On the maximum weight of a planar graph of given order and size

Discrete Applied Mathematics 177 (2014) 101–110

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On the maximum weight of a planar graph of given order and size Andrej Gajdoš, Mirko Horňák ∗ , Peter Hudák, Tomáš Madaras Institute of Mathematics, P.J. Šafárik University, Jesenná 5, 040 01 Košice, Slovakia

article

info

Article history: Received 22 January 2014 Received in revised form 23 May 2014 Accepted 28 May 2014 Available online 14 June 2014 Keywords: Edge weight Graph weight Planar graph

abstract The weight of an edge uv of a graph is defined to be the sum of degrees of the vertices u and v . The weight of a non-empty graph G is the minimum of the weights of edges of G. The paper is concerned with the maximum weight of a planar graph having n vertices and −12n ⌋. m edges. It is shown that if m ≥ 2n + 1, then the maximum weight is at most ⌊ 9m m−2n Moreover, there are infinitely many pairs (n, m) such that the maximum weight is at least −12n ⌊ 9m ⌋ − 1. m−2n © 2014 Elsevier B.V. All rights reserved.

1. Introduction Throughout this paper, we consider non-empty simple (having neither loops nor multiple edges) finite graphs. Let G be such a graph. The degree of a vertex v ∈ V (G) is denoted by deg(v) and a vertex of degree d is called a d-vertex. The weight of an edge uv ∈ E (G) is w(uv) := deg(u) + deg(v) and the weight of G is w(G) := min(w(e) : e ∈ E (G)). For p, q ∈ Z let [p, q] denote the integer interval bounded by p, q, i.e. the set {z ∈ Z : p ≤ z ≤ q}, and by [p, ∞) the integer interval lower bounded by p, i.e. the set {z ∈ Z : p ≤ z }. The research of graph weights can be traced back to the paper [9] of Wernicke who proved (in dual form) that each plane triangulation of minimum degree 5 has the weight at most 11. This result was generalised to polyhedral graphs in the classical paper [8] by Lebesgue (with the upper bound 14) and by Kotzig [7] who proved that the weight of a polyhedral graph is at most 13, and at most 11 in the case of the absence of 3-vertices; both bounds 13 and 11 are best possible. These results initiated a systematic study of structural properties of planar graphs. A survey of results on the weight of planar graphs and related topics can be found in Jendrol’ and Voss [5]. Beyond planar graphs, the study of graph weights was further stimulated by the problem of Erdős which was posed at the Fourth Czechoslovak Symposium on Combinatorics held in 1990 in Prachatice: What is the maximum weight of a graph G on n vertices and m edges? In more general setup, the problem of Erdős can be written in the following way: A graph property is an isomorphism closed subclass of the class I of all  n simple finite graphs, see e.g. Borowiecki, Broere and Mihók [1]. Consider a graph property P . For n ∈ [2, ∞) and m ∈ [1, 2 ] let P (n, m) denote the class of all graphs G ∈ P with |V (G)| = n and |E (G)| = m. If P (n, m) ̸= ∅, the number

w(P , n, m) := max(w(G) : G ∈ P (n, m)) is well defined.

∗

Corresponding author. Tel.: +421 552342561. E-mail addresses: [email protected] (A. Gajdoš), [email protected] (M. Horňák), [email protected] (P. Hudák), [email protected] (T. Madaras). http://dx.doi.org/10.1016/j.dam.2014.05.047 0166-218X/© 2014 Elsevier B.V. All rights reserved.

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A. Gajdoš et al. / Discrete Applied Mathematics 177 (2014) 101–110

First results for the most general property I (corresponding to the original Erdős question) were obtained by Ivančo and Jendrol’ [3], who found among other things the following lower bound for w(I, n, m).

 √ ], a = ⌈ 12 (1 + 1 + 8m)⌉, b = 12 (a2 − a − 2m), h = ⌈ 12 (2n − 1 − (2n − 1)2 − 8m)⌉ 2p and let p, k be integers such that hk + p = m and h(h − 3) < 2p ≤ h(h − 1). Let f (n, m) = h + k + ⌊ h ⌋ and let g (n, m) be Theorem 1. Let n ∈ [2, ∞), m ∈ [1,

n 2

defined by

g ( n, m ) =

2a − 2,  2a − 3,     2a − 4,

if b = 0, if b = 1,

   2a − 5,     2a − 6,

a

if 2 ≤ b ≤ if

2

a 2

or b = 3,



+1≤b≤

a+2



2

or (a, b) = (8, 6),

otherwise.

Then w(I, n, m) ≥ max(f (n, m), g (n, m)).



Note that the original formulation of Theorem 1 included also the condition h + k ≤ n, which can be shown to be redundant (deducible from other assumptions). The authors of [3] conjectured that the lower bound of Theorem 1 is in fact equal to w(I, n, m). Their conjecture was proved by Jendrol’ and Schiermeyer [4]. Theorem 2. If n ∈ [2, ∞), m ∈ [1, (f (n, m), g (n, m)). 

n ], and f (n, m), g (n, m) are functions defined in Theorem 1, then w(I, n, m) = max 2

The generalised problem was completely solved by Horňák, Jendrol’ and Schiermeyer [2] for the property O 2 ‘‘to be a bipartite graph’’:



