A comb can be defined by a handle $H$ and a number of teeths $T_1,T_2,\dots,T_t$ such that:
- $H,T_1,T_2,\dots,T_t \subseteq V$
- $T_j \setminus H \neq \emptyset$ $\,\,\, \forall 1 \leq j \leq t$
- $T_j \cap H \neq \emptyset$ $\,\,\, \forall 1 \leq j \leq t$
- $T_i \cap T_j = \emptyset$ $\,\,\, \forall 1 \leq i < j \leq t$
- $t \geq 3$ and odd
A comb inequality states that:
$x(E(H)) + \sum_{j=1}^{t} x(E(T_j)) \leq |H| + \sum_{j=1}^{t} |T_j| - \frac{3t+1}{2}$
Comb inequalities are valid inequalities for the TSP.
I'm looking for a simple explanation of why these inequalities are valid for any hamiltonian cycle. I'm not looking for a formal proof. A simple, informal and intuitive explanation will work.
Since comb inequalities generalize 2-matching inequalities, perhaps this can be better explained by showing the special case and then extending it to the general case.