By Professor Cun-Quan Zhang

The well-known Circuit Double disguise conjecture (and its various editions) is taken into account one of many significant open difficulties in graph concept due to its shut dating with topological graph conception, integer stream concept, graph coloring and the constitution of snarks. you can actually nation: each 2-connected graph has a kinfolk of circuits protecting each aspect accurately two times. C.-Q. Zhang presents an updated assessment of the topic containing all the recommendations, tools and effects constructed to aid resolve the conjecture because the first e-book of the topic within the Nineteen Forties. it's a worthy survey for researchers already engaged on the matter and a becoming creation for these simply getting into the sector. The end-of-chapter workouts were designed to problem readers at each point and tricks are supplied in an appendix.

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**Sample text**

8 (H¨ agglund [102]) An eulerian (1, 2)-weighted cubic graph (G, w) is called a strong contra pair if, for every e ∈ Ew=2 , both (G, w) and (G − e, w) are contra pairs. Find a strong contra pair. 9 (Fleischner, Genest and Jackson [75] (revised)) If (G, w) is a cyclically 4-edge-connected, (1, 2)-weighted, contra pair with a cyclic 4-edge-cut T and w(T ) = 4, then (G, w) = (G1 , w1 ) ⊗IF J (G2 , w2 ) such that one of {(G1 , w1 ), (G2 , w2 )} is a contra pair. Remark. One may notice that we are only able to show that one of {(G1 , w1 ), (G2 , w2 )} is a contra pair, not both yet.

2 (Alspach, Goddyn and Zhang [4]) Every minimal (1, 2)weighted contra pair must be cubic. 2. Our goal is to verify the nonexistence of high degree vertex. The following is a step-by-step outline of the proof. Step 1. (1-1) The degree of every vertex is either 3 or 4. (1-2) Each degree 4 vertex is a cut-vertex of some component of G[Ew=1 ], and every block of G[Ew=1 ] is a circuit. (1-3) The even subgraph G[Ew=1 ] has a unique circuit decomposition X. Step 2. 2. (2-1) The family X of circuits has precisely two members X1 , X2 of odd lengths and all others (if they exist) are of even length.

If G has a faithful circuit cover F with respect to a weight w : E(G) → Z+ , then the total weight of every edge-cut must be even since, for every circuit C of F and every edge-cut T , the circuit C must use an even number of distinct edges of the cut T . With this observation, the requirements of being eulerian and admissible are necessary for faithful circuit covers. 4 Let G be a bridgeless graph with w : E(G) → Z+ . If w is admissible and eulerian, does G have a faithful circuit cover with respect to w?