IOSR Journal of Mathematics (IOSR-JM) e-ISSN: 2278-5728, p-ISSN: 2319-765X. Volume 13, Issue 6 Ver. IV (Nov. - Dec. 2017), PP 72-79 www.iosrjournals.org

On the Incidence Chromatic Number of Sierpiński Graphs

Handan Akyar1 1(Department of Mathematics, Anadolu University, Turkey) Corresponding Author:Handan Akyar

Abstract : In this paper we consider the incidence coloring of Sierpiński graphs 푆(푛, 푘) and prove that the incidence chromatic number of Sierpiński graphs 푆(푛, 3) is 4 when 푛 > 1. Moreover, an alternative proof for the incidence chromatic number of the complete bipartite graph 퐾푚,푛 is given. Algorithms for coloring incidence graphs of Sierpiński graphs 푆(푛, 3) and complete bipartite graph 퐾푚,푛 are presented explicitly. Keywords: Sierpiński graphs, incidence graph, incidence chromatic number ------Date of Submission: 27-12-2017 Date of acceptance: 13-01-2018 ------

I. Introduction The incidence coloring of a graph was first introduced by Brualdi and Massey (see [1]). An incidence in a graph 퐺 = (푉, 퐸) is an ordered pair (푣, 푒) with 푣 ∈ 푉 and 푒 ∈ 퐸, such that the vertex 푣 and the edge 푒 are incident. The set of all incidences in 퐺 usually denoted by 퐼(퐺). For every vertex 푣, we denote by 퐼푣 the set of all incidences of the form (푣, {푣, 푤}) and by 퐴푣 the set of all incidences of the form (푤, {푤, 푣}). The incidences (푣, 푒) ∈ 퐼(퐺) and (푤, 푓) ∈ 퐼(퐺) are called adjacent if one of the following holds:

i. 푣 = 푤, ii. 푒 = 푓, iii. The edge {푣, 푤} equals 푒 or 푓. A 푘-incidence coloring of a graph 퐺 is a mapping 휎 from 퐼(퐺) to a set 푋 of 푘 different colors such that adjacent incidences are assigned different colors. The incidence chromatic number 휒푖 (퐺) of 퐺 is the smallest number 푘 such that 퐺 admits a 푘-incidence coloring. Certainly, it is not easy to find the incidence chromatic number of an arbitrary graph, but many results on incidence coloring were obtained by several researchers, including especially Brualdi and Massey (see [1]). Brualdi and Massey proved that the following results hold for a graph 퐺 with maximum degree Δ(퐺):

휒푖 (퐺) ≥ Δ(퐺) + 1, (1) 휒푖 (퐺) ≤ 2Δ(퐺), (2) 휒푖 (퐾푛 ) = 푛, (푛 ≥ 2) (3) 휒푖 (퐾푚,푛 ) = 푚 + 2, (푚 ≥ 푛 ≥ 2) (4)

휒푖 (푇) = Δ(푇) + 1 (foreverytree푇oforder푛 ≥ 2). (5) Brualdi and Massey also conjectured that 휒푖 (퐺) ≤ Δ(퐺) + 2, but in [2] Guiduli showed that 휒푖 (퐺) ≥ Δ(퐺) + Ω(logΔ(퐺)), which means that this conjecture is invalid. Guiduli also proved that 휒푖 (퐺) ≤ Δ(퐺) + 푂(logΔ(퐺)) and showed that the problem of incidence coloring is a special case of directed star arboricity which was introduced by Algor and Alon (see [3]). The problem of determining incidence chromatic number of a graph has been extensively studied by many authors, and is still a fruitful area of research in graph theory (see [1, 2, 4—8] and references there in)Although Brualdi and Massey’s result, given by equation (4), on complete bipartite graph 퐾푚,푛 is correct, their proof is sadly inaccurate (see [7]). In [7] a correction for the proof of equation (4) is proposed without any interpretation. In the following theorem, we provide an alternate proof for equality (4).

