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# Chromatically unique bipartite graphs with low 3-independent ...

Chromatically unique bipartite graphs with low 3-independent ...

## 114 F.M. Dong et al. /

114 F.M. Dong et al. / Discrete Mathematics 224 (2000) 107–124= |{{Q 1 ;Q 2 }|Q 1 ;Q 2 ∈ (G); |Q 1 | =3; |Q 2 | =2; |Q 1 ∩ Q 2 |61}|− |{{Q 1 ;Q 2 }|Q 1 ;Q 2 ∈ (G); |Q 1 | =3; |Q 2 | =2; |Q 1 ∩ Q 2 | =1}|= t(s − 2) − 2p 4 (G ′ ): (5)There are exactly ( t 2 ) sets {Q 1;Q 2 }, where Q 1 ;Q 2 ∈ (G), with |Q 1 | = |Q 2 | =3and Q 1 ≠ Q 2 . Thus |Q 1 ∩ Q 2 |62 for such Q 1 ;Q 2 . The number of sets {Q 1 ;Q 2 } with|Q 1 | = |Q 2 | = 3 and |Q 1 ∩ Q 2 | =2 is p 4 (G ′ ), because(i) |Q 1 ∩ Q 2 | =2 i Q 1 ∪ Q 2 induces a path P 4 in G ′ and(ii) for each path P 4 in G ′ , there is only one set {Q 1 ;Q 2 } with |Q 1 | = |Q 2 | = 3 suchthat Q 1 ∪ Q 2 induces this path P 4 .Similarly, the number of sets {Q 1 ;Q 2 } with |Q 1 | = |Q 2 | = 3 and |Q 1 ∩ Q 2 | =1 isp 5 (G ′ ). Hence|{{Q 1 ;Q 2 }|Q 1 ;Q 2 ∈ (G); |Q 1 | = |Q 2 | =3;Q 1 ∩ Q 2 = ∅}|= |{{Q 1 ;Q 2 }|Q 1 ;Q 2 ∈ (G); |Q 1 | = |Q 2 | =3; |Q 1 ∩ Q 2 |62}|−|{{Q 1 ;Q 2 }|Q 1 ;Q 2 ∈ (G); |Q 1 | = |Q 2 | =3; |Q 1 ∩ Q 2 | =2}|−|{{Q 1 ;Q 2 }|Q 1 ;Q 2 ∈ (G); |Q 1 | = |Q 2 | =3; |Q 1 ∩ Q 2 | =1}|( ) t= − p 4 (G ′ ) − p 5 (G ′ ): (6)2By (4)–(6), the result is obtained.For G ∈ B(p; q; s; t), dene ′′ (G; 4) = ′ (G; 4) − (s(2 p−2 +2 q−2 − 2) + t(2 p−3 +2 q−2 − 2)+(s + t)(s + t − 1)=2 − 3t): (7)Observe that for G; H ∈ B(p; q; s; t); ′′ (G; 4) = ′′ (H; 4) i (G; 4) = (H; 4).Lemma 3.4. For G ∈ B(p; q; s; t); if each component of G ′ is a path; then ′′ (G; 4)=(2 p−3 − 2 q−3 ) 2 (G ′ ;A ′ ) − 3p 4 (G ′ ) − p 5 (G ′ ):Proof. It follows from Lemmas 3.1–3.3.For a graph G with uv ∉ E(G), let G + uv (resp. G · uv) denote the graph obtainedfrom G by adding the edge uv (resp. by identifying u and v). For any vertex setA ⊆ V (G), let G − A denote the graph obtained from G by deleting all vertices in Aand all edges incident to vertices in A.For two disjoint graphs G and H, let G + H denote the graph with vertex setV (G) ∪ V (H) and edge set E(G) ∪ E(H) ∪{xy|x ∈ V (G);y ∈ V (H)}.

F.M. Dong et al. / Discrete Mathematics 224 (2000) 107–124 115Lemma 3.5. For a bipartite graph G=(A; B; E); if uvw is a path in G ′ with d G ′(u)=1and d G ′(v)=2; then for any k¿2;(G; k)=(G + uv; k)+(G −{u; v};k − 1) + (G −{u; v; w};k − 1):Proof. Since P(G; )=P(G + uv; )+P(G · uv; ), we have(G; k)=(G + uv; k)+(G · uv; k):Let x be the vertex in G · uv produced by identifying u and v. Notice that x is adjacentto all vertices in V (G ·uv)−{x; w}. Thus G ·uv+xw=K 1 +(G −{u; v}) and G ·uv·xw=K 1 +(G −{u; v; w}). We also observe that for any graph H; (K 1 + H; k)=(H; k − 1),sinceHenceP(K 1 + H; )=P(H; − 1):(G · uv; k)=(G · uv + xw; k)+(G · uv · xw; k)The lemma is then obtained.= (K 1 +(G −{u; v});k)+(K 1 +(G −{u; v; w});k)= (G −{u; v};k − 1) + (G −{u; v; w};k − 1):Theorem 3.1. Let p; q and s be integers with p¿q¿3 and 06s6q − 1. For everyG ∈ ⋃ 4t=1B(p; q; s; t); if G is 2-connected; then G is -unique.Proof. By Theorem 1.3, B(p; q; s; t) ∩ K −s2(p; q) is-closed for each t¿0. To showthat every 2-connected graph in B(p; q; s; t) is-unique, it suces to show that forevery two graphs G and H in B(p; q; s; t), if G ≁ = H, then (G; 4) ≠ (H; 4) or(G; 5) ≠ (H; 5). Recall that for G; H ∈ B(p; q; s; t); ′′ (G; 4) ≠ ′′ (H; 4) i (G; 4) ≠(H; 4).For each t =1; 2; 3; 4, the graphs in B(p; q; s; t) are named as G t;1 ;G t;2 ;:::, and areshown in a table together with the values ′′ (G t;1 ); ′′ (G t;2 );:::: For each graph G t;i ,if every component of G ′ t;i is a path, then ′′ (G t;i ; 4) can be obtained by Lemma 3.4;otherwise, we must rst nd ′ (G t;i ; 4) by Lemma 3.1, and then nd ′′ (G t;i ; 4) by (7).(1) B(p; q; s; 1): The set B(p; q; s; 1) includes two graphs by Theorem 2.1, G 1;1and G 1;2 (see Table 1). Notice that ′′ (G 1;1 ; 4) ≠ ′′ (G 1;2 ; 4) when p ≠ q. But whenp = q; G 1;1∼ = G1;2 .(2) B(p; q; s; 2): The set B(p; q; s; 2) includes four graphs by Theorem 2.1, G 2;1 ;G 2;2 ;G 2;3 and G 2;4 (see Table 2). Notice that only ′′ (G 2;1 ; 4) is odd. If p¿q, thethree values ′′ (G 2;2 ; 4); ′′ (G 2;3 ; 4) and ′′ (G 2;4 ; 4) are distinct. If p = q, then G 2;2∼ =G 2;3 and we shall show that (G 2;3 ; 5) ¿(G 2;4 ; 5). When p = q, by Lemma 3.5

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