Open Journal of Discrete Mathematics, 2013, 3, 183-191
http://dx.doi.org/10.4236/ojdm.2013.34032 Published Online October 2013 (http://www.scirp.org/journal/ojdm)
Graph Derangements*
Pete L. Clark
University of Georgia, Athens, USA
Email: plclark@gmail.com
Received June 25, 2013; revised July 15, 2013; accepted August 10, 2013
Copyright © 2013 Pete L. Clark. This is an open access article distributed under the Creative Commons Attribution License, which
permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
ABSTRACT
We introduce the notion of a graph derangement, which naturally interpolates between perfect matchings and Hamilto-
nian cycles. We give a necessary and sufficient condition for the existence of graph derangements on a locally finite
graph. This result was first proved by W. T. Tutte in 1953 by applying some deeper results on digraphs. We give a new,
simple proof which amounts to a reduction to the (Menger-Egerváry-König-)Hall(-Hall) Theorem on transversals of set
systems. We also consider the problem of classifying all cycle types of graph derangements on m × n checkerboard
graphs. Our presentation does not assume any prior knowledge in graph theory or combinatorics: all definitions and
proofs of needed theorems are given.
Keywords: Graph Derangement Cycle Perfect Matching
1. Introduction
1.1. The Cockroach Problem
An interesting problem was conveyed to me at the
Normal Bar in Athens, GA. The following is a mathe-
matically faithful rendition, although some of the details
may be misremembered or mildly embroidered.
Consider a square kitchen floor tiled by 255 5
square tiles in the usual manner. Late one night the
house’s owner comes down to discover that the floor is
crawling with cockroaches: in fact each square tile
contains a single cockroach. Displeased, she goes to the
kitchen cabinet and pulls out an enormous can of roach
spray. The roaches sense what is coming and start skitter-
ing. Each roach has enough time to skitter to any
adjacent tile. But it will not be so good for two (or more)
roaches to skitter to the same tile: that will make an
obvious target. Is it possible for the roaches to perform a
collective skitter such that each ends up on its own tile?
This is a nice problem to give undergraduates: it is
concrete, fun, and far away from what they think they
should be doing in a math class.
1.2. Solution
Suppose that the tiles are painted black and white with a
checkerboard pattern, and that the center square is black,
so that there are 13 black squares and 12 white squares.
Therefore there are 13 roaches who start out on black
squares and are seeking a home on only 12 white squares.
It is not possible—no more than for pigeons!—for all 13
cockroaches to end up on different white squares.
1.3. A Problem for Mathematicians
For a grown mathematician (or even an old hand at
mathematical brainteasers), this is not a very challenging
problem, since the above parity considerations will
quickly leap to mind. Nevertheless there is something
about it that encourages further contemplation. There
were several other mathematicians at the Normal Bar and
they were paying attention too. “What about the 66
case?” One of them asked. “It reminds me of the Brou-
wer fixed point theorem,” muttered another1.
One natural follow-up is to ask what happens for
cockroaches on an mn
rectangular grid. The preced-
ing argument works when and are both odd. On
the other hand, if e.g.
m
mn n
2
it is clearly possible
for the cockroaches to skitter, and already there are
several different ways. For instance, we could divide the
rectangle into two dominos and have the roaches on each
domino simply exchanging places. Or we could simply
have them proceed in a (counter)clockwise cycle.
Dominos are a good idea in general: if one of and
is even, then an
m
nmn
rectangular grid may be tiled
1Unfortunately, a connection with Brouwer’s theorem is not achieved
here, but see [1].
*Thanks to Mariah Hamel, Josh Laison, and the Math Department at
Willamette University.
C
opyright © 2013 SciRes. OJDM
P. L. CLARK
184
with dominos, and this gives a way for the roaches to
skitter. Just as above though this feels not completely
satisfactory and one naturally looks for other skittering
patterns: e.g. when , one can have the roaches
in the inner square skittering clockwise as before
and then roaches in the outer ring of the square skittering
around in a cycle: isn’t that more fun? There are many
other skittering patterns as well.
4mn
V
1
v
vv
22
E
I found considerations like the above to be rather
interesting (and I will come back to them later), but for
me the real problem was a bit more meta: what is the
mathematical structure underlying the Cockroach Pro-
blem, and what is the general question being asked about
this structure?
Here we translate the Cockroach Problem into graph-
theoretic terms. In doing so, we get a graph theoretic
problem which in a precise sense interpolates between
two famous and classic problems: existence of perfect
matchings and existence of Hamiltonian cycles. On the
other hand, the more general problem does not seem to
be well known. But it’s interesting, and we present it
here: it is the existence and classification of graph
derangements.
2. Graph Derangements and Graph
Permutations
2.1. Basic Definitions
Let be a simple, undirected graph: that is,
we are given a set of vertices and a set of edges,
which are unordered pairs of distinct elements of . For
, we say that and 2 are adjacent if
and write 12
. In other words, for a set
, to give a graph with vertex set is equivalent to
giving an anti-reflexive, symmetric binary relation on ,
the adjacency relation. A variant formalism is also
useful: we may think of a graph as a pair of sets
,GV
12 V

