## 3 Helix Polygon
### Necessity of \(n-2\) Natural Vertex Guards for Helix
We have \(F(x)=g_{1}(x).(T_{1}(x)+T_{2}(x)+\ldots+T_{l}(x))=True\). Thus, at least one of the expressions \(T_{i}\)s is True. Without loss of generality, it can be named \(T_{1}\) and is written as \(T_{1}=g_{i_{1}}.g_{i_{2}}\ldots g_{i_{m}}\) where \(g_{i_{j}}\)s are some natural guards in \(G\) and \(i_{j}\
eq 1,2\). Since \(T_{1}(x)=True\), we have  
\[F(x)=T_{2}(x) \tag{10}\]0  
Regarding the structure of \(H_{n}\), \(i_{j}\)s are even, because \(\forall i\geq 2:g_{2i}(x)=True\). Now, let \(y\in R_{3}\) be an arbitrary point. Obviously, \(\forall i\geq 2\), \(g_{2i}(y)=True\) which implies \(T_{1}(y)=True\). Then, \(F(y)=True\). However, \(y\
ot\in H_{n}\), which denotes a contradiction. In addition, \(F\) can be manifested as below:  
\[F(x)=T_{2}(x) \tag{10}\]1  
Now note that for all points which are located in the interior (or exterior) of \(H_{n}\), the above formula has the same value with the formula \(g_{1}.(g_{2}+T_{2})\). This fact can be shown easily by considering all cases. Then, \(F=g_{1}.(g_{2}+F_{1})\). Lemma 4: _It is not possible to define \(H_{5}\) with less than 3 natural vertex guards. The formula is \(F=g_{1}. (g_{2}+g_{5})\)._  
Proof. : Regarding Lemmas 2 and 3, \(F\) can be written as \(g_{1}.(g_{2}+F_{2})\). The edges \(v_{4}v_{5}\) and \(v_{3}v_{5}\) should have at least one guard on their endpoints. The optimal possibility is to locate a guard on \(v_{5}\) as their intersection point. Clearly, \(g_{1}. (g_{2}.g_{5})\) defines \(H_{5}\). \[F(x)=T_{2}(x) \tag{10}\]2 Regions \(R_{1}\) and \(R_{3}\) can be distinguishable without the existence of \(g_{2}\) in the formula. **Lemma 5**. : _Every arbitrary set of natural vertex guards \(G\) which defines \(H_{n}\) contains \(g_{3}\) which is a natural guard on \(v_{3}\). The final formula is in the form of \(F=g_{1}. (g_{2}+g_{3}.F_{3})\) where \(F_{3}\) is a Boolean expression of \(G-\{g_{1},g_{2},g_{3}\}\)._  
Proof. : Let \(G\) be an arbitrary set of natural guards which defines \(H_{n}\) by Boolean formula \(F\) for \(n\geq 6\). Suppose a contradiction in which \(g_{3}\) does not belong to \(G\). By Lemma 3, \(F=g_{1}.(g_{2}+F_{2})\). Assume that \(F_{2}=T_{1}+T_{2}+\ldots+T_{l}\), where \(T_{l}\)s are multiplication of natural guards in \(G\). Consider two regions, \(R_{3}\) and \(R_{4}\), as shown in \[F(x)=T_{2}(x) \tag{10}\]3 Let \(x\in R_{4}\) be an arbitrary point. We have \(F(x)=F_{2}(x)=True\). So, at least one of \(T_{l}\)s is True. Without loss of generality, we call it \(T_{1}\), which can be expressed as follows:  
\[F(x)=T_{2}(x) \tag{10}\]4  
where \(g_{i_{j}}\)s are natural guards in \(G\) and \(i_{j}\
eq 1,2,3\). Since \(T_{1}(x)=True\), we have  
\[F(x)=T_{2}(x) \tag{10}\]5  
Therefore, \(i_{j}\)s are even. Now, let \(y\in R_{3}\) be an arbitrary point. This point casts light that for all \(i\geq 2\), \(g_{2i}(y)=True\), which implies \(T_{1}(y)=True\). Then, \(F(y)=True\). However, \(y\
otin H_{n}\), showing a contradiction. So, \(g_{3}\in G\). In addition, \(F\) can be written as below:  
\[F(x)=T_{2}(x) \tag{10}\]6  
It can be easily obtained from equivalency of \(g_{1}. (g_{2}+F_{2})\) and \(g_{1}. (g_{2}+g_{3}.F_{3})\) for all points with respect to \(H_{n}\). **Lemma 6**. : _It is not possible to define \(H_{5}\) with fewer than three natural vertex guards. The formula is \(F=g_{1}. (g_{2}+g_{3}.g_{6})\)._  
Proof. : Considering Lemma 5, \(H_{6}\) can be defined by \(F=g_{1}.(g_{2}+g_{3}.F_{3})\). Since \(v_{4}v_{6}\) and \(v_{6}v_{5}\) are two edges of \(H_{6}\), it is required to place at least one guard on one of the endpoints of these two edges. The optimal placement is to place a guard on \(v_{6}\). Obviously, \(H_{6}\) can be defined by \(F=g_{1}.(g_{2}+g_{3}.g_{6})\). \[F(x)=T_{2}(x) \tag{10}\]7 Regions \(R_{3}\) and \(R_{4}\) can be distinguishable without the existence of \(g_{3}\) in the formula. **Lemma 7**. : _Every arbitrary set of natural vertex guards \(G\) which defines \(H_{n}\), for all \(n\geq 7\), contains \(g_{4}\), a natural guard on \(v_{4}\).