AbstractA graph G is said to be well-covered if every maximal independent set of vertices has the same cardinality. A planar (simple) graph in which each face is a triangle is called a triangulation. It is the aim of this paper to prove that there are no 5-connected planar well-covered triangulations

Finbow, A.

Hartnell, B.

Nowakowski, R.

Plummer, Michael D.

Elsevier - Publisher Connector 

On well-covered triangulations: Part I 

Discrete Applied Mathematics 132 (2004) 97–108
www.elsevier.com/locate/dam
On well-covered triangulations: Part I
A. Finbowa , B. Hartnella , R. Nowakowskib , Michael D. Plummerc
aDepartment of Mathematics and Computing Science, Saint Mary’s University, Halifax, Nova Scotia,
Canada B3H 3C3
bDepartment of Mathematics, Dalhousie University, Halifax, Nova Scotia, Canada B3H 4H8
cDepartment of Mathematics, Vanderbilt University, Nashville, Tennessee 37240, USA
Received 24 June 2001; received in revised form 30 May 2002; accepted 4 September 2002
Abstract
A graph G is said to be well-covered if every maximal independent set of vertices has the same
cardinality. A planar (simple) graph in which each face is a triangle is called a triangulation. It
is the aim of this paper to prove that there are no 5-connected planar well-covered triangulations.
? 2003 Elsevier B.V. All rights reserved.
Keywords: Independent set; Well-covered; Triangulation
1. Introduction
In this paper all graphs will be ;nite and simple.
In 1969, the fourth author ;rst proposed the study of graphs in which each maximal
independent set of vertices has the same size and suggested that the name well-covered
be applied to them [9]. Although it is now well known that the independent set problem
is NP-complete for graphs in general (cf. [5]), for certain interesting subfamilies of
graphs, such as those called claw-free, the problem becomes polynomially solvable
(cf. [8,12]). Clearly, the independent set problem has a polynomial solution for the
class of well-covered graphs, but how does one recognize this class? It was shown
independently by ChvBatal and Slater [2] and by Sankaranarayana and Stewart [11] that
the recognition problem for well-covered graphs is co-NP-complete. In contrast, if the
E-mail addresses: art.;nbow@stmarys.ca (A. Finbow), bert.hartnell@stmarys.ca (B. Hartnell),
rjn@mathstat.dal.ca (R. Nowakowski), plummemd@ctrvax.vanderbilt.edu (M.D. Plummer).
0166-218X/$ - see front matter ? 2003 Elsevier B.V. All rights reserved.
doi:10.1016/S0166-218X(03)00393-7
98 A. Finbow et al. / Discrete Applied Mathematics 132 (2004) 97–108
graphs are claw-free, then the recognition problem becomes polynomial. (See [13,14].)
For more comprehensive treatments of well-covered graphs, see [3,10].
A widely studied subclass of graphs are those which are maximal planar and which
are commonly called (planar) triangulations. Any triangulation (larger than a single
triangle) must have vertex connectivity 3, 4 or 5 since, by Euler’s theorem it must
contain a vertex of degree 3, 4 or 5. Lebesgue [7], Kotzig [6], Borodin [1] and Jendrol’
[4] have extensively investigated what kind of con;gurations must always exist in
any triangulation. In the next section, these results are used to prove that there is no
5-connected planar well-covered triangulation. Planar well-covered triangulations which
are 3- and 4-connected will be treated in a subsequent paper.
Note that in this paper all graphs are simple and if v is a vertex of a graph, N [v]
will denote the closed neighborhood of the vertex v; namely, N [v] = N (v) ∪ {v}.
2. There exists no 5-connected well-covered triangulation
In order to prove the main result of this section, as well as other results to follow,
we shall make extensive use of the work of Kotzig [6], Borodin [1] and Jendrol’ [4]
on the structure of triangulations. After Jendrol’, we shall call a triangle with vertex
degrees a, b and c an (a; b; c)-triangle. The results of the above three authors which
we shall need can be summarized as follows. (See [4, Theorem 4].)
