Journal ArticleDOI

# A note on independent sets in trees

01 Jan 1988-SIAM Journal on Discrete Mathematics (Society for Industrial and Applied Mathematics)-Vol. 1, Iss: 1, pp 105-108

TL;DR: A simple graph-theoretical proof that the largest number of maximal independent vertex sets in a tree with n vertices is given by m( T), a result first proved by Wilf.

AbstractWe give a simple graph-theoretical proof that the largest number of maximal independent vertex sets in a tree with n vertices is given by $m( T ) = \begin{cases} 2^{k - 1} + 1& {\text{if }} n = 2k, \\ 2^k & {\text{if }} n = 2k + 1, \end{cases}$ a result first proved by Wilf [SIAM J. Algebraic Discrete Methods, 7 (1986), pp. 125–130]. We also characterize those trees achieving this maximum value. Finally we investigate some related problems.

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### Summary

• The authors also characterize those trees achieving this maximum value.
• Finally the authors investigate some related problems.
• Key words, independent vertices, trees, extremal graphs AMS(MOS) subject classifications.

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SIAM
J.
Disc.
MATH.
Vol.
1,
No.
1,
February
1988
(C)
1988
Society
for
Industrial
and
Applied
Mathematics
012
A
NOTE
ON
INDEPENDENT
SETS
IN
TREES*
BRUCE
E.
SAGANf
Abstract.
We
give
a
simple
graph-theoretical
proof
that
the
largest
number
of
maximal
independent
vertex
sets
in
a
tree
with
n
vertices
is
given
by
2
k-
+
if
n
2k,
m(T)
2
if
n
2k
+
1,
a
result
first
proved
by
Wilf
[SIAM
J.
Algebraic
Discrete
Methods,
7
(I
986),
pp.
125-130].
We
also
characterize
those
trees
achieving
this
maximum
value.
Finally
we
investigate
some
related
problems.
Key
words,
independent
vertices,
trees,
extremal
graphs
AMS(MOS)
subject
classifications.
05C35, 05C30,
05C70
1.
Introduction.
Herbert
Wilf
[5]
was
the
first
to
answer
the
following
question:
What
is
the
largest
number
of
maximal
independent
vertex
sets
in
a
tree
with
n
vertices?
His
proof
had
an
algebraic
flavor
and
was
somewhat
complicated.
Subsequently
Daniel
Cohen
[1]
was
able
to
provide
a
graph-theoretical
proof,
but
one
which
was
still
fairly
complex
in
view
of
the
simplicity
of
the
bound
(see
Theorem
3
below).
The
purpose
of
this
note
is
to
give
a
simple
graph-theoretical
demonstration
of
this
result
which,
in
addition,
completely
characterizes
all
trees
achieving
the
maximum
value.
J.
Griggs
and
C.
Grinstead
[2]
independently
found
a
straightforward
proof
which
is
similar
to
ours
in
some
respects
but
differs
in
others.
2.
Maximizing
independent
sets.
We
begin
with
some
preliminary
definitions
and
lemmas.
For
any
concepts
that
are
not
defined,
the
reader
can
consult
Harary’s
book
[4].
Given
a
graph,
G,
let
V(G)
be
the
vertex
set
of
G
and
let
v(G)
IV(G)[
where
denotes
cardinality.
Recall
that
a
vertex
u
V(G)
is
called
an
endpoint
if
deg
u
1.
We
will
say
that
a
vertex
v
V(G)
is
penultimate
if
v
is
not
an
endpoint
and
v
is
adjacent
to
(at
least)
deg
v
endpoints.
Note
that
v
is
adjacent
to
deg
v
endpoints
if
and
only
if
v
is
the
center
of
the
star
Kt.
LEMMA
1.
Every
finite
tree
T
with
v(T)
>-_
3
has
a
penultimate
vertex.
Proof.
The
next-to-last
vertex
on
any
diameter
must
be
penultimate.
If
v
is
penultimate
in
T,
then
T
v
consists
of
deg
v
isolated
vertices
and
one
other
component
called
the
penultimate
component
P
(if
v
is
the
center
of
a
star,
choose
any
fixed
component
as
the
penultimate
one).
Now
let
End
v
{
w
Plw
is
adjacent
to
v}
so
that
V(T)
V(P)
U
{
v}
U
End
v
where
denotes
disjoint
union.
Call
a
set
I
_
V(G)
independent
if
no
two
vertices
of
I
are
adjacent
in
G.
Now
let
M(G)
{I
_
V(G)II
is
independent
and
maximal},
i.e.,
if
!
M(G)
then
there
is
no
independent
set
J
with
I
J.
Also
set
m(G)
IM(G)i.
We
wish
to
find
the
maximum
value
of
m(T)
over
all
trees
T
with
v(T)
n.
First,
however,
we
need
an
upper
bound.
Received
by
the
editors
April
28,
1986;
accepted
for
publication
(in
revised
form)
March
24,
1987.
This
research
was
supported
in
part
by
a
NATO
post-doctoral
grant
administered
by
the
National
Science
Foundation.
Department
of
Mathematics,
University
of
Pennsylvania,
Philadelphia,
Pennsylvania
19104-6395.
Present
address,
Department
of
Mathematics,
Michigan
State
University,
East
Lansing,
Michigan
48824.
105
Downloaded 05/09/14 to 35.8.72.212. Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php

