Star covers and star partitions of double-split graphs

Abstract

A graph that is isomorphic to the complete bipartite graph $K_{1,r}$ for some $r\ge 0$ is called a star. A collection $\mathcal {C} = \{V_1, \ldots , V_k\}$ of subsets of the vertex set of a graph $G = (V, E)$ is called a star cover of G if each set in the collection induces a star and has $V_1\cup \ldots \cup V_k = V$. A star cover $\mathcal {C}$ of a graph $G = (V, E)$ is called a star partition of G if $\mathcal {C}$ is also a partition of V. The problem Star Cover takes a graph G as input and asks for a star cover of G of minimum size. The problem Star Partition takes a graph G as input and asks for a star partition of G of minimum size. From Shalu et al. (Discrete Appl Math 319:81–91, 2022), it follows that both these problems are NP-hard even for bipartite graphs. In this paper, we show that both Star Cover and Star Partition have $O(n^7)$ time exact algorithms for double-split graphs. Proving that our algorithms indeed have running time $\varOmega (n^7)$ necessitates the construction of an intricate infinite family of double-split graphs meeting several requirements. Other contributions of the paper are a simple linear time recognition algorithm for double-split graphs and a useful succinct matrix representation for double-split graphs.

The proof of Theorem 1 and the C program of Example 8

In this appendix, we prove Theorem 1. Following it, we present the C program that was used for checking the correctness of Example 8.

We now state and prove Theorem 1.

Theorem 6

There is an infinite family of double-split graphs such that each graph $G_r = (A_r \cup B_r, E_r)$ in the family is free of compatible edge pairs as well as compatible edge triples spanning either four or five non-edges but has a compatible edge triple spanning six non-edges.

Proof

We construct the infinite family of double-split graphs using the $9\times 6$ core L given in Fig. 14 in a particular way. Indeed, for each $r \ge 1$, we define a double-split graph $G_r$ by using a double-split adjacency matrix $M_r$ whose core $L_r$ is given in Fig. 18. We emphasize that L in each $L_r$ stands for the matrix L given in Fig. 14.

Fig. 18

The cores of certain double-split adjacency matrices

For instance, the double-split graphs $G_1$ and $G_2$ correspond to the double-split adjacency matrices $M_1$ and $M_2$ below:

$$\begin{aligned} M_1 = \begin{bmatrix} L_1 &{} \overline{L_1}\\ \overline{L_1} &{} L_1 \end{bmatrix} = \begin{bmatrix} \begin{array}{l} L \\ \overline{L} \\ \end{array} &{} I_{18} &{} \begin{array}{l} \overline{L} \\ L \\ \end{array} &{} \overline{I_{18}} \\ &{}\\ \begin{array}{l} \overline{L} \\ L \\ \end{array} &{} \overline{I_{18}} &{} \begin{array}{l} { L} \\ \overline{L} \\ \end{array} &{} I_{18} \\ \end{bmatrix} \text{ and } M_2 = \begin{bmatrix} L_2 &{} \overline{L_2}\\ \overline{L_2} &{} L_2 \end{bmatrix} = \begin{bmatrix} \begin{array}{l} L \\ \overline{L} \\ L \\ \overline{L} \end{array} &{} I_{36} &{} \begin{array}{l} \overline{L} \\ L \\ \overline{L} \\ L \end{array} &{} \overline{I_{36}} \\ &{}\\ \begin{array}{l} \overline{L} \\ L \\ \overline{L} \\ L \end{array} &{} \overline{I_{36}} &{} \begin{array}{l} L \\ \overline{L} \\ L \\ \overline{L} \end{array} &{} I_{36} \\ \end{bmatrix} \end{aligned}$$

More generally, for each $r\ge 1$, the double-split graph $G_r = (A_r\cup B_r, E_r)$ has $M_r$ defined above as its double-split adjacency matrix. It may be noted that the matrix $M_r$ has dimension $(36r)\times (12+36r)$ since its core $L_r$ has dimension $(18r)\times (6+18r)$. In the ensuing discussion, for clarity of exposition, we denote the parts $A_r$ and $B_r$ of $G_r$ simply by A and B, respectively. Thus, we have $|A| = |A_r| = 2p = 36r$ and $|B| = |B_r| = 2q = 12 + 36r$.

