Complexity of Steiner Tree in Split Graphs - Dichotomy Results

Necessity: Suppose $\Delta_{G}^{I}\geq 3$, and let $v\in C$ has at least 3 neighbours, say $x,y,z\in I$. Then the set $\{v,x,y,z\}$ forms a claw in $G$ with $v$ as its centre vertex. It follows that if $G$ is claw-free, then $\Delta_{G}^{I}\leq 2$. Now suppose $\Delta_{G}^{I}=2$. Let $u\in C$ such that there exist vertices $x,y\in I,\{x,y\}\subseteq N_{G}^{I}(u)$. We assume on the contrary that there exist $v\in C,v\neq u$ such that $N_{G}^{I}(u)\cap N_{G}^{I}(v)=\emptyset$. Since $C$ is a clique, $\{u,v\}\in E(G)$. It follows that vertices $\{u,x,y,v\}$ forms a claw in $G$ with $u$ as its centre, a contradiction. This proves Condition 2, and completes the proof of the forward direction.
Sufficiency: On the contrary assume that $G$ is not claw free. No claw in $G$ can have its centre vertex in the set $I$, since for any $v$ in $I$, the set $N_{G}(v)\subseteq C$ and hence induces a clique in $G$. So every claw in $G$ has its centre vertex in the set $C$. Consider a claw with the vertex set $\{v,x,y,z\}$, with the centre $v$ being in $C$. No two of the other three vertices of the claw can be in $C$, because then there would be an edge between them. So at most one of $\{x,y,z\}$ is in $C$, and the rest (of which there are at least two) are in $I$. It follows that if $G$ contains a claw, then $\Delta_{G}^{I}\geq 2$. Equivalently, if $\Delta_{G}^{I}\leq 1$ then $G$ is claw-free. Finally, consider the case where $\Delta_{G}^{I}=2$. Suppose the vertex set $\{v,x,y,z\}$ induces a claw in $G$, with its centre vertex being $v$. Then $v$ is in $C$, and at least two of $\{x,y,z\}$ are in $I$, as we argued above. Since $\Delta_{G}^{I}=2$ we get that exactly two of $\{x,y,z\}$, say $x$ and $y$, are in $I$. Then $z$ is in $C$, and $\{x,z\},\{y,z\}\notin E(G)$. It follows that $N^{I}_{G}(v)\cap N^{I}_{G}(z)=\emptyset$ which is a contradiction to Condition 2. Therefore, our assumption that there exist a claw in $G$ is wrong, and this completes the sufficiency. Therefore, the theorem follows.$\hfill\squareforqed$

Since $\Delta_{G}^{I}=2$, let there exist a vertex $v\in C$ such that $d_{G}^{I}(v)=2$. On the contrary, assume that $|I|>3$, that is, $\{a,b,c,d\}\subseteq I$ such that $N_{G}^{I}(v)=\{a,b\}$. Let $X=N_{G}(a)$ and $Y=N_{G}(b)$ as shown in Figure 1. If there exist a vertex $t\in C$ such that $t\notin X$, and $t\notin Y$, then vertices $\{a,b,v,t\}$ induces a claw. Therefore, $C=X\cup Y$. If $X\subseteq Y$, then $C\cup\{b\}$ induces a larger clique, which is a contradiction to the assumption on the maximality of clique $C$. Therefore, $X\not\subseteq Y$ and similarly, $Y\not\subseteq X$. It follows that, $X-Y\neq\emptyset$ and $Y-X\neq\emptyset$. For every vertex $v\in X\cap Y$, $\{v,c\}\notin E(G)$ and $\{v,d\}\notin E(G)$ otherwise, $N_{G}^{I}(v)\cup\{v\}$ induces $K_{1,3}$.