Theorem 2.1. Each 5-connected planar triangulation of order at least 7ve contains
an (a; b; c)-triangle, where
(i) a= 5, b= 5, 56 c6 7, or
(ii) a= 5, b= 6, c = 6.
First we remind the reader that if a graph is 5-connected, it has a unique embedding
in the plane and that any face may be considered to be the outer or in;nite face.
We further note that in a 5-connected triangulation, no 4-cycle is induced. Finally we
remind the reader of the following well known (see [10]) but very useful observation
when working with well-covered graphs: if G is a well-covered graph and I is an
independent set of vertices of G, then G − N [I ] is also well-covered.
A technical lemma was found to facilitate the proof of the main result.
Denition 2.2. The 5-tuple (F; L; B; C; d) is called an F-con7guration in the graph G
provided that F is a subgraph of G, B ∪ C ∪ {d} ⊆ V (F), B ∩ C = ∅, and there is an
integer k¿ 2 such that the following are true:
Card(B) = k and {b1; b2; : : : ; bk}= B ⊆ N (d) ∩ V (F); (2.1)
Card(C) = k and {c1; c2; : : : ; ck}= C ⊆ V (F); (2.2)
N [C] ⊆ V (F); (2.3)
A. Finbow et al. / Discrete Applied Mathematics 132 (2004) 97–108 99
L ⊆ F is independent in V (G) and is a maximal independent set in V (F); (2.4)
B ⊆ L; (2.5)
(L− {bi}) ∪ {ci} is independent in V (G) and a maximal independent set in V (F)
for each i∈{1; 2; : : : ; k}; (2.6)
(L− B) ∪ {d} ∪ (C − {c1; c2}) is independent in V (G) and a maximal independent
set in V (F): (2.7)
We note that (2.1), (2.5) and (2.7) combine to yield
N (d) ∩ L= B (2.8)
and that (2.6) together with B ∩ C = ∅ implies that
L ∩ C = ∅: (2.9)
Lemma 2.3. A 5-connected well-covered planar triangulation does not contain an
F-con7guration.
Proof. Suppose for a contradiction that (F; L; B; C; d) is an F-con;guration in a 5-
connected well-covered planar triangulation G. Let I be a maximal independent set in
V (G) containing L such that |I ∩ N (d)| is a minimum.
Observe that for each j∈{1; 2; : : : ; k}, Ij =(I −{bj})∪{cj} is independent by (2.3)
and (2.6). Further, since G is well-covered and since the cardinality of I and Ij are
equal, Ij must be a maximal independent set in G.
As a result, we note that
if v∈G − F then v∈N [I − {bj}] for each j∈{1; 2; : : : ; k}: (3.1)
We claim that J = (I − N (d)) ∪ {d} ∪ (C − {c1; c2}) is a maximal independent
set in G. Note that (2.3) combined with (2.7) imply that J is independent; and (2.7)
combined with (2.8) imply that N [J ] ⊇ F .
To show that J is a maximal independent set in G assume that v∈G − F and that
v ∈ N [J ]. Then we note that v is not adjacent to d, but that v is adjacent to one or
more elements of I (by the maximality of I), and all of these elements must be in
N (d) (by the construction of J ).
However, if v were adjacent to 1 and 2 ∈N (d) ∩ I , then G would contain the
induced 4-cycle 1d2v (dv is not in G and 12 ∈ G since 1 and 2 are in the
independent set I), which as noted above is impossible. Hence, v is adjacent to exactly
one element of N (d) ∩ I .
By (3.1) v is adjacent to no element of B, so v is adjacent to exactly one ∈ (N (d)∩
I)−F . But then J ′=(I−{})∪{v} is an independent set in G with the same cardinality
100 A. Finbow et al. / Discrete Applied Mathematics 132 (2004) 97–108
Fig. 1.
as I and hence by the well-covered property of G, J ′ is a maximal independent set in
G containing L, but such that |J ′ ∩ N (d)|¡ |I ∩ N (d)|. This contradicts the minimum
intersection property of I . This shows that no such v exists and hence that J is a
maximal independent set in G.
