We start by the following identity: \(\mathbf{1} _ { k = 0 } \equiv 2 ^ k \pmod{ 2 }\).

Let \(V _ 1, V _ 2, E _ 1, E _ 2\) denote the golden vertexes, silver vertexes, golden edges and silver edges, respectively. \(V _ 1 \sqcup V _ 2 = V, E _ 1 \sqcup E _ 2 = E\).

So the answer is

\[\begin{aligned} f &= \sum \limits _ { V _ 1 \sqcup V _ 2 = V, E _ 1 \sqcup E _ 2 = E } \mathbf{1} _ { \{ u: u \in V _ 2, \forall e \in N( u ), e \in E _ 1 \} = \varnothing } \mathbf{1} _ { \{ e = ( u, v ): e \in E _ 1, u, v \in V _ 2 \} = \varnothing } \\ &\equiv \sum \limits _ { V _ 1 \sqcup V _ 2 = V, E _ 1 \sqcup E _ 2 = E } \sum \limits _ { V _ 3 \subset V _ 2 } \mathbf{1} _ { \forall u \in V _ 3, \forall e \in N( u ), e \in E _ 1 \ } \mathbf{1} _ { \{ e = ( u, v ): e \in E _ 1, u, v \in V _ 2 \} = \varnothing } \pmod{ 2 } \end{aligned} (\star)\]

Consider when we fixed \(V _ 1 \sqcup V _ 2 = V, V _ 3 \subset V _ 2\), the value of \(g = \sum \limits _ { E _ 1 \sqcup E _ 2 = E } \mathbf{1} _ { \forall u \in V _ 3, \forall e \in N( u ), e \in E _ 1 \ } \mathbf{1} _ { \{ e = ( u, v ): e \in E _ 1, u, v \in V _ 2 \} = \varnothing }\).

• \(\forall e \in E[ V _ 1 ] \cup E[ V _ 1, V _ 2 \setminus V _ 3 ]\), \(e\) can in either \(E _ 1\) or \(E _ 2\), both situations affect nothing about others.
• \(\forall e \in E[ V _ 3 ] \cup E[ V _ 3, V _ 2 \setminus V _ 3 ]\), \(e\) can in neither \(E _ 1\) nor \(E _ 2\).
• \(\forall e \in E[ V _ 2 \setminus V _ 3 ]\), \(e\) must in \(E _ 2\).
• \(\forall e \in E[ V _ 1, V _ 3 ]\), \(e\) must in \(E _ 1\).

So \(g = 2 ^ { \# E[ V _ 1 ] + \# E[ V _ 1, V _ 2 \setminus V _ 3 ] } \cdot 0 ^ { \# E[ V _ 3 ] + \# E[ V _ 3, V _ 2 \setminus V _ 3 ] } \equiv \mathbf{1} _ { \# E[ V _ 1 ] = \# E[ V _ 1, V _ 2 \setminus V _ 3 ] = \# E[ V _ 3 ] = \# E[ V _ 3, V _ 2 \setminus V _ 3 ] = 0 } \pmod{ 2 }\).

Bring into \((\star)\), we get \(f \equiv \sum \limits _ { V _ 1 \sqcup V _ 3 \sqcup V _ 4 = V } \mathbf{1} _ { \# E[ V _ 1 ] = \# E[ V _ 1, V _ 4 ] = \# E[ V _ 3 ] = \# E[ V _ 3, V _ 4 ] = 0 } \pmod{ 2 }\), where \(V _ 4\) denote \(V _ 2 \setminus V _ 3\).

Let \(F = \{ ( V _ 1, V _ 3, V _ 4 ): V _ 1 \sqcup V _ 3 \sqcup V _ 4 = V, \# E[ V _ 1 ] = \# E[ V _ 1, V _ 4 ] = \# E[ V _ 3 ] = \# E[ V _ 3, V _ 4 ] = 0 \}\). Then \(f \equiv |F| \pmod{ 2 }\).

Consider the mapping \(\varphi: F \to F\) defined by \(\varphi( V _ 1, V _ 3, V _ 4 ) = ( V _ 3, V _ 1, V _ 4 )\). Easy to notice:

• \(\forall ( V _ 1, V _ 3, V _ 4 ) \in F\), \(\varphi( \varphi( V _ 1, V _ 3, V _ 4 ) ) = ( V _ 1, V _ 3, V _ 4 )\).
• \(\forall ( V _ 1, V _ 3, V _ 4 ) \in F\), \(\varphi( V _ 1, V _ 3, V _ 4 ) = ( V _ 1, V _ 3, V _ 4 )\) iff \(V _ 1 = V _ 3\) iff \(V _ 1 = V _ 3 = \varnothing, V _ 4 = V\) is uniquely determined.

So that \(|F| \equiv 1 \pmod{ 2 }\). Which leads to the final answer \(f \equiv 1 \pmod{ 2 }\).