EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
2025, Vol. 18, Issue 2, Article Number 5942
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

The Total Choosability with Neighbor Sum
Distinguishing Properties of Planar Graphs That Are

Devoid of C5 Adjacent to C3

Pongpat Sittitrai1,4,Kittikorn Nakprasit2,5, Patcharapan Jumnongnit3,∗

1 Futuristic Science Research Center, School of Science, Walailak University,
Nakhonsithammarat, 80161, Thailand
2 Department of Mathematics, Faculty of Science, Khon Kaen University, 40002,
Khon Kaen, Thailand
3 Division of Mathematics, School of Science, University of Phayao, Phayao, 56000,
Thailand
4 Research Center for Theoretical Simulation and Applied Research in Bioscience and
Sensing, Walailak University, Nakhon Si Thammarat 80160, Thailand
5 Centre of Excellence in Mathematics, MHESI, Bangkok 10400, Thailand

Abstract. Let G be a graph such that a proper total coloring ψ : V (G) ∪ E(G) −→ N. Let s(x)
denote the sum of colors assigned to x and those incident edges of x. A coloring ψ is a neighbor
sum distinguishing total coloring if s(x) ̸= s(y), whenever xy is an edge in G. Let L be a k-list
assignment of a graph G if L is a function, say L : V (G) ∪ E(G) → 2N such that |L(t)| = k for all
t in the set V (G) ∪ E(G). If G has a total coloring such that α(x) is the element of L(x) for all x
in the set V (G)∪E(G), then we call α total-L-coloring. Moreover, a total-L-coloring α is called a
neighbor sum distinguishing total-L-coloring if s(x) ̸= s(y) where xy is an edge in G. Let Ch′′∑(G)
be the minimum number k such that G has such coloring for every k-list assignment L.
In this work, we present that for a graph G has Ch′′∑(G) ≤ max{10,∆(G) + 3} if G is a planar
graph that are devoid of C5 adjacent to C3.

2020 Mathematics Subject Classifications: 05C10

Key Words and Phrases: Neighbor sum distinguishing total coloring, coloring, discharging
method

1. Introduction

Simple, undirected, finite graphs are considered in all cases. When referring to the
maximum degree, face set, edge set, and vertex set of a graph G, respectively, we use

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v18i2.5942

Email addresses: pongpat.sittitrai@gmail.com (P. Sittitrai),
kitnak@hotmail.com (K. Nakprasit),patcharapan.ju@up.ac.th (P. Jumnongnit)

https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0)



P. Sittitrai, K. Nakprasit, P. Jumnongnit / Eur. J. Pure Appl. Math, 18 (2) (2025), 5942 2 of 7

the notation ∆(G), F (G), E(G), and V (G), respectively. If the boundary of two faces is
shared, then two faces are said to be adjacent.

Let a proper total coloring ψ : V (G) ∪ E(G) −→ N and s(x) denote the sum of colors
assigned to x and those incident edges of x. A coloring ψ is a neighbor sum distinguishing
total coloring (abridged as nsdt-coloring) if s(x) ̸= s(y), whenever xy is an edge in G. The
neighbor sum distinguishing total chromatic number of G, say tndi∑(G) is the minimum
number such that G has a nsdt-coloring.

Piĺsniak and Woźniak [1] introduced neighbor sum distinguishing total coloring and
obtained tndi∑(G) for cycles, bipartite graphs, cubic graphs, and complete graphs. Ad-
ditionally, the authors posed the following conjecture.

[1] If |G| ≥ 2, then tndi∑(G) ≤ ∆(G) + 3.
The conjecture is verified for K4-minor free graphs by Li, Liu, and Wang [2] and planar

graphs with maximum degree at least 13 by Zhang and Li [3].
Later, the result of Zhang and Li [3] was improved by IC-planar graphs with a maxi-

mum degree of 13 by Song et al. [4] and planar graphs with a maximum degree at least
11 by Qu et al. [5]. For other classes, Wang, Ma, and Han [6] and Ge, Li, and Xu [7]
confirmed the conjecture by planar graphs without 3-cycles with maximum degree at least
7 and planar graphs without 5-cycles with maximum degree at least 7, respectively.