Let $u_1v_1, \ldots , u_{18r}v_{18r}$ be the edges of $G_r[A]$ and let $x_1y_1, \ldots , x_6y_6, z_1w_1, \ldots , z_{18r}w_{18r}$ be the non-edges of $G_r[B]$. Then, without loss of generality, we shall assume that the double-split matrix $M_r$ corresponds to these orderings. So, we have that rows of $M_r$ correspond to the ordered set $A = A_r = \{u_1, \ldots , u_{18r}, v_1, \ldots , v_{18r}\}$ and its columns correspond to the ordered set $B = B_r = \{x_1, \ldots , x_6, z_1, \ldots , z_{18r}, y_1, \ldots , y_6, w_1, \ldots , w_{18r}\}$. Let $U = U_r = \{u_1, u_2, \ldots , u_{18r}\}$ and $V = V_r = \{v_1, v_2, \ldots , v_{18r}\}$.

The double-split matrix $M_r$ (and hence the graph $G_r$) we consider is based specifically on the matrix L of Fig. 14. A careful inspection of this L reveals that the double-split graph $G_r$ (which is based on $M_r$) has the three intersections $N_A(x_1)\cap N_A(y_2)$, $N_A(x_3)\cap N_A(y_4)$ and $N_A(x_5)\cap N_A(y_6)$ equal to $A_r$. This implies that, for any $r\ge 1$, $\{(x_1,x_2), (x_3,x_4), (x_5,x_6)\}$ is a compatible edge triple of $G_r$ spanning six non-edges of G[B], namely $x_1y_1, \ldots , x_6y_6$.

We also note that, for any $r\ge 1$, no two columns of the core $L_r$ of the matrix $M_r$ are either equal or complement to each other. This implies that no two columns of $M_r$ that correspond to a pair of adjacent vertices from set $B = B_r$ are either equal or complement to each other. But this means that the set $B = B_r$ does not have any adjacent pair of vertices $\alpha$ and $\beta$ for which $N_A(\alpha )$ and $N_A(\beta )$ are either equal or disjoint. Thus, from the equivalence of statements (4), (5) and (7) of Lemma 3, we have that $G_r$ has no compatible edge pairs for any $r\ge 1$.

If $G_r$ has no compatible edge triples spanning four non-edges, then, since it is free of compatible edge pairs, by Lemma 8, it has no compatible edge triples spanning five non-edges either. So, to complete the proof of the theorem, it suffices to prove that $G_r$ has no compatible edge triples spanning four non-edges.

Since $G_r$ does not have any compatible edge pairs, by Lemma 7, it has no compatible edge triples spanning four non-edges if and only if it has no sure edge triples spanning four non-edges. Thus, we complete the proof of the theorem by just showing that $G_r$ has no sure edge triples spanning four non-edges.

For the purpose of showing that $G_r$ has no sure edge triples spanning four non-edges, we first analyze the nature of intersections corresponding to different types of edges in G[B]. To begin with, we observe the following:

• 1. $N_A(z_i)\cap N_A(z_j) = V \setminus \{v_i,v_j\}$ for any $1\le i < j \le 18r$. This follows because $N_A(z_i) = \{u_i\} \cup (V {\setminus } \{v_i\})$ for each $1\le i \le 18r$.

• 2. $N_A(w_i)\cap N_A(w_j) = U \setminus \{u_i,u_j\}$ for any $1\le i < j \le 18r$. This follows because $N_A(w_i) = (U {\setminus } \{u_i\})\cup \{v_i\}$ for each $1\le i \le 18r$.

• 3. $N_A(z_i)\cap N_A(w_j) = \{u_i,v_j\}$ for any $1\le i, j \le 18r$ with $i\ne j$. This follows because $N_A(z_i)$ and $N_A(w_j)$ share only $u_i$ and $v_j$ when $1\le i, j \le 18r$ and $i\ne j$.