Theorem 1.1For all 푚 ≥ 푛 ≥ 2, 휒푖 (퐾푚,푛 ) = 푚 + 2.

DOI: 10.9790/5728-1306047279 www.iosrjournals.org 72 | Page The Incidence Chromatic Number of Sierpiński Graphs

Proof. It is clear that 푚 + 1 colors are not sufficient to color 퐼(퐾푚,푛 ) (see [1]). So we have 푚 + 2 ≤

휒푖 (퐾푚,푛 ) and to complete the proof, it remains to be shown that 퐾푚,푛 has an incidence coloring with 푚 + 2 colors.Since for each 푚 ≥ 푛 ≥ 2, 퐾푚,푛 is a subgraph of 퐾푚,푚 , the inequality 휒푖 (퐾푚,푛 ) ≤ 휒푖 (퐾푚,푚 ) holds.

Hence, it is enough to show that 휒푖 (퐾푚,푚 ) ≤ 푚 + 2.

Then let us consider the complete bipartite graph 퐾푚 ,푚 with partition sets 푉1 = {푤1, 푤2, … , 푤푚 } and 푉2 =

{푢1, 푢2, … , 푢푚 }, and color it using the following method:

If (푤푖 , {푤푖 , 푢푗 }) ∈ 퐼푤푖 ⊂ 퐼(퐾푚 ,푚 ) then 푚 if푗 = 푚 − 푖 + 1,

휎1((푤푖 , {푤푖 , 푢푗 })) = 푚 + 1 if푗 = 푚and푖 ≠ 1, 푗 otherwise.

If (푢푗 , {푢푗 , 푤푖 }) ∈ 퐴푤푖 ⊂ 퐼(퐾푚,푚 ) then 푚 + 2 푖푓푗 = 푚 − 푖 + 1,

휎2((푢푗 , {푢푗 , 푤푖 })) = 푚 + 1 푖푓푗 ≠ 푚and푖 = 1, 푚 − 푖 + 1 otherwise, where 푖, 푗 ∈ {1,2, … , 푚}.

By means of 휎1 and 휎2 we define a new function 휎: 퐼(퐾푚,푚 ) → {1,2, … , 푚 + 2} by 휎 ((푣, 푒)) if푣 ∈ 푉 , 휎((푣, 푒)) = 1 1 휎2((푣, 푒)) if푣 ∈ 푉2.

We shall now show that adjacent incidences are assigned different colors by 휎. Let (푤푖 , {푤푖 , 푢푗 }) ∈ 퐼푤푖 be any incidence element then it has three type of neighbors:

1. (푤푖 , {푤푖 , 푢푟 }) (1 ≤ 푟 ≤ 푚, 푟 ≠ 푗), using the definition of 휎1, we have 휎((푤푖 , {푤푖 , 푢푗 })) ≠ 휎((푤푖 , {푤푖 , 푢푟 })).

2. (푢푘 , {푢푘 , 푤푖 }) (1 ≤ 푘 ≤ 푚), by virtue of the definition of 휎1 and 휎2, we get 휎((푤푖 , {푤푖 , 푢푗 })) ≠

휎((푢푘 , {푢푘 , 푤푖 })).

3. (푢푗 , {푢푗 , 푤푝 }) (1 ≤ 푝 ≤ 푚), in the same way, because of the definition of 휎1 and 휎2, we obtain

휎((푤푖 , {푤푖 , 푢푗 })) ≠ 휎((푢푗 , {푢푗 , 푤푝 })).

Therefore, all neighbors of (푤푖 , {푤푖 , 푢푗 }) ∈ 퐼푤푖 are colored different from itself.Similarly, let

(푢푗 , {푢푗 , 푤푖 }) ∈ 퐴푤푖 be any incidence element. If we interchange the partition sets 푉1 and 푉2, then one can easily conclude that (푢푗 , {푢푗 , 푤푖 }) and its all neighbors are assigned different colors. This completes the proof.In the following example, we demonstrate how to color incidence graph of a complete bipartite graph by the method given in the proof of Theorem 1.1.

