Let \(G=(L,R;E)\) be a bipartite graph with color classes \(L\) and \(R\) and edge set \(E\). A set of two bijections \(\{\varphi_1 , \varphi_2\}\), \(\varphi_1 , \varphi_2 :L \cup R \to L \cup R\), is said to be a \(3\)-biplacement of \(G\) if \(\varphi_1(L)= \varphi_2(L) = L\) and \(E \cap \varphi_1^*(E)=\emptyset\), \(E \cap \varphi_2^*(E)=\emptyset\), \(\varphi_1^*(E) \cap \varphi_2^*(E)=\emptyset\),  where \(\varphi_1^*\), \(\varphi_2^*\) are the maps defined on \(E\), induced by \(\varphi_1\), \(\varphi_2\), respectively.  We prove that if \(|L| = p\), \(|R| = q\), \(3 \leq p \leq q\), then every graph \(G=(L,R;E)\) of size at most \(p\) has a \(3\)-biplacement

Beata Orchel

Edyta Leśniak

Lech Adamus

Let G = (L,R;E) be a bipartite graph with color classes L and R and edge set E. A set of two bijections {&phi;1, &phi;2}, &phi;1, &phi;2 : L &cup; R &rarr; L &cup; R, is said to be a 3-biplacement of G if [formula], where &phi;*/1, &phi;*/2 are the maps defined on E, induced by &phi;1, &phi;2, respectively. We prove that if &zwnj;L&zwnj; = p, &zwnj;R&zwnj; = q, 3 &le; p &le; q, then every graph G = (L, R; E) of size at most p has a 3-biplacement

Adamus, L.

Leśniak, E.

Orchel, B.

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3-biplacement of bipartite graphs

