## 4 Identifying Subproblem Solutions and Defining the Recursive Formulas
By Definition 7, \(f(C)\geq x\) for every \(C\in C_{T}(G_{\alpha})\), and \(f(C)\geq y\) for every \(C\in C(G_{\alpha}[TS(G_{\alpha})])\). Note that \(C_{T}(G_{\alpha})=C(G_{\alpha}[TS(G_{\alpha})])\). Therefore, \(f(C)\geq\max\{x,y\}\) for every \(C\in C_{T}(G_{\alpha})\) and \(C\in C(G_{\alpha}[TS(G_{\alpha})])\). The function \(f\) is a \(\Gamma_{p,1}^{d,\ell}(w,w)\)-clique transversal function of \(G_{\alpha}\). We have  
\[\tau_{p,1}^{d,\ell}(w,w)(G_{\alpha})\leq\tau_{p,1}^{d,\ell}(x,y)(G_{\alpha}).\]  
Conversely, let \(f\) be a \(\Gamma_{p,1}^{d,\ell}(w,w)\)-clique transversal function of \(G_{\alpha}\). By Definition 7, we obtain that \(f(C)\geq w\geq x\) for every \(C\in C_{T}(G_{\alpha})\), and \(f(C)\geq w\geq y\) for every \(C\in C(G_{\alpha}[TS(G_{\alpha})])\). The function \(f\) is a \(\Gamma_{p,1}^{d,\ell}(x,y)\)-clique transversal function of \(G_{\alpha}\). Thus,  
\[\tau_{p,1}^{d,\ell}(x,y)(G_{\alpha})\leq\tau_{p,1}^{d,\ell}(w,w)(G_{\alpha}).\]  
Following the discussion above, we know that \(\tau_{p,1}^{d,\ell}(x,y)(G_{\alpha})=\tau_{p,1}^{d,\ell}(w,w)(G_{\alpha})\). Next, let us consider the equation for \(\tau_{p,1}^{d,\ell}(w,w)(G_{\alpha})\). Let \(w_{min}=p\cdot\omega(G)\), and \(w_{max}=(p+(\ell-1)\cdot d)\cdot\omega(G)\). Let \(w_{1}\) and \(w_{2}\) be integers such that \(w_{min}\leq w_{1},w_{2}\leq w_{max}\), and \(w_{1}+w_{2}=w\). Let \(f_{1}\) be \(\Gamma_{p,1}^{d,\ell}(w_{1},w_{1})\)-clique transversal function of \(G_{\alpha_{1}}\) and \(f_{2}\) be \(\Gamma_{p,1}^{d,\ell}(w_{2},w_{2})\)-clique transversal function of \(G_{\alpha_{2}}\). By Definition 7, \(f_{1}(C)\geq w_{1}\) for every \(C\in C_{T}(G_{\alpha_{1}})\) and \(C\in C(G_{\alpha_{1}}[TS(G_{\alpha_{1}})])\). Similarly, \(f_{2}(C)\geq w_{2}\) for every \(C\in C_{T}(G_{\alpha_{2}})\) and \(C\in C(G_{\alpha_{2}}[TS(G_{\alpha_{2}})])\). Let \(f\) be a function of \(G_{\alpha}\) such that \(f(v)=f_{1}(v)\) for every \(v\in V(G_{\alpha_{1}})\), and \(f(v)=f_{2}(v)\) for every \(v\in V(G_{\alpha_{2}})\). A maximal clique \(C\in C_{T}(G_{\alpha})\) is the union of a clique \(C_{1}\in C(G_{\alpha_{1}}[TS(G_{\alpha_{1}})])\) and a clique \(C_{2}\in C(G_{\alpha_{2}}[TS(G_{\alpha_{2}})])\). Hence, \(f(C)=f_{1}(C_{1})+f_{2}(C_{2})\geq w_{1}+w_{2}=w\). Note that \(C_{T}(G_{\alpha})=C(G_{\alpha}[TS(G_{\alpha})])\). We obtain \(f(C)=f_{1}(C_{1})+f_{2}(C_{2})\geq w_{1}+w_{2}=w\) for every \(C\in C(G_{\alpha}[TS(G_{\alpha})])\). The function \(f\) is a \(\Gamma_{p,1}^{d,\ell}(w,w)\)-clique transversal function of \(G_{\alpha}\). Consequently,  
\[\tau_{p,1}^{d,\ell}(w,w)(G_{\alpha})\leq\tau_{p,1}^{d,\ell}(w_{1},w_{1})(G_{ \alpha_{1}})+\tau_{p,1}^{d,\ell}(w_{2},w_{2})(G_{\alpha_{2}}).\]  
Conversely, let \(f\) be a \(\Gamma_{p,1}^{d,\ell}(w,w)\)-clique transversal function of \(G_{\alpha}\). Let \(f_{1}\) be a function of \(G_{\alpha_{1}}\) such that \(f_{1}(v)=f(v)\) for every \(v\in V(G_{\alpha_{1}})\), and let \(f_{2}\) be a function of \(G_{\alpha_{2}}\) such that \(f_{2}(v)=f(v)\) for every \(v\in V(G_{\alpha_{2}})\). Recall that \(C_{T}(G_{\alpha})=C(G_{\alpha}[TS(G_{\alpha})])=C_{X}(G_{\alpha})\). We know that \(C_{T}(G_{\alpha_{1}})=C(G_{\alpha_{1}}[TS(G_{\alpha_{1}})])\) and \(C_{T}(G_{\alpha_{2}})=C(G_{\alpha_{2}}[TS(G_{\alpha_{2}})])\). Furthermore, a maximal clique \(C\in C_{T}(G_{\alpha})\) is the union of a maximal clique \(C_{1}\) of \(G_{\alpha_{1}}[TS(G_{\alpha_{1}})]\) and a maximal clique \(C_{2}\) of \(G_{\alpha_{2}}[TS(G_{\alpha_{2}})]\). Thus, \(f_{1}(C_{1})+f_{2}(C_{2})=f(C_{1})+f(C_{2})=f(C)\geq w\). Let \(\hat{C}_{1}\in C(G_{\alpha_{1}}[TS(G_{\alpha_{1}})])\) and \(\hat{C}_{2}\in C(G_{\alpha_{2}}[TS(G_{\alpha_{2}})])\) such that \(f(\hat{C}_{1})\) and \(f(\hat{C}_{2})\) are the smallest values in \(\{f(C_{1})\mid C_{1}\in C(G_{\alpha_{1}}[TS(G_{\alpha_{1}})])\}\) and \(\{f(C_{2})\mid C_{2}\in C(G_{\alpha_{2}}[TS(G_{\alpha_{2}})])\}\), respectively. Then, the clique \(\hat{C}=\hat{C}_{1}\cup\hat{C}_{2}\) is a maximal clique in \(G_{\alpha}\). Let \(f(\hat{C}_{1})=x_{1}\) and \(f(C_{2})=x_{2}\).