Edge omission in cubic graph
Task number: 4047
Let \( G \) be a cubic graph without bridges. Prove it for each its edge there is a perfect matching that avoids it.
Remove the edge from the graph and check Tutte’s condition, e.g. by a contradiction.
Note that at least 3 edges lead from the odd components.
Let \( G \) be a cubic graph without bridges. Cubic graphs satisfy Tutte’s condition, that is, for every \( S \subseteq V (G) \) holds that \( | S | \ge C_o (G, S) \), where \( C_o (G, S) \) stands for the number of odd components of the graph \( G-S \).
When we remove the edge \( e \), we assume for a contradiction that \( G-e \) does not have a perfect matching and therefore Tutte’s condition does not apply to it and there is a set \( S \subseteq V (G) \) such that \( | S | < C_o (G-e, S) \)
If \( e \) had both ends in \( S \) or both ends in the same component, then \( C_o (G, S) = C_o (Ge, S) \) and we get the contradiction immediately. So necessarily \( e \) connects two odd components of \( G-e \). They become one even by returning \( e \) in \( G \). Thus \( C_o (G-e, S) = C_o (G, S) +2 \).
These two components are not only connected to each other by the egde \( e \), but are also connected to the set \( S \). Because in the graph \( G-e \) these are odd components, an odd number of edges should lead into them according to the parity principle. Since the \( G \) graph has no bridges, there are at least three edges connecting each odd component (in both \( G \) and \( G-e \)). Similarly, at least two edges lead to even components. Thus, at least four edges connect the merged component in \( G \). In total of at least \( 3 C_o (G, S) +4 \) edges lead to \( S \).
Conversely, in \( G-e \), it holds that \( | S | < C_o (G-e, S) \). Both sides are integers, so \( 3 | S | < 3 C_o (G-e, S) -2 \).
Because \( G \) is cubic, vertices of \( S \) are incident with at most \( 3 | S | \) edges. Therefore, for the number of edges \( \#e \) incident with \( S \) in the graph \(G\) the following inequality holds: \[ 3 C_o (G, S) +4 \le \#e \le 3 | S | <3 C_o (G-e, S) -2 = 3 (C_o (G, S) +2) -2 = 3 C_o (G, S) +4, \] which is the desired contradiction.