, ±2 · 11, ±4 · 11})
0
10
9
8
7
6
110
99
1
88
2 77
66
55
3
44
33
4
22
11
5
0
0 1 2 3 4 5 6 7 8 9 10
Hamilton decomposition of C AY Zp2 ; S
For example consider C AY (Z112 ; {±1, ±1 · 11, ±2 · 11, ±4 · 11})
0
10
9
8
7
6
110
99
1
88
2 77
66
55
3
44
33
4
22
11
5
0
0 1 2 3 4 5 6 7 8 9 10
Hamilton decomposition of C AY Zp2 ; S
For example consider C AY (Z112 ; {±1, ±1 · 11, ±2 · 11, ±4 · 11})
0
10
9
8
7
6
110
99
1
88
2 77
66
55
3
44
33
4
22
11
5
0
0 1 2 3 4 5 6 7 8 9 10
Hamilton decomposition of C AY Zp2 ; S
For example consider C AY (Z112 ; {±1, ±1 · 11, ±2 · 11, ±4 · 11})
0
10
9
8
7
6
110
99
1
88
2 77
66
55
3
44
33
4
22
11
5
0
0 1 2 3 4 5 6 7 8 9 10
Hamilton decomposition of C AY Zp2 ; S
For example consider C AY (Z112 ; {±1, ±1 · 11, ±2 · 11, ±4 · 11})
0
10
9
8
7
6
110
99
1
88
2 77
66
55
3
44
33
4
22
11
5
0
0 1 2 3 4 5 6 7 8 9 10
Hamilton decomposition of C AY Zp2 ; S
For example consider C AY (Z112 ; {±1, ±1 · 11, ±2 · 11, ±4 · 11})
0
10
9
8
7
6
110
99
1
88
2 77
66
55
3
44
33
4
22
11
5
0
0 1 2 3 4 5 6 7 8 9 10
Hamilton decomposition of C AY Zp2 ; S
For example consider C AY (Z112 ; {±1, ±1 · 11, ±2 · 11, ±4 · 11})
0
10
9
8
7
6
110
99
1
88
2 77
66
55
3
44
33
4
22
11
5
0
0 1 2 3 4 5 6 7 8 9 10
Hamilton decomposition of C AY Zp2 ; S
For example consider C AY (Z112 ; {±1, ±1 · 11, ±2 · 11, ±4 · 11})
0
10
9
8
7
6
110
99
1
88
2 77
66
55
3
44
33
4
22
11
5
0
Hamilton decomposition of C AY Zp2 ; S
For example consider C AY (Z112 ; {±1, ±1 · 11, ±2 · 11, ±4 · 11})
0
10
9
8
7
6
110
99
1
88
2 77
66
55
3
44
33
4
22
11
5
0
Hamilton decomposition of C AY Zp2 ; S
For example consider C AY (Z112 ; {±1, ±1 · 11, ±2 · 11, ±4 · 11})
0
10
9
8
7
6
110
99
1
88
2 77
66
55
3
44
33
4
22
11
5
0
0
10
1
9
2
⇒
3
8
7
4
6
5
Theorem: Connected C AY Zp2 ; S is Hamilton decomposable.
Proof. It suffices to show that C AY Zp2 ; {±1, ±`1 p, ±`2 p, . . . ± `k p} ,
where 0 < `1 < `2 < · · · < `k < (p − 1/2) is Hamiliton decomposable.
There are 3 stages.
0
`1
−`2
`3
`3
S
⇒
⇒
−`2
S
`1
I. Genesis.
II. Mitosis.
0
III. Evolution.
References
I B. Alspach, Research Problem 59, Discrete Math. 50 (1984), 115.
I B. Alspach, J.-C. Bermond and D. Sotteau, Decompositions into cycles I:
Hamiliton decompositions, in Cycles and Rays, edited by G. Hahn, G. Sabidussi,
and R. Woodrow, Kluwer Acad. Publ., Dordrecht, (1990) 9–18.
I J.-C. Bermond, Hamilton decomposition of graphs, directed graphs, and
hypergraphs, Advances in Graph Theory, Annals of Discrete Math. 3 (1978),
21–28.
I J.-C. Bermond, O. Favaron, and M. Maheo, Hamiltonian decomposiiton of Cayley
graphs of degree four, J. Combin. Theory Ser. B 46, (1990), 142–153.
I M. Dean, Hamilton cycle decomposition of 6-regular circulants of odd order,
J. Combin. Des. 15 (2007), 91–97. 69B (1965), 147–153.
I J. Liu, Hamiltonian decompositions of Cayley graphs on abelian groups, Discrete
Math. 131 (1994), 163–171.
I J. Liu, Hamiltonian decompositions of Cayley graphs on abelian groups of odd
order, J. Combin. Theory Ser. B 66 (1996), 75–86.
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Amusing e-mail
Mail - We are done! - 4/30/13
From: Brian Alspach
<[email protected]>
To: [email protected]
Cc: [email protected]
Yesterday morning I awoke with a new idea on how to finish our
problem. When I arrived at the university, Don already was
there so I showed him what I was trying to do. I then told Don I
really wanted to finish a SIAM book review and why didn’t he
play with it. At 3 pm I had finished a first draft of the book
review and was ready to look at this approach with Don. He
was looking somewhat smirky and I quickly discovered
why. He had nicely refined my idea and we had a proof. What I
like is that it generalizes one thing I had done for the one
vertical length case. Here is a thumbnail sketch. .....
As always, Brian