Then by Example \[2K3\], we have $\tilde{\mathcal L}_{\varepsilon}(f)=2{\rm lk}(f)$. ]{}

\[K5K33\_2\] [ Let $G$ be $K_{5}$ or $K_{3,3}$ labeled and oriented as illustrated in Figure \[2K3K5K33\], and let $f$ be an embedding of $G$ in $S^3$. Let $\varepsilon$ be the isomorphism from $L(G)$ to ${\mathbb Z}$ defined by $\varepsilon(e_1,e_3)=1$ for $G=K_5$ and $\varepsilon(c_1, c_3)=1$ for $G=K_{3,3}$. Then it follows that $\tilde{\mathcal L}_{\varepsilon}(f)=\sum_{a\cap b=\emptyset}\varepsilon(a,b)\ell(f(a),f(b))$, where the value of $\varepsilon(a,b)$ for an arbitrary pair of edges is given in Example \[K5\] if $G=K_{5}$ and in Example \[K33\] if $G=K_{3,3}$. ]{}

\[K6gs\]

Consider $K_{6}$, labeled and oriented as in Figure \[K6Simon\], and let $f$ be an embedding of $K_{6}$ in $S^3$. For any pair of disjoint edges $a$ and $b$ in $K_6$, we define $\varepsilon(a,b)$ as follows:

$$\varepsilon(x_i,x_j) = 
    \begin{cases}
        3 &\text{if } x_i \text{ and } x_j \text { are anti-parallel in Figure~\ref{K6Simon}}
\\
        2 & \text{if } x_i \text{ and } x_j \text { are neither parallel nor anti-parallel in Figure~\ref{K6Simon}}
        \end{cases}$$

$$\varepsilon(y_i,y_j) = 
    \begin{cases}
        0 &\text{if } y_i \text{ and } y_j \text { are anti-parallel in Figure~\ref{K6Simon}}
\\
        -1 & \text{if } y_i \text{ and } y_j \text { are neither parallel nor anti-parallel in Figure~\ref{K6Simon}}
        \end{cases}$$

$$\varepsilon(x_i,z_j) = 
    