12
vv E
EV
,vv
,
V
v
VV
,VE
and an incidence relation on V. Namely, for E
x
V and , eE
x
is incident to if e
e
, or,
less formally, if
x
is one of the two vertices comprising
the endpoints of . If one knows the incidence relation
as a subset of then one knows in particular for
each the pair of vertices
12 which are
incident to and thus one knows the graph .
e
V
E
,GVE

,E
E
vV
eE
,vv G
For and , the degree of is the
number of edges which are incident to . A degree zero
vertex is isolated; a degree one vertex is pendant.
v
v
A graph is finite if its vertex set is finite (and hence its
edge set is finite as well). A graph is locally finite if
every vertex has finite degree.
If and
GV
,E
GV
are graphs with
, we say is an edge subgraph of : it has
the same underlying vertex set as G and is obtained
from by removing some edges. If GV and
E
E
G
G G
,E
,GVE

are finite graphs with V and
V
E
equal the set of all elements of linking two vertices
in
E
V
, we say G
is an induced subgraph of . G
For a graph
,GVE, a subset
X
V is in-
dependent if for no 12
,
x
xX
do we have 12
x
x.
Example 2.1: For , we define the checker-
board graph ,mn. Its vertex set is
,mn
R

mn1,,, 1,,
and we decree that

,
2
,
121
x
xyy if
11
xy 2 2
1xy
.
Example 2.2: More generally, for and
1
n
,,mm
n

1,,
n
mm
R
we may define an n-dimensional an-
alogue of . Its vertex set is
,mn
R
11,
n
i,i
m
, and we decree that
11
,,
mm
,,
x
xyy if 1
Example 2.3: For we define the cylinder
checkerboard graph ,mn. This is a graph with the
same vertex set as and having edge set consisting
of all the edges of ,mn together with
1
m
ii
ixy

1n
.
2,
C
,mn
RR
m

,,1
x
nx
for all 1
x
m
. We put ,1m
, the cycle graph.
The checkerboard graph is an edge subgraph of
.
m
C
,mn
R
,2
mn
C
,
Example 2.4: For we define the torus
checkerboard graph ,mn. Again it has the same vertex
set as ,mn and contains all of the edges of
together with the following ones: for all 1
mn
C
T
R,mn
R
x
m
,
,
x
n,1x and for all 1
y
n,

1,y,my
T
.
The cylinder checkerboard graph ,mn is an edge
subgraph of the torus checkerboard graph .
C
,mn
For a graph
,EGV and
x
V, we define the
neighborhood of x as

x
NyVx y. More
generally, for any subset
X
V we define the neigh-
borhood o f X as
such that
x
xX
NNXy VxXxy
 .
Remark 2.5: Although ,
x
x
NX and
NX need
not be disjoint. In fact, iff

XNX
X
is an
independent set.
Now the “cockroach skitterings” that we were asking
about on can be formulated much more generally.
Let
,mn
R
,GVE be a graph. A graph derangement of
is an injection
G:
f
VV such that
f
vv
for
all vV
. Let be the set of all graph derange-
ments of .
DerG
G
Example 2.6: If n
GK
is the complete graph on the
vertex set
1, ,nn, then a graph permutation of
is nothing else than a permutation of
G
n. A
derangement of n
K
is a derangement in the usual sense,
i.e., a fixed-point free permutation. Derangements exist
iff and the number of them is asymptotic to >1n!n
e
as .
n
It is natural to also consider a slightly more general
definition.
Copyright © 2013 SciRes. OJDM
P. L. CLARK 185
For a graph , a graph permutation of
is an injection
,GVE
:
G
f
VV such that for all vV
,
either

f
vv

or
f
v
vV
v
G

. Let be the set
of all graph permutations of . Thus a graph derange-
ment is precisely a graph permutation which is fixed
point free: for all ,
PermG
f
vv.
Lemma 1 Let be a graph.
,GV
E
1) If has an isolated vertex, . GDerG
2) If has two pendant vertices adjacent to a com-
mon vertex, .
G
DerG
Proof. Left to the reader as an exercise to get com-
fortable with the definitions.
Remark 2.7: If is an edge subgraph of G, any
graph derangement (resp. graph permutation)
G
of G
is also a graph derangement (resp. graph permutation) of
.
G
2.2. Cycles and Surjectivity
Given a graph , we wish not only to decide whether
is nonempty but also to study its structure. The
collection of all derangements on a given graph is likely
to be a very complicated object: consider for instance
G
DerG
Der n
K
, which has size asymptotic to !n
e. Just as in the
case of ordinary permutations and derangements, it seems
interesting to study the possible cycle types of graph
derangements and graph permutations on a given graph
. Let us give careful definitions of these.
G
First, let be a set and
V:
f
VV an function. For
, let
m
m
f
f f
be the mth iterate of
f
.
We introduce a relation on as follows: for
V
,
x
yV,
x
y
n iff there are such that ,mn
m
f
xfy
. This is an equivalence relation on : the
reflexivity and the symmetry are immediate, and as for
the transitivity: if
V
,,
x
yzV are such that
x
y
and
then there are abc with yz,,,d
ab
f
xfy
and cd
f
yf
ac
z
c a
, and then
.
bc bd bd
c bbc
f
xf fxf fyfy
f
fy ffz fz