But by (2.8), |I |¿ |J |+ 1 and thus since G is well-covered, the hypothesis that G
contains an F-con;guration is false.
We now turn to the main result of this section.
Theorem 2.4. No 5-connected triangulation is well-covered.
Proof. From the results of Jendrol’ et al. above, we see that every 5-connected trian-
gulation G must contain a triangle of type (5, 5, 5), (5, 5, 6), (5, 5, 7), or (5, 6, 6).
We proceed to treat each of these possibilities in sequence.
Step 1: First we will show that G contains neither a (5, 5, 5)-triangle nor a (5, 5, 7)-
triangle.
Suppose, to the contrary, that G does contain either a (5, 5, 5)-triangle or a (5, 5, 7)-
triangle. Then G will contain as an induced subgraph the basic 9-vertex con;guration
(if it contains the (5, 5, 5)-triangle a1a2a3) or the basic 11-vertex con;guration (if it
contains the (5, 5, 7)-triangle a1a2a3) shown in Fig. 1(i) and (ii), respectively.
In either case, denote this con;guration by F .
Observe that planarity and 5-connectedness imply that F is an induced subgraph of G
(for example the edge zw cannot be in G for otherwise zwa1a3 forms an induced 4-cycle
in G.) Then the set L = {x1; x2; z} in case (i) (respectively, the set L = {x1; x2; z1; z2}
in case (ii)) is independent in G. But then setting X = {x1; x2}, and A= {a1; a2}, we
see that (F; L; X; A; w) is an F-con;guration in G in violation of Lemma 2.3. Thus, the
hypothesis that G contains either a (5, 5, 5)-triangle or a (5, 5, 7)-triangle is false,
completing step 1.
Step 2: Next we will show that G contains no (5, 5, 6)-triangle.
A. Finbow et al. / Discrete Applied Mathematics 132 (2004) 97–108 101
Fig. 2.
Suppose, to the contrary, that G does contain a (5, 5, 6)-triangle a1a2a3. Then G
must contain the 10-vertex con;guration H of Fig. 2
As in the case of the subgraph F in step 1, H is an induced subgraph of G. We
begin by noting that by the 5-connectivity and maximal planarity, vertices yi and zi
must share a common neighbor pi ∈G−H , for i=1 and 2 (see Fig. 3). Note that the
absence of separating 4-cycles implies that p1 = p2, and that neither edge p1z2 nor
edge p2z1 is in G. Furthermore, the edge p1p2 is not present, for if it were, then in
the 4-cycle p1z1z2p2, one of the diagonals p2z1 or p1z2 would have to be in G (since
no 4-cycle in G is induced). Finally note that the edge x1x2 is not in G since H is
induced.
Claim 1. Either the set {p1; z2; x1; x2} or the set {p2; z1; x1; x2} is independent.
To prove Claim 1, ;rst note that the edges x1x2; x1z2, and z1x2 are not in G since
H is induced. Hence it will suMce to show that
For some i in {1; 2}; neither the edge x1pi nor the edge x2pi is in G: (4.1)
Suppose that G contains neither the edge x1p1 nor the edge x2p2. Then, since by
planarity one of the edges x1p2 or x2p1 is not in G, (4.1) follows.
Next suppose that G contains exactly one of the edges x1p1 or x2p2. Without loss
of generality assume that the edge x1p1 is not in G and the edge x2p2 is in G. We
obtain (4.1) in this case by establishing that the edge p1x2 is also not in G.