The conjecture is also shown to be true for planar graphs with a better bound,
tndi∑(G) ≤ ∆(G) + 2, by [4, 8, 9].

Let L be a k-list assignment of a graph G if L is a mapping, say L : V (G)∪E(G) → 2N

where |L(t)| = k for each t ∈ V (G)∪E(G). If G has a total coloring such that α(t) ∈ L(t)
for all t in the set V (G) ∪ E(G), then we call α total-L-coloring. Moreover, a total-L-
coloring α is called a neighbor sum distinguishing total-L-coloring if s(x) ̸= s(y) where xy
is an edge in G. Let Ch′′∑(G) be the minimum number k such that G has such coloring for
every k-list assignment L, which is called the neighbor sum distinguishing total choosability
of G.

Qu et al. [10] proved that Ch′′∑(G) ≤ ∆(G) + 3 for every planar graph G with

∆(G) ≥ 13. Yao et al. [11] studied Ch′′∑(G) of d-degenerate graphs.
Conjecture 1 has been verified for several classes of planar graphs as follows: planar

graphs without adjacent 3-cycles with maximum degree at least 8 by [12], planar graphs
without adjacent 6-cycles with maximum degree at least 7 [13], and planar graphs without
4-cycles adjacent to 3-cycles with maximum degree at least 7 by [14].

More results about the neighbor sum distinguishing total choosability for planar graphs
can be seen in [3, 15–19].

In this paper, we give Theorem 1 to confirm the list version of Conjecture 1 on planar
graphs without neighboring 5-cycles and 3-cycles. The theorem also improves the results
of Wang, Ma, and Han [6] (on planar graphs without 3-cycles) and Ge, Li, and Xu [7] (on
planar graphs without 5-cycles).

2. Notations

All subscripts are modulo l unless stated otherwise for the remainder of the paper.



P. Sittitrai, K. Nakprasit, P. Jumnongnit / Eur. J. Pure Appl. Math, 18 (2) (2025), 5942 3 of 7

An l-vertex (face), an l+-vertex (face), or an l−-vertex (face) is a vertex (a face) which
has degree l, at least l, or at most l, respectively. Moreover, we call f a (d1, d2, . . . , dl)-face
if f is an l-face where its incident vertices have degree d1, d2, . . . , dl in clockwise order.

If f1 and f2 are adjacent 3-faces with a common incident vertex v, then we call a vertex
v a rough vertex of f1 and f2.

For an l-face f , we let f1, . . . , fl adjacent faces of f in clockwise order. An adjacent
k-face fi of f is called a thin adjacent k-face of f if fi−1 or fi+1 is a 4+-face.

3. Helpful tools

Consider a minimal planar graph G with Ch′′∑(G) > ∆(G) + 3.

We denote the plane embedding of the graph obtained by removing all 2−-vertices of
G by H

′
.

Lemma 1. (Claim 1, Observation 1, and Claim 4 in [20]) The graph H
′
satisfies the

following properties:
(i) A graph H

′
has no 2−-vertices.

(ii) Every 3-vertex have to be adjacent to a 5+-vertex.
(iii) Only a (3, 5+, 5+)-face or a (4+, 4+, 5+)-face is a 3-face of a graph H

′
.

Lemma 2. (Lemma 7 in [14]) The graph H
′
satisfies the following properties: A 5-vertex

is not adjacent to two 3-vertices.

From now on, we impose an additional condition that a planar graph G has no the
adjacency of 3-cycles and 5-cycles. The additional condition implies the following lemma.

Lemma 3. Faces and vertices in H
′
satisfy the following properties.

(i) For l ∈ {4, 5}, an l-face is not adjacent to any 3-faces.
If f is a 3-face, then f is not adjacent to l-face for l ∈ {4, 5}.
(ii) Let fi, fi+1, and fi+2 be three consecutive incident faces of an l-vertex v. If l ≥ 4,

then one of fi, fi+1, and fi+2 is a 4+-face.
(iii) Any 3-face is not adjacent to three 3-faces.