√

n−

n2 −4m 2



, s∗ = a∗ b∗ − m, p∗ = min(s∗ , 2), w ∗ = a∗ + a∗  √ √ b∗ − p∗ , r0 = n2 − 4m, r1 = (n − 1)2 − 4m, and r1′ = n2 − 4m − 4. 1. If r0 is an integer, then w(O 2 , n, m) = n = w ∗ . 2. If r0 is not an integer and (exactly) one of r1 , r1′ is, then w(O 2 , n, m) = n − 1 = w ∗ + 1. 3. If r0 , r1 , r1′ are not integers, then w(O 2 , n, m) = w ∗ .  Theorem 3. Let n ∈ [2, ∞), m ∈ [1, ⌊ ⌋⌈ ⌉], a = n 2

n 2

∗

, b∗ =

m

In this paper we are interested in analysing w(T3 , n, m), where T3 denotes the graph property ‘‘to be a planar graph’’. If m ∈ [1, 2n − 3], it is easy to determine w(T3 , n, m), see Propositions 1 and 2. For m ≥ 2n + 1 we bound w(T3 , n, m) from above in Theorem 4. In addition, we show in Theorem 5 that the bound given by Theorem 4 is close to be the best possible. 2. Optimum weight for planar graphs

˜ is an embedding of a planar graph G to a sphere (or, equivalently, to a plane), we denote by V (G˜ ), E (G˜ ) and F (G˜ ) the If G ˜ respectively. As usual, we do not distinguish elements of V (G) and V (G˜ ) (and vertex set, the edge set and the face set of G, ˜ ) as well) that correspond to each other (unless explicitly stated otherwise). of E (G) and E (G ˜ ) := {(u, v), (v, u) : uv ∈ E (G˜ )} be the set of all oriented edges of G. ˜ Consider a fixed ω, one of the two possible Let O(G ˜ An oriented edge (u, v) ∈ O(G˜ ) is incident to a face f ∈ F (G˜ ) (with respect to ω), orientations of the sphere σ carrying G. provided that the (non-oriented) edge uv is incident to f and u precedes v if the curve uv is oriented according to ω. Let O(f ) ˜ ) that are incident to the face f . be the set of all oriented edges in O(G A sequence (v1 , v2 , . . . , vk ) of vertices of f is an f -trail of length k − 1 if (vi , vi+1 ) ∈ O(f ) for each i ∈ [1, k − 1] and (vi , vi+1 ) ̸= (vj , vj+1 ) for each i, j ∈ [1, k − 1], i ̸= j. An angle of f is such an f -trail τ = (u, v, w) that for any ε > 0 the points u, w of the sphere σ can be joined by an arc Aε lying in f with the distance between each point of Aε and the set uv ∪ vw (of the points of the edges of the trail τ ) smaller than ε . A facial trail of f is such an f -trail (v1 , v2 , . . . , vk ) that (vi−1 , vi , vi+1 ) is an angle of f for each i ∈ [2, k − 1]. The degree of f is the length of a longest facial trail of f ; we denote it by deg(f ), and, similarly as in the case of vertices, a face of degree d is called a d-face. Each facial trail of f of length deg(f ) is closed, its first and last vertex are equal. Let F (f ) ˜ ). be the set of all facial trails of f that are of length deg(f ). Clearly, |F (f )| = deg(f ) for any face f ∈ F (G  ˜ ˜ ˜ If We denote by A(f ) the set of all angles of f and by A(G) := f ∈F (G˜ ) A(f ) the set of all angles of G (angles of faces of G). ˜ the vertex v is its centre. Let A(v) be the set of all angles of G˜ with the centre v . From the definitions it (u, v, w) is an angle of G, ˜ )| = 2|E (G˜ )|, |O(f )| = deg(f ) = |A(f )| for any f ∈ F (G˜ ) and |A(v)| = deg(v) for any v ∈ V (G˜ ). immediately follows that |O(G Proposition 1. If n ∈ [2, ∞) and m ∈ [1, n − 1], then w(T3 , n, m) = m + 1 = w(I, n, m). Proof. By Theorem 2 we have h = 1, p = 0, and w(I, n, m) = m + 1. The inequality w(T3 , n, m) ≤ w(I, n, m) is trivial, hence this proposition follows from the fact that the planar graph K1,m ∪ (n − m − 1)K1 of order n and size m has the weight m + 1. 

A. Gajdoš et al. / Discrete Applied Mathematics 177 (2014) 101–110

103

Proposition 2. If n ∈ [2, ∞) and m ∈ [n, 2n − 3], then w(T3 , n, m) = ⌈ m ⌉ + 2 = w(I, n, m). 2 Proof. In this case Theorem 2 yields h = 2, p ∈ [0, 1], and w(I, n, m) = ⌈ m ⌉ + 2. If m is even, m = 2l, the graph K2,l ∪ 2 (n − l − 2)K1 ∈ T3 (n, m) has the weight l + 2 = ⌈ m2 ⌉ + 2. Adding to the above graph a heavy non-edge joining two vertices of degree l leads to an optimum graph with m = 2l + 1.  Our main theorem bounds the number w(T3 , n, m) from above provided that m ≥ 2n + 1. Theorem 4. If n ∈ [7, ∞) and m ∈ [2n + 1, 3n − 6], then w(T3 , n, m) <

