By Okamura H.

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**Extra resources for 2-reducible cycles containing three consecutive edges in (2k + 1)-edge-connected graphs**

**Sample text**

Y1 N ~) = k + 2 + e and X1 fq Y ~ X1, contrary to the minimality of X 1. Thus X1 - T and X2 - T are independent sets. In (A) and (B), if x e X1 vl, then e(x, X 2 ) > ~ + 1, and so IXxl = 2 since e(v~,X2)> 2, R 2 = (x) and e(v~) = k + 1, which contradicts e(v~)= k in (A). 11). In (T), we can choose P* such that IE(P*)N O(X1)I = 1. For otherwise, starting from Ux along P*, let z be the first vertex in X1, then X 2 ÷ v~ separates z from u2. Then there are disjoint D~, D2 c_ V(G) - (X2 + vl) such that V(G) = Dx O D 2 (3 X 2 (J {01 }, e(D1, D2) --- 0, z 6 D1 and u 2 ~ D 2.

Acknowledgment. The author would like to thank the referee for his helpful comments. References 1. : On a theorem of Mader. Discrete Mathematics 101, 49-57 (1992) 2. : Counterexamples to a conjecture of Mader about cycles through specified vertices in n-edge-connected graphs. Graphs and Comb. 8, 253-258 (1992) 3. : A reduction method for edge-connectivity in graphs. Ann. Discrete Math. 3, 145-164 (1978) 4. : Paths in graphs, reducing the edge-connectivity only by two. Graphs and Comb. I, 81-89 (1985) 5.

Then there are disjoint D~, D2 c_ V(G) - (X2 + vl) such that V(G) = Dx O D 2 (3 X 2 (J {01 }, e(D1, D2) --- 0, z 6 D1 and u 2 ~ D 2. 8). Thus Xx = DIUD2U{vl} = {vt,v2,z}. 6) e(z, X 2 ) < ~ and e(z, v l) < ~, a contradiction. 11), V(P*) = {u~, x, y, u2} for some x ~ X2 and y e X 1 . 23), there is (k + 3)-set Z with Z f ) ( T U {x,y})-(vl, u2, x}. e(X1 tq Z) >_ k + 4, and e(X1 tq Z) > k + 4 since X1 and X2 are minimal (k + 3)-sets, contrary to Lemma 2. 25), V(G) - T is an independent set. 22). 6).