In the first case there is a board with dimensions 2*2 with 4 squares inside it of 1*1 dimension. The first square is empty i.e. it is a missing square in the upper left corner, so the L-shaped arrangement for this case will be LR and it will be filled at the remaining coordinates as it has the corner for the L-shaped arrangement at the lower right. Similarly, In the second case a missing square is there in the upper right corner so the L-shaped arrangement for this case will be LL and it will be filled at the remaining coordinates as it has the corner for the L-shaped arrangement at the lower left, in the third case a missing square is there in the lower right corner so the L-shaped arrangement for this case will be UL and it will be filled at the remaining coordinates as it has the corner for the L-shaped arrangement at the upper left and in the fourth case a missing square is there in the lower left corner so the L-shaped arrangement for this case will be UR and it will be filled at the remaining coordinates as it has the corner for the L-shaped arrangement at the upper right

We will use x coordinate and y coordinate to record the location of the missing square. It is convenient to treat the origin (0, 0) as the lower left corner of the board. import java.util.Scanner;

public class TrominoTiling {
private static String[][] board;

public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.print("Please enter size of board (need to be 2^n and 0 to quit): ");

int boardSize = scanner.nextInt();
if (boardSize == 0) {
scanner.close();
return;
}

System.out.print("Please enter coordinates of missing square (separated by a space): ");
int xMissing = scanner.nextInt();
int yMissing = scanner.nextInt();

// Initialize the board
initializeBoard(boardSize);

// Mark the initially missing square
board[yMissing][xMissing] = “MS”;

// Fill the board with trominoes
fillBoard(0, 0, boardSize, xMissing, yMissing);

// Print the result
for (int y = boardSize - 1; y >= 0; y–) { // Start from the top for display
for (int x = 0; x < boardSize; x++) {
System.out.print(board[y][x] + “\t”);
}
System.out.println();
}

scanner.close();
}

private static void initializeBoard(int boardSize) {
board = new String[boardSize][boardSize];
for (int i = 0; i < boardSize; i++) {
for (int j = 0; j < boardSize; j++) {
board[i][j] = " "; // Initialize with spaces
}
}
}

private static void fillBoard(int x, int y, int size, int xMissing, int yMissing) {
// Base condition: If the board size reaches 2x2, place a tromino around the missing square
if (size == 2) {
// Mark the current batch of trominoes with a unique identifier or simple labels based on the spec
String trominoMarker = “UR”; // Default placement, can vary based on conditions below
for (int i = 0; i < size; i++) {
for (int j = 0; j < size; j++) {
if (x + i != xMissing || y + j != yMissing) {
// Here, adjust the marker based on the quadrant of the missing tile
if (xMissing == x && yMissing == y) trominoMarker = “UR”; // Lower Left
else if (xMissing == x + 1 && yMissing == y) trominoMarker = “UL”; // Lower Right
else if (xMissing == x && yMissing == y + 1) trominoMarker = “LR”; // Upper Left
else if (xMissing == x + 1 && yMissing == y + 1) trominoMarker = “LL”; // Upper Right

board[y + j][x + i] = trominoMarker;
}
}
}
return;
}

int halfSize = size / 2;
// Determine the quadrant of the missing square relative to the current board section
// Recursively fill each quadrant
// Adjust xMissing and yMissing for each recursive call to be within the new quadrant bounds
int newXMissing = xMissing < x + halfSize ? xMissing : x + halfSize - 1;
int newYMissing = yMissing < y + halfSize ? yMissing : y + halfSize - 1;

// Quadrant 1 - Lower Left
fillBoard(x, y, halfSize, xMissing, yMissing);
// Quadrant 2 - Lower Right
fillBoard(x + halfSize, y, halfSize, xMissing >= x + halfSize ? xMissing : newXMissing, yMissing);
// Quadrant 3 - Upper Left
fillBoard(x, y + halfSize, halfSize, xMissing, yMissing >= y + halfSize ? yMissing : newYMissing);
// Quadrant 4 - Upper Right
fillBoard(x + halfSize, y + halfSize, halfSize, xMissing >= x + halfSize ? xMissing : newXMissing, yMissing >= y + halfSize ? yMissing : newYMissing);
}

} Please enter size of board (need to be 2^n and 0 to quit): 4
Please enter coordinates of missing square (separated by a space): 1 3
LL MS UR UR
LL LL UR UR
UR UR UR UR
UR UR UR UR the above code is giving this output which is incorrect. The MS is placed at the correct position but the other positions are incorrect. The program must fill the other squares with the L-shaped arrangement that I have already specified. What you need to do is to decide which L shape to be put in the remaining squares after the missing square is placed correctly at its position. The correct and expected output when the similar input is given should be Please enter size of board (need to be 2^n and 0 to quit): 4
Please enter coordinates of missing square (seperated by a space): 1 3
LL MS UR UR
LL LL LR UR
LL LR LR LR
LL LL LR LR where the other positions except MS are filled with LL,LR,UL,UR according to the L-shaped arrangement that I had specified earlier. The remaining squares should be filled with these values. The chess board (with a missing square) should be covered by an arrangement of trominoes (an L-shaped arrangement of three squares; refer to the four cases that I had specified earlier) such that all trominoes are confined to the boundaries of the chess board, and no tromino overlaps another.