Indeed if p1x2 were in G, then let I be a maximal independent set in G containing
{p1; a2; x1; z2}. Now since G is well-covered, the set (I − {z2; a2}) ∪ {y2} fails to be
102 A. Finbow et al. / Discrete Applied Mathematics 132 (2004) 97–108
Fig. 3.
maximal and hence there exists a vertex z adjacent to z2, but not adjacent to y2 nor to
any other vertex in I −{z2}. Also since G is well-covered, the set (I −{x1; a2})∪{a1}
fails to be maximal and hence there exists a vertex x adjacent to x1, but to no other
vertex in I (see Fig. 4). But x and z are not adjacent by planarity and hence J = (I −
{a2; x1; z2}) ∪ {x; z; a1; y2} is independent. Since |J | = |I | + 1, this contradicts the fact
that G is well-covered. Hence, the edge p1x2 is not in G and (4.1) is established in
this case.
Finally suppose that G contains both the edge x1p1 and the edge x2p2. Since G
is maximal planar and since p1 and p2 are not adjacent, x1 and p1 share a common
neighbor u1 in G − H and by symmetry x2 and p2 share a common neighbor u2 in
G − H . Note that planarity and 5-connectivity prevent the ui from being any of the
vertices in {a1; a2; a3; y1; y2; p1; p2; x1; x2; z1; z2; w}.
Now if u1 =u2 or if the edge u1u2 is not in G, then let I be a maximal independent
set in G containing {u1; u2; y1; y2} and let J = (I − {y1; y2}) ∪ {a3}. Since I and
J are maximal independent sets of diNerent sizes, this contradicts the fact that G is
well-covered. Hence, u1 = u2 and the edge u1u2 is in G as shown in Fig. 5.
Further, neither u1 nor u2 can be adjacent to w, for suppose edge u1w is in G.
Then by 5-connectivity, there are no vertices external to the 4-cycle u1wx2u2. But then
G−N [{p1; p2}] would contain as a component the subgraph induced by {a1; a2; a3; w}
which is not well-covered. Hence G is not well-covered, a contradiction.
Now, for i=1; 2; let qi be the external common neighbor of pi and zi. Observe that
qi= zj and qi=uj, where i and j range over {1; 2}, are all excluded by 5-connectivity.
A. Finbow et al. / Discrete Applied Mathematics 132 (2004) 97–108 103
Fig. 4.
We further observe that if q1 = q2 or if the edge q1q2 is not in G, then G −
N [{q1; q2; w}] would contain as a component the subgraph of G induced by {y1; y2; a3}
which is not well-covered. Hence again G is not well-covered, a contradiction. Hence
q1 = q2 and the edge q1q2 ∈E(G), as shown in Fig. 6.
But now by planarity, one of the edges u1q2 or q1u2 is not in G; without loss of
generality suppose edge u1q2 is not in G. Note that then {u1; q2; w} is independent in
G and G−N [{u1; q2; w}] contains as a component the graph induced by {y2; a3; y1; z1}
which is not well-covered. Therefore again G is not well-covered, a contradiction.
Hence Claim 1 is proved.
Without loss of generality, suppose that the set K = {p1; x1; x2; z2} is independent.
Let F be the subgraph of G depicted in Fig. 3 and set X = {x1; x2}, and A =
{a1; a2}. Since w can be adjacent to neither p1 nor z2, we see that (F; K; X; A; w) is an
F-con;guration in G in violation of Lemma 2.3. Thus, the hypothesis that G contains
a (5, 5, 6)-triangle is false completing step 2.
Step 3: Next we will show that G contains no (5, 6, 6)-triangle.
Suppose, to the contrary, that G does contain a (5, 6, 6)-triangle a3a2a1. Then G
must contain the 11-vertex con;guration H of Fig. 7.
As in the case of the subgraph F in step 1, H is an induced subgraph of G. We
begin by noting that by 5-connectivity and maximal planarity, vertices wi and xi must
104 A. Finbow et al. / Discrete Applied Mathematics 132 (2004) 97–108
Fig. 5.
share a common neighbor pi ∈G−H , for both i=1 and 2. Furthermore, vertices z and
wi must have a common neighbor ri ∈G−H , for i= 1 and 2. Note that it is possible
that r1 = r2, but since separating 4-cycles are forbidden, ri = p1 = p2 = ri, for i = 1
and 2 (see Fig. 8).
