## 4 Network Having Minimum \(\xi_{i}\)-Value, \(i=1,2\) w.r.t \(\mathcal{B}_{n}^{d}\)
Using this fact, one has \(\xi_{1}(\aleph)\geqslant\xi_{1}(K_{n-t,t})>\xi_{1}(K_{n-t,t})=4n\) and \(\xi_{2}(\aleph)\geqslant\xi_{2}(K_{n-t,t})>\xi_{2}(K_{n-t,t})=4t(n-t)\), where \(e\) is any link in \(K_{n-t,t}\). Thus, we get our desired result. Theorem 3: _Assume that \(\aleph\) belongs to \(\mathscr{B}_{n}^{d}\) with the minimum \(\xi_{1}\)-value. If \(d\geqslant 3\), then \(\aleph\cong\aleph(n,d)\) for odd \(d\), where \(\aleph(n,d)\) is already defined._  
**Proof. ** We opt \(\aleph\in\mathscr{B}_{n}^{d}\) such that its \(\xi_{1}\)-value is as small as possible. Let \(v_{0}v_{1}\ldots v_{d}\) is the diametrical path. Thereby, we partition \(V_{\aleph}\) as \(V_{0}\cup V_{1}\cup\cdots\cup V_{d}\). To complete the proof, we need the following claim. **Claim 1. ** For odd \(d\), one has  
\[|V_{0}|=|V_{1}|=\cdots=|V_{\frac{d+1}{2}-1}|=|V_{\frac{d+1}{2}+1}|=\cdots=|V_ {d-1}|=|V_{d}|=1,\big{|}|V_{\frac{d-1}{2}}|-|V_{\frac{d+1}{2}}|\big{|}\leq 1. \tag{5}\]  
**Proof of Claim 1. ** Note that \(|V_{0}|=\{v_{0}\}\) and \(|V_{d}|=\{v_{d}\}\). Here, we only need to prove that \(|V_{1}|=1\) holds. In the same way, one can show that \(|V_{2}|=\cdots=|V_{\frac{d+1}{2}-1}|=|V_{\frac{d+1}{2}+1}|=\cdots=|V_{d-1}|=1\), we omit the procedure here. Since, for \(d=3\), the desired result is trivial. In what follows we choose the case \(d\geqslant 5\), for odd \(d\). In the case \(|V_{1}|\geqslant 2\), then we opt any \(u\in V_{1}\) and let \(\aleph^{\prime}=\aleph-uv_{0}+\{ux:x\in V_{4}\}\). Here, \(\{v_{0}\}\cup(V_{1}\setminus\{u\})\cup V_{2}\cup(V_{3}\cup\{u\})\cup V_{4} \cup\ldots\cup\{v_{d}\}\) is the node partition of \(\aleph^{\prime}\); the choice of \(\aleph\) as well as in view of Lemma 4 i.e, for two of the neighbour blocks in \(V_{\aleph^{\prime}}\) induces the complete bipartite subnetwork and \(|V_{d}|=1\) for \(d\geqslant 5\). By considering the construction of \(\aleph\) and \(\aleph^{\prime}\), it is easy to verify that \(\varepsilon(u)\geqslant\varepsilon^{\prime}(u)+1\), \(\varepsilon(x)=\varepsilon^{\prime}(x)\) for every \(x\in V_{\aleph}\setminus\{u\}\). This gives  
\[\xi_{1}(\aleph)-\xi_{1}(\aleph^{\prime}) =\sum_{x\in V(\aleph)}\varepsilon^{2}(x)+\varepsilon^{2}(u)-\sum_{ x\in V(\aleph^{\prime})}\varepsilon^{\prime 2}(x)-\varepsilon^{\prime 2}(u)\] \[=\varepsilon^{2}(u)-\varepsilon^{\prime 2}(u)\] \[\geqslant\varepsilon^{2}(u)-(\varepsilon(u)-1)^{2}\] \[=2\varepsilon(u)-1\] \[= 4t+4(n-t)\]0  
The last inequality (6), follows by \(d\geqslant 5\) and \(\varepsilon(u)\geqslant 4\). _i.e._\(\xi_{1}(\aleph^{\prime})<\xi_{1}(\aleph)\), which contradicts our selection of \(\aleph\). Hence, \(|V_{1}|=1\). In a similar manner one can also prove that \(|V_{2}|=\cdots=|V_{\frac{d-1}{2}-1}|=|V_{\frac{d+1}{2}+1}|=\cdots=|V_{d-1}|=1\). Next we show that if \(d\) is odd, then \(||V_{\frac{d-1}{2}}|-|V_{\frac{d+1}{2}}||\leq 1\). Without loss of generality, we assume that \(|V_{\frac{d-1}{2}}|\geqslant|V_{\frac{d+1}{2}}|\). Then it suffices to show that \(|V_{\frac{d-1}{2}}|-|V_{\frac{d+1}{2}}|\leq 1\). If this is not true, then \(|V_{\frac{d-1}{2}}|-|V_{\frac{d+1}{2}}|\geqslant 2\). Choose \(w\in V_{\frac{d-1}{2}}\), let  
\(\aleph^{*}=\aleph-\{wx:x\in V_{\frac{d-2}{2}}\cup V_{\frac{d+1}{2}}\}+\{wy:y\in( V_{\frac{d-1}{2}}\setminus\{w\})\cup V_{\frac{d+2}{2}}\}\). Then the node partition of \(\aleph^{*}\) is \(\{v_{0}\}\cup V_{1}\cup V_{2}\cup\ldots\cup V_{\frac{d-2}{2}}\cup(V_{\frac{d-1 }{2}}\setminus\{w\})\cup(V_{\frac{d+1}{2}}\cup\{w\})\cup V_{\frac{d+1}{2}} \cup\ldots\cup\{v_{d}\}\) and every two adjacent blocks in \(V_{\aleph^{*}}\) induce a complete bipartite network. Based on the constructions of \(\aleph\) and \(\aleph^{*}\), it is straightforward to see that \(\varepsilon^{2}(v)={\varepsilon^{*}}^{2}(v)\) for every \(v\in V_{\aleph}\). Thus  
\[= 4t+4(n-t)\]1 \[= 4t+4(n-t)\]2  
This shows a class of networks such that \(a+b=n-(d-1)\) where \(a=|V_{\frac{d-1}{2}}|\), \(b=|V_{\frac{d+1}{2}}|\), which contradicts the option of \(\aleph\). Hence, this completes the proof of Claim 1.