Suppose we have a graph **G**. Say **B** a fundamental basis of the cycle space of **G**. Say *LP* a linear programming problem where there is a variable for each vertex of **G**, each variable can take value $\geq 0$, for each odd cycle of **B** we add to *LP* the constraint $x\_{a} + x\_{b} + x\_{c} + ... + x\_{i} \geq k$ where $x\_{a},x\_{b},x\_{c},...,x\_{i}$ are the verteces of the cycle and $k$ is the number of vertices of the cycle. The objective function of *LP* is $\min\sum\limits\_{i=1}^{n}{x\_{i}}$.


We say **S** an optimal solution of *LP*. Can we say that each vertex, whose variable takes a value $\gt 0$ in **S**, is a vertex of at least a minimum [odd cycle transversal](https://en.wikipedia.org/wiki/Odd_cycle_transversal) of **G**?



Odd cycle transversal
**No.** Let $G$ be the graph obtained by gluing a $3$-cycle $abc$ and a $5$-cycle $cdefg$ together at vertex $c$. Then $(x\_a, x\_b, x\_c, x\_d, x\_e, x\_f, x\_g)=(0,0,3,1,1,0,0)$ is an optimal solution of the LP. However, neither $d$ nor $e$ are contained in a minimum odd cycle transversal of $G$, since $\{c\}$ is the unique minimum odd cycle transversal of $G$. 


The answer is also **no** to the updated question in the comment below. Here is a counterexample for both versions. Let $C\_n$ be an odd cycle with $n \geq 5$. Fix $e=ab \in E(C\_n)$ and let $T=C\_n - e$. Let $G$ be the graph obtained from $C\_n$ by adding all edges $f$ such that $T \cup f$ contains an even cycle. Then, the fundamental basis of $G$ with respect to $T$ contains exactly one odd cycle, namely $C\_n$. Thus, $x=(\frac{1}{n}, \dots, \frac{1}{n})$ is an optimal solution to the revised LP (see comment below) and $x=(1, \dots, 1)$ is an optimal solution to the original LP. However, it is not true that every vertex of $G$ is in a minimum odd cycle transversal, because the only minimum odd cycle transversals of $G$ are $\{a\}$ and $\{b\}$. 