Therefore, the vertices $c,d$ can have adjacency in two disjoint sets $X-Y$ or $Y-X$.
Case 1: $N_{G}(c)\cap X\neq\emptyset$ and $N_{G}(d)\cap Y\neq\emptyset$. Edge $\{u,c\}\in E(G)$ where $u\in X-Y$ and $\{z,d\}\in E(G)$ where $z\in Y-X$. Observe that $\{a,z\},\{c,z\}\notin E(G)$ otherwise $N_{G}^{I}(z)\cup\{z\}$ induces $K_{1,3}$. Similarly, $\{d,u\}\notin E(G)$ otherwise $N_{G}^{I}(u)\cup\{u\}$ induces $K_{1,3}$. From the discussion, it follows that the vertices $\{u,a,c,z\}$ induces a claw, which is a contradiction. Similar argument holds for $N_{G}(c)\cap Y\neq\emptyset$ and $N_{G}(d)\cap X\neq\emptyset$.
Case 2: $N_{G}(c)\subseteq X$ and $N_{G}(d)\subseteq X$. Let $\{c,u\},\{d,w\}\in E(G)$ such that $u,w\in X-Y$. Note that there exist at least one vertex $z\in Y-X$. If $\{c,z\}\notin E(G)$, then the vertices $\{u,a,c,z\}$ induces a claw. If $\{c,z\}\in E(G)$, then the vertices $\{w,a,d,z\}$ induces a claw as, $\{d,z\}\notin E(G)$. The argument is symmetric for $N_{G}(c)\subseteq Y$ and $N_{G}(d)\subseteq Y$.
Cases 1 and 2 give a contradiction to the fact that $G$ is claw free. Therefore, our assumption that $|I|>3$ is wrong, and hence, the lemma follows. ∎

Necessity: On the contrary, let us assume there exist $v\in C$ such that $N_{G}^{I}(u)\cap N_{G}^{I}(v)=\emptyset$. Since $d_{G}^{I}(u)=3$, vertices $\{u,v\}\cup N_{G}^{I}(u)$ induces a $K_{1,4}$, which is a contradiction and the necessary condition follows.
Sufficiency: On the contrary, assume that $G$ is not $K_{1,4}$ free and there exists a $K_{1,4}$ induced on $\{u,v,w,x,y\}$ with $u$ as the centre vertex. No $K_{1,4}$ in $G$ can have its centre vertex in the set $I$, since for any $u$ in $I$, the set $N_{G}(u)$ is a subset of the set $C$ and hence induces a clique in $G$. So every $K_{1,4}$ in $G$ has its centre vertex in the set $C$ particularly, $u\in C$. Since $G$ is a $3$-split graph, $d_{G}^{I}(u)=3$. This implies that there exist at least one vertex of $K_{1,4}$, say $v\in C$, and $u\in V_{3}$. It follows that $N_{G}^{I}(u)\cap N_{G}^{I}(v)=\emptyset$, which is a contradiction and the sufficiency follows. This completes the proof of the lemma. ∎

By Lemma 3, for every $u\in C$, $N_{G}^{I}(v)\cap N_{G}^{I}(u)\neq\emptyset$. This implies that for every $u\in C$, $\{v_{1},v_{2},v_{3}\}\cap N_{G}^{I}(u)\neq\emptyset$. It follows that $N_{G}(v_{1})\cup N_{G}(v_{2})\cup N_{G}(v_{3})=C$. ∎

If $M$ is a connected graph, then the minimum Steiner set in $G$ corresponds to the minimum edge cover in $M$. For any graph $M$ with maximum matching $P$, the cardinality of minimum edge cover is $|V(M)|-|P|$. Therefore, a minimum Steiner set $S$ is such that $|S|=|V(M)|-|P|=|I|-k$. If $M$ is not connected, let $C_{1},C_{2},\ldots,C_{r}$ be the components such that $C_{1},C_{2},\ldots,C_{i},i\leq r$ are non-trivial components with at least one edge and $C_{i+1},C_{i+2},\ldots,C_{r}$ are trivial ones. For components $C_{1},C_{2},\ldots,C_{i}$, we find the maximum matching $P$ where $k=|P|$ and $Q\subseteq C$ be the corresponding vertex set of the matching $P$. Clearly, $|N_{G}^{I}(Q)|=2|Q|=2|P|=2k$. Let $Q^{\prime}$ be the corresponding clique set of $I\backslash N_{G}^{I}(Q)$. From the definition of the corresponding clique set, $|Q^{\prime}|\leq|I\backslash N_{G}^{I}(Q)|$. Note that, there does not exist two vertices $x,y\in I\backslash N_{G}^{I}(Q)$ such that $N_{G}(x)\cap N_{G}(y)\neq\emptyset$, otherwise it contradicts the maximality of $P$. Since there does not exist the possibility to have two such vertices $x,y\in I\backslash N_{G}^{I}(Q)$, it follows that $|Q^{\prime}|=|I\backslash N_{G}^{I}(Q)|$ and the graph induced on $V(G)\backslash N_{G}^{I}(Q)$ is a $1$-split graph. Therefore, $|Q^{\prime}|=|I|-2k$, and $I\setminus N_{G}^{I}(Q)\subseteq N_{G}^{I}(Q^{\prime})$. It follows that the set $S=Q^{\prime}\cup Q$ forms a Steiner set of $G$ and $|S|=|I|-2k+k=|I|-k$. ∎

