The join of null graph Om and complete graph Kn, denoted by S(m; n), is called a complete split graph.&nbsp;In this paper, we characterize unique list colorability of the graph G = S(m; n). We shall prove that G is uniquely 3-list colorable graph if and only if m&gt;=4, n&gt;=4 and m + n&gt;=10, m(G)&gt;=4 for every&nbsp;1&lt;=m&lt;=5 and n&gt;=6.The join of null graph Om and complete graph Kn, denoted by S(m; n), is called a complete split graph.&nbsp;In this paper, we characterize unique list colorability of the graph G = S(m; n). We shall prove that G is uniquely 3-list colorable graph if and only if m&gt;=4, n&gt;=4 and m + n&gt;=10, m(G)&gt;=4 for every&nbsp;1&lt;=m&lt;=5 and n&gt;=6

Xuan Hung, Le

English

Universidad Nacional de Trujillo: Publicaciones Científicas

Journal homepage http://revistas.unitru.edu.pe/index.php/SSMMSELECCIONES MATEMÁTICASUniversidad Nacional de TrujilloISSN: 2411-1783 (Online)2021; Vol. 8(1): 120-124.Uniquely List Colorability of Complete Split GraphsLe Xuan Hung∗Received, Apr. 10, 2021 Accepted, Jun. 30, 2021How to cite this article:Xuan Hung L. Uniquely List Colorability of Complete Split Graphs. Selecciones Matemáticas. 2021;8(1):120–124.http://dx.doi.org/10.17268/sel.mat.2021.01.11AbstractThe join of null graph Om and complete graph Kn, denoted by S(m,n), is called a complete split graph.In this paper, we characterize unique list colorability of the graph G = S(m,n). We shall prove that Gis uniquely 3-list colorable graph if and only if m ≥ 4, n ≥ 4 and m + n ≥ 10, m(G) ≤ 4 for every1 ≤ m ≤ 5 and n ≥ 6.Keywords . Chromatic number, list- chromatic number, uniquely list colorable graph, complete split graph.1. Introduction. All graphs considered in this paper are finite undirected graphs without loops ormultiple edges. If G is a graph, then V (G), E(G) (or V , E in short) and G will denote its vertex-set, itsedge-set and its complementary graph, respectively. The set of all neighbours of a subset S ⊆ V (G) isdenoted by NG(S) (or N(S) in short). Further, for W ⊆ V (G) the set W ∩NG(S) is denoted by NW (S).If S = {v}, then N(S) and NW (S) are denoted shortly by N(v) and NW (v), respectively. For a vertexv ∈ V (G), the degree of v (resp., the degree of v with respect toW ), denoted by deg(v) (resp., degW (v)), is|NG(v)| (resp., |NW (v)|). The subgraph ofG induced byW ⊆ V (G) is denoted byG[W ]. The null graphsand complete graphs of order n are denoted by On and Kn, respectively. Unless otherwise indicated, ourgraph-theoretic terminology will follow [1].Let G1 = (V1, E1), G2 = (V2, E2) be two graphs such that V1 ∩ V2 = ∅. Their union G = G1 ∪ G2has, as expected, V (G) = V1 ∪ V2 and E(G) = E1 ∪ E2. Their join defined is denoted G1 + G2 andconsists of G1 ∪G2 and all edges joining V1 with V2.A graph G = (V,E) is called a split graph if there exists a partition V = I ∪ K such that G[I] andG[K] are null and complete graphs, respectively. We will denote such a graph by S(I ∪K,E). The join ofnull graph Om and complete graph Kn, Om +Kn = S(m,n), is called a complete split graph.The notionof split graphs was introduced in 1977 by Földes and Hammer [7]. A role that split graphs play in graphtheory is clarified in [7] and in [3], [4], [15], [17], [20], [21], [22]. These graphs have been paid attentionalso because they