9m−12n . m−2n

Proof. We proceed by way of contradiction. For that purpose suppose that the set



G := G ∈ T3 (n, m) : n ≥ 7, 2n + 1 ≤ m ≤ 3n − 6, w(G) ≥

9m − 12n



m − 2n

˜ = (V˜ , E˜ , F˜ ) of a graph (of counterexamples to the statement of Theorem 4) is non-empty and consider an embedding G ˘G ∈ T3 (n, m) to a sphere satisfying |V˜ | = min(|V (G)| : G ∈ G) =: ν and |E˜ | = max(|E (G)| : G ∈ G, |V (G)| = ν). So, G˘ minimises the number of vertices (the main criterion), and then maximises the number of edges in a counterexample (the ˘ and let G˘ i be the ith component of G, ˘ i ∈ [1, c ]. There is j ∈ [1, c ] side criterion). Let c be the number of components of G ˘ j )| ≥ 2|V (G˘ j )| + 1, for otherwise with |E (G c 

˘ )| = m = |E (G

|E (G˘ i )| ≤ 2

i=1

c 

|V (G˘ i )| = 2|V (G˘ )| = 2n,

i=1

˘ i ) ≥ w(G˘ ) for each i ∈ [1, c ], we have G˘ j ∈ G, which (using the main criterion) shows that G˘ is a contradiction. Since w(G connected. ˜ := O(G˜ ) and A˜ := A(G˜ ). Putting ρ := |E˜ |/|V˜ | we obtain Let us define for simplicity O 2<

2n + 1 n

≤ρ≤

3n − 6

< 3.

n

(1)

˜ is a connected plane graph, from Euler’s formula |V˜ | − |E˜ | + |F˜ | = 2 it follows As |E˜ | = ρ|V˜ | and G  8 − 7ρ 2

v∈V˜

+

 9ρ − 10

+

2

e∈E˜

 8 − 9ρ

= 8 − 9ρ.

2

f ∈F˜

(2)

The number of angles in A˜ with centre v ∈ V˜ is equal to deg(v), hence

 8 − 7ρ 2

v∈V˜

=

8 − 7ρ

  v∈V˜ (u,v,w)∈A˜

2 deg(v)



=

8 − 7ρ



f ∈F˜ (u,v,w)∈A(f )

2 deg(v)

.

(3)

˜ | = 2|E˜ | and |O(f )| = |A(f )| for any f ∈ F˜ , we have also Since |O  9ρ − 10 2

e∈E˜

=

 9ρ − 10

=

4

(u,v)∈O˜



9ρ − 10



f ∈F˜ (u,v)∈O(f )

4

=





f ∈F˜ (u,v,w)∈A(f )

9ρ − 10 4

.

(4)

Thus, putting c (f ) :=

8 − 9ρ 2



+



8 − 7ρ 2 deg(v)

(u,v,w)∈A(f )

+

9ρ − 10



4

,

f ∈ F˜ ,

(5)

from (2) to (4) and from ρ ∈ (2, 3) (see (1)) we obtain



c (f ) = 8 − 9ρ < −10 < 0.

(6)

f ∈F˜

Let S denote the set of all finite sequences of positive integers and let q : S → Q be the mapping defined by q(d1 , d2 , . . . , dk ) :=

8 − 9ρ 2

+

k   8 − 7ρ i =1

2di

+

9ρ − 10 4



.

(7)

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A. Gajdoš et al. / Discrete Applied Mathematics 177 (2014) 101–110

The order of summands in the definition (5) is not important. Therefore, if deg(f ) = k, (v1 , v2 , . . . , vk+1 ) ∈ F (f ) and di := deg(vi ) for i ∈ [1, k], from (5) and (7) we see that c (f ) = q(d1 , d2 , . . . , dk ).

(8)

˜ the sequence (d1 , d2 , . . . , dk ) involved in (8) satisfies As vi vi+1 ∈ E, di + di+1 ≥

9|E˜ | − 12|V˜ |

9ρ − 12

=

|E˜ | − 2|V˜ |

> 15,

ρ−2

i = 1, 2, . . . , k

(9)

is decreasing in the interval (2, ∞), = 15 for ρ ∈ (2, 3). Let Sρ denote the set of all sequences in S of length at least 2, in which the sum

(with dk+1 := d1 ), where the last inequality comes from the fact that the function and so

9ρ−12 ρ−2

>

9·3−12 3 −2

9x−12 x −2

9ρ−12

of any two neighbouring terms is at least ρ−2 , and the last term is considered to be neighbouring the first term. Claim 1. If k ∈ [4, ∞), (d1 , d2 , . . . , dk ) ∈ Sρ and min(d1 , d2 , . . . , dk ) ≥ 2, then q(d1 , d2 , . . . , dk ) ≥

(k−4)(9ρ−8) 8

.

Proof. With dk+1 := d1 we have k   8 − 7ρ

2di

i=1

+

9ρ − 10

 =

4

k   8 − 7ρ

+

4di

i=1

8 − 7ρ 4di+1

+

9ρ − 10



4

.

We will be done by showing that, for each i ∈ [1, k], s(di , di+1 ) :=

8 − 7ρ 4di

+

8 − 7ρ

≥

4di+1

12 − 9ρ 8

,

(10)

because then q(d1 , d2 , . . . , dk ) ≥

8 − 9ρ 2

 +k

12 − 9ρ 8

+

9ρ − 10



4

=

(k − 4)(9ρ − 8) 8

.

Since 8 − 7ρ < −6 < 0, the inequality in (10) is equivalent to 1 di

+

1

≤

di+1

9ρ − 12 14ρ − 16

.

