Existence of Decompositions
of Multipartite Graphs,
some Gregarious and some Fair
Chris Rodger
Auburn University
Graph Decompositions
• Partition the edges of one of your favourite
graphs G into copies of another of your
favorite graphs H.
Perhaps G = K8 and H = P3 =
Since K8 has
28 edges,
you made a
bad choice!
Graph Decompositions
• Partition the edges of one of your favourite
graphs G into copies of another of your
favorite graphs H.
Now you have a
Perhaps G = K8 and H = P4 chance!
Can’t the top
left vertex
be the
middle of
some path?
Graph Decompositions
• Partition the edges of one of your favourite
graphs G into copies of another of your
favorite graphs H.
Perhaps G = K8 and H is a hamilton cycle.
Another bad
choice: each
vertex in K8
has odd
degree!
Graph Decompositions
• Partition the edges of one of your favourite
graphs G into copies of another of your
favorite graphs H.
Perhaps G = K8 - F and H is a hamilton cycle.
Now you
have a
chance to
succeed!
Spouse Avoiding Dinners
Try to find a way for 4 couples to sit at 1 round
table so that each sits next to each other
person exactly once.
Not the spouses!
1
2
3
M1
4
W4
M4
Men
Women
Friday; Saturday; Sunday
M2
W2
M3
W3
W1
Spouse Avoiding Dinners
Try to find a way for 4 couples to sit at 2 tables,
each seating 4 people so that each sits next to
each other person exactly once.
Not the spouses!
1
2
3
4
Friday
Saturday
Men
Women
Sunday
M1
M3
M2
W4
W1
W3
W2
M2
M1
M3
W2
W1
W3
W3
W2
W1
M4
M3
M2
M1
Spouse Avoiding Dinners
Try to find a way for 4 couples to sit at 2
tables, each seating 4 people so that each sits
next to each other person exactly once.
Not the spouses!
1
Men
Women
2
3
4
Friday
Saturday
Sunday
Can you do this so
that each table has 2
men and 2 women?
Must one avoid one’s spouse??
No! You now have an excuse for another dinner!
Cycle systems of graphs other than Kn are
also interesting.
Join vertices in the
same group with λ1
edges and vertices in
different groups with
λ2 edges
λ1 = 2 and λ2 = 1
Pure and Mixed Edges
Graph Decompositions with Two
Associate Classes
Here is my current favourite G:
1
2
K(a,p;λ1,λ2) =
λ1
λ2
a
1
2
p
Cycle Systems with 2 Associate Classes
Maybe you have one big
table!
1
2
There exists a Cap-factorization λ1
(i.e. hamilton decomposition)
a
λ2
of K(a,p;λ1,λ2) if and only if:
1 2
p
1. λ1 (a-1) + λ2a(p-1) is even,
and
K(a,p;λ1,λ2)
2. λ2a(p-1) ≥ λ1.
(Bahmanian, Rodger)
Cycle Systems with 2 Associate Classes
Tables of size 4 are more common!
Suppose a is even.
1
2
There exists a C4-factorization
of K(a,p;λ1,λ2) if and only if
1. 4 divides ap
2. λ1 is even, and
3. If a ≡ 2 (mod 4) then
λ2a(p-1) ≥ λ1.
(Billington, Rodger)
λ1
λ2
a
1 2
K(a,p;λ1,λ2)
p
Why must λ2a(p-1) ≥ λ1?
Suppose a ≡ 2 (mod 4).
Consider one C4-factor.
Every part must contain at
least 2 vertices incident with
mixed edges.
So each C 4
1
factor must
contain at
least p mixed
edges!
2
a =6
K(a,p;λ1,λ2)
1
p
The same
argument
works for
hamilton
cycles.
What about a ≡ 3 (mod 4)?
Looks
difficult
from my
point of
view!
How do the proofs go?
You need different perspectives!
For the hamilton
cycles,
use amalgamations!
For 4-cycles use design
theoretic methods!
