## 3 Helix Polygon
### Necessity of \(n-2\) Natural Vertex Guards for Helix
The final formula is in the form of \(F=g_{1}.(g_{2}+g_{3}. (g_{4}+F_{4}))\), where \(F_{4}\) is a Boolean expression of \(G-\{g_{1},g_{2},g_{3},g_{4}\}\)._  
Proof. : Let \(G\) be an arbitrary set of natural guards defining \(H_{n}\) by Boolean formula \(F\) and \(n\geq 7\). Suppose a contradiction in which \(g_{4}\) does not belong to \(G\). By Lemma 5, \(F=g_{1}.(g_{2}+g_{3}.F_{3})\). It is intended to show that \(F_{3}=g_{4}+F_{4}\). Assume that \(F_{3}=T_{1}+T_{2}+\ldots+T_{t}\), where \(T_{i}\)s are the multiplication of natural guards of \(G\). Consider two regions, \(R_{5}\) and \(R_{6}\), shown in \[F(x)=T_{2}(x) \tag{10}\]8 Let \(x\in R_{5}\) be an arbitrary point, then \(F(x)=F_{3}(x)=True\). So, at least one of \(T_{i}\)s is True. Without loss of generality, it is called \(T_{1}\) which can be expressed as below:  
\[F(x)=T_{2}(x) \tag{10}\]9  
where \(g_{i_{j}}\)s are natural guards and \(i_{j}\
eq 1,2,3,4\). Since \(T_{1}(x)=True\), for all \(j:1\leq j\leq m\), \(g_{i_{j}}(x)=True\). Therefore, \(i_{j}\)s are even. Now, let \(y\in R_{6}\) be an arbitrary point (see Figure 8). It is clear that for all \(i\geq 2\), \(g_{2i}(y)=True\), which implies \(T_{1}(y)=True\). Then, \(F(y)=True\). However, \(y\
otin H_{n}\), indicating a contradiction. So, \(g_{4}\in G\). In addition, \(F\) can be written as below:  
\[T_{2}=T_{2}^{(1)}+T_{2}^{(2)}+\ldots+T_{2}^{(l)} \tag{11}\]0  
It can be easily shown that \(g_{1}. (g_{2}+g_{3}.F_{3})\) is equivalent with \(g_{1}.(g_{2}+g_{3}. (g_{4}+F_{4}))\) for all the points inside or outside of \(H_{n}\). **Lemma 8**. : _Let \(H_{n}\) be defined by \(F=g_{1}.(g_{2}+g_{3}.(g_{4}+F_{4}))\). Formula \(F_{P}=g_{1}^{{}^{\prime}}. (g_{2}^{{}^{\prime}}+g_{3}^{{}^{\prime}}.F_{4}^{c})\) defines the pocket of \(H_{n}\), \(P(H_{n})\), where \(g_{i}^{{}^{\prime}}\)s are defined in Equation (7)._  
Proof. : Let \(x\in P(H_{n})\) and \(y\
otin P(H_{n})\) be two arbitrary points (see Figure 9). Then, it can be demonstrated that \(F_{P}(x)=True\) and \(F_{P}(y)=False\). Additionally, \(g_{2}^{{}^{\prime}}=G_{3}.g_{5}^{c}\) and \(g_{3}^{{}^{\prime}}=g_{4}^{c}\), so \(x\in P(H_{n})\implies g_{1}^{{}^{\prime}}(x)=True\) and \(G_{3}(x)=True\). So  
\[T_{2}=T_{2}^{(1)}+T_{2}^{(2)}+\ldots+T_{2}^{(l)} \tag{11}\]1  
On the other hand, \(x\in P(H_{n})\) implies that \(F(x)=False\) and \(g_{1}(x)=True\). Hence,  
\[T_{2}=T_{2}^{(1)}+T_{2}^{(2)}+\ldots+T_{2}^{(l)} \tag{11}\]2  
\[T_{2}=T_{2}^{(1)}+T_{2}^{(2)}+\ldots+T_{2}^{(l)} \tag{11}\]3 Regions \(R_{5}\) and \(R_{6}\) can be distinguishable without the existence of \(g_{4}\) in the formula. So,  
\[T_{2}=T_{2}^{(1)}+T_{2}^{(2)}+\ldots+T_{2}^{(l)} \tag{11}\]4  
From Equations (12) and (13), \(F_{P}(x)=\mathit{True}\) can be obtained. If \(g^{{}^{\prime}}(y)=\mathit{False}\), for \(y\
otin P(H_{n})\), \(F_{P}(y)=\mathit{False}\) and the process of our calculations has been completed. Assume that \(g^{{}^{\prime}}(y)=\mathit{True}\), then as \(y\
otin P(H_{n})\), consequently \(g_{2}^{{}^{\prime}}(y)=\mathit{False}\) (see Figure 9). If \(g_{3}^{{}^{\prime}}(y)=\mathit{False}\), \(F_{P}(y)=\mathit{False}\). Now, suppose that \(g_{3}^{{}^{\prime}}(y)=\mathit{True}\) (i.e., \(g_{4}(y)=\mathit{False}\)). This assumption implies that \(y\in H_{n}\) and we have \(g_{1}(y)=\mathit{True}\), \(g_{2}(y)=\mathit{False}\), \(g_{3}1(y)=\mathit{True}\) and \(g_{4}(y)=\mathit{False}\). Since \(y\in H_{n}\), \(F_{\left(y\right)}=F_{4}(y)\) and consequently, \(F_{4}(y)=\mathit{True}\). So, we have \(F_{P}(y)=\mathit{False}\). This means that \(F_{P}\) defines \(P(H_{n})\). **Theorem 1**. : \(H_{n}\) _requires at least \(n-2\) natural vertex guards._  
Proof. : We prove this theorem by induction. It is clear that for \(n=4\), \(H_{4}\) is a tetragon and cannot be defined by fewer than two guards. We proved this in corollaries 4 and 6 for \(n-5\). Now, assume that this holds for \(n-1\), and we have to prove it for \(n\) where \(n\geq 7\). Let \(F\) be a Boolean formula which defines \(H_{n}\). With regard to Lemma 7, \(F=g_{1}.(g_{2}+(g_{3}.(g_{4}+F_{4}))\).