Now suppose f is injective: we now call the
-equi-
valence classes cycles. Let
x
V
, and denote the cycle
containing
x
by
x
C. Then:
x
C is finite iff
x
lies in the image of
f
and there
is m
such that m
f
xx
.
x
C is singly infinite iff it is infinite and there are
y
V, m
such that m
f
yx and y is not
in the image of
f
.
x
C is doubly infinite iff it is infinite and every
x
yC lies in the image of
f
.
These cases are mutually exclusive and exhaustive, so
f is surjective iff there are no singly infinite cycles.
Suppose
,GVE is a graph and Perm
f
G. We
can define the cycle type of f as a map from the set of
possible cycle types into the class of cardinal numbers.
When V is finite, this amounts to a partition of in
the usual sense: e.g. the cycle type of roaches skittering
counterclockwise on a
#V
55
grid is
. A graph
permutation is a graph derangement if it has no -cycles.
We will say a graph derangement is matchless if it has
no 2-cycles.
1
1,8,16
Proposition 2 Suppose that a graph G admits a graph
derangement. Then G admits a surjective graph derange-
ment.
Proof. It is sufficient to show that any graph derange-
ment can be modified to yield a graph derangement with
no singly infinite cycles, and for that it suffices to con-
sider one singly infinite cycle, which may be viewed as
the derangement 1nn
on the graph
with
1nn
for all n
,2
12n
. This derangement can be
decomposed into an infinite union of 2-cycles:
12,34, ,n
.
2.3. Disconnected Graphs
Proposition 3 Let be a graph with components G
i
GiI
, and let Perm
f
G
.
1) For all
ii
G,iIf G
.
2) Conversely, given graph permutations i
f
on each
i, G:
i
iI
f
fG
Der
G
Der
i
is a graph permutation. More-
over i
f
GfG
 for all . iI
E
The proof is immediate. Thus we may as well restrict
attention to connected graphs.
2.4. Bipartite Graphs
A bipartition of a graph is a partition
12
of the vertex set such that each i
V is an
independent set. A graph is bipartite if it admits at least
one bipartition.
,GV
VV V
For k
, a k-coloring of a graph
E,GV is a
map
:1,,kCV such that for all ,
x
yV
,
x
yCxCy
2
2
G
. There is a bijective corres-
pondence between -colorings of and bipartitions
of : given a -coloring we define
G
C
VxVCxi
i
, and given a bipartition we define
iCx
if i
x
V
. Thus a graph is bipartite iff it
admits a 2-coloring.
Remark 2.8: For a graph
,GVE
, a map
:0,CV1 is a 2-coloring of G iff its restriction to
each connected component i is a 2-coloring of i. It
follows that a graph is bipartite iff all of its connected
components are bipartite.
G G
Remark 2.9: Any subgraph of a bipartite graph G
is bipartite. Indeed, any 2-coloring of restricts to a
2-coloring of
G
G
G
.
Example 2.10: The cycle graph is bipartite iff
is even.
m
Cm
Copyright © 2013 SciRes. OJDM
P. L. CLARK
186
Corollary 4 Let G be a bipartite graph, and let
DerG
. Then every finite cycle of
has even
length.
Proof. Since a subgraph of a bipartite graph is bipartite,
a bipartite graph cannot admit a cycle of odd finite
degree2.
Example 2.11:
1) The n-dimensional checkerboard graph 1,,
n
mm
admits a graph derangement iff are not all
odd.
R
1,,
n
mm
2) The cylinder checkerboard graphs ,mn all admit
graph derangements. Hence, by Remark 2.7, so do the
torus checkerboard graphs .
C
,mn
3) For odd , the square checkerboard graph ,nn
admits graph permutation with a single fixed point. For
instance, by dividing the square into concentric rings we
get a graph permutation with cycle type
T
nR
1
1,8,16, ,82
n





.
3. Existence Theorems
3.1. Halls’ Theorems
The main tool in all of our Existence Theorems is a truly
basic result of combinatorial theory. There are several (in
fact, notoriously many) equivalent versions, but for our
purposes it will be helpful to single out two different
formulations.
Theorem 5 (Halls Theorem: Transversal Form) Let
be a set and iI be an indexed family of
finite subsets of . The following are equivalent:
V

i
S
V
1) (Hall Condition) For every finite subfamily
J
I,
##
i
iJ
.
J
S
2)
,VI admits a transversal: a subset
X
V
and a bijection :
f
XI such that for all
x
X
,

f
x
x
S.
We will deduce Theorem 5 from the following refor-
mulation.
Theorem 6 (Halls Theorem: Marriage Form) Let
be a bipartitioned graph in which every
has finite degree. The following are equivalent:
12
,,GVVE
vV
1
1) (Cockroach Condition) For every finite subset of
, .
1
V