(11)

Also, from s(di+1 , di ) = s(di , di+1 ) we see that, when bounding s(di , di+1 ) from below, without loss of generality we may suppose di ≤ di+1 . 9ρ−12 If di ≥ 2ρ−4 , then 1 di

+

1

≤

di+1

2 di

≤

4ρ − 8 9ρ − 12

,

in which case the inequality 4ρ − 8

≤

9ρ − 12

9ρ − 12

(12)

14ρ − 16

is sufficient for proving (11). However, the assumption ρ ∈ (2, 3) leads to 9ρ − 12 > 0 and 14ρ − 16 > 0, and so (12) is equivalent to the inequality 0 ≤ 25ρ 2 − 40ρ + 16 = (5ρ − 4)2 . 9ρ−12−ρ di +2di 9ρ−12 9ρ−12 If 2 ≤ di < 2ρ−4 , then from di + di+1 ≥ ρ−2 it follows that di+1 ≥ and ρ−2 1 di

+

1 di+1

≤

1 di

+

ρ−2 9ρ − 12 − ρ di + 2di

=

9ρ − 12 di (9ρ − 12 − ρ di + 2di )

.

9ρ−12

It is left to the reader to show that the function x(9ρ−12−ρ x+2x) is decreasing for 0 < x ≤ 1 di

+

as required.

1 di+1

≤

9ρ − 12 2(9ρ − 12 − 2ρ + 4)

=

9ρ − 12 14ρ − 16



Claim 2. If (d1 , d2 , d3 ) ∈ Sρ and min(d1 , d2 , d3 ) ≥ 3, then q(d1 , d2 , d3 ) > 0.

9ρ−12 . 2ρ−4

Therefore, we obtain

A. Gajdoš et al. / Discrete Applied Mathematics 177 (2014) 101–110

105

Proof. We have q(d1 , d2 , d3 ) =

9ρ − 14 4

+

8 − 7ρ



2

1

+

d1

1 d2

+



1 d3

.

(13)

8−7ρ

Since 2 < 0, when bounding q(d1 , d2 , d3 ) from below we need to bound [1, 3] : dj = min(d1 , d2 , d3 )). If di ≥ d :=

9ρ−12 , 2ρ−4

q(d1 , d2 , d3 ) ≥

1 d1

then

+

9ρ − 14 4

1 d2

+

1 d3

+

8 − 7ρ 2

3 di

≤

≤

6ρ−12 9ρ−12

6ρ − 12

·

9ρ − 12

1 d1

1 d2

+

+

1 d3

from above. Let i := min(j ∈

and

=

−ρ 2 + 10ρ − 8 . 12ρ − 16

In this case the statement of Claim 2 follows from the fact that for ρ ∈ (2, 3) the numerator as well as the denominator of the last fraction is positive. 9ρ−12 If 3 ≤ di < d, then di + dj ≥ ρ−2 for each j ∈ [1, 3] − {i}, hence 1 d1

+

1 d2

+

1 d3

≤

1 di

+2·

ρ−2

9ρ − 12 + ρ di − 2di

=

9ρ − 12 − ρ di + 2di

di (9ρ − 12 − ρ di + 2di )

.

(14)

Putting g (x) :=

9ρ − 12 + ρ x − 2x x(9ρ − 12 − ρ x + 2x)

,

g1 (x) := x2 (ρ − 2)2 + 2x(ρ − 2)(9ρ − 12) − (9ρ − 12)2 , g2 (x) := x2 (9ρ − 12 − ρ x + 2x)2 we obtain g ′ (x) = g1 (d) =

(9ρ−12)2

g1 (x) . g2 (x)

The assumption 3 ≤ x < d leads to g2 (x) > 0 and g1′ (x) > 0. Since g1 (3) = 18(2 − ρ 2 ) < 0 and

> 0, there is a unique zero z of the function g1 (x) in the interval ⟨3, d⟩. Thus, g (x) is decreasing in the inter  2ρ−3 2ρ−4 val ⟨3, z ⟩ and increasing in the interval ⟨z , d⟩. Therefore, max(g (x) : 3 ≤ x < d) ≤ max(g (3), g (d)) = max 3ρ−3 , 3ρ−4 = 2ρ−3 , 3ρ−3

4

and so, using (13) and (14), q(d1 , d2 , d3 ) ≥

9ρ−14 4

8−7ρ 2

+

·

2ρ−3 3ρ−3

=

(ρ−2)(3−ρ) 12(ρ−1)

Claim 3. If (d1 , d2 , d3 ) ∈ Sρ and min(d1 , d2 , d3 ) = 2, then q(d1 , d2 , d3 ) ≥ 9ρ−12 ρ−2

Proof. If di = 2, then 2 + dj ≥ q(d1 , d2 , d3 ) ≥

9ρ−14 4

+

8−7ρ 2



1 2

+2·

ρ−2 7ρ−8



and dj ≥

=

1−ρ . 2

7ρ−8 ρ−2

> 0.