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Opuscula Mathematica  Vol. 28  No. 3  2008Lech Adamus, Edyta Le±niak, Beata Orchel3-BIPLACEMENT OF BIPARTITE GRAPHSAbstract. Let G = (L;R;E) be a bipartite graph with color classes L and R and edge setE. A set of two bijections f'1;'2g, '1;'2 : L [ R ! L [ R, is said to be a 3-biplacementof G if '1(L) = '2(L) = L and E \ '1(E) = ;, E \ '2(E) = ;, '1(E) \ '2(E) = ;, where'1;'2 are the maps dened on E, induced by '1, '2, respectively.We prove that if jLj = p, jRj = q, 3  p  q, then every graph G = (L;R;E) of size atmost p has a 3-biplacement.Keywords: bipartite graph, packing of graphs, placement, biplacement.Mathematics Subject Classication: 05C70.1. INTRODUCTION1.1. BASIC DEFINITIONSThroughout the paper we will only consider nite, undirected graphs without loopsand multiple edges.Let G be a graph with vertex set V (G) and edge set E(G). The cardinality of theset V (G) is called the order of G and is denoted by jGj, while the cardinality of theedge set E(G) is the size of G, denoted by kGk.For a vertex x 2 V (G), N(x;G) denotes the set of its neighbors in G. The degreed(x;G) of the vertex x in G is the cardinality of the set N(x;G). A vertex x of G issaid to be pendent (resp. isolated) if d(x;G) = 1 (resp. d(x;G) = 0).A set of pairwise non-incident edges in a graph G is called a matching.Let G1 and G2 be vertex disjoint graphs. The union G = G1 [G2 is a graph withV (G) = V (G1) [ V (G2) and E(G) = E(G1) [ E(G2). If a graph G is the union of kdisjoint copies of a graph H, then we write G = kH.Let G = (L;R;E) be a bipartite graph with vertex set V (G) = L [ R and edgeset E(G) = E. We denote then L(G) = L and R(G) = R, and we call these sets theleft and right set of bipartition of the vertex set of G.We denote by L(G) (resp. R(G)) the maximum vertex degree in the set L (resp. R).223224 Lech Adamus, Edyta Le±niak, Beata OrchelIf jLj = p and jRj = q, we say that G is a (p;q)-bipartite graph. Kp;q stands forthe complete (p;q)-bipartite graph. Gbipis the complement of G in Kp;q. ThusGbip= (L;R;E0), where E0 consists of all the edges joining L with R which are notin E.1.2. 2-PLACEMENT AND 3-PLACEMENT OF SIMPLE GRAPHSDenition 1. Let G be a simple graph. We say that G is 2-placeable if there existsa bijection ' : V (G) ! V (G) such thatif xy 2 E(G); then '(x)'(y) = 2 E(G):The bijection ' will be called a 2-placement of G.The study of placing problems was initiated by a series of papers published in thelate 1970s. The following theorem, proved by Sauer and Spencer [3], was the rstresult in this area.Theorem A. Let G be a graph of order n. If kGk  n   2, then G is 2-placeable.This theorem can be generalized in a great variety of ways. Wo¹niak and Wojda[5] showed that under the assumptions of Theorem A there exists a 3-placement of agiven graph G, unless G is an exception (see Theorem B below).A 3-placement of a given graph can be dened analogously to a 2-placement.Denition 2. Let G be a simple graph of order n. A graph G is 3-placeable ifthere exist bijections '1;'2 : V (G) ! V (G) such that E(G) \ '1(E(G)) = ;,E(G) \ '2(E(G)) = ;, '1(E(G)) \ '2(E(G)) = ;, where the map 'i dened onE(G) is induced by 'i (i = 1;2), that is 'i(xy) = 'i(x)'i(y).The set f'1;'2g is called a 3-placement of G.Wo¹niak and Wojda proved the following theorem.Theorem B. Let G be a simple graph of order n. If kGk  n   2, then either G is3-placeable or G is isomorphic to K3 [ 2K1 or to K4 [ 4K1.Exhaustive surveys of the results concerning the problems of placing of simplegraphs are given in [1, Chapter 8] and [4]. However, we would like to focus onplacements of bipartite graphs, the so-called biplacements, dened by Fouquet andWojda [2] in 1993.1.3. 