11
##VNV
2) There is a semiperfect matching, that is, an in-
jection 12
:VV
such that for all 1
x
V,
x
x
V.
Proof. We follow [2]. Step 1: Suppose is finite.
We go by induction on . The case 1 is trivial.
Now suppose that 1 and that the result holds
for all bipartitioned graphs with first vertex set of car-
dinality smaller than . It will be notationally con-
venient to suppose that , and we do so.
1
#V
#Vn
n
1
#V
n
Case 1: Suppose that for all , every -
element subset of 1 has at least neighbors. Then
we may match n to any element of 2
V and semi-
perfectly match
1<kn
1kk
V
, 1n1,
into the remaining ele-
ments of by induction.
2
Case 2: Otherwise, for some , , there is a
-element subset 1
Vk1<k
n
k
X
V such that

#NXk
. The
subset
X
may be semiperfectly matched into 2
V by
induction, say via 12
:
X
V
, so it suffices to show that
the Hall Condition still holds on the induced biparti-
tioned subgraph on

X
12
\,V1. Indeed, if not,
then for some h,
\
k
VX
1hn

ha
, there would be an h-
element subset Yving fewer than h neig
bors in
1\VX h-
21
\V
en ), but thX



## ##
#.
NX YNXNYNXNY
kh XY



Step 2: Suppose is infinite. For
1
V1
x
V, endow
x
N with the discrete topology; endow 1
x
xV
NN
with the product topology. Each Nx is finite hence com-
pact, so N is compact by Tychonoff’s Theorem. For any
finite subset 1
X
V, let

,.
Xixy
H
nn NnnxyX
Then X
H
is closed in and is nonempty by Step 1.
Since is compact, there is , and any such
is a semiperfect matching.
N
GX
X
nH
n
Remark 3.1: Theorems 5 and 6 are equivalent results:
Assume Theorem 5. In the setting of Theorem 6, take
2
VV
, 1
I
V
and
x
x
I. The local finiteness
of the graph means each element of is finite, and the
assumed Cockroach Condition is precisely the Hall
Condition, so by Theorem 5 there is
N
2
X
V and a
bijection such that
:fX I
x
x
Xf xN.
Let 1
1
:
f
V
X
. Then for , let
1
Vy
y
x
, so
x
fy yx
.
Assume Theorem 6. In the setting of Theorem 5 take
1
VI
, 2
VV
, and the set of pairs E
,ix such
that i
x
S
. Since each i is finite, the graph is locally
finite, and the assumed Hall Condition is precisely the
Cockroach Condition, so by Theorem 6 there is a semi-
perfect matching
S
:
I
V
. Let

X
fI, and let
be the inverse function. For
:fX I
x
X
, if
xif,
x
i
, so

i
f
x
xSS
.
Remark 3.2: Theorem 5 was first proved for finite
I
by Philip Hall [3]. Eventually it was realized that equi-
valent or stronger versions of P. Hall’s Theorem had
been proven earlier by Menger [4], Egerváry [5] and
König [6]. The matrimonial interpretation was introduced
some years later by Halmos and Vaughan [2]. Never-
theless, with typical disregard for history the most com-
mon name for the finite form of either Theorem 5 or 13
is Halls Marriage Theorem.
1
1>
1,
V
,
2Conversely, a graph with no cycles of odd degree is bipartite, a famous
result of D. Köni
g
.
Copyright © 2013 SciRes. OJDM
P. L. CLARK 187
Remark 3.3: The generalization to arbitrary index sets
was given by Marshall Hall Jr. [7], whence “Halls’
Theorem” (i.e., the theorem of more than one Hall).
Remark 3.4: M. Hall Jr.’s argument used Zorn’s
Lemma, which is equivalent to the Axiom of Choice
(AC). The proof supplied above uses Tychonoff’s Theo-
rem, which is also equivalent to (AC) [8]. However, all
of our spaces are Hausdorff. By examining the proof of
Tychonoff’s Theorem using ultrafilters [9], one sees that
when the spaces are Hausdorff, by the uniqueness of
limits one does not need (AC) but only that every filter
can be extended to an ultrafilter (UL). In turn, (UL) is
equivalent to the fact that every Boolean ring has a prime
ideal (BPIT). (BPIT) is known to be weaker than (AC),
hence Halls’ Theorem cannot imply (AC).
Question 1 Does Halls Theorem imply the Boolean
Prime Ideal Theorem?
Remark 3.5: The use of compactness of a product of
finite, discrete spaces is a clue for the cognoscenti that it
should be possible to find a nontopological proof using
the Compactness Theorem from model theory. The
reader who is interested and knowledgeable about such
things will enjoy doing so. The Compactness Theorem
(and also the Completeness Theorem) is known be equi-
valent to (BPIT).
Example 3.6 ([10], pp. 288-289): Let 1 be the set of
non-negative integers and let 2
V be the set of positive
integers. For all positive integers 1
V
x
V, we decree that
x
is adjacent to the corresponding positive integer in
2 and to no other elements of 2. However, we decree
that 1 is adjacent to every element of . It is clear
that there is no semiperfect matching, because if we
match 0 to any 2, then the corresponding element
1 cannot be matched. But the Cockroach Condition
holds: for a finite subset 1
V
n
V
0V
V
Y
nV
X
V, if 0
X
then
1
#NV 1
#V, whereas if 0
X
then
12
VNV
.
3.2. The First Existence Theorem
Theorem 7 Consider the following conditions on a
graph
,GVE
DerG 
:
(D) .
(H) For all subsets
X
V,