Example 1.2 We can give an incidence coloring of the complete bipartite graph K4,4 using the method presented in the proof of Theorem 1.1 as follows:

휎((푤1, {푤1, 푢1})) = 1 휎((푤2, {푤2, 푢1})) = 1 휎((푤3, {푤3, 푢1})) = 1 휎((푤4, {푤4, 푢1})) = 4,

휎((푤1, {푤1, 푢2})) = 2 휎((푤2, {푤2, 푢2})) = 2 휎((푤3, {푤3, 푢2})) = 4 휎((푤4, {푤4, 푢2})) = 2,

휎((푤1, {푤1, 푢3})) = 3 휎((푤2, {푤2, 푢3})) = 4 휎((푤3, {푤3, 푢3})) = 3 휎((푤4, {푤4, 푢3})) = 3,

휎((푤1, {푤1, 푢4})) = 4 휎((푤2, {푤2, 푢4})) = 5 휎((푤3, {푤3, 푢4})) = 5 휎((푤4, {푤4, 푢4})) = 5,

휎((푢1, {푢1, 푤1})) = 5 휎((푢2, {푢2, 푤1})) = 5 휎((푢3, {푢3, 푤1})) = 5 휎((푢4, {푢4, 푤1})) = 6,

휎((푢1, {푢1, 푤2})) = 3 휎((푢2, {푢2, 푤2})) = 3 휎((푢3, {푢3, 푤2})) = 6 휎((푢4, {푢4, 푤2})) = 3,

휎((푢1, {푢1, 푤3})) = 2 휎((푢2, {푢2, 푤3})) = 6 휎((푢3, {푢3, 푤3})) = 2 휎((푢4, {푢4, 푤3})) = 2,

휎((푢1, {푢1, 푤4})) = 6 휎((푢2, {푢2, 푤4})) = 1 휎((푢3, {푢3, 푤4})) = 1 휎((푢4, {푢4, 푤4})) = 1.

DOI: 10.9790/5728-1306047279 www.iosrjournals.org 73 | Page The Incidence Chromatic Number of Sierpiński Graphs

Figure1: A proper coloring of 퐼(퐾4,4) with 6 colors

In Fig. 1, means that the incidence element (푤1, {푤1, 푢1}) is colored by color 1. The Sierpiński graphs 푆(푛, 푘) (푛, 푘 ≥ 1) is defined on the vertex set {0,1, … , 푘 − 1}푛 , in which two different vertices 푢 = (푢1, … , 푢푛 ) and 푣 = (푣1, … , 푣푛 ) are adjacent if and only if there exists an 푕 ∈ {1, … , 푛} such that i. 푢푡 = 푣푡 , for 푡 = 1, … , 푕 − 1, ii. 푢푕 ≠ 푣푕 , and iii. 푢푡 = 푣푕 and 푣푡 = 푢푕 for 푡 = 푕 + 1, … , 푛.

In the sequel we will use abbreviation 푢1푢2 … 푢푛 for the vertex (푢1, … , 푢푛 ). From the definition of 푛 푛 푘(푘 −1) 푆(푛, 푘), it is clear that |푉(푆(푛, 푘))| = 푘 , |퐸(푆(푛, 푘))| = , 푆(1, 푘) ≅ 퐾 and 푆(2, 푘) ≅ 푃 푘 . For 2 푘 2 푖 ∈ {0,1, … , 푘 − 1} we call vertices of the form 푖푖 … 푖 extreme vertices of 푆(푛, 푘). Obviously, only 푘 of 푘푛 vertices are extreme vertices of 푆(푛, 푘). In Fig. 2, the Sierpiński graphs 푆(2,5) and 푆(3,4) with their vertex labelings are illustrated. Let 푛 ≥ 2, then for 푖 ∈ {0,1, … , 푘 − 1} we define 푖푆(푛 − 1, 푘) be the subgraph of

푆(푛, 푘) induced by the vertices of the form 푖푣2푣3 … 푣푛 . The edge {푖푗푗 … 푗 , 푗푖푖 … 푖} is the unique edge between 푖푆(푛 − 1, 푘) and 푗푆(푛 − 1, 푘) and it is denoted by 푒(푛)푖푗 or 푒(푛)푗푖 (see Fig. 3). For further information on Sierpiński graphs 푆(푛, 푘), please see [9].

