Independence and domination on shogiboard graphs

Abstract

Given a (symmetrically-moving) piece from a chesslike game, such as shogi, and an n × n board, we can form a graph with a vertex for each square and an edge between two vertices if the piece can move from one vertex to the other. We consider two pieces from shogi: the dragon king, which moves like a rook and king from chess, and the dragon horse, which moves like a bishop and rook from chess. We show that the independence number for the dragon kings graph equals the independence number for the queens graph. We show that the (independent) domination number of the dragon kings graph is n − 2 for 4 n 6 and n − 3 for n 7. For the dragon horses graph, we show that the independence number is 2n − 3 for n 5, the domination number is at most n − 1 for n 4, and the independent domination number is at most n for n 5.

Reflections on the n +k dragon kings problem

“…The bound in Proposition 13 is not tight, as we can see in the monodiagonally symmetric 9 + 3 dragon kings problem solution with pawns on squares (4, 4), (5,6), and (6,5) and dragon kings on squares (0, 4), (1, 7), (2,5), (3,3), (4, 0), (4, 6), (5,2), (5,8), (6,4), (6,6), (7,1), and (8, 5).…”

“…For the base case of k = 0, take a 5 × 5 board and place dragon kings on squares (2, 0),(1, 2),(0, 4), (3,3), and (4, 1). These pieces do not attack each other, and squares (0, 0), (4, 0), (4,3), (4,4), (3,4) and (2,4) are empty. We use Lemma 2 for the induction step.…”

“…Proof: First we show by induction on k that if k is even, a centrosymmetric n + k dragon kings problem solution exists for n = 2k + 6. For the base case, we use the 6 × 6 board with dragon kings on squares (0, 1), (1, 3), (2, 5), (3, 0), (4,2), and (5, 4). Then we use Lemma 8 for the inductive step.…”

“…Next we show that the statement is true for n = 2k + 6. For k = 0, take the 6 × 6 board with dragon kings on squares (0, 3), (1, 5), (2, 2), (3, 0), (4,4) and (5,1). For k = 1, take the 8 × 8 board with a pawn on square (5,5) and dragon kings on squares (0, 5), (1, 1), (2,4), (3,6), (4, 2), (5, 0), (5,7), (6,3), and (7,5).…”

“…4. In this case, for n = 11 consider the 11 × 11 board with a pawn on square (5,5) and dragon kings on squares (0, 8), (1, 6), (2, 0), (3,5), (4,1), (5,3), (5,7), (6,9), (7,5), (8,10), (9, 4), and (10, 2). For n = 11 + 4j use Lemma 15 inductively.…”