Proof. Let b(f) be the boundary of a face f .
(i) Let f be a 3-face and g be an l-face where l ∈ {4, 5}. Suppose that f is adjacent to

g. Give b(f) = v1v2v3 and b(g) = v1v2u1 . . . ul−2.
Consider l = 4. One can see that b(f) and b(g) share three vertices; otherwise, a

5-cycle v2u1u2v1v3 is adjacent to a 3-cycle v1v2v3, a contradiction. If v3 = u1 or v3 = u2,
then there is a 2-vertex or parallel edges, a contradiction by Lemma 1 (1) or property of
a graph H ′.

Consider l = 5. One can see that a 5-cycle b(g) is adjacent to a 3-cycle b(f), a
contradiction.

(ii) Let v be an 4+-vertex. Let fi, fi+1, and fi+2 be three consecutive incident faces of
v. Suppose that fi, fi+1, and fi+2 are 3-faces. Give b(fi) = vvivi+1, b(fi+1) = vvi+1vi+2,
and b(fi+2) = vvi+2vi+3. Since each vertex is not a 4+-vertex by Lemma 1 (i), five vertices



P. Sittitrai, K. Nakprasit, P. Jumnongnit / Eur. J. Pure Appl. Math, 18 (2) (2025), 5942 4 of 7

v, vi, vi+1, vi+2, vi+3 should be distinct; otherwise, parallel edges exist, a contradiction.
Then a 5-cycle vvivi+1vi+2vi+3 is adjacent to a 3-cycle vvivi+1, a contradiction.

(iii) Let f be a 3-face adjacent to faces f1, f2, and f3. Suppose that f1, f2, and f3 are
3-face. By Lemma 3 (ii), each incident vertex of f is a 3-vertex. This contradicts Lemma
1 (iii).

4. Main Theorem

Theorem 1. A graph G has Ch′′∑(G) ≤ max{10,∆(G)+3} if G is a planar graph without
C5 adjacent to C3.

Proof. Suppose Theorem 1 to the contrary. Consider a minimal counterexample G.
Recall that H

′
is defined as in the previous section. Discharging method is used to show

that H
′
does not exist. For each z ∈ V (H ′) ∪ F (H ′), we let µ(z) = d(z) − 4. One can

observe that ∑
z∈V (H′)∪F (H′)

µ(z) = −8

by Handshaking lemma and Euler’s formula.
Next, we desire some discharging rules to transfer charge w(x→ y) from x to y for some

x, y ∈ V (H ′)∪F (H ′). In additional, we have a new charge µ∗(z) for each z ∈ V (H ′)∪F (H ′)
after the transfer. Moreover,

∑
z∈V (H′)∪F (H′) µ

∗(z) =
∑

z∈V (H′)∪F (H′) µ(z) = −8. Our

goal is to find the discharging rules that make µ∗(z) ≥ 0 for each z ∈ V (H ′) ∪ F (H ′).
The following are discharging rules.

(R1) For a 3-face f , let g be an adjacent to a l-face g where u is a rough 5+-vertex of f
and g.

(R1.1) Let l = 3, w(u → f) = 1
3 when the incident vertex of f not incident to g is a

4+-vertex.
(R1.2) Let l ≥ 6, w(g → f) = 1

2 if g is a thin adjacent face of f , otherwise w(g →
f) = 1

3 .
(R2) For a 5+-vertex u, w(u→ v) = 1

3 for each its adjacent 3-vertex v.
Now, it is necessary to claim that µ ∗ (z) ≥ 0 follows discharge for each z ∈ V (H ′) ∪

F (H ′).
It is clear that if f is a 4-face or 5-face, then µ∗(f) = 0 or 1 respectively.

CASE 1: Consider a 3-face f .
let f1, f2, and f3 be adjacent faces of f where fi is incident to vi and vi+1.
We consider three cases by Lemma 3 (iii).
- A face f is not incident to a 3-face.
It follows that µ∗(f) ≥ µ(f) + 3× 1

3 = 0 by (R1.2).
- A face f is incident to exactly one 3-face, say f1.
It follows that v1 and v2 are rough vertices of f and f1.
If v1 and v2 are 4-vertices or v3 is a 3-vertex, then f2 and f3 are thin adjacent 6+-faces

of f by Lemma 3 (i) or Lemma 3 (ii), respectively. Then µ∗(f) ≥ µ(f) + 2 × 1
2 = 0 by

(R1.2).



