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

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

120 F.M. Dong et al. /

120 F.M. Dong et al. / Discrete Mathematics 224 (2000) 107–124Table 4. (continued)(c) For s¿5; we have q¿6 and 2 q−4 is a factor of ′′ (G; 4) + 9 for every G ∈ S 5 ;but 4 is not a factor of ′′ (G; 4) + 9 for every G ∈ S 1 ∪ S 2 ∪ S 3 . Hence(S 1 ∪ S 2 ∪ S 3 ; S 5 )=0.(d) For s¿5; we have q¿6; and 2 q−2 is a factor of ′′ (G; 4) + 11 for G ∈ S 1 ; 2 3is not a factor of ′′ (G; 4) + 11 for G ∈ S 2 ; and 2 3 is a factor of ′′ (G; 4)+11but 2 4 is not for G ∈ S 3 . Hence (S 1 ; S 2 ; S 3 )=0.

F.M. Dong et al. / Discrete Mathematics 224 (2000) 107–124 121(e) For s¿6; we have q¿7; and 2 2 is a factor of ′′ (G; 4) for every G ∈ S 6 but itis not for every G ∈ S 4 . Hence (S 4 ; S 6 )=0.By (b)–(e), Claim 1 holds.The remaining work is to compare every two graphs in each S i . Both S 1 and S 4contain only one graph. For S 5 , when p=q; G 4;10∼ = G4;11 ; when p ¿ q; ′′ (G 4;10 ; 4) ≠ ′′ (G 4;11 ; 4). In the following, we shall study the three sets S 2 ; S 3 and S 6 .(4.1) S 3 : When p ¿ q; ′′ (G 4;6 ; 4) ¿ ′′ (G 4;7 ; 4) ¿ ′′ (G 4;8 ; 4). When p = q, wehave G 4;6∼ = G4;8 and by the method used in (8),(G 4;7 ; 5) − (G 4;8 ; 5)= (G 4;7 −{a 4 ;b 4 ;c 4 }; 4) − (G 4;8 −{a 5 ;b 5 ;c 5 }; 4) = −2 q−5 ≠0:(4.2) S 6 : When p¿q, ′′ (G 4;12 ; 4) ¿ ′′ (G 4;13 ; 4) ¿ ′′ (G 4;14 ; 4) ¿ ′′ (G 4;15 ; 4) ¿ ′′ (G 4;16 ; 4):When p = q; G 4;12∼ = G4;16 ;G 4;13∼ = G4;15 and by the method used in (8),(G 4;14 ; 5) − (G 4;15 ; 5) = −2 q−5 ;(G 4;15 ; 5) − (G 4;16 ; 5) = −3 × 2 q−5 ¡ 0:(4.3) S 2 : Observe that ′′ (G 4;3 ; 4) = ′′ (G 4;4 ; 4). When p¿q, ′′ (G 4;2 ; 4) ¡ ′′ (G 4;3 ; 4) ¡ ′′ (G 4;5 ; 4):When p=q; G 4;2∼ = G4;5 and G 4;3∼ = G4;4 . In the following, we shall compare (G 4;4 ; 5)with (G 4;5 ; 5) for p = q, and (G 4;3 ; 5) with (G 4;4 ; 5) for p¿q.By Lemma 3.5, when p = q, by the method used in (8),(G 4;4 ; 5) − (G 4;5 ; 5)= (G 4;4 −{a ′ 2;b ′ 2;c ′ 2}; 4) − (G 4;5 −{a 3 ;b 3 ;c 3 }; 5)= − 2 q−5¡ 0:For G 4;3 and G 4;4 , we prove the following claim:Claim 2. (G 4;3 ; 5) − (G 4;4 ; 5) = 3(2 p−5 − 2 q−5 ).By Lemma 3.5,(G 4;3 ; 5)= (G 4;3 + a 1 b 1 ; 5) + (G 4;3 −{a 1 ;b 1 }; 4) + (G 4;3 −{a 1 ;b 1 ;c 1 }; 4)= (G 4;3 + a 1 b 1 + b 1 c 1 ; 5) + (G 4;3 −{b 1 ;c 1 }; 4) + (G 4;3 −{b 1 ;c 1 ;d 1 }; 4)+ (G 4;3 −{a 1 ;b 1 }; 4) + (G 4;3 −{a 1 ;b 1 ;c 1 }; 4);

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