2-BIPLACEMENT AND 3-BIPLACEMENT OF BIPARTITE GRAPHSDenition 3. Let G = (L;R;E) be a bipartite graph. We say that G is 2-biplaceableif there exists a bijection ' : L [ R ! L [ R such that '(L) = L andif xy 2 E; then '(x)'(y) = 2 E:The bijection ' is called a 2-biplacement of G.3-biplacement of bipartite graphs 225Fouquet and Wojda [2] proved the following theorem, which is an analogue ofTheorem A for bipartite graphs.Theorem C. Let G = (L;R;E) be a (p;q)-bipartite graph such that either p  3,q  3 and kGk  p+q 3, or 2 = p  q and kGk  p+q 2. Then G is 2-biplaceable.The aim of this paper is to nd a sucient condition for a bipartite graph to be3-biplaceable; in other words, nd an analogue of Theorem B for bipartite graphs.By analogy to a 2-biplacement we consider a 3-biplacement of a bipartite graph.Let G = (L;R;E) be a (p;q)-bipartite graph. Then G can be considered as asubgraph of the complete bipartite graph Kp;q.Denition 4. The graph G = (L;R;E) is 3-biplaceable if there exist bijections'1;'2 : L [ R ! L [ R such that '1(L) = '2(L) = L and E \ '1(E) = ;,E \ '2(E) = ;, '1(E) \ '2(E) = ;, where the maps '1;'2 : E ! E(Kp;q) areinduced by '1;'2, respectively (i:e:; 'i(xy) = 'i(x)'i(y) for i = 1;2).The set f'1;'2g is called a 3-biplacement of G.It is easy to see that a (p;q)-bipartite graph G is 3-biplaceable if and only if wecan nd two edge-disjoint copies of G, say Gr and Gb, in the graph Gbip. We then callthe edges of G black, the edges of Gr red, the edges of Gb blue, and there is L(G) =L(Gr) = L(Gb), R(G) = R(Gr) = R(Gb), E(G) \ E(Gr) = ;, E(G) \ E(Gb) = ;,E(Gr) \ E(Gb) = ;.Now we are ready to formulate the main result of this paper.2. MAIN RESULTLet G1 denote a (2,3)-bipartite graph such that kG1k = 2 and L(G1) = 2.Our goal is to prove the following theorem.Theorem 1. Let G = (L;R;E) be a (p;q)-bipartite graph, p  q and q  3. IfkGk  p then either G is 3-biplaceable or G is isomorphic to G1.Proof. We will proceed by induction on p + q.The assertion is easy to check for p  3 and q = 3 (see Fig. 1), and hence for allq  3.1 Gp+q=4 p+q=5p+q=6Fig. 1226 Lech Adamus, Edyta Le±niak, Beata OrchelNow assume that p + q  8, q  p  4, and the theorem holds for all integersp0  1, q0  3, such that p0  q0 and p0 + q0 < p + q.Let G = (L;R;E) be a (p;q)-bipartite graph with p and q as above. Without lossof generality, we can assume that kGk = p. We will show that G is 3-biplaceable.In the proof, we shall consider three cases.Case 1. L(G)  3.Let v 2 L be a vertex such that d(v;G) = L(G). It is evident that there are atleast two isolated vertices, say x and y, in L.We dene a new graph G0 := G n fv;x;yg. G0 is (p0;q0)-bipartite, where p0 =p   3  1, q0 = q  4, p0  q0. Thus G0 6= G1 and kG0k  p   3 = p0. Hence, by theinductive hypothesis, G0 is 3-biplaceable. Let f'01;'02g be a 3-biplacement of G0.We dene a 3-biplacement f'1;'2g of G as follows:'1(v) = x, '1(x) = v, '1(y) = y, '1(w) = '01(w) 8w 2 V (G0),'2(v) = y, '2(x) = x, '2(y) = v, '2(w) = '02(w) 8w 2 V (G0).Case 2. L(G) = 2.Pick v 2 L with d(v;G) = 2. We need to consider several subcases.Subcase 2.1. There is a pendent vertex in L, say x, such that N(x;G)\N(v;G) = ;.Let N(v;G) = fw1;w2g  R, N(x;G) = fw3g  R, and let y be an isolated vertexin L. We have