Figure2:Sierpiński graphs 푆(2,5) and 푆(3,4)

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Figure3:Sierpiński graph 푆(3,3) and its subgraphs 0푆(2,3), 1푆(2,3), 2푆(2,3) and 20푆(1,3)

Sierpiński graphs and their coloring properties have been extensively studied in the literature by some authors. In [10], Parisse showed that 휒(푆(푛, 푘)) = 푘, for 푛, 푘 ∈ ℕ. In [11], Klavžar showed that 휒′(푆(푛, 3)) = 3 and this coloring is unique. This result was extended by Jakovac and Klavžar who showed that for any 푘, 휒′(푆(푛, 푘)) = 푘. They also proved that for any 푛 ≥ 2 and any odd 푘 ≥ 3, 휒′′ (푆(푛, 푘)) = 푘 + 1 (see [12]). The exact value of the total chromatic number of 푆(푛, 푘) was given by Hinz and Parisse and they obtained for any 푘, 푛 ≥ 2, 휒′′ (푆(푛, 푘)) = 푘 + 1 (see [13]).In [14], authors concentrated on Sierpiński graphs and presented certain results on their metric aspects,domination-type invariants with an emphasis on perfect codes, different colorings, and embeddingsinto other graphs.We previously studied the game chromatic number and the game coloring number of Sierpiński graphs, and obtain certain results (see [15]). Namely, the game chromatic number of a graph 퐺 is defined via a two-person finite game. Two players, generally called Alice and Bob, with Alice going first, alternatively color the uncolored vertices of 퐺 with a color from a color set 푋, such that no two adjacent vertices have the same color. Bob wins if at any stage of the game before the 퐺 is completely colored, one of the players has no legal move; otherwise, that is, if all the vertices of 퐺 are colored properly, Alice wins.

The game chromatic number 휒푔 (퐺) of 퐺 is the least number of colors in the color set 푋 for which Alice has a winning strategy.Accordingly, the game coloring number of a graph 퐺 is defined by modifying the rules of the above-mentioned coloring game as follows. The players fix a positive integer 푘, and instead of coloring vertices, only mark an unmarked vertex each turn. Bob wins if at some stage, some unmarked vertex has 푘 marked neighbors, while Alice wins if this never happens. The game coloring number of a graph 퐺 is defined as the least number 푘 for which Alice has a winning strategy on graph 퐺, and it is denoted by the symbol col푔(퐺). Although game chromatic number and game coloring number of the graphs are extensively studied, and many significant results are obtained, there are still many open problems in this subject (please see [15—29] and references therein).In the following theorem, we obtain the game coloring number of Sierpiński graphs 푆(푛, 푘), and particularly the game chromatic number of Sierpiński graphs 푆(푛, 3).

Theorem 1.3 ([15])For any 푛 ≥ 1 and 푘 ≥ 2 푘 if푛 = 1, col (푆(푛, 푘)) = 푔 푘 + 1 if푛 > 1, and 3 if푛 = 1, 휒 (푆(푛, 3)) = 푔 4 if푛 > 1.

We could not prove, but we conjecture that for each 푘 ≥ 1

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푘 if푛 = 1, 휒 (푆(푛, 푘)) = 푔 푘 + 1 if푛 > 1.

1.2 Main Theorem In this section, we state a theorem which gives the exact value of the incidence chromatic number of Sierpiński graphs 푆(푛, 3). 3 if 푛 = 1, Theorem 2.1휒 (푆(푛, 3)) = 푖 4 if 푛 > 1. Proof. If 푛 = 1 then 푆(1,3) ≅ 퐾3, and since for each 푘 ≥ 1, 휒푖 (퐾푘 ) = 푘, we get 휒푖 (푆(1,3)) = 3. Now let 푛 ≥ 2. Clearly, Δ(푆(푛, 3)) = 3 and from (1) we obtain

4 ≤ 휒푖 (푆(푛, 3)). Let 푋 = {0,1,2,3} be the set of colors. If we prove there is a proper coloring 휎 of 퐼(푆(푛, 3)) with colors from 푋, then we complete the proof for 푛 ≥ 2. We prove it by induction on 푛.