106
BRUCE
E.
SAGAN
FIG.
1.
Batons
of
length
O.
LEMMA
2.
Let
T
be
a
tree
and
v
V(
T
be
penultimate
with
corresponding
component
P.
Then
m(T)<=2m(P).
Proof
Let
I
be
a
maximal
independent
set
in
T.
Then
either
End
v
_
I
or
v
e
I
(exclusive
or).
In
the
first
case,
I
End
v
tO
Ip
where
Ip
is
a
maximal
independent
set
of
P.
In
the
second,
I
{
v}
tO
(I
{
w})
where
w
is
the
unique
vertex
of
P
adjacent
to
v
(w
need
not
be
in
I).
[2]
Define
a
baton
of
length
l
as
follows.
Start
with
a
path
L
of
length
I
and
attach
any
number
of
paths
of
length
two
to
L’s
endpoints.
Hence
the
batons
of
length
0
are
just
"extended"
stars
and
the
first
few
are
displayed
in
Fig.
1.
Similarly,
the
batons
of
length
form
a
family
some
of
whose
members
are
shown
in
Fig.
2.
THEOREM
3.
Among
all
labeled
trees
T
with
n
vertices,
the
maximum
value
of
m(T)
is
2
k-+l
ifn=2k,
m(T)=
2
k
ifn=2k+
1.
Furthermore
this
maximum
is
attained
only
by
the
batons
of
length
0
(when
n
is
odd)
or
by
the
batons
of
lengths
and
3
(when
n
is
even).
Proof
Induct
on
n.
The
theorem
can
be
checked
by
hand
for
v(T)
<-
10
using
Lemma
2
and
Harary’s
tables
[4,
pp.
233-234].
(The
author
has
made
this
calculation
and
does
not
recommend
it
to
the
reader.)
Now
let
T
be
a
tree
with
re(T)
maximum
among
all
trees
with
v(T)
n
>
10.
By
Lemma
there
exists
a
penultimate
vertex
v
V(T)
with
corresponding
component
P.
There
are
two
cases
depending
upon
the
parity
of
n.
If
n
is
odd,
n
2k
+
1,
then
consider
the
unique
baton
of
length
0
with
n
vertices,
denoted
Tn.
Since
T2k
/
contains
k
paths
of
length
2,
a
simple
calculation
shows
that
m
(T2k
+
1)
2
k.
Hence
(1)
m(T)>=2
k.
FIG.
2.
Batons
of
length
1.
Downloaded 05/09/14 to 35.8.72.212. Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php