to consider three subcases depending on the degrees of the verticesw1;w2;w3.Subcase 2.1.1. d(w3;G) = 1.Put G0 := G n fv;x;y;w3g. G0 is a (p0;q0)-bipartite graph with p0 = p   3  1,q0 = q   1  3, p0  q0, kG0k = p0. Obviously, G0 is not isomorphic with G1, forotherwise p = 5 and q = 4, which contradicts the assumption p  q. By the inductivehypothesis, there is a 3-biplacement of G0, say f'01;'02g. We dene bijections '1 and'2 in the following way:'1(v) = y, '1(x) = v, '1(y) = x, '1(w3) = w3, '1(w) = '01(w) 8w 2 V (G0),'2(v) = x, '2(x) = y, '2(y) = v, '2(w3) = w3, '2(w) = '02(w) 8w 2 V (G0).f'1;'2g is a 3-biplaceament of G.Subcase 2.1.2. d(w3;G) > 1 and d(w1;G) = d(w2;G) = 1.In the case of p = q = 4, we get one graph only. Obviously, it is 3-biplaceable (seeFig. 2).Thus we can assume that q  5. Then we dene a graph G0 := Gnfv;x;y;w1;w2g,which is (p0;q0)-bipartite with p0 = p 3  1, q0 = q 2  3, p0  q0. Since kG0k = p0,there exists a 3-biplacement of G0, unless G0 = G1.In the case of G0 = G1, the graph G is 3-biplaceable (see Fig. 3).3-biplacement of bipartite graphs 227In the case of G0 6= G1, let f'01;'02g be a 3-biplacement of G0.To get a 3-biplacement f'1;'2g of G, put:'1(v) = y, '1(x) = v, '1(y) = x, '1(w1) = w1, '1(w2) = w2,'1(w) = '01(w) 8w 2 V (G0),'2(v) = x, '2(x) = y, '2(y) = v, '2(w1) = w1, '2(w2) = w2,'2(w) = '02(w) 8w 2 V (G0).vxyw1w2w3Fig. 2vxyw1w2w3Fig. 3Subcase 2.1.3. d(w3;G) > 1; d(w1;G) > 1 or d(w2;G) > 1.These assumptions imply that p  5. It is easy to check that, for q  p = 5, G is3-biplaceable. Therefore, we may assume that q  p  6.Let u1;u2 be isolated vertices in R and G0 := G n fv;x;y;w3;u1;u2g. Again, G0is 3-biplaceable; let f'01;'02g be a 3-biplacement of G0.A set of bijections f'1;'2g such that'1(v) = y, '1(x) = v, '1(y) = x, '1(w3) = u1, '1(u1) = w3, '1(u2) = u2, '1(w) ='01(w) 8w 2 V (G0),'2(v) = x, '2(x) = y, '2(y) = v, '2(w3) = u2, '2(u1) = u1, '2(u2) = w3, '2(w) ='02(w) 8w 2 V (G0),is then a 3-biplacement of G.Subcase 2.2. There is a pendent vertex in L, say x, such that N(x;G) \ N(v;G)6=;.Without loss of generality, we put N(v;G) = fw1;w2g and N(x;G) = fw2g.Consequently, for all z 2 L of degree 2, there is N(z;G)  fw2g, and for all y 2 Lof degree 1, there is N(y;G)  fw1;w2g. Otherwise, we get Subcase 2.1.We have to consider the following subcases.Subcase 2.2.1. For all z 2 L of degree 2, there is N(z;G) = fw1;w2g.In this case all (p;q)-bipartite graphs for p + q = 8;9;10 are 3-biplaceable, whichis easily veriable. Hence we can assume that q  6. If so, there are at least fourisolated vertices in R, say u1;u2;u3;u4.A 3-biplacement f'1;'2g of G is dened as follows:'1(w1) = u1, '1(w2) = u2, '1(u1) = w1, '1(u2) = w2,'1(w) = w 8w 2 V (G) n fw1;w2;u1;u2g,'2(w1) = u3, '2(w2) = u4, '2(u3) = w1, '2(u4) = w2,'2(w) = w 8w 2 V (G) n fw1;w2;u3;u4g.228 Lech Adamus, Edyta Le±niak, Beata OrchelSubcase 2.2.2. There exists z 2 L of degree 2 such that N(z;G) = fw2;w3g andw3 6= w1.It follows that p  5. Moreover, every pendent vertex in L is joined with w2, forotherwise we would get Subcase 2.1. Consequently, all non-isolated vertices in L arejoined with w2.Firstly, suppose that d(w3;G) = 1.A trivial verication shows that the theorem is true for q  p = 5. Therefore, assumethat p  6. Let y1;y2 2 L;u 2 R be isolated vertices in G.Consider a graph G0 := G n fv;x;z;y1;y2;w2;w3;ug. G0 6= G1 and by the inductivehypothesis G0 is 3-biplaceable.A 3-biplacement of G is given by the maps '1;'2 dened