Claim 2. Each of r1 and r2 is adjacent to x3 and thus N (z) − H = {r1; r2}, where
either r1 = r2 or edge r1r2 is in G.
To prove Claim 2, ;rst suppose that the edge r1x3 is not in G. Thus, the set K =
{x1; x2; x3; r1} is independent in G. Let F be the subgraph of G depicted in Fig. 8 and
set F1 =(F ∩N [K])−{p1}, X ={x1; x3}, and A={a1; a3}. Since y1 can be adjacent to
neither r1 nor x2, we see that (F1; K; X; A; y1) is an F-con;guration in G in violation
of Lemma 2.3. Thus, the hypothesis that the edge r1x3 is not in G is false. Similarly,
the hypothesis that the edge r2x3 is not in G is false.
If r1=r2, then by planarity and 5-connectivity, N (z)−H={r1}={r1; r2}. Otherwise,
each r ∈N (z)− (H ∪{r1; r2}) lies outside the 4-cycle r1zr2x3 (that is, is separated from
the vertices a1; a2 and a3) and thus cannot exist. Hence, N (z) − H = {r1; r2} and by
maximal planarity, the edge r1r2 is in G. This completes the proof of Claim 2.
In summary then, the graph G contains a subgraph as shown in Fig. 9 where r1 = r2
is possible.
A. Finbow et al. / Discrete Applied Mathematics 132 (2004) 97–108 105
Fig. 6.
Fig. 7.
106 A. Finbow et al. / Discrete Applied Mathematics 132 (2004) 97–108
Fig. 8.
Fig. 9.
Claim 3. Each vertex in {r1; r2} is adjacent to either p1 or p2.
To prove Claim 3, let us begin by assuming that r1 is adjacent to neither p1 nor p2.
Then the set K = {y1; y2; p1; p2; r1} is independent in G. (Indeed, edges piyj, i = j,
are prohibited by planarity combined with Claim 2, whereas the edges piyi and r1yj
are prohibited by 5-connectivity.)
A. Finbow et al. / Discrete Applied Mathematics 132 (2004) 97–108 107
Fig. 10.
Let F be the subgraph of G depicted in Fig. 9 and set Y = {y1; y2; r1}, and A =
{a1; a2; z}. We note that for i = 1 and 2, edge pix3 would yield the 4-cycle x3pixiyi.
But both of the edges x3xi and piyi are prohibited by the 5-connectivity and hence
this 4-cycle would be induced in G, a contradiction. Thus, we see that (F; K; Y; A; x3)
is an F-con;guration in G in violation of Lemma 2.3.
Thus, the hypothesis that r1 is adjacent to neither p1 nor p2 is false. Similarly, the
hypothesis that r2 is adjacent to neither p1 nor p2 is false. This completes the proof
of Claim 3.
Now we observe that, because of planarity, if r1 = r2, the only way that Claim 3
can be satis;ed is for both edges r1p1 and r2p2 to be in G. Hence, the graph must
108 A. Finbow et al. / Discrete Applied Mathematics 132 (2004) 97–108
contain one of the two con;gurations H1 and H2 shown in Fig. 10 where in H2 the
edge r2p2 may or may not exist.
Let K={a3; p1; p2; z} and note that this set is independent in G. Let I be a maximal
independent set in G containing K and observe that it contains no vertices of Fig. 10
except those in K . Now both the sets Ji = (I − {z; pi}) ∪ {wi}, for i = 1 and 2 are
independent. Since |I | = |Ji| + 1, and G is well-covered, there exists a vertex qi such
that qi = N [Ji] because Ji is not maximal. But qi ∈N [I ] and hence qi is adjacent to pi,
but qi is adjacent to no other vertex in I . However, q1q2 is not in G by planarity and
hence J = (I −{z; p1; p2})∪ {q1; q2; w1; w2} is independent. But |J |= |I |+1, violating
the well-covered property of G and completing the proof of step 3 and thence the
theorem.
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On well-covered triangulations: Part I

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