Further, since the columns of $M_r$ corresponding to $x_i$ and $y_i$, where $1\le i \le 6$, are from columns of r copies of $\begin{bmatrix} L\\ \overline{L} \end{bmatrix}$ and r copies of $\begin{bmatrix} \overline{L} \\ L \end{bmatrix}$, both $N_A(x_i)$ and $N_A(y_i)$ consist of exactly p/2 elements from U and exactly p/2 elements from V. We may also recall that $N_A(z_i) = \{u_i\} \cup (V {\setminus } \{v_i\})$ and $N_A(w_i) = (U {\setminus } \{u_i\}) \cup \{v_i\}$. Thus, we also have the following.

• 4. $N_A(x_i)\cap N_A(z_j) \subseteq (V \cup \{u_j\})$ and $|N_A(x_i)\cap N_A(z_j)| \le p/2 + 1$ for $1 \le i \le 6$ and $1 \le j \le 18r$.

• 5. $N_A(x_i)\cap N_A(w_j) \subseteq (U \cup \{v_j\})$ and $|N_A(x_i)\cap N_A(w_j)| \le p/2 + 1$ for $1 \le i \le 6$ and $1 \le j \le 18r$.

• 6. $N_A(y_i)\cap N_A(z_j) \subseteq (V \cup \{u_j\})$ and $|N_A(y_i)\cap N_A(z_j)| \le p/2 + 1$ for $1 \le i \le 6$ and $1 \le j \le 18r$.

• 7. $N_A(y_i)\cap N_A(w_j) \subseteq (U \cup \{v_j\})$ and $|N_A(y_i)\cap N_A(w_j)| \le p/2 + 1$ for $1 \le i \le 6$ and $1 \le j \le 18r$.

Fig. 19

The graph $G_1$: The number of elements of $U = U_1$ (also $V = V_1$) covered by the intersections $N_A(x_i)\cap N_A(x_j)$, $N_A(y_i)\cap N_A(y_j)$ and $N_A(x_i)\cap N_A(y_j)$

When $r = 1$, by straightforward counting, we learn that each of the intersections $N_A(x_i)\cap N_A(x_j)$, $N_A(y_i) \cap N_A(y_j)$, where $1\le i < j \le 6$ and $N_A(x_i) \cap N_A(y_j)$, where $1\le i, j \le 6$ and $i\ne j$, has at most $p/2 - 2$ elements from $U = U_1$ and at most $p/2 - 2$ elements from $V = V_1$. Thus, when $r = 1$, these sets have size at most $p-4$. The exact values are given in Fig. 19. Also, for larger values of r, the columns of $M_r$ corresponding to $x_i$ and $y_i$, where $1\le i \le 6$, are from columns of r copies of $\begin{bmatrix} L\\ \overline{L} \end{bmatrix}$ and r copies of $\begin{bmatrix} \overline{L} \\ L \end{bmatrix}$. Thus for any $r\ge 1$, we have the following.

• 8. $N_A(x_i)\cap N_A(x_j)$ has at most $p/2 - 2r$ elements from $U = U_r$ and at most $p/2 - 2r$ elements from $V = V_r$. So, $|N_A(x_i)\cap N_A(x_j)| \le p - 4r$ for each $1\le i < j \le 6$.

• 9. $N_A(y_i)\cap N_A(y_j)$ has at most $p/2 - 2r$ elements from $U = U_r$ and at most $p/2 - 2r$ elements from $V = V_r$. So, $|N_A(y_i)\cap N_A(y_j)| \le p - 4r$ for each $1\le i < j \le 6$.

• 10. $N_A(x_i)\cap N_A(y_j)$ has at most $p/2 - 2r$ elements from $U = U_r$ and at most $p/2 - 2r$ elements from $V = V_r$. So, $|N_A(x_i)\cap N_A(x_j)| \le p - 4r$ for each $1\le i, j \le 6$ with $i\ne j$.

Now, using the bounds 1 through 10 above, we prove that $G_r = (A\cup B, E)$ does not have any compatible edge triples spanning four non-edges of G[B]. Since the item numbers 1 through 10 that we make use of in our arguments below will be clear from the type of intersections considered, we avoid explicit references to them.