Consider the base case when 푛 = 2, then we have a proper coloring 휎0 of 퐼(0푆(1,3)) using only 3 colors, since 휒푖 (0푆(1,3)) = 3. We choose 휎0 as follows:

휎0: 퐼(0푆(1,3)) → {0,1,2} ⊂ 푋, 휎0((0푖, {0푖, 0푗})) = 푗 (푖, 푗 = 0,1,2). Let us define a permutation 푝1 from 퐼(0푆(1,3)) to 퐼(1푆(1,3)) 푝1((00, {00,01})) = (11, {11,10}), 푝1((00, {00,02})) = (11, {11,12}), 푝1((01, {01,00})) = (10, {10,11}), 푝1((01, {01,02})) = (10, {10,12}), 푝1((02, {02,00})) = (12, {12,11}), 푝1((02, {02,01})) = (12, {12,10}), and a permutation 푝2 from 퐼(0푆(1,3)) to 퐼(2푆(1,3)) 푝2((00, {00,01})) = (22, {22,21}), 푝2((00, {00,02})) = (22, {22,20}), 푝2((01, {01,00})) = (21, {21,22}), 푝2((01, {01,02})) = (21, {21,20}), 푝2((02, {02,00})) = (20, {20,22}), 푝2((02, {02,01})) = (20, {20,21}). We also define two permutations 푐1, 푐2: 푋 → 푋,

0 1 2 3 0 1 2 3 푐 = and푐 = . 1 0 3 2 1 2 0 1 3 2 Now, composing the functions 푐1, 휎0 and the inverse of 푝1 we get a proper coloring of 퐼(1푆(1,3)) as follows: −1 휎1: 퐼(1푆(1,3)) → {0,2,3} ⊂ 푋, 휎1 = 푐1 ∘ 휎0 ∘ 푝1 . Similarly, we obtain a proper coloring of 퐼(2푆(1,3)) as follows: −1 휎2: 퐼(2푆(1,3)) → {0,1,3} ⊂ 푋, 휎2 = 푐2 ∘ 휎0 ∘ 푝2 .

Figure4: Proper colorings of 퐼(1푆(1,3)) and 퐼(2푆(1,3)) by virtue of proper coloring of 퐼(0푆(1,3))

Finally, we give a proper coloring of incidence elements (푖푗, 푒(2)푖푗 ) = (푖푗, {푖푗, 푗푖})푖, 푗 = 0,1,2, 푖 ≠ 푗 which are not belong to 퐼(0(푆(1,3))), 퐼(1(푆(1,3))) and 퐼(2(푆(1,3))). We color an incidence element

(푖푗, {푖푗, 푗푖})푖, 푗 = 0,1,2, 푖 ≠ 푗 with a color not used in 퐼(푖푆(1,3)). Then, since 3 ∉ ℛ(휎0), we can color incidence element (0푗, {0푗, 푗0})) with color 3. Similarly, since 1 ∉ ℛ(휎1), we can color incidence element (1푗, {1푗, 푗1})) with color 1, and finally since 2 ∉ ℛ(휎2), we can color incidence element (2푗, {2푗, 푗2})) with color 2. Here, ℛ(휎0) denotes the range of 휎0. Hence, we get 휎∗((0푗, {0푗, 푗0})) = 3, 휎∗((1푗, {1푗, 푗1})) = 1, and휎∗((2푗, {2푗, 푗2})) = 2. As a result, by combining all these colorings together, we obtain the following proper coloring of 퐼(푆(2,3)):

DOI: 10.9790/5728-1306047279 www.iosrjournals.org 76 | Page The Incidence Chromatic Number of Sierpiński Graphs

