Moore graphs of d=2 are v-valent and order is v^2+1

Moore graphs of

d

=

2

are

v

-valent and order is

v

2

+

1

Given

a

Moore graph

with

diameter

2 and girth 5, which implies the
existence of cycles.

To prove

every node (vertex) has the same

valency

,

v

say, and the number

n

of nodes (vertices) is

v

2

+

1

.

Lemma:

a key feature of Moore graphs with diameter 2 is that any nodes X
and that are not

adjacent

have exactly one shared neighbour Y.

Proof of lemma:

at least one Y

k

because

distance

XY is not 1 so must be 2, at most one because XY

0

ZY

1

would be a 4-cycle.

∘

∘

∘

Lemma:

every two adjacent nodes B and C lie together on some 5-cycle.

Proof of lemma:

every node (other than B) is at distance 1 or 2 from B, and every node (other than C) at distance 1 or 2 from C. There are no nodes X
at distance 1 from both (BXC would be a 3-cycle). Suppose the graph only has nodes at distance 1 from B and 2 from C (call them A

i

), and nodes at distance 1 from C and 2 from B (call them D

j

). Now no cycle can exist (the only edges are A

i

B, BC, CD

j

; any edge of type AA, AC, , BD, DD would create a 3-or 4-cycle). But Moore graphs have cycles by definition. So there must be at least one node E at distance 2 from both B and C. Let A be the joint neighbour of B and E, and D that of C and E (note A

≠

D, otherwise BAC would be a 3-cycle). Now CDEAB is a 5-cycle with edge BC.

∘

∘

∘

Lemma:

two nodes O and Q at distance 2 have the same valency.

Proof
of lemma:

let P be the unique joint neighbour. Let O have

o

other neighbours
N

i

and let Q have

q

other neighbours R

j

.

No N

i

can be adjacent to P (3-cycle PON

i

) so the unique joint neighbours
of any N

i

and Q must be among the R

j

. Different N

i

and N

i

′

cannot use the same R

j

(4-cycle ON

i

R

j

N

i

′

) so we have

o

⩽

q

.
By the same argument (swapping O and Q, N

i

and R

j

) also

q

⩽

o

so we
have

o

=

q

, both nodes have valency

q

+

1

.

∘

∘

∘

Lemma:

two adjacent nodes O and S have the same valency.

Proof
of lemma:

let OPQRS be the 5-cycle through SO. Calling the valency of Q
again

q

+

1

, both O and S have that same valency by the previous lemma.

∘

∘

∘

Proof of theorem:

the graph is

connected

. Travel from any node to any
other via adjacent ones, the valency stays the same by the last lemma (let’s
keep calling it

q

+

1

).

Now let N and R be any two adjacent nodes. N has

q

other neighbours O

i

and R has

q

other neighbours Q

j

. Call the joint neighbour of O

i

and
Q

j

now P

i

⁢

j

, these

q

2

nodes are all distinct (4-cycles of type
NOPO and/or RQPQ otherwise) and none of them coincide with N, R, the Os or Qs
(3- or 4-cycles otherwise). On the other hand, there are no further nodes
(distance

>

2

from N or R otherwise). Tally:

q

2

+

2

⁢

q

+

2

=

(

q

+

1

)

2

+

1

,
for valency

q

+

1

.

∘

∘

∘

Title

Moore graphs of

d

=

2

are

v

-valent and order is

v

2

+

1

Canonical name

MooreGraphsOfD2AreVvalentAndOrderIsV21

Date of creation

2013-03-22 15:11:30

Last modified on

2013-03-22 15:11:30

Owner

marijke (8873)

Last modified by

marijke (8873)

Numerical id

6

Author

marijke (8873)

Entry type

Proof

Classification

msc 05C75

Generated on Fri Feb 9 07:56:13 2018 by

LaTeXML