We observe that $N_A(z_i)\cap N_A(w_j) = \{u_i,v_j\}$ for each $1\le i < j \le 18r$. But all other types of intersections are of size at most $p-2$. So, it follows that $z_iw_j$, where $1\le i, j \le 18r$ and $i\ne j$, cannot be part of any sure edge triple of $G_r$.

We now also argue that any sure edge triple of $G_r$ spanning four non-edges can have at most one edge from $S = \{z_iz_j, w_iw_j \mid 1 \le i < j \le 18r\}$. Suppose, for contradiction, this is not the case. By definition, any sure edge triple of $G_r$ can have at most one edge spanning any particular pair of non-edges. So, if any sure edge triple spanning four non-edges has two or three edges from S, then their end vertices must be from either (a) three or (b) four non-edges of the form $z_iw_i$, where $1\le i \le 18r$. We now prove that neither of these cases are possible.

Case (a): Suppose $G_r$ has a sure edge triple with vertices from three distinct non-edges $z_iw_i, z_jw_j$ and $z_kw_k$. Without loss of generality, assume that $z_iz_j$ and $w_iw_k$ are part of a sure edge triple of $G_r$. But $[N_A(z_i)\cap N_A(z_j)]\cup [N_A(w_i)\cap N_A(w_k)] = [V {\setminus } \{v_i,v_j\}] \cup [U{\setminus } \{u_i, u_k\}] = A{\setminus } \{v_i,v_j,u_i,u_k\}$. This means that the set of elements of A that are not covered by these two intersections has a pair of adjacent vertices from A, namely $u_i$ and $v_i$, and hence cannot be a subset of $N_A(z)$ for any $z\in B$ as $N_A(z)$ is an independent set for each $z\in B$. This effectively implies $z_iz_j$ and $w_iw_k$ cannot be part of a sure edge triple of $G_r$.

Case (b): Suppose $G_r$ has a sure edge triple with vertices from four distinct non-edges $z_iw_i, z_jw_j, z_kw_k$ and $z_lw_l$. Suppose a sure edge triple contains $z_iz_j$ and $z_kz_l$. Then the union of the two intersections corresponding to these edges leaves all vertices of U uncovered. Since the sure edge triple in consideration spans only four non-edges, the third edge in it must be of the form $w_sw_t$, where $s\in \{i,j\}$ and $t\in \{k,l\}$. Also the intersection corresponding to an edge of the form $w_sw_t$ can cover only $p-2$ vertices of U. But $|U|=p$. Thus, $z_iz_j$ and $z_kz_l$ cannot be part of any sure edge triple. Similarly, we conclude that $w_iw_j$ and $w_kw_l$ are not part of any compatible edge triple. So, without loss of generality, assume that a sure edge triple contains $z_iz_j$ and $w_kw_l$. In this case, the union of the two intersections corresponding to these edges leave the vertices in $\{v_i,v_j,u_k,u_l\}$ uncovered. Since the sure edge triple in consideration spans only four non-edges, the third edge in it must be of the form $z_sw_t$, where $s\in \{k,l\}$ and $t\in \{i,j\}$. But the intersection corresponding to an edge of the form $z_sw_t$ can cover at most two vertices. So, this case also cannot arise.

Therefore, any sure edge triple of $G_r$ spanning four non-edges can have at most one edge from S.

For convenience, we call any edge of G[B] with their end vertices from the set $\{x_1,\ldots , x_6,$ $y_1,\ldots , y_6\}$ old, call any edge of G[B] with their end vertices from the set $\{z_1,\ldots , z_{18r},$ $w_1,\ldots , w_{18r}\}$ new and call any edge of G[B] with one vertex $\{x_1,\ldots , x_6, y_1,\ldots , y_6\}$ and one vertex from $\{z_1,\ldots , z_{18r}, w_1,\ldots , w_{18r}\}$ mixed. Thus, from the preceding arguments, we have that any sure edge triple spanning four non-edges can have at most one new edge and it has the form $z_iz_j$ or $w_iw_j$. This implies the following cases for any sure edge triple spanning four non-edges in terms of the number of new, old and mixed edges.