##
X
NX.
(H) For all finite subsets
X
V,
##
X
NX.
1) Then (D) (H) (H).
 
2) If is locally finite, then (H) (D) and thus
(D) (H) (H).
G
 
Proof. a) (D) (H): If
DerG
and
X
V,
then :VV
is an injection with

X
NX
.
Thus
##
X
NX
. (H) (H) is immediate.
G
3) (H) (D) if is locally finite: for each
x
V, let
x
x, and let SN

x
x
V. Since G is
locally finite,
IS
I
is an indexed family of finite subsets of
. By assumption, for any finite subfamily
V
J
I,
#### :
x
i
xJ iJ
J
NJNS



this is the Hall Condition. Thus by Theorem 5, there is
X
V and a bijection such that for all :fX V
x
X
,

f
x
xN
, i.e.,

x
fx. Let
1:
f
VX
. Then for all
y
V,
yf y

y
, so DerG
.
Remark 3.7: Example 3.3 shows (H) need not imply
(D) without the assumption of local finiteness. The graph
with vertex set and such that every
x
is
adjacent to every integer satisfies (H) but not
(H).
>nx
Lemma 8 Let be a locally finite graph which
violates the cockroach condition: there is a finite subset
G
X
VG such that

#>#
X
NX
YX
. Then there is an
independent subset such that
#>#YNY.
Proof. Let be the subset of all vertices which
are not adjacent to any element of
YX
X
, so is an in-
dependent set. Put 1
Y
#mY
, 2, and
\mX
Y#
12
nmm #X
 . By hypothesis,
#<NX n12
m m
; since

\\
X
YNXNY, we
find
12
#<NYmm2 1
Combining Proposition 2, Theorem 7 and Lemma 8
we deduce the following result.
<#m Ym .
Theorem 9 (First Existence Theorem)
For a locally finite graph, the following are equivalent:
1) For every finite independent set
X
in , G
##
X
NXG.
2) admits a surjecti ve gra ph dera ngement.
Remark 3.8: Theorem 9 was first proved by W.T.
Tutte ([11], 7.1). Most of Tutte’s paper is concerned with
related—but deeper—results on digraphs. The result
which is our Theorem 9 appears at the end of the paper
and is proven by passage to an auxiliary digraph and
reduction to previous results. Perhaps because this was
not the main focus of [11], Theorem 9 seems not to be
well known. In particular, our observation that one need
only apply Theorem 5 appears to be new.
3.3. Bipartite Existence Theorems
A matching on a graph is a subset
such that no two edges in share a common
vertex. A matching is
perfec t if every vertex of
is incident to exactly one edge in .
,VGE
E
G
A graph permutation is dyadic if all of its cycles have
length at most 2.
Proposition 10 Let be a graph. G
1) Matchings of correspond bijectively to dyadic
graph permut a t i ons .
G
2) Under this bijection perfect matchings correspond
to dyadic graph derangements.
Proof. 1) Let be a matching. We define
E
SymV
as follows: if
x
incident to the edge
Copyright © 2013 SciRes. OJDM
P. L. CLARK
188

,exy, we put
x
y
. Otherwise we put
x
x
.
This is well-defined since by definition every vertex is
incident to at most one edge and gives rise to a dyadic
graph permutation. Conversely, to any dyadic graph per-
mutation SymV
, let
X
V be the subset of ver-
tices which are not fixed by
and put
,
xX
x
x
. Then
and

are mutually inverse.
2) A matching is perfect iff every vertex
x
V
is incident to an edge of iff the permutation
E
is
fixed-point free.
,,GVV
12
,VVE
,,E
Let 12 be a bipartitioned graph. A semi-
perfect matching on G is a matching such
that every vertex of 1 is incident to exactly one ele-
ment of . Thus a subset is a perfect match-
ing on iff it is a semiperfect matching on
both and on

.
E
V
VV
G
,,VVE
12 21
A semiderangement of
12 is an in-
jective function
,,VGV E
21
:VV
such that

x
x
for all
1
x
V
.
But in fact we have defined the same thing twice: a
semiderangement of a bipartitioned graph is nothing else
than a semiperfect matching.
Theorem 11 (Semiderangement Existence Theorem)
Consider the following conditions on a bipartitioned
graph :

E
12
(SD) There is an injection
,,GVV
1
V
2
:V
such that for all
1
x
V,

x
x
.
(H) For every finite subset 1
J
V,
##
J
NJ.
Then (SD) (H), and if G is locally finite, (H)
(SD).
Proof. This is precisely Theorem 6 stated in the
language of semiderangements.
Theorem 12 (Königs Theorem) Let
12
,,VE
2
V
GV
11
:V
be a bipartitioned graph, which need not be locally finite.
Suppose there is a semiderangement
V
of
and a semiderangement 221
12
,,VVE
:V
of
21 . Then admits a perfect matching, i.e., a
bijection such that
,,VVE
1
:fV
G
2
V

x
fx for all
1
x
V.
Proof. Let 12
. Then 12 is a
graph derangement. The injection
VVV
:V
V

partitions V into
finite cycles, doubly infinite cycles, and singly infinite
cycles. For each cycle C which is finite or doubly
infinite, 11 2
:CV CV
 and 221
:CV CV

are bijections. Thus if there are no singly infinite cycles,
taking 1
f
we are done. If C is a singly infinite
cycle, it has an initial vertex 1
x
. If 11
x
V, then
211
:CV CV
2
is surjective; the problem occurs if
1
x
V: then 1
x
does not lie in the image of 1
. We
have 1