1−ρ . 2

for each j ∈ [1, 3] − {i}, which leads to the inequality



Now let Sρ∗ denote the set of all sequences (d1 , d2 , . . . , dk ) ∈ Sρ that satisfy the following two conditions: (C1) min(d1 , d2 , . . . , dk ) = 2 ⇒ k ≥ 4; (C2) |{i ∈ [1, k] : di > 1}| ≥ |{i ∈ [1, k] : di = 1}| + 3. From (C2) we see that all sequences in Sρ∗ are of length at least 3. Claim 4. If (d1 , d2 , . . . , dk ) ∈ Sρ∗ , tj = |{i ∈ [1, k] : di = j}| for both j ∈ [1, 2], and t1 t2 = 0, then q(d1 , d2 , . . . , dk ) ≥ 0. Proof. We proceed by induction on the number t1 . Let first t1 = 0. If k ≥ 4, the required inequality follows from Claim 1. If k = 3, then, because of (C1), min(d1 , d2 , d3 ) ≥ 3, and, by Claim 2, q(d1 , d2 , d3 ) > 0. Next suppose that t1 ≥ 1, and if (d′1 , d′2 , . . . , d′l ) ∈ Sρ∗ satisfies t1′ = t1 − 1 and t1′ t2′ = 0, where tj′ = |{i ∈ [1, l] : d′i = j}| for both j ∈ [1, 2], then q(d′1 , d′2 , . . . , d′l ) ≥ 0. Consider the sequence {di }i∈Z , the cyclic extension of the sequence (d1 , d2 , . . . , dk ), in which di = dj if i ≡ j (mod k). With i ∈ Z then (di+1 , di+2 , . . . , di+k ) ∈ Sρ∗ and q(di+1 , di+2 , . . . , di+k ) = q(d1 , d2 , . . . , dk ); thus, without loss of generality we may suppose that d1 = 1. For the sequence (d′1 , d′2 , . . . , d′l ) = (d3 , d4 , . . . , dk ) we have t1′ = t1 − 1 (see (9)), t2′ = t2 = 0, min(d′1 , d′2 , . . . , d′l ) ̸= 2 and

|{i ∈ [1, l] : d′i > 1}| − |{i ∈ [1, l] : d′i = 1}| = |{i ∈ [1, k] : di > 1}| − |{i ∈ [1, k] : di = 1}| ≥ 3, so that (d′1 , d′2 , . . . , d′l ) ∈ Sρ∗ . Therefore, by the induction hypothesis, q(d3 , d4 , . . . , dk ) =

8 − 9ρ 2

+

k   8 − 7ρ i =3

2di

+

9ρ − 10 4



≥ 0.

(15)

106

A. Gajdoš et al. / Discrete Applied Mathematics 177 (2014) 101–110 9ρ−12

8ρ−10

From 1 + d2 ≥ ρ−2 it follows that d2 ≥ ρ−2 , and so 2   8 − 7ρ

2di

i=1

+

9ρ − 10



4

≥

8 − 7ρ 2

+

9ρ − 10 2

+

8 − 7ρ 2

·

ρ−2 9ρ 2 − 14ρ + 4 = >0 8ρ − 10 16ρ − 20

√

(notice that both roots of the polynomial 9x2 − 14x + 4, i.e. 7±9 13 , are smaller than 2), which together with (15) proves that q(d1 , d2 , . . . , dk ) > 0.  Claim 5. If a face of F˜ is incident to a d-vertex for some d ∈ [1, 2], then it is not incident to a (3 − d)-vertex. Proof. Suppose that f ∈ F˜ is incident to a d-vertex xd for both d ∈ [1, 2]. There is a neighbour y of x2 that is distinct from the ˆ = G˜ + x1 y; unique neighbour of the vertex x1 . Both vertices x1 and y are incident to the face f , hence there is a plane graph G 9ρ−12 ˆ ˜ ˆ ˆ ˜ ˆ ˜ )|, each such graph satisfies w(G) ≥ w(G) ≥ ρ−2 , which means that G ∈ G. However, |V (G)| = |V (G)| and |E (G)| > |E (G

˜ which contradicts the assumptions on the counterexample G.



A problematic face is a 3-face f1 = xyz1 that is incident to a 2-vertex z1 and vertices x, y of degree greater than 2 (see (9)); the edge xy is the basis of the problematic face f1 . Consider the maximal (i.e. non-extendable) simple sequence of 2-vertices (z1 , z2 , . . . , zp ) such that fi := xzi yzi+1 ∈ F˜ for each i ∈ [1, p − 1] and call it the 2-sequence of the face f1 . It is easy to see that then the face comp(f ) ∈ F˜ (called compensating for f1 ) that is incident to the vertices x, y, and zp , but distinct from fp , is of degree at least 4; indeed, otherwise F˜ = {fi : i ∈ [1, p]} ∪ {comp(f1 )} and ρ = |E˜ |/|V˜ | = (2p + 1)/(p + 2) < 2, a contradiction. Moreover, since the sequence (z1 , z2 , . . . , zp ) is non-extendable, either deg(comp(f1 )) ≥ 5 or the fourth vertex of the face comp(f1 ) (distinct from x, y, zp ) is of degree at least 3. Claim 6. There are at most two problematic faces sharing the same compensation face in F˜ . Moreover, if comp(f1 ) = comp(f1′ )

for distinct problematic faces f1 , f1′ ∈ F˜ , then f1 and f1′ share the same basis and the face comp(f1 ) is incident to exactly two 2-vertices. Proof. Let f1 = xyz1 , f1′ = x′ y′ z1′ ∈ F˜ be distinct problematic faces with the bases xy, x′ y′ and the 2-sequences (z1 , z2 , . . . , zp ) and (z1′ , z2′ , . . . , zr′ ) such that comp(f1 ) = comp(f1′ ). Suppose that {x, y} ̸= {x′ , y′ }, say x′ ̸∈ {x, y}. Since both vertices zp and

ˆ = G˜ + zp x′ . Each such graph satisfies w(Gˆ ) ≥ w(G˜ ) and Gˆ ∈ G. Then, x′ are incident to comp(f1 ), there is a plane graph G