4-cycle systems of K(a,p;λ1,λ2)
There exists a 4-cycle system of K(a,p;λ1,λ2) if and only if
1. Each vertex has even degree,
2. The number of edges is divisible by 4, 1
2
3. If a = 2 then
λ1
• λ2 > 0,
and
a
λ2
• λ1 ≤ 2(p-1) λ2
12
p
4. If a = 3 then
K(a,p;λ1,λ2)
• λ2 > 0, and
• λ1 ≤ 3(p-1) λ2/2
if λ2 is even, and
• λ1 ≤ 3(p-1) λ2/2 - (p-1)/9
if λ2 is odd.
(Hung Lin Fu, Rodger)
For 3-cycles: Fu, Rodger, Sarvate
For block designs: Bose and Shimamoto – 1952!
Gregarious Graph Decompositions
with Two Associate Classes
Given a partition of the vertices of G, a graph
decomposition of G into copies of H is said to be
gregarious if the vertices of each copy of H “are
shared as evenly among the parts as possible”.
(Other versions: “come from as many parts as
possible”; “are all in different parts”. )
Gregarious 3-cycle systems (also GDDs)
with Two Associate Classes
• 3- cycles are also copies of K3
• The most interesting case is when p = 2.
1
2
λ1
a
λ2
1 2
p
Gregarious 3-cycle systems (also GDDs)
with Two Associate Classes
• 3- cycles are also copies of K3
• The most interesting case is when p = 2,
• but we allow the parts to have different sizes.
0
0
In general:
# pure edges ≥ # (mixed edges)/2
So:
λ1(m(m-1) + n(n-1)) ≥ λ2mn
with equality if and only if:
m-1
n-1
the GDD is gregarious.
Decompose G = Kn v2 Km
0
0
Consider the gregarious case where
λ1 = 1 and λ2 = 2.
The necessary condition
λ1(m(m-1) + n(n-1)) ≥ λ2mn
becomes
1. (m-n)2 = m + n,
m-1
and all vertices must have even degree, so
2. m and n are both odd.
n-1
Try this gregarious case: G = Kn v2 Km
1. (m-n)2 = m + n,
2. m and n are both odd.
Strangely enough, this means that:
m = ((m-n)2 + (m – n))/2, and
n = ((m-n)2 - (m-n))/2.
0
m-1
0
n-1
Notice that m-n is even, but is not divisible by 4.
So m-n = 4x+2 for some x, so we can write:
m = (4x+3)(2x+1), and
n = (4x+1)(2x+1).
Four Ingredients: #1
There exists a near 1-factorization of K2x+1 (a partition of
the edges into sets of x independent edges).
5 more of these rotations provides the partition
of the edges into near 1-factors.
Four Ingredients: #2
There exists a 1-factorization of Hi v Hj,
where Hi and Hj are hamilton cycles.
The join, v, is the union
of a 1-factor that is an
isomorphism from Hi to Hj,
Hi
Hj
Four Ingredients: #2
There exists a 1-factorization of Hi v Hj, where
Hi and Hj are hamilton cycles.
The join, v, is the union of:
1. a 1-factor that is an
isomorphism from Hi to Hj, and
2. a regular bipartite graph.
2
1
Hi
3
2
1
2
3
3
3
1
1
3
2
Hj
1
2
Give the hamilton cycles
the “same” coloring using
3 colors.
Colour the 1-factor with
the “missing” color.
Properly color the remaining bipartite graph.
Four Ingredients: #3
There exists a frame M(4x+1, 2x+1):
• Partition the vertices of G = K(4x+1)(2x+1) into
2x+1 parts Pi of size 4x+1.
• Partition the edges into sets, each of which
induces a 1-factor of G – V(Pi) for some i.
This looks like an expanded near 1-factorization
in some sense!
Partition the edges into sets, each of which
induces a 1-factor of K(4x+1)(2x+1) – V(Pi) for some i.
x=2
Use a near 1-factorization of K5: first near 1-factor.
Put a
1-factorizn
of H1 v H1
on each
pair of
copies of
the graph
in purple.
Use a near 1-factorization of K5: second near 1-factor.
Put a
1-factorizn
of H2 v H2
on each
pair of
copies of
the graph
in green.
Use a near 1-factorization of K5:three more near 1-factors.
So far
we have:
Put a
1-factorizn
of Hi v Hj
on each
pair of
the other
three
copies of
the graph.
The 5th near 1-factor:
the 1-factor of K(4x+1)(2x+1) – V(P2) is a little different.