221
x
xV
,
312 2
x
x
V, and so forth. So
we can repair matters by defining 1
on 24 by
21
,,xx
x
x, 43
x
x, and so forth. Doing this on every
singly infinite cycle with initial vertex lying in V2 a
bijection 1. Moreover, since
2
:fV V2
is a semi-
derangement and
21 2nn
x
x
for all n
, we
have
2212nnn
f
xx
V
xf
1
V
, so is a bijection.
Remark 3.9: Suppose and 2 are sets and
112
V
:V
and 221
:VV
are injections between
them. If we apply Theorem 12 to the bipartitioned graph
on
12
,VV in which
121
x
VyV xy
  or
2
y
x
, we get a bijection 12
: this is the
celebrated Cantor-Bernstein Theorem. As a proof of
Cantor-Bernstein, this argument was given by Gyula
(“Julius”) König [12] and remains to this day one of the
standard proofs. His son Dénes König explicitly made
the connection to matching in infinite graphs in his
seminal text [13].
:fV V
Theorem 13 (Second Existence Theorem) Let
12
,,VEGV
G
G
G
be a locally finite bipartitioned graph.
The following are equivalent:
1) admits a perfect matching.
2) admits a dyadic graph derangement.
3) admits a gr ap h de rangem en t .
4) For every subset
J
V,

##
J
NJ.
Proof. 1)
2) by Proposition 10.
2) 3) is immediate.
3) 4) is the same easy argument we have already
seen.
4) 1): By Theorem 11, we have semiderange-
ments 112
:VV
and 1
V
22 :V
. By Theorem 12
this gives a perfect matching.
Remark 3.10: The equivalence 1) 4) above is due
to R. Rado [14].
3.4. An Equivalence
Theorems 9 and 13 are “equivalent” in the sense that
they were proved using equivalent formulations of Halls’
Theorem (together with, in the case of Theorem 13, a
Cantor-Bernstein argument). In this section we will show
their equivalence in a stronger sense: each can be rapidly
deduced from the other.
Assume Theorem 9, and let be a
locally finite bipartitioned graph satisfying the Cock-
roach Condition: for all finite independent subsets
12
12
,,VEGV
J
V
V#NJ  , . Then Theorem 9 applies to
yield a surjective graph derangement f. A cycle
admits a dyadic graph derangement iff it is not finite of
odd length; since G is bipartite, no cycle in f is finite of
odd length. Thus decomposing f cycle by cycle yields a
dyadic graph derangement.
#JC
Assume Theorem 13, and let
,GVE be a locally
finite graph satisfying the Hall Condition: for all finite
independent subsets
J
V#NJ, . Let #J
22
,,VE
V
21
GV
VV be the bipartite double of : we put
12
G
. For
x
V
, let 1
x
(resp. 2
x
) denote the
copy of
x
in (resp. ). For every
1
V2
V
,exyE,
Copyright © 2013 SciRes. OJDM
P. L. CLARK 189
we give ourselves edges

12 122
,,,
x
yyxE
GG
2
f
. Then
2 is locally finite bipartitioned, and it is easy to see
that the Hall Condition in implies the Cockroach
Condition in 2. By Theorem 13, 2 admits a dyadic
graph derangement . From we construct a graph
derangement of : for
G
G
f2
fG
x
V1
, let
x
be the
corresponding element of 1; let V

221
y
fxy, and let
be the element of corresponding to 2. Then we
put
y V

f
xy. It is immediate to see that is a graph
derangement of . It need not be surjective, but no
problem if it isn’t: apply Proposition 2.
f
G
Remark 3.11: Our deduction of Theorem 9 from
Theorem 13 is inspired by an unpublished manuscript of
L. Levine [15].
Remark 3.12: Recall that Theorem 13 implies the
Cantor-Bernstein Theorem. The above equivalence thus
has the following curious consequence: Cantor-Bernstein
follows almost immediately from the transversal form of
Halls’ Theorem.
3.5. Dyadic Existence Theorems
One may ask if there is also an Existence Theorem for
dyadic derangements in non-bipartite graphs. The answer
is yes, a quite celebrated theorem of Tutte. A later result
of C. Berge gives information on dyadic graph per-
mutations.
Let be a graph. For a subset
,GVE
X
V, we
denote by the induced subgraph with vertex set
.
\GX
\VX
Theorem 14 (Third Existence Theorem) For a finite
graph , TFAE:
,GVE
G
1) has a dya dic gra p h dera ngement.
2) For every subset
X
V, the number of connected
components of with an odd number of vertices is
at most
\GX
#
X
.
Proof. See [16].
Remark 3.13: Theorem 14 can be generalized to
locally finite graphs: see [17].
A maximum matching of a finite graph G is a
matching such that is maximized among all
matchings of . Thus if admits a perfect matching,
a matching is a maximum matching iff it is perfect,
whereas in general the size of a maximum matching
measures the deviation from a perfect matching in .
#
GG
G
For a finite graph
H
, let

odd
H
be the number of
connected components of
H
with an odd number of
vertices.
Theorem 15 (Berges Theorem [18]) Let be any
finite graph. The size of a maximum matching in is
GG
1#odd(\)#
min
2
GXV
BXGXV
 
