in every case, (a) and (b) can not both happen. In a similar way we show that (a) and (c) can not both happen. However, cases (b) and (c) can both happen.

On the other hand, any $6$-cycle contains exactly three pairs of diametrically opposite vertices and every $4$-cycle $u_1,u_2,u_3,u_4,u_1$ has exactly $\deg(u_1) + \deg(u_2)+\deg(u_3)+\deg(u_4) - 8$ exiting edges. Therefore, $$p_3(G) - f(G) - 3|C_6(G)|$$ represents the number of unordered pairs of distinct vertices $u,v$ for which there exists a path of length three between them.

However, it can happen that $d(u,v) < 3$. In such a case we consider two possibilities:

-   $d(u,v)=1$:\
    In this case, $u$ and $v$ lie on a $4$-cycle $C=x,y,v,u,x$, see Figure \[siti\]. Obviously, it holds $d_C(u,v)=1$ and $C$ is the only $4$-cycle with such property (otherwise we obtain $4$-cycles with two common edges or a 4-cycle and a $6$-cycle with three common edges).

    ![\[siti\] $4$-cycle in case $(i)$.](siti.eps)

-   $d(u,v)=2$:\
    In this case, $u$ and $v$ lie on a $5$-cycle $C=x,y,v,a,u,x$ such that $a \in V(G) \setminus \lbrace x,y,u,v \rbrace$, see Figure \[sitii\]. Obviously, it holds $d_C(u,v)=2$ and $C$ it the only $5$-cycle with such property (otherwise we obtain 5-cycles with three common edges 