P. Sittitrai, K. Nakprasit, P. Jumnongnit / Eur. J. Pure Appl. Math, 18 (2) (2025), 5942 5 of 7

If v1 or v2 is a 5+-vertex or v3 is a 4+-vertex, then v1 or v2 gives charge 1
3 to f by

(R1.1). Combining with (R1.2), we have µ∗(f) ≥ µ(f) + 3× 1
3 = 0.

- A face f is incident to two 3-faces, say f1 and f2.
Note that v2 is incident to three consecutive 3-faces. Then v2 is a 3-vertex by Lemma

3 (ii). Moreover, v1 and v3 are 5+-vertices by Lemma 1 (iii). Similarly, a vertex incident
to f1 or f2 but not incident to f is a 5+-vertex. Then each of v1, v2, and f3 gives charge
1
3 to f by (R1). Thus µ∗(f) ≥ µ(f) + 3× 1

3 = 0.
CASE 2: Consider a 6+-face.

Let f be an l-face where l ≥ 6. We let f1, . . . , fl adjacent faces of f in clockwise order.
For the convenience of calculating µ∗(f), we redistribute the charge that was transferred
from f as follows.

First, we give w(f → fi) =
1
3 for each fi where fi transfer its charge

1
6 revived form f

to fi−1 and fi+1. One can see that the process is as described in (R1.2).
- If fi is a 3-face being not a thin adjacent face of f , then w(f → fi) ≥ 1

3 .
- If fi is a thin adjacent 3-face of f , then w(f → fi) ≥ 1

3 + 1
6 = 1

2 .
- If fi is a 4+-face, then w(f → fi) ≥ 1

3 − 2× 1
3 = 0.

Hence, µ∗(f) ≥ µ(f)− l × 1
3 = l − 4− l × 1

3 = l × 2
3 − 4 ≥ 0 as desired.

CASE 3: Consider a 3-vertex v.
By Lemma 1(ii), v is not adjacent to any 4−-vertices. Thus µ∗(v) ≥ µ(v) + 3× 1

3 = 0
by (R2)
CASE 4: Consider a 5-vertex v.

By Lemma 3 (ii), a vertex v is a rough vertex of at most two of its incident faces.
It follows from Lemma 2 that a vertex v is adjacent to at most one 3-vertex. Hence,
µ∗(v) ≥ µ(v)− 3× 1

3 = 0 by (R1.1) and (R2).
CASE 5: consider a 6+-vertex v.

Let v be a l-vertex where k ≥ 6.
We let v1, . . . , vl adjacent vertices of v in clockwise order. For the convenience of

calculating µ∗(v), we redistribute the charge that was transferred from v as follows. First,
we give w(v → vi) =

1
3 for each vi. Now, we consider a 3-face f bounded by vi, vi+1, and

v in two situations. If (1) vi is not a 3-vertex, and (2) vi+1, vi+2, and v bound the same
3-face then reduce w(v → vi) to 0 and transfer charge 1

3 to a 3-face f instead. If (1) vi+1 is
a 4+-vertex and (2) or vi−1, vi, and v bound the same 3-face, then adjust w(v → vi+1) to
0 and carry charge 1

3 from v to f instead. Note that Lemma 3 (ii) implies two previously
mentioned situations cannot happen simultaneously. Consequently, each of a 3-vertex vi
receives 1

3 from v as by (R2) and each of a 3-face that v is a rough vertex as in (R1.1)
receive 1

3 from v. Additionally, we get µ∗(v) ≥ µ(v)− l× 1
3 = l− 4− l× 1

3 = l× 2
3 − 4 ≥ 0

as desired.
This completes the proof.



P. Sittitrai, K. Nakprasit, P. Jumnongnit / Eur. J. Pure Appl. Math, 18 (2) (2025), 5942 6 of 7

Acknowledgements

This work was supported by Walailak University under the New Researcher Develop-
ment scheme (Contract Number WU67234)

The second author is (partially) supported by the Centre of Excellence in Mathematics,
Ministry of Higher Education, Science, Research, and Innovation, Thailand.

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