as:'1(v) = z, '1(x) = x, '1(z) = v, '1(y1) = y1, '1(y2) = y2, '1(w2) = u, '1(w3) = w3,'1(u) = w2, '1(w) = '01(w) 8w 2 V (G0),'2(v) = y1, '2(x) = x, '2(z) = y2, '2(y1) = v, '2(y2) = z, '2(w2) = w3, '2(w3) = u,'2(u) = w2, '2(w) = '02(w) 8w 2 V (G0),where f'01;'02g is a 3-biplacement of G0.Secondly, suppose that d(w3;G)  2.It follows that d(w1;G)  2, for if not, we would replace w1 with w3, and get thecase proved above. Since all non-isolated vertices in L are joined with w2, thend(w2;G)  5.We conclude that q  p  9 and there are at least three isolated verticesin L and six isolated vertices in R. Let us denote by y1;y2;y3 isolated ver-tices in L and by u1;u2;u3;u4 isolated vertices in R. Consider a graph G0 :=G n fv;x;z;y1;y2;y3;w2;w3;u1;u2;u3;u4g. As p  9, there is G0 6= G1. Thus G0has a 3-biplacement, say f'01;'02g.A 3-biplacement f'1;'2g of G is dened below:'1(v) = z, '1(x) = x, '1(z) = v, '1(yi) = yi for i = 1;2;3, '1(w2) = u1,'1(w3) = u2, '1(u1) = w2, '1(u2) = w3, '1(u3) = u3, '1(u4) = u4, '1(w) = '01(w)8w 2 V (G0),'2(v) = y1, '2(x) = x, '2(z) = y2, '2(y1) = v, '2(y2) = z, '2(y3) = y3,'2(w2) = u3, '2(w3) = u4, '2(u1) = u1, '2(u2) = u2, '2(u3) = w2, '2(u4) = w3,'2(w) = '02(w) 8w 2 V (G0).Subcase 2.3. There are no pendent vertices in L.It follows that all vertices in L are of degree 0 or 2. Three subcases need to beconsidered.Subcase 2.3.1. There are no pendent vertices in R.Then we dene sets:A := fw 2 L : d(w;G) = 2g, B := fw 2 L : d(w;G) = 0g,C := fw 2 R : d(w;G)  2g, D := fw 2 R : d(w;G) = 0g.We have A  L, B  L, jAj = jBj (since kGk = p) and C  R, D  R,jCj  jAj; jCj  jBj; jCj  jDj.3-biplacement of bipartite graphs 229It is easy to see that G is 3-biplaceable (see Fig. 4).ABCDFig. 4Subcase 2.3.2. There are no vertices in R of degree greater than 1.Set N(v;G) = fw1;w2g  R. There is d(w1;G) = d(w2;G) = 1. We deduce thatthere are at least two isolated vertices in L, say y1;y2, and, apart from v, at least oneother vertex of degree 2, say x.It is a simple matter to show that G is 3-biplaceable in the case of q  p = 4.Therefore, we assume that p  5 and apply the inductive hypothesis to the graphG0 := G n fv;x;y1;y2;w1;w2g. We extend bijections '01 and '02 of a 3-biplacement ofG0 to '1 and '2, maps of a 3-biplacement of G, in the following way:'1(v) = x, '1(x) = v, '1(y1) = y1, '1(y2) = y2, '1(w1) = w1, '1(w2) = w2,'1(w) = '01(w) 8w 2 V (G0),'2(v) = y2, '2(x) = y1, '2(y1) = x, '2(y2) = v, '2(w1) = w1, '2(w2) = w2,'2(w) = '02(w) 8w 2 V (G0).Subcase 2.3.3. There is a vertex of degree 2 in L such that one of its neighbors hasdegree 1 and the other has degree at least 2.Without loss of generality, we can choose our v to be this vertex. Put N(v;G) =fw1;w2g with d(w1;G) = 1; d(w2;G)  2.It follows that there exists a vertex x 2 L such that N(x;G) = fw2;w3g, w3 6= w1,and there exist isolated vertices, say y1;y2 2 L and u 2 R.The case of q  p = 4 is left to the reader. We assume that q  p  5. In fact,since every non-isolated vertex in L has degree 2 and kGk = p, it implies that p  6.Let G0 := G n fv;x;y1;y2;w1;w2;ug. If G0 = G1, then G is one of the two graphswhich are 3-biplaceable, which is easy to check. If G0 6= G1, then by the inductivehypothesis there exists a 3-biplacement f'01;'02g of G0.A 3-biplacement of G is given by the maps '1;'2 dened as'1(v) = x, '1(x) = v, '1(y1) = y1, '1(y2) = y2, '1(w1) = w1, '1(w2) = u,'1(u) = w2, '1(w) = '01(w) 8w 2 V (G0),'2(v) = y1, '2(x) = y2, '2(y1) = v, '2(y2) = x, '2(w1) = w2, '2(w2) = w1,'2(u) = u, '2(w) = '02(w) 8w 2 