Internat. J. Math. & Math. Sci.
VOL. 11 NO. 4 (1988) 825-827
825
A NOTE ON THE VERTEX-SWITCHING RECONSTRUCTION
I. gRAS(OV
School of Mathematlcal Sciences
Tel-Avlv University
Te1-Aviv
Israel
(Received October 7, 1987 and in revised form November 15, 1987)
ABSTRACT.
Bounds on the maximum and minimum degree of a graph establlshing its
reconstructibility from the vertex switching are given.
It is also shown that any
disconnected graph with at least five vertices is reconstructible.
KEY WORDS AND PHRASES. Vertex-Swltchlng, Reconstruction.
1980 AMS SUBJECT CLASSIFICATION CODE. 05C06.
I.
INTRODUCTION.
A switching
Gv of
a graph G at vertex v is a graph obtained from G by deletlng
all edges incident to v and inserting all possible edges to v which are not in G.
Since switching is a commutative operation, i.e.,
(Gv) u
(Gu)v,
be naturally extended to arbitrary subsets of the vertex set V(G).
for all A c
the definition can
Thus, G
A is defined
V(G).
The Vertex-Switching Reconstruction Problem, proposed by Stanley [1], asks:
G uniquely determined up to isomorphism by the set (deck
),
{Gv}v
Is
V(G) ?
If the answer is "yes" then G is called reconstructlble.
It was shown in [1] that any graph G with n
IV(G)[
=
0(mod 4) is reconstructible.
It seems that a little is known about the case n
0(mod 4). However, Stanley
pointed out [1], that the degree sequence of a graph, and consequently, the number of
4. Bounds on the number of edges in a
edges easily reconstructlble, provided n
graph, e(G), establishing its reconstructibillty was given [2]. Namely:
n(n 2)
e(G) i [,
4
n
2
--], n 4.
4
As might be expected, in virtue of the last result, G is reconstructlble if it
Here we will prove
the last claim (Theorem 2) and show that for sufficiently large n a graph is recon6) > 0.gn, where A and 6 are the maximum and the minimum
structible if max (A,n
degree of G respectively. Actually, we prove a little more, namely:
has a vertex of degree not close to n/2 or if G is disconnected.
I. KRASIKOV
826
2.
MAIN RESULTS.
(n-l) n(n-1)
If min (n
THEOREM I.
2n/2-3
then G is reconstructible.
0(mod 4). We
In virtue of the quoted result of Stanley, we may assume n
will consider a graph G as a spanning subgraph of a fixed copy of the complete graph
K
The switching equivalence class G of G is the set of all H c K isomorphic to G
PROOF.
n
such that H
n
V(G).
for some switching A
G
A
For each subgraph g
G, let
c
p(G*
g) be the number of those elements of G
which
contain a fixed copy of g.
First we show that G is reconstructible if
p(G*
s(g
where s(H
F)
Kn)
g)s (g
2
n/2-2
G)
(2.1)
is the number of the
subgraphs of F isomorphic to H.
Observe that
IG*Is(g
p(G*
G)
D
On the other hand, consider the set S.
1
7.1Sil 21G*
Observe that
since G
A and
that for a nonreconstructible graph
Kn).
g)s (g
{A
A6
G
G
V(G)\A,
G,
IS4il
(2.2)
IAI
i}.
It is known
are identical.
([2], Corollary 2.4).
>_-
Thus,
2i
if G is not reconstructible then
2IG*{
>
In/2 )
x:
2n/2_i
(2.3)
2i
.
Comparing (2.2) and (2.3), we get that (2.1) is enough for the reconstructibility of G.
Let now g be a star
KI,
Observe that p(G
switching, possibly preserving a fixed copy of K
s(g
n)
K
n
KI, )
is
A
& 2 since the only proper
V(K
1
Hence, by (2.1), G is reconstructible if n
Furthermore,
<
Now, to complete the proof, one has to consider the complementary graph G, which is
reconstructible iff G is.
Now we will prove that disconnected graphs are reconstructible.
First we need
the following simple lemma:
LEMMA I.
for any
PROOF.
such that
obtaln G
Suppose that nonisomorphic graphs G and H have the same deck. Then
H.
there is uEV(G), 9
u, such that G
9u
Since the decks of G and H are equal then there is a bijection :V(G)
V(H)
be an isomorphism. Choosing u =h(())we
Let
H().
H()
EV(G)
G
u
h
H.
Moreover, since G
G
G, then
(9).
827
VERTEX-SWITCHING RECONSTRUCTION
COROLLARY i.
Let n
deg () + deg (u)
PROOF.
n or n
If G
and G,
u, have the same deck then
u
and u are adjacent in G or are not.
2, depending on whether
Let e(,u) be the number of edges between
deg ()
COROLLARY 2.
PROOF.
4.
+ deg (u)
2(n
2e(,u)
and u.
2)
2 <. 6
+
A
.<
n.
We omit the details.
This easily follows from Lemma 1 and Corollary I.
Any disconnected graph is reconstructible, provided n
THEOREM 2.
e(H) then
e(G)
=
2.
n
4 then n
If G is not reconstructible and n
Since
4.
Assume the contrary. Then there are two nonisomorphic graphs G and H
4, and, say, G is disconnected. Denote by C a minimal connected
component of G. First we show that G has exactly two connected components and C =I.
PROOF.
with the same deck, n
K6+
be a vertex of the minimal degree in C, and let u be such a vertex that
We claim that either u
() 6 or G is regular of degree n -2 Indeed,
2
Let
G
9u
H.
otherwise,
CI
>.
max
(deg () + I, deg (u) + I)
which contradicts the minimality of C.
>
n/2,
Furthermore, if G is regular then again
u are in different components since, otherwise, the degree sequences of G and
different.
Now it follows by Corollary I, deg(9) + deg (u)
n -2.
Gu
Therefore, G has
exactly two components, C is regular, and A >. n/2.
A
Let us show that C is just
I. Since all vertices of degree are in C, we have
deg (9)
+
K6+
1 <- ICI
-< n
A
i.
+
A <- n
Hence, applying Corollary 2, we get
n
Thus, deg ()
2 <. 6
+
6, deg (u)
A
=<
deg (9)
A, and C
Finally, G/ z-G since deg ()
which.is a contradiction.
-= K6+I.
ICl
2.
u6 C and dam (u)
A
ICI
This completes the proof.
REFERENCES
I.
STANLEY, R.P.
2.
(1985), 132-138.
KRASIKOV, I. and RODITTY, Y.
Reconstruction from vertex switching. J.
Combin... Theory (B)
38,
Balance equations for reconstruction problems.
48 (1987), 458-464.
Archly der Mathematik, Vol.
and
are