Finding the labeled graph $M$ of $G$, incurs $O(n)$ effort where $n=|V(G)|$. Maximum matching $P$ of $M$ can be found in $O(n^{\frac{3}{2}})$ time. Note that the corresponding vertex set $Q$ of $P$ can be found in linear time. Similarly, the corresponding clique set also can be obtained in linear time. Therefore, the overall running time for finding the Steiner set is $O(n^{\frac{3}{2}})$ and OPT Steiner tree in any $2$-split graph is polynomial-time solvable. ∎

Recall from Corollary 1 that, $G^{2}$ is a $l$-split graph for some $0\leq l\leq 2$. On the contrary, let $\alpha(M)\geq 3$. Let vertices $\{a,b,c,d,e,f\}\subseteq V(M)$ be those vertices participating in the matching of size at least $3$ such that $\{u,v,w\}\subseteq C^{1}$ and $\{a,b\}\subseteq N^{I}_{G^{1}}(u)$, $\{c,d\}\subseteq N^{I}_{G^{1}}(v)$, $\{e,f\}\subseteq N^{I}_{G^{1}}(w)$ as shown in Figure 2. Clearly, from Lemma 3, $N_{G^{1}}^{I}(x)\cap N_{G^{1}}^{I}(u)\neq\emptyset$. Similarly, $N_{G^{1}}^{I}(x)\cap N_{G^{1}}^{I}(v)\neq\emptyset$ and $N_{G^{1}}^{I}(x)\cap N_{G^{1}}^{I}(w)\neq\emptyset$. We consider the following scenario.

Suppose $\{g,v\}\in E(G^{1})$ and $\{g,u\}\notin E(G^{1})$. Since $N_{G^{1}}^{I}(x)\cap N_{G^{1}}^{I}(u)\neq\emptyset$, without loss of generality, $\{h,u\}\in E(G^{1})$. Observe that $\{u,v\}\cup N_{G^{1}}^{I}(v)$ induces a $K_{1,4}$. Therefore, $\{g,u\}\in E(G^{1})$. Similar argument holds true for $w$ and $\{g,w\}\in E(G^{1})$. Since the clique $C^{1}$ is maximal, $g$ is not adjacent to all vertices of $C^{1}$, and therefore there exist $y\in C^{1}$ such that $\{g,y\}\notin E(G^{1})$. Clearly, $N_{G^{1}}^{I}(y)\cap N_{G^{1}}^{I}(u)\neq\emptyset$, $N_{G^{1}}^{I}(y)\cap N_{G^{1}}^{I}(v)\neq\emptyset$ and $N_{G^{1}}^{I}(y)\cap N_{G^{1}}^{I}(w)\neq\emptyset$. Observe that in $G^{2},d_{G^{2}}^{I}(y)=3$, and $G^{2}$ is not a $l$-split graph, $l\leq 2$. This is a contradiction to Corollary 1. It follows that our assumption $\alpha(M)\geq 3$ is wrong and therefore, $\alpha(M)\leq 2$. This completes the proof of the lemma. ∎

Observe that $V_{3}\neq\emptyset$ as $G^{1}$ is $3$-split. For any $v\in V_{3}$, $G^{2}$ is the graph induced on $V(G^{1})\backslash N^{I}_{G^{1}}(v)$. By Corollary 1, $G^{2}$ is a $l$-split graph, $l\leq 2$. Let $S^{2}$ be the minimum Steiner set of $G^{2}$ such that $N_{G^{2}}^{I}(S^{2})=I^{2}$ and $|I^{2}|=|I^{1}|-3$. If $M$ is the labeled graph of $G^{2}$, then by Lemma 6, $\alpha(M)\leq 2$. Let $\{\{a,b\},\{c,d\}\}\subseteq E(M)$ be the matching edges of a maximum matching in $M$. Observe that there exist two vertices $v_{1},v_{2}\in C^{2}$ such that $N_{G^{2}}^{I}(v_{1})=\{a,b\}$, $N_{G^{2}}^{I}(v_{2})=\{c,d\}$. Notice that for each vertex $w\in I^{2}\backslash\{a,b,c,d\}$, there exist a vertex $u\in C^{2}\backslash\{v_{1},v_{2}\}$ in $S^{2}$ such that $\{w,u\}\in E(G^{2})$. The graph induced on $V(G^{2})\backslash\{a,b,c,d\}$ is a $1$-split graph, $|S^{2}|\geq|I^{2}|-4+2$ and it follows that $|S^{2}|\geq|I^{1}|-3-2=|I^{1}|-5$. It can be concluded that $|S|\geq|I^{1}|-5$ as $|S|\geq|S^{2}|$. This completes the proof of the lemma. ∎