.
We immediately deduce the following result on dyadic
graph permutations.
Corolla ry 16 Let be a finite graph. Then the least
number of fixed points in a dyadic graph permutation of
is
G
G#2
G
VB
.
3.6. Matchless Graph Derangements
Let
,GVE be a finite graph with . Then a
graph permutation of cycle type is called a Hamil-
tonian c y c l e (or Hamiltonian circuit).
#Vn

n
From the perspective of graph derangements it is clear
that Hamiltonian cycles lie at the other extreme from
dyadic derangements and permutations. Here much less
is known than in the dyadic case: there is no known
Hamiltonian analogue of Tutte’s Theorem on perfect
matchings, and in place of Berge’s Theorem we have the
following open question.
Question 2 Given a finite graph G, determine the
largest integer such that admits a graph per-
mutation of type nG
,1,,1n. Equivalen tly, deter mine the
maximum order of a vertex subgraph of admitting a
Hamiltonian cycle.
G
A general graph derangement is close to being a
partition of the vertex set into Hamiltonian cycles, except
that if x and y are adjacent vertices, then sending x to y
and y to x meets the requirements of a graph derange-
ment but does not give a Hamiltonian subcycle because
the edge from x to y is being used twice. Let us say a
graph derangement is matchless if each cycle has length
at least 3. The above Existence Theorems lead us na-
turally to the following question.
Question 3 Is there an Existence Theorem for match-
less graph derangements?
As noted above, there is no known Existence Theorem
for Hamiltonian cycles. Since a Hamiltonian cycle is a
graph derangement of a highly restricted kind, one might
hope that Question 3 is somewhat more accessible.
4. Checkerboards Revisited
4.1. Universal and Even Universal Graphs
Let G be a graph on the vertex set
1, ,nn
such
that DerG
. As in §2, it is natural to inquire
about the possible cycle types of graph derangements
(and also graph permutations) of G. We say G is uni-
versal if for every partition of there is a graph
derangement of with cycle type . For instance, the
complete graph
.2
pn
pG
n
K
is (rather tautologously) universal.
If and G is bipartite, by Corollary 4 G is not
universal, because the only possible cycle types are even.
Thus for graphs known to be bipartite the more interest-
ing condition is that every possible even partition of
5n
n
G
occurs as the cycle type of a graph derangement of :
we call such graphs even un iversal.
Copyright © 2013 SciRes. OJDM
P. L. CLARK
190
4.2. On the Even Universality of Checkerboard
Graphs
Proposition 17 For all n
, the checkerboard graph
is
even universal.
2,n
RProof. This is an easy inductive argument which we
leave to the reader.
Proposition 18 Let be an even number. Then: 4n
1) If are even numbers greater than 2 such
1,,
k
aa
k
that , then there is no graph derangem ent
1
4
i
ia

3n
of of cycle type .
3,n
2) It follows that is not even universal.
R

1,,,4
k
aa
3,n
Proof. 1) A -cycle on any checkerboard graph must
be a -square. By symmetry, we may assume that
the -square is placed so as to occupy portions of
the top two rows of ,mn. The two vertices immediately
underneath the square cannot be part of any Hamiltonian
cycle in the complement of the square, so any graph
derangement containing a -cycle must also contain a
-cycle directly underneath the -cycle. Thus the cycle
type is excluded.
R
4
4
4
22

22
aa
n
R
4
3n
2 4
1
2) Since is even, is even and at least ,
so there are even partitions of of the above form:
if is divisible by 4;
otherwise.
,,,
k
12
, 4
3n
4, 4,,4
n
6, 4,4,
Proposition 19 If is odd and is divisible by
, then admits no graph derangement of cycle
type and is therefore not even universal.
m n
4,mn
R
4,, 44,
Proof. Left to the reader.
Example 4.1
: There are
3,4
G12 11
2
p

 even
partitions of 12. By Proposition 19, we cannot realize
the cycle types , by graph derangements.
We can realize the other nine:

8, 4
4, 4,4
 
 
12,10, 2,8, 2, 2,6, 6,6, 4, 2,6, 2, 2,2,
4, 4, 2, 2,4, 2, 2, 2, 2,2, 2, 2,2, 2,2.
Example 4.2 : Of the
4,4
G16 22
2
p

 even parti-
tions of , we can realize :
16 20

  
16,14, 2,12, 4,12, 2, 2,10, 4, 2,10, 2, 2,2,
8,8,8, 6, 2,8, 4, 4,8, 4, 2, 2,


8, 2, 2, 2,6, 6, 2, 2,6, 4, 4, 2,6, 4, 2, 2, 2,
6, 2, 2,2, 2,2,4,4,4, 4,4, 4, 4,2, 2,
 