ˆ )| = |V˜ | and |E (Gˆ )| > |E˜ |, a contradiction. however, |V (G From the definition it evidently follows that each edge in E˜ is the basis for at most two problematic faces in F˜ . The above consideration then immediately shows that more than two problematic faces cannot share the same compensation face (otherwise the total number of vertices in their bases, and hence in the common compensation face would be at least three). Assume now that xy = x′ y′ . If the face comp(f1 ) is incident to a 2-vertex u ̸∈ {zp , zr′ }, then at least one of the neighbours ˆ = G˜ + zp v yields a similar contradiction as above. of the vertex u, say v , must be out of {x, y}, and so any plane graph G



The compensation number of a face f ∈ F˜ is defined as cn(f ) := |{f1 ∈ F˜ : comp(f1 ) = f }|. From Claim 6 we know that cn(f ) ∈ [0, 2] for each f ∈ F˜ .

˜ and cn(comp(f1 )) = 1, then c (f1 ) + c (comp(f1 )) > 0. Claim 7. If f1 is a problematic face of G Proof. Let k := deg(comp(f1 )). If k = 4, there is the (unique) sequence (v1 , v2 , v3 , v4 ) ∈ F (comp(f1 )) with deg(v2 ) = 2, 9ρ−12

7ρ−8

deg(v4 ) ≥ 3 and deg(vi ) ≥ ρ−2 − 2 = ρ−2 , i = 1, 3, which leads to c (comp(f1 )) ≥ ρ−2

2 · 7ρ−8



=

7ρ−8 . 12

Therefore, by Claim 3, c (f1 ) + c (comp(f1 )) >

If k ≥ 5, then, by Claims 1 and 3, c (f1 ) + c (comp(f1 )) >

1−ρ 2

1−ρ 2

+

+

7ρ−8 12

(k−4)(9ρ−8) 8

= ≥

ρ−2 12

8−9ρ 2

+ 9ρ − 10 +

8−7ρ 2

1 2

+

1 3

+

> 0.

5ρ−4 8

> 0.



Claim 8. If cn(f ) = 2 and comp(f1 ) = comp(f1′ ) = f for faces f , f1 , f1′ ∈ F˜ , f1 ̸= f1′ , then c (f ) + c (f1 ) + c (f1′ ) > 0. Proof. Let xy be the common basis of the faces f1 , f1′ , and let z (z ′ , respectively) be the last vertex of the 2-sequence of f1

˜ Suppose without loss of generality that (of f1′ ). Since deg(f ) > 4, at least one of the vertices x, y must be a cut-vertex of G. ˜ By Claim 6 there are exactly two 2-vertices incident to f . Therefore, there is the (unique) sequence x is a cut-vertex of G. (v1 , v2 , . . . , vk+1 ) ∈ F (f ) with v1 = vk+1 = x (where, clearly, k ≥ 7), deg(v2 ) = 2 and v3 = y. Let di := deg(vi ) for i ∈ [1, k + 1].

A. Gajdoš et al. / Discrete Applied Mathematics 177 (2014) 101–110

107

˜ then deg(v4 ) = 2 and v1 = v5 = vk+1 . Thus, (d1 , d2 , d3 , d4 ), (d5 , d6 , . . . , dk ) ∈ Sρ∗ , and so, by If y is not a cut-vertex of G, Claims 1 and 2, q(d1 , d2 , d3 , d4 ) = q(d5 , d6 , . . . , dk ) =

8 − 9ρ 2

+

4   8 − 7ρ

2di

i =1

8 − 9ρ 2

+

k   8 − 7ρ

2di

i =5

+

+

9ρ − 10



4 9ρ − 10 4



≥ 0, ≥ 0.

As a consequence we obtain c (f ) = q(d1 , d2 , . . . , dk ) =

8 − 9ρ 2

+

k   8 − 7ρ

2di

i=1

+

9ρ − 10 4

 ≥

9ρ − 8 2

,

6 + 2 · 1−ρ = 7ρ− > 0. 2 2 ˜ there is l ∈ [6, k − 4] such that vl = v3 , deg(vl+1 ) = 2 and vl+2 = vk+1 . Similarly as above in this If y is a cut-vertex of G, 9ρ−8 case we have (d1 , d2 , d3 , dl+1 ), (d4 , d5 , . . . , dl ), (dl+2 , dl+3 , . . . , dk ) ∈ Sρ∗ , which yields c (f ) ≥ 2 · 2 and c (f )+ c (f1 )+ c (f1′ ) ≥ 8ρ − 7 > 0. 

and then, using Claim 3, c (f ) + c (f1 ) + c (f1′ ) ≥

9ρ−8 2

Claim 9. If a face f ∈ F˜ is not problematic, (v1 , v2 , . . . , vk+1 ) ∈ F (f ) and di = deg(vi ), i = 1, 2, . . . , k, then (d1 , d2 , . . . , dk )

∈ Sρ∗ .

