# Optimal Constructions of Low-Hit Zone Frequency-Hopping Sequence Set Based on _m_-Sequence
## 2 Preliminaries
### \(q\)-Ary \(m\)-Sequences and Its Decimated Sequence
\begin{split} H(\mathbf{s}_{\mathbf{c}_{\alpha}},\mathbf{s}_{ \mathbf{c}_{\beta^{\prime}}}\tau)&=\sum_{t=0}^{(q^{n}-1)/d-1}h[( \mathbf{S}_{\boldsymbol{\pi}}(t),\mathbf{S}_{\boldsymbol{\pi}}(t+1),..., \mathbf{S}_{\boldsymbol{\pi}}(t+k-1)),(0,0,...,0)]\\ &=N(\mathbf{S}_{\boldsymbol{\pi}},(0,0,...,0))\end{split}  \tag{37}\]  
The following cases exist here. Case 1: \(\alpha=\beta\). According to (36) and Lemma 3, \(\mathbf{S}_{\boldsymbol{\pi}}\) is a \(d\)-decimated sequence of \(L^{(\tau d)}(e)-e\) which is still a \(q\)-ary \(m\)-sequence of degree \(n\) for any \(\tau\
eq 0\). It follows that \(H(\mathbf{s}_{\mathbf{c}_{\alpha}},\mathbf{s}_{\mathbf{c}_{\beta}},\tau)= \frac{q^{n-k}-1}{d}\) from Lemma 4. Thus, the MPHAC of \(\mathbf{S}_{\mathbf{c}}\) is  
\[H_{d}(\mathbf{S}_{\mathbf{c}})=\frac{q^{n-k}-1}{d}. \tag{38}\]  
Case 2: \(\alpha\
eq\beta\). Case 2. 1: \(\alpha^{\prime}=\beta^{\prime}\). \(\alpha_{0}\
eq\beta_{0}\) because of \(\alpha\
eq\beta\) and \(\alpha^{\prime}=\beta^{\prime}\). Therefore, \(\mathbf{S}_{\boldsymbol{\pi}}\) is still a \(d\)-decimated sequence of \(L^{((\tau+(\alpha_{0}-\beta_{0})(L_{z}+1))d)}(e)-e\). When \(\tau=(\beta_{0}-\alpha_{0})(L_{z}+1)\), \(\mathbf{S}_{\boldsymbol{\pi}}\) is a sequence in which all elements are zero. Thus, \(H(\mathbf{s}_{\mathbf{c}_{\alpha}},\mathbf{s}_{\mathbf{c}_{\beta^{\prime}}}, \tau)=\frac{q^{n}-1}{d}\). When \(\tau\
eq(\beta_{0}-\alpha_{0})(L_{z}+1)\), \(\mathbf{S}_{\boldsymbol{\pi}}\) is still a \(q\)-ary \(m\)-sequence of degree \(n\) for any \(\tau\) from Lemma 3. In this case, \(H(\mathbf{s}_{\mathbf{c}_{\alpha}},\mathbf{s}_{\mathbf{c}_{\beta^{\prime}}}, \tau)=\frac{q^{n-k}-1}{d}\). Case 2. 2: \(\alpha^{\prime}\
eq\beta^{\prime}\). \(\mathbf{S}_{\boldsymbol{\pi}}\) is a \(d\)-decimated sequence of \(L^{(\alpha^{\prime}-\beta^{\prime}+(\tau+(\alpha_{0}-\beta_{0})(L_{z}+1))d)}(e)-e\) which is still a \(q\)-ary \(m\)-sequence of degree \(n\) for any \(\tau\) according to Lemma 3. From Lemma 4, we have \(H(\mathbf{s}_{\mathbf{c}_{\alpha}},\mathbf{s}_{\mathbf{c}_{\beta^{\prime}}} \tau)=\frac{q^{n-k}-1}{d}\). Thus, the MPHCC of \(\mathbf{S}_{\mathbf{c}}\) is equal to  
\[H_{c}(\mathbf{S}_{\mathbf{c}})=\frac{q^{n-k}-1}{d} \tag{39}\]  
for \(\tau\leq L_{z}\). To sum up, \(\mathbf{S}_{\mathbf{c}}\) is an LHZ FHS set, and the MPHC is  
\[H(\mathbf{S}_{\mathbf{c}})=\frac{q^{n-k}-1}{d} \tag{40}\]  
for \(\tau\leq L_{z}\). Then, input all the parameters of \(\mathbf{S_{c}}\) into Peng-Fan bound (4), and we have  
\[\lambda_{right} =\left[\frac{\left\lfloor\frac{q^{n}-1}{d(L_{z}+1)}\right\rfloor d( L_{z}+1)-q^{k}}{\left\lfloor\frac{q^{n}-1}{d(L_{z}+1)}\right\rfloor d(L_{z}+1)-1} \cdot\frac{q^{n}-1}{dq^{k}}\right] \tag{41}\] \[=\left[\frac{q^{n}-1}{dq^{k}}-\frac{q^{k}-1}{\frac{q^{q}-1}{d(L_{z }+1)}\rfloor d(L_{z}+1)-1}\cdot\frac{q^{n}-1}{dq^{k}}\right]\] \[=\left[\frac{q^{n-k}-1}{d}-\frac{(q^{k}-1)(q^{n}-\left\lfloor\frac {q^{n}-1}{d(L_{z}+1)}\right\rfloor d(L_{z}+1))}{\left(\left\lfloor\frac{q^{n} -1}{d(L_{z}+1)}\right\rfloor d(L_{z}+1)-1\right)dq^{k}}\right].\]  
It is easy to check that \(\frac{(q^{k}-1)(q^{n}-\left\lfloor\frac{q^{n}-1}{d(L_{z}+1)}\right\rfloor d( L_{z}+1))}{\left(\left\lfloor\frac{q^{n}-1}{d(L_{z}+1)}\right\rfloor d(L_{z}+1) -1\right)dq^{k}}>0\). And let \(q^{n}-1=\left\lfloor\frac{q^{n}-1}{d(L_{z}+1)}\right\rfloor d(L_{z}+1)+R\), \(0\leq R\leq L_{z}\leq\frac{q^{n}-1}{2d}-1\). We have  
\[q^{n} -\left\lfloor\frac{q^{n}-1}{d(L_{z}+1)}\right\rfloor d(L_{z}+1) -(\left\lfloor\frac{q^{n}-1}{d(L_{z}+1)}\right\rfloor d(L_{z}+1)-1)d \tag{42}\] \[=q^{n}+d-(d+1)\left\lfloor\frac{q^{n}-1}{d(L_{z}+1)}\right\rfloor d (L_{z}+1)\] \[=q^{n}+d-(d+1)(q^{n}-1-R)\] \[\leq q^{n}+d-(d+1)\frac{(2d-1)q^{n}+1}{2d}\] \[H_{c}(\mathbf{S}_{\mathbf{c}})=\frac{q^{n-k}-1}{d} \tag{39}\]0  
Therefore, \(\frac{(q^{k}-1)(q^{n}-\left\lfloor\frac{q^{n}-1}{d(L_{z}+1)}\right\rfloor d( L_{z}+1))}{\left(\left\lfloor\frac{q^{n}-1}{d(L_{z}+1)}\right\rfloor d(L_{z}+1)- 1\right)dq^{k}}<1\).