A
NOTE
ON
INDEPENDENT
SETS
IN
TREES
107
FIG.
3.
G
and
G2.
Now
v(P)
=<
n
2
2k-
1.
However,
if
v(P)
<
2k-
1,
then
by
Lemma
2
and
induction
(for
n
at
least
7)
we
have
(2)
m(T)<=2m(P)<2
2
k-=
2
k,
which
contradicts
(1).
Thus
v(P)
2k-
which
implies
that
deg
v
2,
and
End
v
{
u}
for
a
single
vertex
u.
Furthermore
P
T2-
since
if
it
does
not,
induction
applies
which
yields
m(P)
<
2
(this
baton
is
the
unique
tree
attaining
the
maximum
value).
But
then
(2)
holds
as
before,
a
contradiction
unless
P
T2_
.
Putting
all
these
facts
together,
we
see
that
T
consists
of
a
tree
T2_
with
a
path
of
length
two
w-v-u
attached
to
some
w
V(T2k-
).
This
leaves
only
three
possibilities
for
T:
T
/
,
G
or
G,
where
G
and
G
are
given
in
Fig.
3.
To
eliminate
G1
and
G2
as
possibilities,
consider
a
second
penultimate
vertex
v’
as
shown.
If
p’
is
the
corresponding
component,
then
for
n
_
9
we
have
P’
q
Tz,-.
Invoking
Lemma
2
again
we
see
that
m(Gi)
<=
2m(P’)
<
2.2
-
2
for
1,
2
and
so
neither
Gi
maximizes
re(T).
For
n
even,
n
2k,
exactly
the
same
line
of
reasoning
as
in
the
odd
case
can
be
used.
It
follows
that
the
only
possibilities
for
T
are
those
obtained
by
attaching
a
path
of
length
2
to
a
baton
of
length
or
3
on
2k
2
vertices.
Hence
T
is
either
a
baton
of
length
or
3
and
re(T)
2
k-
+
or
T
is
one
of
the
five
graphs
H,
H5
shown
in
Fig.
4.
Note
that
in
H2
(respectively
H3)
we
require
that
deg
c
_
3
(respectively
deg
d
>-
2)
so
that
the
graph
does
not
degenerate
into
a
baton
of
length
3
(respectively
1).
It
is
easy
to
verify
that
because
n
>
10
we
can
find
in
each
of
these
five
graphs
a
second
penultimate
vertex
v’
such
that
P’
is
not
a
baton
of
length
or
3.
It
follows
from
FIG.
4.
H
through
Hs.
Downloaded 05/09/14 to 35.8.72.212. Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php