휎: 퐼(푆(2,3)) → 푋 = {0,1,2,3}, 휎0((푖푗, {푖푗, 푘푙})) if (푖푗, {푖푗, 푘푙}) ∈ 퐼(0푆(1,3)), 휎 ((푖푗, {푖푗, 푘푙})) if (푖푗, {푖푗, 푘푙}) ∈ 퐼(1푆(1,3)), 휎((푖푗, {푖푗, 푘푙})) = 1 휎2((푖푗, {푖푗, 푘푙})) if (푖푗, {푖푗, 푘푙}) ∈ 퐼(2푆(1,3)), 휎∗((푖푗, {푖푗, 푘푙})) otherwise , where 푖, 푗, 푘, 푙 ∈ {0,1,2}.

Now, suppose that the induction hypothesis holds for 푛 where 푛 ≥ 2, that is, 휒푖 (푆(푛, 3)) = 4 for 푛 ≥ 2, and we want to prove that it also holds for 푛 + 1. By the induction hypothesis, we have a proper coloring

휎0: 퐼(0푆(푛, 3)) → 푋 of 퐼(0푆(푛, 3)) using 4 colors. Similar to the above strategy we define the permutation

푝1: 퐼(0푆(푛, 3)) → 퐼(1푆(푛, 3)), (0푢2푢3 … 푢푛+1, {0푢2푢3 … 푢푛+1, 0푣2푣3 … 푣푛+1}) → ′ ′ ′ ′ ′ ′ ′ ′ ′ (1푢2푢3 … 푢푛+1, {1푢2푢3 … 푢푛+1, 1푣2푣3 … 푣푛+1}) such that for all 푖 = 2,3, … , 푛 + 1

1 if푢푖 = 0, 1 if푣푖 = 0, ′ ′ 푢푖 = 0 if푢푖 = 1, and푣푖 = 0 if푣푖 = 1, 2 if푢푖 = 2, 2 if푣푖 = 2. Likewise, we define the permutation

푝2: 퐼(0푆(푛, 3)) → 퐼(2푆(푛, 3)), (0푢2푢3 … 푢푛+1, {0푢2푢3 … 푢푛+1, 0푣2푣3 … 푣푛+1}) → ′ ′ ′ ′ ′ ′ ′ ′ ′ (2푢2푢3 … 푢푛+1, {2푢2푢3 … 푢푛+1, 2푣2푣3 … 푣푛+1}) such that for all 푖 = 2,3, … , 푛 + 1

2 if푢푖 = 0, 2 if푣푖 = 0, ′ ′ 푢푖 = 1 if푢푖 = 1, and푣푖 = 1 if푣푖 = 1, 0 if푢푖 = 2, 0 if푣푖 = 2. We define two permutation functions 푐1, 푐2: 푋 → 푋, 0 1 2 3 if푛isodd, 푐 = 0 3 2 1 1 0 1 2 3 if푛iseven; 1 0 2 3 and 0 1 2 3 if푛isodd, 푐 = 0 1 3 2 2 0 1 2 3 if푛iseven. 2 1 0 3 Now, composing the functions 푐1, 휎0 and the inverse of 푝1 we get a proper coloring of 퐼(1푆(푛, 3)) as follows: −1 휎1: 퐼(1푆(푛, 3)) → 푋, 휎1 = 푐1 ∘ 휎0 ∘ 푝1 . Analogously, we obtain a proper coloring of 퐼(2푆(푛, 3)) as follows: −1 휎2: 퐼(2푆(푛, 3)) → 푋, 휎2 = 푐2 ∘ 휎0 ∘ 푝2 .