Case	New	Mixed	Old
1	0	0	3
2	0	1	2
3	0	2	1
4	0	3	0
5	1	0	2
6	1	1	1
7	1	2	0

Case 1: In this case, a compatible edge triple for $G_r$ spanning four non-edges implies a compatible edge triple spanning four non-edges for the graph G defined in Fig. 14, which is a contradiction. So, this case is not possible.

Case 2: Two intersections corresponding to any two old edges leave at least 4r vertices of U and at least 4r vertices of V uncovered. But any intersection corresponding to a mixed edge includes at most p/2 vertices from U and one vertex from V or vice versa. So, this case is not possible.

Case 3: Two intersections corresponding to any pair of mixed edges and one intersection corresponding to any old edge can cover at most $2(p/2+1) + (p-4r) < 2p$ vertices since $r \ge 1$. So, this case is not possible.

Case 4: Three intersections corresponding to any three mixed edges can cover at most $3(p/2+1) < 2p$ vertices; the inequality follows since $p\ge 18$. So, this case is not possible.

Case 5: One new edge requires two non-adjacent pairs of G[B] of the form $z_iw_i$, $1\le i \le 18r$. Also two old edges require, by the definition of a sure edge triple, three non-adjacent pairs of G[B] of the form $x_iy_i$, $1\le i \le 6$. But this means that the sure edge triple spanning four non-edges has vertices from five non-edges of G[B], a contradiction. So, this case is not possible.

Case 6: The intersection corresponding to any old edge leaves at least $p/2 + 2r$ vertices of U and at least $p/2 + 2r$ vertices of V uncovered. Also the intersection corresponding to any new edge covers $p-2$ vertices from either V or U. This leaves behind at least $p/2 + 2r$ vertices of A uncovered. But any intersection corresponding to a mixed edge covers at most $p/2 + 1$ vertices of A. So, this case is not possible.

Case 7: One intersection corresponding to a new edge and two intersections corresponding to two distinct mixed edges can potentially cover $(p-2) + 2(p/2+1) = 2p$ vertices of A. But this will be possible only if these three intersections are disjoint. So we check if three such disjoint intersections are possible. Any intersection corresponding to a new edge consists of exactly $p-2$ vertices of either U or V. Without loss of generality, assume that the sure edge triple includes the new edge $w_jw_k$. Then the corresponding intersection equals $U{\setminus } \{u_j, u_k\}$. Also any intersection corresponding to a mixed edge covers at most p/2 vertices from U and at most one vertex from V or vice versa. If there is any intersection that covers p/2 vertices of V and $u_j$, then it necessarily corresponds to a mixed edge of the form $x_iz_j$ for some $1\le i\le 6$. So, we shall also assume that the sure edge triple contains the mixed edge $x_iz_j$. Each $N_A(y_j)$ covers exactly p/2 elements of V and, among these, only $N_A(y_i)$ covers all those elements of V that are left uncovered by $N_A(x_i)$. So, it follows that the other mixed edge in the sure edge triple must necessarily equal $y_iz_k$ as the corresponding intersection only can possibly cover the remaining p/2 elements of V and $u_k$. But this means that the sure edge triple spans only three non-edges, namely $x_iy_i, z_jw_j$ and $z_kw_k$. But a sure edge triple, by definition, must span at least four non-edges. So, this case also is not possible.

This completes our proof of the fact that $G_r$ has no sure edge triples spanning four non-edges. Thus, since $G_r$ has no compatible edge pairs, from Lemmas 7 and 8, it follows that $G_r$ has no compatible edge triples spanning either four or five non-edges as already remarked.

Hence, for each $r\ge 1$, the graph $G_r$ has neither compatible edge pairs nor compatible edge triples spanning either four or five non-edges but has a compatible edge triple spanning six non-edges. $\square$

We now present the C program that we used for checking the correctness of Example 8. On inputting the $9\times 6$ core L given in Fig. 14, it computes the corresponding $18\times 12$ double-split adjacency matrix M and verifies that the associated double-split graph does not have any compatible edge triples spanning four non-edges.