The Number of Minimum Roman and Minimum Total Dominating Sets for Some Chessboard Graphs

In this paper, both the roman domination number and the number of minimum roman dominating sets are found for any rectangular rook’s graph. In a similar fashion, the roman domination number and the number of minimum roman dominating sets are found on the square bishop’s graph for odd board sizes. Also found are the number of minimum total dominating sets associated with the light-colored squares when and same for the dark-colored squares when


Introduction
The m n × rook's graph, denoted by γ denotes the roman domination number on the square, n n × bishop's graph. For more on the roman domination parameter in general, including some work for specific families of graphs, see [2]- [9]. Likewise, given a graph ( ) , G V E = , a set of vertices is said to be a total dominating set if every vertex in V is adjacent to at least one vertex in the set. The minimum cardinality among all total dominating sets is known as the total domination number, denoted γ for the bishop's graph. On the bishop's graph, a set can also be said to be a total dominating set if and only if every square is attacked. The total domination number for the bishop's graph was determined to be ( ) [10]. This paper's findings also imply that for ( ) 1 mod12 n ≡ (with 1 n > ), the total domination number of the subgraph associated with the light-colored squares is ( ) for the subgraph associated with the dark-colored squares. These findings are central to the results in Section 4 of this paper. For more on the total domination number, a good book on the current literature is [11]. Theorem 1. ( ) 2  2 1 a a n n + − + . But since this quadratic has a minimum at 1 a n = − , then

Rook's Graph Results for Roman Domination
a b a n a n n n n n r n R n γ ≤ − , then place twos in every square along one of the main diagonals-save for one of those squares. In this square, plant a one. This provides a roman dominating set of size 2 1 n − .  Proof: To begin our proof we will first define a subgraph G of the m n × rook's graph similar to the ones first introduced in [12] [13] for the queen's graph, with the vertex set taken to correspond to the two entries. Our subgraph G differs from the definition given for the queen's graph in that there are no diagonal edges for G on the rook's graph, just column and row edges. Given the Figure 1. Let the pawn in the lower-left hand corner represent a one entry and the rooks represent two entries. Then the above is a minimum roman dominating set for the standard 8×8 board.
subgraph of vertices associated with our two entries as our vertex set, two vertic-  , place m two entries, each having a different row. It is then easy to see that any square will be in the same row with at least one two entry, and thereby be roman dominated. Figure 2 will illustrate.  Theorem 3. The number of minimum Roman Dominating sets for the square rook's graph is ( ) ! n n , where n is the length of both of our dimensions.
Proof: First note it follows that we must have exactly 1 n − two entries, and the lone, one entry placed on our n n × board, since our quadratic in Theorem one had its only minimum at 1 a n = − , leaving us with the sole, one entry to provide our count of 2 1 n − for the roman dominating set. It also follows that each entry can't share the same row or column, for if either any single pair of two entries share a row (or column, without loss of generality) then we'd have ( ) ( ) 2 1 2 2 r n R n n γ ≥ − + = , a contradiction. Also note no two entry can be adjacent to our lone one entry, for if so we can then obtain a roman dominating set with less cardinality than our proven minimum by changing the associated one entry to a zero entry, thereby a contradiction.
Making no distinction between the 1 n − two entries and the sole one entry, we note we can associate these n objects with independent vertices on our board Open Journal of Discrete Mathematics    Next consider again, as in Theorem 2, the inequality , which is a contradiction. Thus for our second class of solutions it follows that we must have 1 j = , 0 c = , and 0 r = or we obtain a similar contradiction for our lower bound. It then also follows we have m j − two entries which are placed independently, and 2 one entries that are placed in the two remaining squares in our row which contains no two entry.
is the number of such sets, we will first place the one entries. Note for this assignment there are m rows to choose from that will lack a two entry, and 1 2 ways to choose the two one placements, given any row without a two entry. It follows that since we can have no one entry that is adjacent to a two entry since we would have redundancy-for the set would not need the adjacent one entry as opposed to a zero entry-that we can eliminate the two columns and one row with ones in them, and consider the placement of our twos. Note our placement of ones can be placed exactly 2 1 2 Given any placement of our 2 one entries for the case where 1 j = , we have exactly 1 m − rows and 1 m − columns on which to assign our two entries, after the one entries have been assigned. Since this can be accomplished exactly ways, then we have our total number of Roman dominating sets in our second class as . Summing those in both the classes of Roman dominating sets proves the theorem. Figure  Proof: In similar way as in Theorems 3 and 4, we see that we are forced to have exactly m two entries, with one placed in each row. Since there are m rows, and n ways to pick a square in each row, then we have our total of m n ways to pick our m two entries. 