Proof. Let us call a subwalk (vi , vi+1 , . . . , vj ) of a walk W = (v1 , v2 , . . . , vk+1 ) with 1 ≤ i ≤ j ≤ k atomic if j ≡ i (mod 2), vi+2l = vi for each l ∈ [0, j−2 i ], di+2l+1 = 1 for each l ∈ [0, j−2i−2 ] and min(di−1 , di , dj+1 ) > 1; here we work again with the cyclic extension of the sequence (d1 , d2 , . . . , dk ) used in the proof of Claim 4. Clearly, either all walks in F (f ) are in Sρ∗ or all walks in F (f ) are out of Sρ∗ . Therefore, without loss of generality we may suppose that p

p

W = (v11 , v21 , . . . , va11 , v12 , v22 , . . . , va22 , . . . , v1 , v2 , . . . , vapp ), where (v1l , v2l , . . . , val l ) is an atomic subwalk of W , l = 1, 2, . . . , p. From the definition of an atomic subwalk we see that the sequence

(deg(v1l ), deg(v2l ), . . . , deg(val l )), contains

al +1 2

terms greater than 1 and

l = 1, 2, . . . , p al −1 2

˜ is a tree (with at most two terms 1. Moreover, p ≥ 3, because otherwise G

< number of terms in the non-leaf vertices) and ρ = sequence D = (d1 , d2 , . . . , ) so the condition (C2) is satisfied for D. If there is i ∈ [1, k] such that di = 2, then, by Claim 5, no term of D is equal to 1, and min(d1 , d2 , . . . , dk ) = 2. Since the face f is not problematic, we have k ≥ 4, and so the condition (C1) is satisfied for D, too. Thus, D ∈ Sρ∗ .  n−1 2, a contradiction. As a consequence, the difference between the n dk that are greater than 1, and the number of 1’s in D, is at least 3, and

We now define the basic partition B of the set F˜ in the following way: Consider a face f ∈ F˜ .

• If cn(f ) = 0 and f is not problematic, then {f } ∈ B . • If cn(f ) = 1 and comp(f1 ) = f , then {f , f1 } ∈ B . • If cn(f ) = 2 and comp(f1 ) = f = comp(f1′ ) with f1 ̸= f1′ , then {f , f1 , f1′ } ∈ B . Claim 10. If B ∈ B , then



b∈B

c (b) ≥ 0.

Proof. If B = {f }, the face f is not problematic, and so, by Claims 9 and 4, b∈B c (b) = c (f ) ≥ 0. If B = {f , f1 }, where comp(f1 ) = f , then cn(f ) = 1, and so, by Claim 7, b∈B c (b) = c (f ) + c (f1 ) > 0. Finally, if B = {f , f1 , f1′ }, where cn(f ) = 2 and comp(f1 ) = comp(f1′ ) = f , then Claim 8 yields b∈B c (b) = c (f ) + c (f1 ) + c (f1′ ) > 0. 



We are now ready to finish the proof of our main theorem. Since B is a partition of the set F˜ , from Claim 10 it follows that

 f ∈F˜

c (f ) =



c (b) ≥ 0,

B∈B b∈B

which represents a contradiction to (6) and proves Theorem 4.



108

A. Gajdoš et al. / Discrete Applied Mathematics 177 (2014) 101–110

−12n Since w(T3 , n, m) ≤ w(I, n, m), Theorem 4 is useful only for pairs (n, m) with 9m < w(I, n, m). To identify (almost m−2n all of) them note first that, for m ∈ [2n − 2, 3n − 6], Theorem 2 yields h = 3, p ∈ [1, 3], and

m

w(I, n, m) = m − 2

3

+3≥

m+8 3

.

(16) 9(2n+l)−12n

(n+2l)+8

Having in mind (16) and Theorem 4, we are interested in determining those l ∈ [1, n − 6], for which (2n+l)−2n < , or, 3 equivalently, l2 +(2n−19)l−18n > 0. If n ≥ 18, then the discriminant D(n) of the quadratic equation x2 +(2n−19)x−18n = 0 satisfies D(n) = 4n2 − 4n + 361 < (2n + 4)2 . Therefore, the inequality l > 21 (19 − 2n + 2n + 4) is sufficient for usefulness of the bound given by Theorem 4; the corresponding interval for m is then [2n + 12, 3n − 6]. If n ≥ 18 and m = 2n + l, l ≥ 12, then w(T3 , n, m) < 6n + 9, and, because of (16), l

w(I, n, m) − w(T3 , n, m) >

2n + l + 8 3

 −

6n l





+9 =n

2 3

−

6



l

+

l − 19 3

≥

n 6

−

7 3

=

n − 14 6

.

Thus, we have proved: Proposition 3. The set {w(I, n, m) − w(T3 , n, m) : n ∈ [2, ∞), m ∈ [1, 3n − 6]} is unbounded.



3. Construction The quality of Theorem 4 can be judged using the next result. Theorem 5. There exists an infinite number of pairs (n, m) with n ∈ [7, ∞) and m ∈ [2n + 1, 3n − 6] such that



9m − 12n



m − 2n

− 1 ≤ w(T3 , n, m) ≤



9m − 12n m − 2n



.

