Home » Subject » Graph Theory » Basic G.T. » Cycles
Importance: High ✭✭✭
Author(s):
Subject: Graph Theory
» Basic Graph Theory
» » Cycles
Keywords: cover
cycle
Eulerian
Recomm. for undergrads: no
Posted by: mdevos
on: March 7th, 2007
Conjecture   If $ G $ is a 6-edge-connected Eulerian graph and $ P $ is a 2-transition system for $ G $, then $ (G,P) $ has a compaible decomposition.

Definition: Let $ G $ be an Eulerian graph and for every vertex $ v $, let $ P(v) $ be a partition of the edges incident with $ v $. We call $ P $ a transition system. If every member of $ P(v) $ has size at most $ k $ (for every $ v $), then we call $ P $ a $ k $-transition sytem. A compatible decomposition of $ (G,P) $ is a list of edge-disjoint cycles $ C_1,\ldots,C_n $ with union $ G $ so that every $ C_i $ contains at most one edge from every member of $ P(v) $.

Let $ G $ be a graph and let $ G' $ be the graph obtained from $ G $ by replacing each edge $ e $ of G by two edges $ e',e'' $ in parallel. Let $ P $ be the 2-transition system of $ G $ with $ {e',e''} \in P(v) $ whenever $ e' $ and $ e'' $ are incident with $ v $. Now, $ G' $ is an Eulerian graph and every compatible decomposition of $ (G',P) $ gives a cycle double cover of $ G $. Since the cycle double cover conjecture can be reduced to graphs which are 3-edge-connected, the above conjecture would imply the cycle double cover conjecture.

We define a transition system $ P $ to be admissable if every member of $ P(v) $ contains no more than half of the edges in any edge-cut. It is easy to see that if there is a compatible decomposition of $ (G,P) $, then $ P $ must be admissable. The converse of this is not true; There is an admissable 2-transition system of the graph $ K_5 $ which does not admit a compatible decomposition. Recently, G. Fan and C.Q. Zhang [FZ] have proved that $ (G,P) $ does have a compatible decomposition whenever $ P $ is admissable and $ G $ has no $ K_5 $ minor. This result imporoved upon an earlier theorem of Fleischner and Frank [FF]. Very recently, I have proved a weak version of the above conjecture, by showing that $ (G,P) $ also has a compatible decomposition when P is a 2-transition system and G is 80-edge-connected. I'd quite like to see an improvement on this bound. Here is a related conjecture.

Conjecture  (Sabidussi)   Let $ W $ be an Euler tour of the graph $ G $. If $ G $ has no vertex of degree two, then there is a cycle decomposition of $ G $, say $ F $, so that no two consecutive edges of $ W $ are in a common circuit of $ F $.

If $ W $ is given by $ v_1,e_1,v_2,e_2,...,e_{m-1},v_m $ then we may form a 2-transition system $ P $ by putting $ \{e_{i-1},e_i\} $ in $ P(v_i) $ for every $ i $ (working modulo $ m $). Now a compatible decomposition of $ (G,P) $ is precisely a cycle decomposition of $ G $ satisfying the above conjecture. Thus, Sabidussi's conjecture is equivalent to the assertion that $ (G,P) $ has a compatible decomposition whenever $ G $ has no vertex of degree two and $ P $ is a 2-transition system which comes from an Euler tour.

Let $ G $ be a directed Eulerian graph and for every vertex $ v $, let $ P(v) $ be a partition of the edges incident with $ v $ into pairs so that every in-edge is paired with an out-edge. We define a compatible decomposition to be a decomposition of $ G $ into directed circuits so that every directed circuit contains at most one edge from every member of $ P(v) $. Our current techniques don't seem to shed any light on the problem of finding compatible decompositions for Eulerian digraphs. Next I pose a very basic question which is still open.

Problem  (DeVos)   Does there exist a fixed integer $ k $ such that $ (G,P) $ has a compatible decomposition whenever $ G $ is a $ k $-edge-connected directed Eulerian graph and $ P $ is a 2-transition system?
add new comment

Reply

Comments are limited to a maximum of 1000 characters.
More information about formatting options