## 4 Identifying Subproblem Solutions and Defining the Recursive Formulas  
This section explores the methodology for addressing the \((p,d,\ell,b)\)-clique transversal problem by decomposing it into smaller, more manageable subproblems using the recursive definition of distance-hereditary graphs. This section demonstrates how the optimal solution can be systematically constructed by combining the solutions to these subproblems. It also introduces the recursive formulas that forms the foundation of this approach, guiding the reader through the step-by-step process of building the complete solution. In this case, \(C_{T}(G)\
eq\emptyset\) and \(C_{\overline{T}}(G)\
eq\emptyset\). By Definition 7, a \(\Gamma_{p,2}^{d,\ell}(b,b,z)\)-clique transversal function of \(G\) is a \((p,d,\ell,b)\)-clique transversal function of \(G\). Thus,  
\[\tau_{C}(p,d,\ell,b)(G)\leq\min_{p\cdot\omega(G)\leq z\leq(p+(\ell-1)\cdot d) \cdot\omega(G)}\Big{\{}\tau_{p,2}^{d,\ell}(b,b,z)(G)\Big{\}}.\]  
Conversely, let \(C\in C(G[TS(G)])\) and \(f\) be a \((p,d,\ell,b)\)-clique transversal function of \(G\). Clearly, \(|C|\leq\omega(G)\) and \(p\cdot\omega(G)\leq f(C)\leq(p+(\ell-1)\cdot d)\cdot\omega(G)\). Let \(f(C)=z\). The function \(f\) is a \(\Gamma_{p,2}^{d,\ell}(b,b,z)\)-clique transversal function of \(G\). Hence,  
\[\min_{p\cdot\omega(G)\leq z\leq(p+(\ell-1)\cdot d)\cdot\omega(G)}\Big{\{}\tau_ {p,2}^{d,\ell}(b,b,z)(G)\Big{\}}\leq\tau_{C}(p,d,\ell,b)(G).\]  
Based on the above discussion, Statement (3) therefore holds. Lemma 6: _Suppose that \(G\) is a distance-hereditary graph and \(G_{\alpha}\) is a subgraph of \(G\) consisting of only one vertex \(v\). Let \(p\in\{0,-1\}\), and let \(d,\ell,x,y,z\) be integers such that \(d,\ell\in\mathbb{Z}^{+}\) are fixed, and \(p\cdot\omega(G)\leq x,y,z\leq(p+(\ell-1)\cdot d)\cdot\omega(G)\). Then, \(\tau_{p,0}^{d,\ell}(x,y)(G_{\alpha})\), \(\tau_{p,1}^{d,\ell}(x,y)(G_{\alpha})\), and \(\tau_{p,2}^{d,\ell}(x,y,z)(G_{\alpha})\) can be computed as follows:_  
1. \(\tau_{p,0}^{d,\ell}(x,y)(G_{\alpha})=\infty\)_;_
2. \(\tau_{p,2}^{d,\ell}(x,y,z)(G_{\alpha})=\infty\)_;_
3. _Let_ \(i=\lceil(\max\{x,y\}-p)/d\rceil\cdot d+p\)_. Then,_ \[\tau_{p,1}^{d,\ell}(x,y)(G_{\alpha})= \begin{cases}p&\text{if }\max\{x,y\}\leq p,\\ i&\text{if }p<\max\{x,y\}\leq p+(\ell-1)\cdot d,\\ \infty&\text{if }p+(\ell-1)\cdot d<\max\{x,y\}.\end{cases} \]  
Proof. : The graph \(G_{\alpha}\) consists of only one vertex \(v\). By Theorem 1, we have \(TS(G_{\alpha})=\{v\}\). Then, \(C_{\overline{T}}(G_{\alpha})=\emptyset\), and \(C(G_{\alpha})=C_{T}(G_{\alpha})=C(G_{\alpha}[TS(G_{\alpha})])=\{\{v\}\}\). By Definition 7, we obtain \(\tau_{p,0}^{d,\ell}(x,y)(G_{\alpha})=\infty\) and \(\tau_{p,2}^{d,\ell}(x,y,z)(G_{\alpha})=\infty\). Statements (1) and (2) therefore hold. We now consider Statement (3). Let \(f\) be a \(\Gamma_{p,1}^{d,\ell}(x,y)\)-clique transversal function \(f\) of \(G_{\alpha}\). Clearly, \(f(v)\in\{p,p+d,\ldots,p+(\ell-1)\cdot d\}\). 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 \(C_{T}(G_{\alpha})=C(G_{\alpha}[TS(G_{\alpha})])=\{\{v\}\}\). Therefore, \(f(v)\geq\max\{x,y\}\). We consider the following cases. **Case 1:**\(\max\{x,y\}\leq p\). Obviously, \(\tau_{p,1}^{d,\ell}(x,y)(G_{\alpha})=f(v)=p\). **Case 2:**\(p<\max\{x,y\}\leq p+(\ell-1)\cdot d\). Let  
\(i=\lceil(\max\{x,y\}-p)/d\rceil\cdot d+p\). Then,  
\[ \begin{array}{ll}i&=\lceil(\max\{x,y\}-p)/d\rceil\cdot d+p\\ &\geq((\max\{x,y\}-p)/d)\cdot d+p\\ &=(\max\{x,y\}-p)+p\\ &=\max\{x,y\}.\end{array} \]  
Assume that there exists an integer \(j<i\) such that \(j\in\{p,p+d,\ldots,p+(\ell-1)\cdot d\}\) and \(j\geq\max\{x,y\}\). However, this leads to a contradiction as follows:  
\[ \begin{array}{ll}j&\leq i-d\\ &=\lceil(\max\{x,y\}-p)/d\rceil\cdot d+p-d\\ &<((\max\{x,y\}-p)/d+1)\cdot d+p-d\\ &=(\max\{x,y\}-p+d)+p-d\\ &=\max\{x,y\}.\end{array} \]  
Hence, \(\tau_{p,1}^{d,\ell}(x,y)(G_{\alpha})=i=\lceil(\max\{x,y\}-p)/d\rceil\cdot d+p\).