Proof. Let G2p with p ∈ [0, ∞) denote the plane map consisting of two caps formed by six pentagons and 2p annular layers formed by five hexagons (a special nanotubical fullerene, see Kardoš et al. [6]). The dual G∗2p of G2p is a plane triangulation in which pentagonal caps of G2p correspond to wheels W5 of six 5-vertices and annular layers of G2p are transformed into (vertex disjoint) cycles C5 of five 6-vertices (see Fig. 1 where G∗2 is depicted). We have |V (G∗2p )| = 10p + 12, |E (G∗2p )| = 30p + 30, |F (G∗2p )| = 20p + 20 and it is easy to see that G∗2p has a perfect matching M2p with all edges joining vertices of the same degree. (In Fig. 1 bold edges form a perfect matching of G∗2 .) Now let H2p be the Kleetope of G∗2p (obtained by inserting a 3-vertex in the interior of each face of G∗2p ); we have then |V (H2p )| = 30p + 32 and |E (H2p )| = 90p + 90. The triangulation H2p contains, besides 3-vertices, only twelve 10-vertices and 2p 12-vertices. Finally, for a t ∈ [0, ∞), build up over each edge of M2p joining (in H2p ) two d-vertices, d ∈ {10, 12}, a ‘‘tower’’ by adding the total number t + 12 − d of paths of length 2 with central 2-vertices. The resulting plane graph H2p,t satisfies

|V (H2p,t )| = |V (H2p )| + |M2p | · t + 6 · 2 = 5p(t + 6) + 6t + 44, |E (H2p,t )| = |E (H2p )| + |M2p | · 2t + 6 · 2 · 2 = 10p(t + 9) + 12t + 114. Moreover, H2p,t has only edges of weights 2 + (t + 12) = t + 14, 3 + (t + 12) = t + 15 and (t + 12) + (t + 12) = 2t + 24, so that w(H2p,t ) = t + 14. With n2p,t := |V (H2p,t )| and m2p,t := |E (H2p,t )| we obtain, by Theorem 4,

w(T3 , n2p,t , m2p,t ) <

9m2p,t − 12n2p,t m2p,t − 2n2p,t

= t + 15 +

Thus, if 30p + 26 > 10t + 108, or, equivalently, p >



9m2p,t − 12n2p,t



m2p,t − 2n2p,t

t 3

+

10t + 108 30p + 26 41 , 15

.

then

− 1 = t + 14 = w(H2p,t ) ≤ w(T3 , n2p,t , m2p,t ) ≤ t + 15 =

and (n2p,t , m2p,t ) : t ∈ [0, ∞), p ∈



 t 3

+

41 15



,∞





9m2p,t − 12n2p,t



m2p,t − 2n2p,t

is an infinite set of pairs (n, m) having the desired property.



Corollary 1. The set {w(T3 , n, m) : n ∈ [2, ∞), m ∈ [1, ∞)} is unbounded. Proof. Since t +3 ≥ t + 14. 

t 3

+

41 15



for any t ∈ [0, ∞), the set under consideration contains the number w(T3 , n2t +6,t , m2t +6,t ) ≥

A. Gajdoš et al. / Discrete Applied Mathematics 177 (2014) 101–110

109

Fig. 1. The plane triangulation G∗2 .

4. Open problems We conclude our paper by formulating two problems. The first one is motivated by the equality in (16). Problem 1. Improve the upper bound w(T3 , n, m) ≤ w(I, n, m) = m − 2⌈ m ⌉ + 3 for m ∈ [2n − 2, 2n + 12]. 3 Kotzig’s Theorem yields w(T3 , n, 3n − 6) ≤ 13 for each n ∈ [3, ∞). Indeed, any graph G ∈ T3 (n, 3n − 6) is a maximal planar graph. Thus, either G = K3 and w(G) = 4 ≤ 13, or G is 3-connected and w(G) ≤ 13 by Kotzig’s Theorem. This fact shows the correctness of the following definition, in which n ∈ [3, ∞) and t ∈ [0, ∞): f (n, t ) := min(m0 ∈ [1, 3n − 6] : (∀m ∈ [m0 , 3n − 6]) w(T3 , n, m) ≤ t + 13). From the definition it immediately follows f (n, t + 1)≤ f (n, t ).  that 16 n , 3n − 6 for t ∈ [2, ∞). To see it notice that the function Our results show that f (n, t ) ≤ min 2tt + +5

 2t +16   variable x is decreasing in the interval (2n, ∞). Therefore, if m ∈ n , 3n − 6 , then, by Theorem 4, t +5  16  9 2tt + n − 12n 9m − 12n 5  ≤  2t ++16 w(T3 , n, m) < m − 2n n − 2n t +5 ≤

9(2t + 16) − 12(t + 5) 2t + 16 − 2(t + 5)

9x−12n x−2n

of the

= t + 14,

and so w(T3 , n, m) ≤ t + 13. The inequality t ≥ 2 required above is a consequence of the inequality holding for t ∈ [0, 1].

n > 3n − 6

 2t +16  t +5

Problem 2. Improve the upper bound f (n, t ) ≤ 3n − 6 for t ∈ [0, 1]. Acknowledgements The authors are grateful for a support by the Science and Technology Assistance Agency under the contract No. APVV0023-10 (M. Horňák, T. Madaras) and by the grant VEGA 1/0652/12 (M. Horňák, P. Hudák, T. Madaras). References [1] [2] [3] [4] [5]

M. Borowiecki, I. Broere, P. Mihók, Minimal reducible bounds for planar graphs, Discrete Math. 212 (2000) 19–27. M. Horňák, S. Jendrol’, I. Schiermeyer, On maximum weight of a bipartite graph of given order and size, Discuss. Math. Graph Theory 33 (2013) 147–165. J. Ivančo, S. Jendrol’, On extremal problems concerning weights of edges of graphs, in: Sets, Graphs and Numbers, North-Holland, 1992, pp. 399–410. S. Jendrol’, I. Schiermeyer, On a max–min problem concerning weights of edges, Combinatorica 21 (2001) 351–359. S. Jendrol’, H.-J. Voss, Light subgraphs of graphs embedded in the plane—a survey, Discrete Math. 313 (2013) 406–421.

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