V (G0).Case 3. L(G) = 1.By the assumption kGk = p, all vertices in L are pendent.We shall consider three subcases depending on the maximum vertex degree in theset R.230 Lech Adamus, Edyta Le±niak, Beata OrchelSubcase 3.1. R(G) = 1.The theorem is evident in this case, since the edges of G dene a matching pK1;1.Subcase 3.2. R(G)  3.It is easily seen that the theorem is true for q  5. For this reason, assume thatq  6. Let u be a vertex in R such that d(u;G) = R(G) and let v1;v2;v3 beneighbors of u. There are at least two isolated vertices in R, say w1;w2. We dene agraph G0 := G n fw1;w2;u;v1;v2;v3g. Obviously, G0 6= G1, since all vertices in L arependent. Consequently, we may dene a 3-biplacement f'1;'2g of G as follows:'1(w1) = u, '1(w2) = w2, '1(u) = w1, '1(vi) = vi for i = 1;2;3,'1(w) = '01(w) 8w 2 V (G0),'2(w1) = w1, '2(w2) = u, '2(u) = w2, '2(vi) = vi for i = 1;2;3,'2(w) = '02(w) 8w 2 V (G0),where f'01;'02g is a 3-biplacement of G0.Subcase 3.3. R(G) = 2.In this case, we have to consider the two situations: either there is a pendentvertex in R or all non-isolated vertices in R are of degree 2.Subcase 3.3.1. There is a pendent vertex in R, say w1.If q  5, then G is 3-biplaceable, which is easy to check. Assume that q  6. Letw2 2 R be of degree 2 and let u be an isolated vertex in R. Let N(w1;G) = fv1gand N(w2;G) = fv2;v3g. We may apply the inductive hypothesis to the graphG0 := G n fw1;w2;u;v1;v2;v3g. Again, G0 6= G1 and, in consequence, G0 has a3-biplacement, say f'01;'02g.A 3-biplacement f'1;'2g of G is dened below:'1(w1) = w2, '1(w2) = u, '1(u) = w1, '1(vi) = vi for i = 1;2;3,'1(w) = '01(w) 8w 2 V (G0),'2(w1) = u, '2(w2) = w1, '2(u) = w2, '2(vi) = vi for i = 1;2;3,'2(w) = '02(w) 8w 2 V (G0).Subcase 3.3.2. There are no pendent vertices in R.A trivial verication shows that in the cases of p + q = 8;9;10;11 the theoremis true. For q  p  6, we dene a graph G0 := G n fw1;w2;u;v1;v2;v3;v4g, wherew1;w2 2 R are vertices of degree 2, u is an isolated vertex in R, v1;v2 and v3;v4 areneighbors of w1 and w2, respectively. G0 is 3-biplaceable, hence so is G: put f'1;'2gto be:'1(w1) = w2, '1(w2) = u, '1(u) = w1, '1(vi) = vi for i = 1;2;3;4,'1(w) = '01(w) 8w 2 V (G0),'2(w1) = u, '2(w2) = w1, '2(u) = w2, '2(vi) = vi for i = 1;2;3;4,'2(w) = '02(w) 8w 2 V (G0),where f'01;'02g is a 3-biplacement of G0.3-biplacement of bipartite graphs 231AcknowledgementsThe research was partially supported by the AGH University of Science and Technologygrant No 1142004.REFERENCES[1] B. Bollobás, Extremal Graph Theory, Academic Press, London, 1978.[2] J.L. Fouquet, A.P. Wojda, Mutual placement of bipartite graphs, Discrete Math. 121(1993), 8592.[3] N. Sauer, J. Spencer, Edge disjoint placement of graphs, J. Combin. Theory Ser. B 25(1978), 295302.[4] M. Wo¹niak, Packing of graphs, Dissertationes Mathematicae CCCLXII, Warszawa,1997.[5] M. Wo¹niak, A.P. Wojda, Triple placement of graphs, Graph Combin. 9 (1993), 8591.Lech Adamusladamus@wms.mat.agh.edu.plAGH University of Science and TechnologyFaculty of Applied Mathematicsal. Mickiewicza 30, 30-059 Cracow, PolandBeata Orchelorchel@uci.agh.edu.plAGH University of Science and TechnologyFaculty of Applied Mathematicsal. Mickiewicza 30, 30-059 Cracow, PolandReceived: September 25, 2007.Revised: February 15, 2008.Accepted: March 1, 2008.

3-biplacement of bipartite graphs

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3-biplacement of bipartite graphs

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