108
BRUCE
E.
SAGAN
induction
and
from
v(P’)
n
2
2k
2
that
we
have
m(P’)
<=
(2
k-
2
q_
1)
2
k
-2.
Hence
by
Lemma
2
m(Hi)
<=
2m(P’)
=<
2.2
k-
2
2
k-
for
1,
...,
5
which
is
less
than
the
value
obtained
for
the
batons.
This
finishes
the
proof
of
the
theorem.
Call
a
tree
T
extrema!
if
m(T)
is
a
maximum
as
compared
to
all
other
trees
with
the
same
number
of
vertices.
Let
e(n)
be
the
number
of
extremal
trees,
up
to
the
labeling,
on
n
vertices.
COROLLARY
4.
The
number
of
extremal
trees
on
n
vertices
is
given
by
e(n)={lk
ifn=2k+
1,
ifn=2k.
Proof
It
is
a
simple
matter
to
count
the
number
of
batons
of
the
appropri-
ate
sizes.
3.
Remarks.
Finding
the
minimum
value
of
m(T)
is
quite
easy.
THEOREM
5.
The
minimum
value
of
m(
T)
over
all
trees
with
n
vertices,
n
>=
2,
is
m(
T
)
2.
Furthermore
the
unique
tree
(up
to
relabeling)
achieving
this
minimum
is
the
star
gl,n-
.
Proof.
If
v(T)
>-
2,
then
for
any
edge
vw
there
is
a
maximal
independent
set
con-
taining
v
and
a
different
one
containing
w.
Hence
m(T)
>=
2
and
dearly
m(K,n-)
2.
If
T
q:
K,n_
,
then
T
contains
a
path
u-v-w-x.
This
forces
m(T)
>=
3
since
a
third
maximal
independent
set
containing
u
and
x
also
exists,
t--1
Once
one
has
determined
the
lower
and
upper
bounds,
b
and
B,
respectively,
for
a
graphical
invariant
fl(G)
one
looks
for
an
interpolation
theorem.
Such
a
result
has
the
following
form:
For
all
integers
z
satisfying
b
<-
z
=<
B
there
is
a
graph
G
with/(G)
z.
Unfortunately
there
is
no
interpolation
theorem
for
m(T)
since
when
v(T)
9
we
have
2
=<
m(T)
=<
16
but
there
is
no
tree
with
m(T)
15.
We
should
compare
our
proof
of
Theorem
3
with
that
of
Griggs
and
Grinstead
mentioned
in
the
Introduction.
They
also
begin
by
proving
Lemmas
and
2.
Then,
however,
they
use
the
lemmas
to
show
that
the
maximum
value
of
rn
(F)
over
all
forests
F
with
v(F)
n
is
achieved
precisely
when
F
is
a
one-factor
(possibly
with
an
isolated
vertex).
By
carefully
amalgamating
the
components
of
the
one-factor,
they
finally
obtain
the
bound
and
extremal
graphs
for
trees.
Following
the
dictum
that
once
something
is
proved
for
trees
it
should
be
extended
to
all
connected
graphs,
one
is
led
to
pose
the
following
question:
What
is
the
maximum
value
of
m(G)
over
all
connected
graphs
G
with
v(G)
n?
Griggs,
Grinstead
and
Guichard
[3]
have
answered
this
query.
Another
proof
has
been
found
independently
by
Fiiredi.
Acknowledgments.
I
would
like
to
thank
Mihily
Hujter
for
pointing
out
the
proof
of
Lemma
1.
I
also
thank
the
referee
for
suggestions
that
considerably
improved
the
exposition.
REFERENCES
1]
D.
COHEN,
Counting
stable
sets
in
trees,
in
Seminaire
Lotharingien
de
combinatoire,
10
session,
R.
KtJnig,
ed.,
Institute
de
Recherche
Math6matique
Avanc6e
pub.,
Strasbourg,
France,
1984,
pp.
48-
52.
[2]
J.
GRIGGS
AND
C.
GRINSTEAD,
private
communication.
[3]
J.
GRIGGS,
C.
GRINSTEAD
AND
D.
GUICHARD,
The
number
of
maximal
independent
sets
in
a
connected
graph,
prepdnt.
[4]
F.
HARARY,
Graph
Theory,
Addison-Wesley,
Reading,
MA,
1969.
[5]
H.
WILF,
The
number
of
maximal-independent
sets
in
a
tree,
SIAM
J.
Algebraic
Discrete
Methods,
7
(1986),
pp.
125-130.
Downloaded 05/09/14 to 35.8.72.212. Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php
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### Additional excerpts

• ...Sagan [26] finally presented an elegant proof, in which trees attaining the upper bound were also found (as did Griggs and Grinstead [7] independently)....

[...]

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### Cites result from "A note on independent sets in trees..."

• ...Quite a lot of similar results are given in the graph-theoretic literature as well: for instance, Hedman [4] studies the (essentially equivalent) problem of maximizing the number of cliques in graphs with a given maximal clique size, and Wilf [21] gives the largest number of maximal independent vertex sets of a tree on n vertices, see also [15]....

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##### References
More filters

Book
01 Jan 1969

15,318 citations

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TL;DR: The maximum number of maximal independent sets which a connected graph on n vertices can have is determined, and the extremal graphs are completely characterize, thereby answering a question of Wilf.
Abstract: We determine the maximum number of maximal independent sets which a connected graph on n vertices can have, and we completely characterize the extremal graphs, thereby answering a question of Wilf.

80 citations