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By Claus Kress

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An ) for which (a1 , a2 , . . , an ) = (0, 0, . . , 0). 5) rank(C) = n + ai . i=1 Proof. Let C = (ci j ), with rows C 1 , . . , Cr , and suppose that c1 C1 + · · · + cr Cr = 0. Our multigraph contains vertices h and i joined by two edges, say e j and ek . 2 we know that, without loss of generality, ch j = chk = 1, ci j = −cik = 1 and chl = cil = 0 for all l = j, k. It follows that ch = ci = 0. 3, we find that c1 = c2 = · · · = cr = 0. The lemma follows. 4. 8. Suppose that H is a connected graph with n vertices and m edges.

Forbidden subgraphs The √ value τ is the largest limit point of , and the second largest limit point the limit point τ is obtained when T = K 1,2 , is − 3. 23, √ and the next limit point − 3 is obtained when T = K 1,3 . All graphs with least eigenvalue greater than τ , and all graphs with least eigenvalue greater than √ − 3, have been determined in [CvSt]. It is well known that for a graph H with at least five vertices, the automorphism groups of H and L(H ) are isomorphic [Whi]. 3. 9]. 24. Let G = L(H ; a1 , a2 , .

3 [CvDS2]. 1. Proof. Suppose, by way of contradiction, that there exist two different partitions of E(G) satisfying the given conditions. 2, the GCPs other than cliques are the same in each partition. For each such GCP, add an edge between each pair of partners. 6. More generally, a proper cover of a generalized line graph is determined completely by one of its GCPs. Indeed, starting from some fixed GCP, we can recognize the types of its vertices. Next, if we pick from it any vertex of atype which has neighbours external to the GCP, then this vertex along with the external neighbours forms a new GCP, and so forth.

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Reflections on the Iudicare Limb of the Grave Breaches Regime by Claus Kress

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