have connection with packing and knapsack problems [5], with the matroid theory [8],with Boolean functions [18], with the analysis of parallel processes in computer programming [11] and withthe task allocation in distributed systems [12]. Many generalizations of split graphs have been made. Thenewest one is the notion of bisplit graphs introduced by Brandstädt et al. [2].Let G = (V,E) be a graph and λ is a positive integer.A λ-coloring of G is a bijection f : V (G) → {1, 2, . . . , λ} such that f(u) 6= f(v) for any adjacentvertices u, v ∈ V (G). The smallest positive integer λ such that G has a λ-coloring is called the chromaticnumber of G and is denoted by χ(G). We say that a graph G is n-chromatic if n = χ(G).Let (Lv)v∈V be a family of sets. We call a coloring f of G with f(v) ∈ Lv for all v ∈ V is a listcoloring from the lists Lv . We will refer to such a coloring as an L-coloring. The graph G is called λ-list-colorable, or λ-choosable, if for every family (Lv)v∈V with |Lv| = λ for all v, there is a coloring of Gfrom the lists Lv . The smallest positive integer λ such that G has a λ-choosable is called the list-chromaticnumber, or choice number of G and is denoted by ch(G).∗HaNoi University for Natural Resources and Environment 41 A, Phu Dien Road, Phu Dien precinct, North Tu Liem district,Hanoi, Vietnam. (lxhung@hunre.edu.vn).120Xuan Hung L.- Selecciones Matemáticas. 2021; Vol. 8(1): 120-124 121Let G be a graph with n vertices and suppose that for each vertex v in G, there exists a list of k colorsLv , such that there exists a unique L-coloring for G, then G is called a uniquely k-list colorable graph ora UkLC graph for short. The idea of uniquely colorable graph was introduced independently by Dinitz andMartin [6] and by Mahmoodian and Mahdian [16] (Mahmoodian and Mahdian have obtained some resultson the uniquely k-list colorable complete multipartite graphs).The list coloring model can be used in the channel assignment. The fixed channel allocation schemeleads to low channel utilization across the whole channel. It requires a more effective channel assignmentand management policy, which allows unused parts of channel to become available temporarily for otherusages so that the scarcity of the channel can be largely mitigated [24]. It is a discrete optimization problem.A model for channel availability observed by the secondary users is introduced in [24].There have been many interesting and insightful research results on these issues for different graphclasses (see [9], [13], [14], [16]). However, these are still issues that have not been resolved thoroughly,so much more attention is needed. In this paper, we shall characterize unique list colorability of the graphG = S(m,n). Namely, we shall prove that G is uniquely 3-list colorable graph if and only if m ≥ 4, n ≥ 4and m+ n ≥ 10, m(G) ≤ 4 for every 1 ≤ m ≤ 5 and n ≥ 6.2. Preliminaries. If a graph G is not uniquely k-list colorable, we also say that G has property M(k).So G has the property M(k) if and only if for any collection of lists assigned to its vertices, each of size k,either there is no list coloring for G or there exist at least two list colorings. The least integer k such thatG has the property M(k) is called the m-number of G, denoted by m(G). This conception was originallyintroduced