The Roman Domination Number on the Bishop's Graph and the Count for Odd n
Proof: We will begin the proof by providing a roman dominating set of cardinality 2 1 n − , namely by associating twos with every square in the center-most row-save for the left-most square in this row which is associated with a one. All other squares are associated with zeros. This provides a roman dominating set of cardinality 2 1 n − for odd n, thus providing the upper bound of To see the rest of the proof we must describe all our possible sets of minimum roman dominating sets by breaking our n n × board up into six regions, and applying the pigeonhole principle to eliminate certain placements. Then, the placements that remain will be shown to all have roman cardinality of 2 1 n − , whereas the eliminated placements will all have roman cardinality of greater than 2 1 n − . We will then show independence among all our one and two entries, beyond the limitation of regional placement. Then the count will be given of all sets of roman dominating sets with these properties which will be shown to be sufficient for a minimum roman dominating set on the bishop's graph.
First take the origin to be the center of the center-most square, and each square identified by the coordinates at the center of its square. Then, begin by labeling the squares that lie on sum diagonals having the sum of the coordinates exclusively between Region five is the region whose squares have sums of the coordinates associated with these squares either strictly less than 1 2 n − , or strictly greater than ( ) We will now define a set of variables used to lay out an argument that "pigeon-holes" the placement of two entries, by contradiction, to Regions one and four solely-namely 2 n − two entries in Region one, and 1 two entry in Region four. Also, a lone one entry will be placed in Region four. Define 1 R , 2 R , 3 R , Open Journal of Discrete Mathematics a R R > + or 6 3 a R R > + , then that are all on a common diagonal. This is a contradiction since the placement of an additional two entry (as opposed to the minimum of three, one entries needed to saturate these squares) in this diagonal, beyond those placed, would further decrease our roman cardinality. Note also that it is not possible for either 5 2 a R R > + or 3 6 a R R > + , for if either, without loss of generality, then R R a + = without loss of generality), note we have all four Region four squares unsaturated. Note also we're limited to only at most a single two entry in Regions two, three, or four without the cardinality of our Roman dominating set exceeding 2 1 n − . Also note the only squares adjacent to these Region four squares are in Regions two, three, and four and that a placement of an additional two entry in Regions two or three only saturates exactly two of these Region four squares. It then follows that we must place an additional two entry in Region four. This leaves us with exactly one unsaturated Region four square. Since no additional two entries are allowed for the Roman dominating set to be of minimum cardinality, we associate a one entry with this square. However, we still have at least two unsaturated squares in Region six which implies the set is not a Roman dominating set.
Next assume 2  Next let us define a subgraph G similar to the subgraph G found in both [12] [13], except we are defining it for the bishop's graph and not the queen's graph. So given the total dominating set of vertices associated with the occupied squares, we define G to contain only the set of vertices associated with the total dominating set. Two vertices are then adjacent in G if and only if their corresponding squares (which are occupied by bishops) are attacked along unoccupied squares. Note if there is at most two bishops in any given diagonal there is no difference between G and the subgraph induced by the vertices associated with the occupied squares. It is only when there are three or more bishops in a given diagonal that a difference arises. For example, note that the subgraph induced by vertices associated with occupied squares in the same diagonal will form a complete graph, whereas in G these same vertices form a path. Then define p d and n d as the number of positive-sloping and negative-sloping diagonal edges for the graph G. It then follows that either a or b is zero, since if not we would have a light-colored square in region one not under attack. Without loss of generality, by symmetry, take 0 a = . It also follows that there is no empty negative sloping diagonal that has the sum of its coordinates exclusively between plus and minus 1 2 n − . It then follows that Next we will show that the unsaturated squares remaining in Region three af-  To count the number of minimum total dominating sets we will note that every diagonal, both positive and negative, that contains a square in Region one must be occupied. To see this first note every square in Region one must be dominated twice, once along a negative diagonal and once along a positive diagonal, or else we'd have unsaturated squares in either Region two or Region three (depending upon whether we had an open negative or positive diagonal).
Also note that there must be exactly 1 3 n − row edges in G and the same number of column edges in G with all the edges between vertices associated with bishops in Region one.
We begin the count by selecting the 1 3 n − positive diagonals that contain exactly two bishops and the 1 3 n − negative diagonals that contain exactly two bishops. Note there are exactly 1 2 n − choose 1 6 n − ways to choose the positive diagonals and the same number of ways to choose the negative diagonals. Thus the result is squared.
Next, given the assignment of two adjacent bishops to the positive and negative diagonals in such a way, we note there are

Conclusion
Certainly, a similar method might provide the roman domination number for the square bishop's graph and the count on the number of minimum roman dominating sets, when n is even. The partition of squares into the six regions indicated in this paper might also provide some use for finding the counts of the remaining cases for the number of minimum total and paired dominating sets on the square bishop's graph. See both A303144 and A303141 in the Online Encyclopedia of Integer Sequences for some of the values on these counts for very small board sizes. Note the available count for 7 n = obtained in this paper matches the one found for 7 n = in the OEIS. It should also be noted that the case solved for the count of the minimum total dominating sets for the light squares when ( )

Conflicts of Interest
The author declares no conflicts of interest regarding the publication of this paper.