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Reports of the Academy of Sciences of the USSR
- Volume 164, No. 4
MATHEMATICS
V. G. VIZING
AN ESTIMATE OF THE EXTERNAL STABILITY NUMBER OF A GRAPH
(Presented by Academician L. V. Kantorovich, 13 III 1965)
Only finite undirected graphs without loops or multiple edges are considered \((^1)\). A subset \(T\) of the vertices of a graph is called externally stable if every vertex not belonging to \(T\) is adjacent to at least one vertex of \(T\). The least cardinality of an externally stable set of a graph \(G\) is denoted by \(\beta(G)\) and is called the external stability number of the graph \(G\). In the present note the relation between the external stability number and the number of edges in an \(n\)-vertex graph is studied. Analogous problems for the internal stability number and the chromatic number were solved in \((^{2-4})\). We shall denote by \(\sigma(x)\) the degree of the vertex \(x\), and by \(\sigma(G)\) the maximum degree of a vertex of the graph \(G\).
Lemma 1. If \(G\) is an \(n\)-vertex graph, then \(\sigma(G) \leq n - \beta(G)\).
Lemma 2. If \(G\) is a connected \(n\)-vertex graph, where \(n \geq 2\), then
\(\beta(G) \leq [n/2]\).*
A graph \(G\) will be called critical if either it is complete, or if, upon joining by an edge two of its nonadjacent vertices, one obtains a graph whose external stability number is equal to \(\beta(G)-1\).
We shall call the gluing of vertices \(x\) and \(y\) of a graph \(G\) the operation of replacing them by one vertex adjacent to those and only those vertices which were adjacent to at least one of the vertices \(x\) and \(y\).
Lemma 3. If a graph \(H\) is obtained from a critical graph \(G\) by gluing two nonadjacent vertices, then \(\beta(H)=\beta(G)-1\).
We shall denote by \(f(n,k)\) the greatest number of edges that a connected \(n\)-vertex graph with external stability number \(k\) can have. In what follows, when we write \(f(n,k)\), we shall, without special reservation, mean that \(k \leq [n/2]\) for \(n \geq 2\) (Lemma 2) and \(k=1\) for \(n=1\).
We shall denote by \(C(n,k)\) a connected \(n\)-vertex graph with \(f(n,k)\) edges and with \(\beta(C(n,k))=k\). Obviously, the graph \(C(n,k)\) is critical.
Theorem 1.
\[ f(n,1)=\frac{n(n-1)}{2}, \qquad f(n,2)=\frac{n(n-2)}{2}, \]
\[ f(n,k)=(n-k+1)(n-k)/2 \quad \text{for } k \geq 3. \]
Proof. \(f(n,1)=n(n-1)/2\), since \(C(n,1)\) is the complete \(n\)-vertex graph.
\(f(n,2) \leq [n(n-2)/2]\), since, by Lemma 1, \(\sigma(C(n,2)) \leq n-2\). On the other hand, for \(n \geq 4\) it is easy to construct a connected \(n\)-vertex graph
* The brackets \([\,]\) denote the integer part of a number.
with \([n(n-2)/2]\) edges and with external stability number 2; for even \(n\) such a graph is obtained by deleting from the complete \(n\)-vertex graph a matching consisting of \(n/2\) edges; for odd \(n\), by deleting from the complete \(n\)-vertex graph a matching of \((n-1)/2\) edges and one edge issuing from a vertex not incident with any edge of this matching. Consequently,
\[
f(n,2)=[n(n-2)/2].
\]
Let us prove the equality
\[
f(n,k)=(n-k+1)(n-k)/2
\]
for \(k\geq 3\).
First,
\[
f(n,k)\geq (n-k+1)(n-k)/2,
\]
since for any natural \(n\) and \(k\), \(3\leq k\leq [n/2]\), one can construct a connected \(n\)-vertex graph \(H\) with \((n-k+1)(n-k)/2\) edges and with \(\beta(H)=k\): the vertices of \(H\) are
\[
p_1,p_2,\ldots,p_k,\ q_1,q_2,\ldots,q_{n-k};
\]
the vertex \(p_i\) is adjacent to \(q_i\) for \(1\leq i\leq k-1\); \(p_k\) is adjacent to all vertices \(q_l\) for \(l\geq k\); the vertices \(q_1,q_2,\ldots,q_{n-k}\) are pairwise adjacent. Note that from the inequality just proved it follows easily that
\[
\sigma(C(n,k))\geq n-2k+2
\]
for \(k\geq 3\).