by Mahmoodian and Mahdian in [16].Lemma 2.1 ([16]). Each UkLC graph is also a U(k − 1)LC graph.Lemma 2.2 ([16]). The graph G is UkLC if and only if k < m(G).Lemma 2.3 ([16]). A connected graphG has the propertyM(2) if and only if every block ofG is eithera cycle, a complete graph, or a complete bipartite graph.Lemma 2.4 ([16]). For every graph G we have m(G) ≤ |E(G)|+ 2.Lemma 2.5 ([16]). Every UkLC graph has at least 3k − 2 vertices.For example, one can easily see that the graph S(2, 2) has the property M(3) and it is U2LC, som(S(2, 2)) = 3.Proposition 2.1. Let G = S(m,n) be a UkLC graph with k ≥ 2. Then(i) m ≥ 2;(ii) k < m2−m+42 ;(iii) k ≤⌊m+n+23⌋. Proof: (i) If m = 1 then G is a complete graph Kn+1. Lemma 2.3, G has theproperty M(2), a contradiction.(ii) It is not difficult to see that |E(G)| = m(m−1)2 . By Lemma 2.4, we havem(G) ≤ |E(G)|+ 2 = m2 −m+ 42.By Lemma 2.2, we have k < m2−m+42 .(iii) Assertion (iii) follows immediately from Lemma 2.5.Let G = S(m,n) be a UkLC graph with V (G) = I ∪ K,G[I] = Om, G[K] = Kn,m ≥ 2, n ≥1, k ≥ 3. SetI = {u1, u2, . . . , um},K = {v1, v2, . . . , vn}.Suppose that, for the given k-list assignment L:Lui = {ai,1, ai,2, . . . , ai,k} for every i = 1, . . . ,m,Lvi = {bi,1, bi,2, . . . , bi,k} for every i = 1, . . . , n,there is a unique k-list color f :f(ui) = ai,1 for every i = 1, . . . ,m,f(vi) = bi,1 for every i = 1, . . . , n.Proposition 2.2.(i) |f(I)| ≥ 2;(ii) |f(I)| ≤ m− 2, where m ≥ 4.Proof: (i) For suppose on the contrary that |f(I)| = 1, then a1,1 = a2,1 = . . . = am,1 = a. SetH = G − I , it is not difficult to see that H is a complete graph Kn. We assign the following lists L′v forthe vertices v of H:If a ∈ Lv then L′v = Lv \ {a},If a /∈ Lv then L′v = Lv \ {b}, where b ∈ Lv and b 6= f(v).It is clear that |L′v| = k − 1 ≥ 2 for every v ∈ V (H). By Lemma 2.3, H has the property M(2). So byLemma 2.1, H has the propertyM(k−1). It follows that with lists L′v , there exist at least two list colorings122 Xuan Hung L.- Selecciones Matemáticas. 2021; Vol. 8(1): 120-124for the vertices v of H . So it is not difficult to see that with lists Lv , there exist at least two list coloringsfor the vertices v of G, a contradiction.(ii) For suppose on the contrary that |f(I)| ≥ m− 1. We consider separately two cases.Case 1: |f(I)| = m− 1.Without loss of generality, we may assume that a1,1 = a2,1 and ai,1 6= aj,1 for every i, j ∈ {2, . . . ,m}, i 6=j. Set graph G′ = (V ′, E′), withV ′ = I ∪K,E′ = (E(G) ∪ {uiuj |i, j = 1, 2, . . . ,m; i 6= j}) \ {u1u2}.It is clear that G′ is complete split graph S(2,m+ n− 2) with V (G′) = I ′ ∪K ′, whereI ′ = {u1, u2},K ′ = {u3, u4, . . . , um, v1, v2, . . . , vn}.Since a1,1 = a2,1, it is not difficult we have got a contradiction.Case 2: |f(I)| = m.In this case, ai,1 6= aj,1 for every i, j ∈ {1, 2, . . . ,m}, i 6= j. Set graph G′′ = (V ′′, E′′), withV ′′ = I ∪K,E′′ = E(G) ∪ {uiuj |i, j = 1, 2, . . . ,m; i 6= j}.It is clear that G′′ is a complete graph Km+n. By Lemma 2.3, G′′ has the property M(2), so with lists Lv ,there exist at least two list colorings for the vertices v of G′′. Since V (G) = V (G′′), it is not difficult tosee that with lists Lv , there exist at least two list colorings for the vertices v of G, a contradiction.3. Main Results. We need the following Lemmas 3.1–3.9 to prove our results.Lemma 3.1. (i) m(S(1, n)) = 2 for every n ≥ 1;(ii) m(S(r, 1)) = 2 for every r ≥ 1;(iii) m(S(2, n)) = 3 for every n ≥ 2. Proof: (i) It is clear that S(1, n) is a complete graph for everyn ≥ 1, by Lemma 2.3, m(S(1, n)) = 2 for every n ≥ 1.