On the contrary assume that there exist a minimum Steiner set $S\subseteq C^{1}$ such that $|S|\leq|I^{1}|-5$.
Suppose that $S\cap V_{3}=\emptyset$. Note that $V_{3}\neq\emptyset$, say $u\in V_{3}$ and for every vertex $z\in S$, $N_{G^{1}}^{I}(z)\cap N_{G^{1}}^{I}(u)\neq\emptyset$ as per Lemma 3. i.e., for every $z\in S$, there exist an edge $\{z,i\}\in E(G^{1})$, where $i\in N_{G^{1}}^{I}(u)$. The graph $G^{2}=I^{2}+C^{2}$ induced on $V(G^{1})\backslash N_{G^{1}}^{I}(u)$ is a $l$-split graph, $l\leq 2$ by Corollary 1. Consider the Steiner set $S^{2}\subseteq C^{2}$ of $G^{2}$ such that $N_{G^{2}}^{I}(S^{2})=I^{2}$. Note that $|S^{2}|=|I^{2}|$ as $G^{2}$ is a $l$-split graph, $l\leq 2$ and for each vertex $w\in S^{2}$, $N_{G^{1}}^{I}(w)\cap N_{G^{1}}^{I}(u)\neq\emptyset$. Notice that $|I^{2}|=|I^{1}|-3$ and $|S|\geq|S^{2}|$ implies that $|S|\geq|I^{1}|-3$. This shows that $S\cap V_{3}=\emptyset$ is not possible.
Next we shall consider the scenario $S\cap V_{3}\neq\emptyset$. Consider the $l$-split graph, $l\leq 2$ $G^{2}$ induced on $V(G^{1})\backslash N_{G^{1}}^{I}(u)$ where $u\in S\cap V_{3}$. Let the labeled graph of $G^{2}$ be $M$. For $S^{{}^{\prime}}=S\backslash\{u\}$ and $V^{{}^{\prime}}=I^{1}\backslash N_{G^{1}}^{I}(u)$ note that $N^{I}_{G^{2}}(S^{{}^{\prime}})=V^{{}^{\prime}}$. Clearly, $|S^{{}^{\prime}}|=|S|-1\leq|I^{1}|-5-1$ and $|V^{{}^{\prime}}|=|I^{1}|-|N_{G^{1}}^{I}(u)|=|I^{1}|-3$. We now claim that there exist at least $3$ vertices say $\{v_{1},v_{2},v_{3}\}\subseteq S^{{}^{\prime}}$ such that $d^{I}_{G^{2}}(v_{i})=2$, $i=1,2,3$ and $N^{I}_{G^{2}}(v_{i})\cap N^{I}_{G^{2}}(v_{j})=\emptyset$, $1\leq i\neq j\leq 3$. Suppose if there exist at most two vertices $v_{1},v_{2}\in S^{{}^{\prime}}$ such that $d_{G^{2}}^{I}(v_{1})=d_{G^{2}}^{I}(v_{2})=2$ and $N^{I}_{G^{2}}(v_{1})\cap N^{I}_{G^{2}}(v_{2})=\emptyset$, then observe that $|V^{{}^{\prime}}|\leq|S^{{}^{\prime}}|+2$. It follows that $|V^{{}^{\prime}}|\leq|I^{1}|-6+2=|I^{1}|-4$, which is a contradiction as $|V^{{}^{\prime}}|$ is $|I^{1}|-3$. Therefore, there exist at least $3$ vertices $v_{1},v_{2},v_{3}\in S^{{}^{\prime}}$ such that $d^{I}_{G^{2}}(v_{i})=2$, $i=1,2,3$ and $N^{I}_{G^{2}}(v_{i})\cap N^{I}_{G^{2}}(v_{j})=\emptyset$, $1\leq i\neq j\leq 3$. Consider the labeled graph $M$ of $G^{2}$. There exist $\{a,b,c,d,e,f\}\subseteq I^{2}$ such that $\{a,b\}=N_{G^{2}}^{I}(v_{1})$, $\{c,d\}=N_{G^{2}}^{I}(v_{2})$, $\{e,f\}=N_{G^{2}}^{I}(v_{3})$. It follows that $\{a,b\}$, $\{c,d\}$, $\{e,f\}$ forms a matching of size $3$ in $M$ which is a contradiction to Lemma 6. Therefore our assumption is wrong and $|S|\geq|I^{1}|-4$. This completes the proof. ∎