To complete the proof of the theorem it remains to show that
\[
f(n,k)\leq (n-k+1)(n-k)/2
\]
for \(k\geq 3\). We prove this by induction on \(k\), beginning with \(k=3\).
Let \(x\) be a vertex of maximum degree in the graph \(C(n,3)\); \(\Gamma(x)\) is the set of vertices adjacent to \(x\). Since \(\sigma(x)\geq n-4\), by Lemma 1 only two cases are possible:
\[
\sigma(x)=n-4
\]
and
\[
\sigma(x)=n-3.
\]
Let \(\sigma(x)=n-4\); let \(x_1,x_2,x_3\) be vertices of the graph \(C(n,3)\) not adjacent to \(x\). Since \(\beta(C(n,3))=3\), in \(\Gamma x\) there is not a single vertex adjacent simultaneously to the vertices \(x_1,x_2,x_3\), and none of the vertices \(x_1,x_2,x_3\) is nonadjacent to the other two. Hence
\[
\sigma(x_1)+\sigma(x_2)+\sigma(x_3)\leq 2(n-4)+2=2(n-3).
\]
Consequently, the number of edges of the graph \(C(n,3)\) is not greater than
\[
2(n-3)+(n-3)(n-4)/2=(n-2)(n-3)/2.
\]
Now let \(\sigma(x)=n-3\); \(x_1\) and \(x_2\) are vertices not adjacent to \(x\). Denote by \(L_1\) and \(L_2\) the subsets of vertices of \(\Gamma x\) adjacent respectively to \(x_1\) and \(x_2\). Obviously, \(x_1\) and \(x_2\) are nonadjacent; \(|L_1|>0\); \(|L_2|>0\)*, \(L_1\cap L_2=\varnothing\); whichever two vertices are taken—one from \(L_1\) and the other from \(L_2\)—there is in \(\Gamma x\) a vertex adjacent to neither of these two. Hence it is not hard to infer that
\[
\sigma(x_1)+\sigma(x_2)=|L_1|+|L_2|
\]
and that the subgraph generated by the vertices of \(\Gamma(x)\) has no more than
\[
(n-3)(n-4)/2-|L_1|-|L_2|
\]
edges. Therefore,
\[
f(n,3)\leq |L_1|+|L_2|+(n-3)(n-4)/2-|L_1|-|L_2|+n-3
=(n-2)(n-3)/2.
\]
Suppose now that for all \(k\), \(3\leq k<k_0\), it has been proved that
\[
f(n,k)\leq (n-k+1)(n-k)/2,
\]
and suppose, contrary to the assertion of the theorem, that
\[
f(n,k_0)>(n-k_0+1)(n-k_0)/2.
\]
Then in the graph \(C(n,k_0)\), for any two nonadjacent vertices \(x'\) and \(x''\), there is a third vertex adjacent to both of them, since otherwise, by “gluing” \(x'\) and \(x''\), we would obtain a connected \((n-1)\)-vertex graph with number of edges \(>f(n-1,k_0-1)\), whose external stability number is equal to \(k_0-1\), which contradicts the induction hypothesis.
Let \(y\) be a vertex of maximum degree in the graph \(C(n,k_0)\), \(\Gamma y\) the set of vertices adjacent to \(y\); let \(y_1,y_2,\ldots,y_j\) be vertices not adjacent to \(y\). Since
\[
\sigma(C(n,k_0))\geq n-2k_0+2,
\]
we have
\[
j\leq 2k_0-3.
\]
Consider two cases.
Case 1. \(\sigma(y)\geq n-2k_0+3\). Then \(j\leq 2k_0-4\). Choose from the set of vertices
\[
y_1,y_2,\ldots,y_j
\]
the maximum number of disjoint pairs, each of which consists of two nonadjacent vertices. Since \(j\leq 2k_0-4\), the number of such pairs \(r\leq k_0-2\). Let \(v_i\) be a vertex adjacent to both vertices of the \(i\)-th pair \((1\leq i\leq r)\). If \(r=k_0-2\), then the vertices
\[
y,\ v_1,\ v_2,\ldots,v_r
\]
form an externally stable set of the graph \(C(n,k_0)\), which contradicts
\[
\beta(C(n,k_0))=k_0.