(ii) It is clear that S(r, 1) is a complete bipartite graph for every r ≥ 1, by Lemma 2.3, m(S(r, 1)) = 2for every r ≥ 1.(iii) By Lemma 2.3, G = S(2, n) is U2LC for every n ≥ 2.It is not difficult to see that |E(G)| = 1. By Lemma 2.4, m(S(2, n)) ≤ 3 for every n ≥ 2.Thus, m(S(2, n)) = 3 for every n ≥ 2.Lemma 3.2 ([9]). m(S(3, n)) = 3 for every n ≥ 2;Lemma 3.3 ([9]). For every r ≥ 2, m(S(r, 3)) = 3.Lemma 3.4 ([10]). Graphs S(5, 4) and S(4, 4) have property M(3).Lemma 3.5 ([19]). The graph S(4, 5) has property M(3).Lemma 3.6. G = S(4, n) has the property M(4) for every n ≥ 2; Proof: Let G = S(4, n) is acomplete split graph with V (G) = I ∪K,G[I] = O4, G[K] = Kn, n ≥ 2. SetI = {u1, u2, u3, u4},K = {v1, v2, . . . , vn}.For suppose on the contrary that graph G = S(4, n) is U4LC. So there exists a list of 4 colors Lv foreach vertex v ∈ V (G), such that there exists a unique L-coloring f for G. By (i) and (ii) of Proposition2.2, |f(I)| = 2.Let f(I) = {a, b}. Set graph H = G− I , it is not difficult to see that H is a complete graph Kn. Weassign the following lists L′v for the vertices v of H:(a) If a, b ∈ Lv then L′v = Lv \ {a, b},(b) If a ∈ Lv, b /∈ Lv then L′v = Lv \ {a, c}, where c ∈ Lv and c 6= f(v),(c) If a /∈ Lv, b ∈ Lv then L′v = Lv \ {b, c}, where c ∈ Lv and c 6= f(v),(d) If a, b /∈ Lv then L′v = Lv \ {c, d}, where c, d ∈ Lv, c 6= d and c, d 6= f(v).It is clear that |L′v| = 2 for every v ∈ V (H). By Lemma 2.3, H has the property M(2). It follows that withlists L′v , there exist at least two list colorings for the vertices v of H . So it is not difficult to see that withlists Lv , there exist at least two list colorings for the vertices v of G, a contradiction.Lemma 3.7 ([25]). (i) For every n ≥ 2, S(5, n) has the property M(4);(ii) If n ≥ 5 then m(S(5, n)) = 4.Lemma 3.8 ([23]). For every m ≥ 1, k ≥ 2, S(m, 2k − 3) has the property M(k).Lemma 3.9 ([23]). For every n ≥ 1, k ≥ 2, S(2k − 3, n) has the property M(k).Now we prove our results.Theorem 3.1. The graph G = S(m,n) is uniquely 3-list colorable graph if and only if m ≥ 4, n ≥ 4and m+ n ≥ 10.Proof: Firrst we prove the necessity. Suppose that G = S(m,n) is U3LC. If m < 4 or n < 4 thenby Lemma 3.8 and Lemma 3.9, it is not difficult to see that G has the property M(3), a contradiction.Xuan Hung L.- Selecciones Matemáticas. 2021; Vol. 8(1): 120-124 123Therefore, m ≥ 4 and n ≥ 4. It follows that m+ n ≥ 8. If m+ n = 8 then m = 4 and n = 4, by Lemma3.4, G has property M(3), a contradiction. If m+ n = 9 then (m,n) ∈ {(4, 5), (5, 4)}, by Lemma 3.4 andLemma 3.5, G has property M(3), a contradiction. Thus, m+ n ≥ 10.Now we prove the sufficiency. Suppose thatm ≥ 4, n ≥ 4 andm+n ≥ 10. Let V (G) = I∪K,G[I] =Om, G[K] = Kn, I = {u1, u2, . . . , um},K = {v1, v2, . . . , vn}. We prove G is U3LC by induction onm+ n. If m+ n = 10, then we consider separately three cases.