4, 4, 2, 2,2, 2,4, 2, 2, 2, 2,2, 2,2, 2, 2,2, 2,2, 2, 2,2.
We cannot realize:
10,6 ,6,6,4 .
Indeed, the only order 6 cycle of a checkerboard graph
is the checkerboard subgraph. Removing any 6-
cycle leaves one of the corner vertices pendant and hence
not part of any cycle of order greater than .
23
2
Example 4.3
3,6
G: There are 18
2

30p even


partitions of . We can realize these 23 of them:
18
 

18, 16,2, 14,2,2, 12,6, 12,4,
12, 2, 2, 2,10,6, 2,10, 4, 2, 2,
2,
 

10,2,2,2,2 ,8,8,2 , 8,6,2,2 , 8,2,2 ,
8, 2,2,2,2, 2, 2,6,6,6,6,6,4,2,
,4,2
 

6, 6, 2, 2, 2,6, 4, 4, 2, 2,6, 4, 2, 2,,
6, 2, 2,2, 2,2, 2,4, 4, 4,2, 2,2,
2, 2

4, 4, 2, 2, 2, 2, 2,4, 2, 2, 2,2,2, 2, 2
2, 2, 2, 2, 2, 2, 2, 2, 2.
,
We cannot realize the following ones:


14, 4,10,8,10, 4, 4,8, 6, 4
8, 4, 4, 2,6, 4, 4,4,4,4, 4,4
,
, 2.
Of these, all but
10,8 and
4, 4,24, 4, are ex-
cluded by Proposition 18, and we leave it to the reader to
check that these two “exceptional” cases cannot occur.
Example 4.5
4,5
G: There are 20
2

42p even


partitions of . We can realize of them. We
cannot realize:
20 39

8,8, 4,8, 4, 4, 4,4, 4, 4, 4,
4.
No
8,8, 4: to place a 4-cycle in 4,5 without
leaving a pendant vertex, we must place it in a corner,
without loss of generality the upper left corner. The only
-cycle which can fit into the remaining lower left
corner is the recantagular one, and this leaves a pendant
vertex at the lower right corner.
G
8
No
8, 4, 4, 4: Any placement of three -cycles in
4,5 gives two of them in either the top half or the
bottom half, and without loss of generality the top half.
Any such placement leaves a pendant vertex in the top
row.
4
G
No
4, 4, 4, 4,4: This follows from Proposition 19.
Alternately, the argument of the previous paragraph
works here as well.
Example 4.6 (
4,6
G): There are 24
12

77p even


partitions of . They can all be realized by graph
derangements: is even universal.
24
G4,6
By analyzing the above examples, we found the fol-
lowing additional families of excluded cycle types. The
proofs are left to the reader.
Proposition 20 Let 21nk
be an odd integer
Copyright © 2013 SciRes. OJDM
P. L. CLARK
Copyright © 2013 SciRes. OJDM
191
[7] M. Hall Jr., “Distinct Representatives of Subsets,” Bulle-
tin of the American Mathematical Society, Vol. 54, No.
10, 1948, pp. 922-926.
http://dx.doi.org/10.1090/S0002-9904-1948-09098-X
greater than 1. Then 4,n admits no matchless graph
derangemen t with or more -cycles.
G
1
2k
4
Proposition 21 For any non-negative integer ,
admits no graph derangement of the cycle type
. Thus these graphs are not even universal.
k
4,6 4k
G
6,6,,6, 4[8] J. L. Kelley, “The Tychonoff Product Theorem Implies
the Axiom of Choice,” Fundamenta Mathematicae, Vol.
37, No. , 1950, pp. 75-76.
Proposition 22 For any odd integer and
mk,
admits no graph derangement of the cycle type
. Thus these graphs are not even universal.
,4 2mk
G
6, 4,,[9] H. Cartan, “Théorie des Filtres,” Comptes Rendus de
lAcadémie des Sciences, Paris, Vol. 205, 1937, pp. 595-
598.
4
In particular, for ,mn to be even universal, it is
necessary that m and n both are even. Proposition 21
shows that having m and n both even is not sufficient;
however, we have found no examples of excluded even
cycle types of ,mn when are both even and at
least . Such considerations, and others, lead to the
following conjecture, made by J. Laison in collaboration
with the author.
R
[10] C. J. Everett and G. Whaples, “Representations of Se-
quences of Sets,” American Journal of Mathematics, Vol.
71, No. 2, 1949, pp. 287-293.
http://dx.doi.org/10.2307/2372244
R,mn
6[11] W. T. Tutte, “The 1-Factors of Oriented Graphs,” Pro-
ceedings of the American Mathematical Society, Vol. 4,
No. 6, 1953, pp. 922-931.
Conjecture 23 There is a positive integer such
that for all , the checkerboard graph is
even universal.
0
N
2,k
R[12] J. König, “Sur la Theorie des Ensembles,” Comptes Ren-
dus de lAcadémie des Sciences, Paris, Vol. 143, 1906,
pp. 110-112.
0
N,kl2l
[13] D. König, “Theorie der Endlichen und Unendlichen Gra-
phen. Kombinatorische Topologie der Streckenkom-
plexe,” Chelsea Publishing Co., New York, 1950.
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