Finally, we give a proper coloring of incidence elements

(푖푗푗 … 푗, 푒(푛 + 1)푖푗 ) = (푖푗푗 … 푗, {푖푗푗 … 푗 , 푗푖푖 … 푖}), 푖, 푗 = 0,1,2, 푖 ≠ 푗 which are not belong to 퐼(0(푆(푛, 3))), 퐼(1(푆(푛, 3))) and 퐼(2(푆(푛, 3))). By the coloring strategy we always have a color from 푋 which is not used in 퐼(푖1푖2 … 푖푛 푆(1,3)) where 푖1, 푖2, … , 푖푛 ∈ {0,1,2}. Thus, we can color an incidence element (푖푗푗 … 푗, {푖푗푗 … 푗 , 푗푖푖 … 푖})푖, 푗 = 0,1,2, 푖 ≠ 푗 with a color not used in 퐼(푗푗 … 푗푖푆(1,3)). Hence, we get 휎∗((0푗푗 … 푗, {0푗푗 … 푗 , 푗00 … 0})) = 3, 휎∗((1푗푗 … 푗, {1푗푗 … 푗 , 푗11 … 1})) = 1, 휎∗((2푗푗 … 푗, {2푗푗 … 푗 , 푗22 … 2})) = 2 when 푛 is odd, and 휎∗((푖00 … 0, {푖00 … 0 , 0푖푖 … 푖})) = 0, 휎∗((푖11 … 1, {푖11 … 1 , 1푖푖 … 푖})) = 1,

DOI: 10.9790/5728-1306047279 www.iosrjournals.org 77 | Page The Incidence Chromatic Number of Sierpiński Graphs

휎∗((푖22 … 2, {푖22 … 2 , 2푖푖 … 푖})) = 2 when 푛 is even. As a result, by combining all these colorings together, we obtain the following proper coloring of 퐼(푆(푛 + 1,3)): 휎: 퐼(푆(푛 + 1,3)) → 푋 = {0,1,2,3}, 휎 ((푢 … 푢 , {푢 … 푢 , 푣 … 푣 })) if 0 1 푛+1 1 푛+1 1 푛+1 (푢 … 푢 , {푢 … 푢 , 푣 … 푣 }) ∈ 퐼(0푆(푛, 3)), 1 푛+1 1 푛+1 1 푛+1 휎1((푢1 … 푢푛+1, {푢1 … 푢푛+1, 푣1 … 푣푛+1})) if

휎((푢1 … 푢푛+1, {푢1 … 푢푛+1, 푣1 … 푣푛+1})) = (푢1 … 푢푛+1, {푢1 … 푢푛+1, 푣1 … 푣푛+1}) ∈ 퐼(1푆(푛, 3)), 휎 ((푢 … 푢 , {푢 … 푢 , 푣 … 푣 })) if 2 1 푛+1 1 푛+1 1 푛+1 (푢1 … 푢푛+1, {푢1 … 푢푛+1, 푣1 … 푣푛+1}) ∈ 퐼(2푆(푛, 3)), ∗ 휎 ((푢1 … 푢푛+1, {푢1 … 푢푛+1, 푣1 … 푣푛+1})) otherwise. This completes the proof.

Although, for any 푛 and 푘, it is not easy to use the technique, presented in the proof of Theorem 2.1, to determine the incidence chromatic number of Sierpiński graphs 푆(푛, 푘), we conjecture that 푘 if푛 = 1, 휒 (푆(푛, 푘)) = 푖 푘 + 1 if푛 > 1.

II. Conclusion In this study,we focus on Sierpińskigraphsandprove that the incidence chromatic number of Sierpiński graphs 푆(푛, 3)is 4 when푛 > 1. We also presentaproof for the incidence chromatic number of the complete bipartite graph 퐾푚,푛 . We have given algorithms for coloring incidence graphs of Sierpińskigraphs푆(푛, 3) and complete bipartite graph 퐾푚,푛 . Further, we conjecture that the incidence chromatic number of Sierpiński graphs 푆(푛, 푘) is 푘 + 1 when 푛 > 1.

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Handan Akyar. ―On the Incidence Chromatic Number of Sierpiński Graphs." IOSR Journal of Mathematics (IOSR-JM) 13.6 (2017): 72-79.

DOI: 10.9790/5728-1306047279 www.iosrjournals.org 79 | Page