(i) m = 4 and n = 6.We assign the following lists for the vertices of G:Lu1 = {1, 3, 4}, Lu2 = {1, 7, 8}, Lu3 = {2, 5, 6}, Lu4 = {2, 7, 8};Lv1 = {1, 2, 3}, Lv2 = {1, 2, 4}, Lv3 = {1, 2, 5}, Lv4 = {1, 2, 6}, Lv5 = {1, 2, 7}, Lv6 = {1, 2, 8}.A unique coloring f of G exists from the assigned lists:f(u1) = 1, f(u2) = 1, f(u3) = 2, f(u4) = 2;f(v1) = 3, f(v2) = 4, f(v3) = 5, f(v4) = 6, f(v5) = 7, f(v6) = 8.(ii) m = 5 and n = 5.We assign the following lists for the vertices of G:Lu1 = {1, 4, 5}, Lu2 = {1, 3, 6}, Lu3 = {2, 3, 7}, Lu4 = {2, 4, 5}, Lu5 = {2, 6, 7};Lv1 = {1, 2, 3}, Lv2 = {1, 2, 4}, Lv3 = {1, 2, 5}, Lv4 = {1, 2, 6}, Lv5 = {1, 2, 7}.A unique coloring f of G exists from the assigned lists:f(u1) = 1, f(u2) = 1, f(u3) = 2, f(u4) = 2, f(u5) = 2;f(v1) = 3, f(v2) = 4, f(v3) = 5, f(v4) = 6, f(v5) = 7.(iii) m = 6 and n = 4.We assign the following lists for the vertices of G:Lu1 = {1, 3, 5}, Lu2 = {1, 4, 5}, Lu3 = {2, 3, 6}, Lu4 = {2, 3, 4}, Lu5 = {2, 4, 6}, Lu6 = {2, 5, 6};Lv1 = {1, 2, 3}, Lv2 = {1, 2, 4}, Lv3 = {1, 2, 5}, Lv4 = {1, 2, 6}.A unique coloring f of G exists from the assigned lists:f(u1) = 1, f(u2) = 1, f(u3) = 1, f(u4) = 2, f(u5) = 2, f(u6) = 2;f(v1) = 3, f(v2) = 4, f(v3) = 5, f(v4) = 6.Now let m + n > 10 and assume the assertion for smaller values of m + n. We consider separatelytwo cases.Case 1: m ≥ 5.Set G′ = G− um = S(m− 1, n). By the induction hypothesis, for each vertex v in G′, there exists alist of 3 colors L′v , such that there exists a unique f′ for G′. We assign the following lists for the vertices ofG:Lum = L′um−1 , Lv = L′v if v ∈ V (G′).A unique coloring f of G exists from the assigned lists:f(um) = f′(um−1), f(v) = f ′(v) if v ∈ V (G′).Case 2: n ≥ 5.Set G′ = G − vn = S(m,n − 1). By the induction hypothesis, for each vertex v in G′, there exists alist of 3 colors L′v , such that there exists a unique f′ for G′. We assign the following lists for the vertices ofG:Lvn = {f ′(vn−1), f ′(vn−2), t} with t /∈ f ′(G′), Lv = L′v if v ∈ V (G′).A unique coloring f of G exists from the assigned lists:f(vn) = t, f(v) = f ′(v) if v ∈ V (G′).Corollary 3.1. m(S(4, n)) = 4 for every n ≥ 6. Proof: It follows from Theorem 3.1 and Lemma3.6.Theorem 3.2. m(S(r, n)) ≤ 4 for every 1 ≤ r ≤ 5 and n ≥ 6. Proof: It follows from Lemma 3.1 toLemma 3.7.ORCID and LicenseLe Xuan Hung https://orcid.org/0000-0003-4560-2892This work is licensed under the Creative Commons - Attribution 4.0 International (CC BY 4.0)References[1] Behzad M, Chartrand G. Introduction to the theory of graphs. Bodton: Allyn and Bacon; 1971.[2] Brandstädt A, Hammer P, Le V, Lozin V. Bisplit graphs. Discrete Math. 1005; 299:11–32.[3] Chen B, Fu H, Ko M. Total chromatic number and chromatic index of split graphs. J. Combin. Math. Combin. Comput. 1995;17:137–146.[4] Burkard R, Hammer P. A note on hamiltonian split graphs. J. Combin. Theory Ser. 1980; B 28:245–248.124 Xuan Hung L.- Selecciones Matemáticas. 2021; Vol. 8(1): 120-124[5] Chvatal V, Hammer P. Aggregation of inequalities in integer programming. Annals Disc. Math. 1977; 1:145–162.[6] Dinitz J, Martin W. The stipulation polynomial of a uniquely list colorable graph. Austran. J. Combin. 1995; 11:105–115.[7] Földes S, Hammer P. Split graphs. In: Proc. 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Uniquely List Colorability of Complete Split Graphs

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Uniquely List Colorability of Complete Split Graphs

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