Four Ingredients: #4
Lemma: There exists a decomposition of
(x+1)K2x+1 into copies of S(x) =
22 2
x=3
So we get 4K7
1
2
33 3
11 1
3
Each vertex is
the “hub” of
exactly x copies
of S(x).
Recall we are considering: G = Kn v2 Km
Necessary Conditions:
1. (m-n)2 = m + n,
2. m and n are both odd.
So m-n = 4x+2 for some x, so :
m = (4x+3)(2x+1), and
n = (4x+1)(2x+1).
0
4x+2
0
m-1
0
n-1
4x
0
0
Km
2x
0
Kn
v2
2x
Ingredient
z
Construction
0
0
z+d
d
z z+d+1
2x
2x
Near 1factorisation Each edge on the left forms a K3 with vertex “d”.
of K2x+1
0
4x+2
0
2x 2x+1
4x
Ingredient
Construction
0
2x+1
2x
4x+2
0
2x 2x+1
4x
0
4x+2
4x+2
4x+1
4x+1
4x
2x
(2x+2)K2x+1
into S(x)’s
Color the edges so that between each of the
vertices each color 2x+1 … 4x+2 is on an edge.
Ingredient
Construction
0
v
v
v
2x
Use a frame: the right hand side is (4x+3)-regular
Near 14x+2
0
2x 2x+1
4x
factorisation 0
Ingredient
Construction
0
v
d
z z+d
v
v
2x
0
4x+2
0
2x 2x+1
4x
The Result of all this:
There exists a gregarious 3-cycle system of
Km v2 Kn
if and only if
1. (m-n)2 = m + n,
2. m and n are both odd.
(El-Zanati, Punnim and Rodger)
That took a lot of muscle to attack just
a little bit of the problem
λ1 = 1
λ2 = 2
Gregarious p-cycles
There exists a resolvable gregarious p-cycle
system of K(n,p;0,λ2) if and only if:
1. If p is even then n is even, and
2. If p = 3 then n ≠ 2, 6
(Billington, Hoffman, Rodger)
1
2
0
λ2
n
Condition 2 can be removed
if you don’t want resolvability.
1 2
p
Gregarious 4-cycles
There exists a resolvable gregarious 4-cycle
system of K(r, s, t) if and only if:
• r, s, and t are all even
• s(r + t) – rt ≥ 8, and
• r(s + t) – st ≥ 8 or r(s + t) – st = 0.
(Billington and Hoffman)
r
They also consider gregarious
decompositions into paths of length 2.
s
t
Other Gregarious Cycle Results
• Gregarious k-cycle systems of complete
equipartite graphs with k in {6,8} and p ≥ k
(Billington, Hoffman, Smith)
• Cyclic gregarious 6-cycle systems of complete
equipartite graphs with p ≥ k
(Cho and Gould)
• Gregarious k-cycle systems of complete
equipartite graphs with p ≥ k when k is prime
(Smith)
Fair Hamilton Decompositions
There exist a hamilton decomposition of the complete
multipartite graph K(n, p) in which the edges between each
pair of parts are shared fairly (as evenly as possible) between
the hamilton cycles
if and only if n(p-1) is even.
(Leach and Rodger)
Fair holey1-factorizations of K(n,p)?
Let’s try K(3,5). So you have 15 holey 1-factors (colors).
Is this a holey 1-factor? No! There are 3 edges
between some pairs of parts
Yes! Is it fair?
and 0 between others.
These unfair decompositions are often easy to make
using direct products of symmetric idempotent latin
squares.
