## 3 Helix Polygon
### Necessity of \(n-2\) Natural Vertex Guards for Helix  
In this section, we will prove that it is impossible to define a Helix polygon with fewer than \(n-2\) natural vertex guards. Lemma 2: Every arbitrary set of natural vertex guards \(G\) which defines \(H_{n}\) contains \(g_{1}\), a natural vertex guard on \(v_{1}\). The final formula is in the form of \(F=F_{1}.g_{1}\), where \(F_{1}\) is a Boolean expression of \(G-\{g_{1}\}\). Proof: Let \(G\) be an arbitrary set of natural guards which defines \(H_{n}\) by Boolean formula \(F\). Suppose for a contradiction that \(g_{1}\) does not belong to \(G\). Since \(v_{1}v_{2}\) and \(v_{1}v_{3}\) are edges of \(H_{n}\), \(G\) should contain two natural guards on \(v_{2}\) and \(v_{3}\) which are called \(g_{2}\) and \(g_{3}\), respectively. So, \(F\) can be written in the general form \(F=g_{2}.g_{3}.T_{1}+g_{2}.T_{2}+g_{3}.T_{3}\) where \(T_{i}\)are Boolean formulas which do not contain \(g_{1}\), \(g_{2}\) and \(g_{3}\). This will result in a contradiction, mentioned below. Consider two regions, \(R_{1}\) and \(R_{2}\), as shown in Figure 5. Let \(x\) be an arbitrary point inside \(R_{1}\) or \(R_{2}\). So, we have \(g_{2}(x)=True\) and \(g_{3}(x)=False\). So,  
\[F(x)=T_{2}(x) \tag{10}\]  
Furthermore, note that we can expand \(T_{2}\) in the following general form:  
\[T_{2}=T_{2}^{(1)}+T_{2}^{(2)}+\ldots+T_{2}^{(l)} \tag{11}\]  
where \(T_{2}^{(l)}\)'s are the multiplication of some natural guards in \(G\). Let \(x\in R_{1}\) be an arbitrary point. So, from Equation (10), it is implied that  
\[x\in R_{1}\implies T_{2}(x)=F(x)=True\]  
Therefore, at least one of the expressions of \(T_{2}\) should be True. Without loss of generality, it can be called \(T_{2}^{(1)}\) and is written as \(T_{2}^{(1)}=g_{i_{1}}\cdot g_{i_{2}}\cdots\cdot g_{in}\), where \(g_{i_{j}}\) is the natural guards of \(G\) and \(j=1,2,3\). Since \(T_{2}^{(1)}(x)=True\), we have  
\[\forall j:1\leq j\leq m:g_{i_{j}}(x)=True\]  
Regarding the structure of \(H_{in}\), none of indices \(i_{j}\) can be odd. This is because we know that  
\[\forall i\geq 1:g_{2i+1}(x)=False\]  
Now, let \(y\in R_{2}\) be an arbitrary point. Due to the structure of \(H_{in}\), it is inferred that for all \(\forall y\in R_{2}\), we have  
\[\forall i\geq 2:g_{2i}(y)=True.\]  
So, \(T_{2}^{(1)}(x)=True\) and from Equation (11), \(T_{2}(y)=True\) is obtained and consequently \(F(y)=T_{2}(y)=True\) (due to Equation (10). Nevertheless, \(y\
otin H_{in}\), which is a contradiction. With regard to the existence of \(g_{1}\) in \(G\), \(F\) can be written in the form of \(F=g_{1}.T_{1}+T_{2}\), where \(T_{2}\) does not contain \(g_{1}\). Indeed, \(g_{1}.T_{1}+g_{1}.T_{2}\) defines \(H_{in}\) as well. Let \(x\in H_{in}\), then \(g_{1}(x)=True\) and \(g_{1}(x).T_{1}(x)+g_{1}(x).T_{2}(x)=g_{1}(x).T_{1}(x)+T_{2}(x)=F(x)=True\). If \(y\
otin H_{in}\) and \(g_{1}(y)=True\), then \(g_{1}(y).T_{1}(y)+g_{1}(y).T_{2}(y)=g_{1}(y).T_{1}(y)+T_{2}(y)=F(y)=False\). On the other hand, if \(g_{1}(y)=False\), then \(g_{1}(y).T_{1}(y)+g_{1}(y).T_{2}(y)=False\). So, \(F=g_{1}. (T_{1}+T_{2})\) defines \(H_{in}\). In other words, \(F\) can be expressed as \(F=g_{1}.F_{1}\). Figure 5: Regions \(R_{1}\) and \(R_{2}\) can be distinguishable without the existence of \(g_{1}\) in the formula. **Lemma 3**. : _Every arbitrary set of natural vertex guards \(G\) defining \(H_{n}\) contains \(g_{2}\) (i.e., a natural guard on \(v_{2}\)). The final formula is in the form of \(F=g_{1}. (g_{2}+F_{2})\), where \(F_{2}\) is a Boolean expression of \(G-\{g_{1},g_{2}\}\)._  
Proof. : Let \(G\) be an arbitrary set of natural guards which defines \(H_{n}\) by Boolean formula \(F\), for \(n\geq 4\). Suppose a contradiction in which \(g_{2}\) does not belong to \(G\). From Lemma 2, we can write  
\[F=g_{1}.(T_{1}+T_{2}+\ldots+T_{l})\]  
where \(T_{i}\)s are Boolean expression of natural guards of \(G\). Consider two regions, \(R_{1}\) and \(R_{3}\), as shown in Figure 6. Let \(x\in R_{1}\) be an arbitrary point.