\]
Now let \(r\leq k_0-3\). Since the vertices among \(y_1,y_2,\ldots,y_j\) that did not enter into the selected pairs are pairwise adjacent, then, taking one of them and the vertices
\[
y,\ v_1,\ v_2,\ldots,v_r,
\]
we
* If \(A\) is a finite set, then by \(|A|\) is denoted the number of its elements.
again obtain an externally stable set of the graph \(C(n,k_0)\) having no more than \(k_0-1\) vertices, which again is impossible.
Case 2. \(\sigma(y)=n-2k_0+2\). Then \(j=2k_0-3\). If in \(\Gamma y\) there is a vertex adjacent to at least three vertices among \(y_1,y_2,\ldots,y_{2k_0-3}\), or if among the vertices \(y_1,y_2,\ldots,y_{2k_0-3}\) there is a vertex adjacent to at least two others, then we easily arrive at case 1. If the indicated possibilities are excluded, then
\[
\sigma(y_1)+\sigma(y_2)+\cdots+\sigma(y_{2k_0-3})\le
2(n-2k_0+2)+2k_0-4=2n-2k_0.
\]
Therefore the number of edges of the graph \(C(n,k_0)\) must be no more than
\[
\bigl[(n-2k_0+3)(n-2k_0+2)+2n-2k_0\bigr]/2
\le (n-k_0+1)(n-k_0)/2,
\]
which contradicts the assumption \(f(n,k_0)>(n-k_0+1)(n-k_0)/2\).
Theorem 1 is proved.
Corollary 1. If a connected \(n\)-vertex graph \(G\) has \(m\) edges \((m\ge n-1)\), then
\[
1\le \beta(G)\le
\begin{cases}
\min\left\{\left[\dfrac n2\right],
\left[\dfrac{1+2n-\sqrt{8m+1}}2\right]\right\},
& \text{if } m\le \dfrac{(n-2)(n-3)}2,\\[1.2em]
\min\left\{\left[\dfrac{n^2-2m}2\right],2\right\},
& \text{if } m> \dfrac{(n-2)(n-3)}2.
\end{cases}
\]
Let \(n\) and \(k\) be positive integers such that \(k\le n\). Denote by \(g(n,k)\) the greatest number of edges in an \(n\)-vertex graph with external stability number \(k\), and by \(M(n,k)\) an \(n\)-vertex graph with \(g(n,k)\) edges and with \(\beta(M(n,k))=k\).
Theorem 2.
\[
g(n,k)=
\begin{cases}
\dfrac{n(n-1)}2, & \text{for } k=1,\\[1.2em]
\left[\dfrac{(n-k+2)(n-k)}2\right], & \text{for } k\ge 2.
\end{cases}
\]
For fixed \(n\) and \(k\), the graphs \(M(n,k)\) are isomorphic.
Proof. The theorem is obvious for \(k=1\) and \(k=2\). By virtue of
\[
g(n,k)\le g(n+1,k+1)
\]
we have
\[
g(n,k)\ge g(n-k+2,2)=\left[\frac{(n-k+2)(n-k)}2\right]
\]
for \(k\ge 2\). Hence it follows that for \(k\ge 3\) the graph \(M(n,k)\) is disconnected. Indeed, if \(k>[n/2]\), then this follows from Lemma 2, while if \(3\le k\le [n/2]\), then from the inequality
\[
g(n,k)\ge \left[\frac{(n-k+2)(n-k)}2\right]>(n-k+1)(n-k)/2=f(n,k).
\]
Therefore, for \(k\ge 3\), the graph \(M(n,k)\) cannot have a connected component whose external stability number is \(\ge 3\). Further, by a simple count of edges one can verify that the graph \(M(n,k)\) does not have two connected components each of which differs from an isolated vertex. Consequently, for \(k\ge 3\) the graph \(M(n,k)\) consists of \(k-2\) isolated vertices and a graph isomorphic to \(M(n-k+2,2)\). Theorem 2 is proved.
Corollary 2. If an \(n\)-vertex graph \(H\) has \(m\) edges, then
\[
\max\{n-m,1\}\le \beta(H)\le [\,n+1-\sqrt{1+2m}\,].
\]
Let us note in conclusion that in the class of all \(n\)-vertex graphs with \(m\) edges (analogously in the class of \(n\)-vertex connected graphs with \(m\) edges \((m\ge n-1)\)) there exists a graph with external stability number equal to any preassigned integer lying within those limits indicated by Corollary 2 (respectively, Corollary 1).
Institute of Mathematics
Siberian Branch of the Academy of Sciences of the USSR
Received
25 II 1965
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