1
4
7
10
13
2
5
8
11
14
3
6
9
12
15
Unfair Frames
1
4
2
5
3
4
2
5
3
1
2
5
3
1
4
5
3
1
4
2
3
1
4
2
5
Unfair Frames
4
4
2
5
3
5
3
1
1
4
2
5
5
3
1
3
1
4
2
2
Unfair Frames
4
4
2
2
5
5
5
3
1 2 3
2 3 1
3 1 2
3
1 2 3
2 3 1
3 1 2
4
5
3
3
1 2 3
2 3 1
3 1 2
1 2 3
2 3 1
3 1 2
4
2
2
Unfair Frames
10
11
12
4
5
6
13
14
15
7
8
9
11
12
10
5
6
4
14
15
13
8
9
7
12
10
11
6
4
5
15
13
14
9
7
8
10
11
12
13
14
15
7
8
9
1
2
3
11
12
10
14
15
13
8
9
7
2
3
1
12
10
11
15
13
14
9
7
8
3
1
2
4
5
6
13
14
15
1
2
3
10
11
12
5
6
4
14
15
13
2
3
1
11
12
10
6
4
5
15
13
14
3
1
2
12
10
11
13
14
15
7
8
9
1
2
3
4
5
6
14
15
13
8
9
7
2
3
1
5
6
4
15
13
14
9
7
8
3
1
2
6
4
5
7
8
9
1
2
3
10
11
12
4
5
6
8
9
7
2
3
1
11
12
10
5
6
4
9
7
8
3
1
2
12
10
11
6
4
5
Unfair Frames
4
5
6
7
8
9
10 11 12 13 14 15
1
10
11
12
4
5
6
13
14
15
7
8
9
2
11
12
10
5
6
4
14
15
13
8
9
7
3
12
10
11
6
4
5
15
13
14
9
7
8
1
2
3
4
10
11
12
13
14
15
7
8
9
1
2
3
5
11
12
10
14
15
13
8
9
7
2
3
1
6
12
10
11
15
13
14
9
7
8
3
1
2
7
4
5
6
13
14
15
1
2
3
10
11
12
8
5
6
4
14
15
13
2
3
1
11
12
10
9
6
4
5
15
13
14
3
1
2
12
10
11
10
13
14
15
7
8
9
1
2
3
4
5
6
11
14
15
13
8
9
7
2
3
1
5
6
4
12
15
13
14
9
7
8
3
1
2
6
4
5
13
7
8
9
1
2
3
10
11
12
4
5
6
14
8
9
7
2
3
1
11
12
10
5
6
4
15
9
7
8
3
1
2
12
10
11
6
4
5
Which symbol produces
these holey 1-factors?
1
4
7
10
13
2
5
8
11
14
3
6
9
12
15
1
4
7
10
13
2
5
8
11
14
3
6
9
12
15
Symbol 1
Fair holey1-factorizations of K(n,p)
Let’s try K(3,5) again. So you have 15 holey 1-factors (colors).
We can make these using amalgamations!
(Erzurumlouglu and Rodger)
Color 1 = Symbol 1
1
4
7
10
13
2
5
8
11
14
3
6
9
12
15
Fair Frames
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15
1
2
3
1
4
1
5
1
6
7
1
1
8
1
9
1
10
1
11
1
12
1
13
1
14
15
1
Fair holey1-factorizations of K(n,p)
Let’s try K(3,5) again. So you have 15 holey 1-factors (colors).
Here is another holey 1-factor. Is it fair?
Which cells would it fill in the quasigroup we are making?
Color 1 = Symbol 1
Color 2 = Symbol ?
1
4
7
10
13
2
5
8
11
14
3
6
9
12
15
Fair Frames
1
2
3
10 11 12 13 14 15
4
5
6
7
8
9
1
7
10
13
4
14
12
6
15
8
5
9
11
2
14
8
11
10
5
15
9
4
13
6
7
12
3
12
15
9
13
11
6
14
7
5
4
8
10
4
7
14
12
1
10
13
15
2
9
3
11
8
5
10
8
15
14
2
11
7
13
3
9
12
1
6
13
11
9
12
15
3
1
8
14
2
10
7
7
4
10
13
1
14
12
5
3
15
11
6
2
8
14
5
11
10
2
15
13
6
1
12
4
3
9
12
15
6
13
11
3
2
14
4
10
1
5
10
6
9
14
15
7
1
5
13
2
8
3
4
11
15
4
7
2
13
8
3
6
14
1
5
9
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8
13
5
9
3
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15
1
4
7
2
6
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6
4
3
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2
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12
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8
1
7
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12
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6
4
1
3
5
2
15
11
12
10
8
1
7
2
3
5
4
9
6
Fair holey1-factorizations of K(n,p)
What would an amalgamation of a fair K(3,5) holey 1factorization look like?
1
4
13
2
5
14
3
6
15
It’s K4!
Thanks for listening!
Time for Coffee!!