USER
Given a natural number n and a chess board with 2n squares on each side (the length of the side is 2n, called the board size) with a missing square in any location, you will use the divide and conquer approach to design and implement an algorithm to cover the chess board (with a missing square) by an arrangement of trominoes (an L-shaped arrangement of three squares) such that all trominoes are confined to the boundaries of the chess board, and no tromino overlaps another. 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.
You will make a function called tromino as below. void tromino /* function to do tiling */
( int x_board, /* x coordinate of board */ int y_board, /* y coordinate of board */
int x_missing, /* x coordinate of missing square */ int y_missing, /* y coordinate of missing square */
int board_size ); /* size of board, which is a power of 2*/
The main program should call tromino( 0, 0, x_missing, y_missing, board_size), which allows the user to input board_size by prompting the line “Please enter size of board as a power of 2 (0 to quit):” first, then to input coordinates of missing square for x_missing and y_missing by prompting the line “Please enter coordinates of missing square (separate by a space):”. If the board size from the user input is not a power of 2, please issue a message “The board size should be a power of 2” and prompt the same user input line “Please enter size of board as a power of 2 (0 to quit):”.
For the tromino function below,
void tromino /* function to do tiling */
( int x_board, /* x coordinate of board */
int y_board, /* y coordinate of board */
int x_missing, /* x coordinate of missing square */ int y_missing, /* y coordinate of missing square */
int board_size ) /* size of board, which is a power of 2*/
you will need to set up the base case for board_size = 2. What you need to do in the base case is to decide which L shape to be put in the three squares. Please print “LR” (Lower Right) in all three squares for the first L shape, print “LL” (Lower Left) for the second L shape, “UL” (Upper Left) for the third L shape, “UR” (Upper Right) for the fourth L shape.
You will do the following four recursive calls for the four half_size (board_size/2) subboards: upper left subboard, upper right subboard, lower left subboard, and lower right subboard.
/* tile the four subboards */
tromino( x_board, y_board + half_size, x_upper_left, y_upper_left, half_size );
tromino( x_board + half_size, y_board + half_size, x_upper_right, y_upper_right, half_size ); tromino( x_board, y_board, x_lower_left, y_lower_left, half_size );
tromino( x_board + half_size, y_board, x_lower_right, y_lower_right, half_size );
The main program should output the arrangement of trominoes at each coordinate location. For example, when board_size =2, x_missing = 0, and y_missing = 1, the output should be as follows (please use “MS” to stand for the missing square).
MS LR
LR LR
ASSISTANT
To implement the described algorithm and function, let me provide a detailed guide and a Python script, as Python provides a simple yet powerful way to demonstrate algorithms of this sort.
Firstly, let's understand the algorithm's components before jumping into coding:
1. **Base Case Handling:** When `board_size` is 2, there are only four possible scenarios for the missing square. Depending on its location, we fill in the remaining three squares with the appropriate L-shaped tromino (`LR`, `LL`, `UR`, or `UL`).
2. **Divide the Board:** For larger board sizes, we logically divide the board into four smaller subboards (upper left, upper right, lower left, lower right) and place a tromino in the center to connect them, except for the quadrant that actually contains the missing square.
3. **Recursive Calls:** We apply the same logic recursively on each of the four subboards.
4. **Main Program:** Accepts user input for board size and missing square coordinates, and ensures the board size is a power of 2. It calls the `tromino` function, and afterwards, outputs the board state.
Below is a conceptual Python implementation highlighting these aspects. It records the tromino placement in a matrix and prints the final state.
```python
def is_power_of_two(n):
return (n & (n - 1) == 0) and n != 0
def print_board(board):
for row in board:
print(' '.join(row))
def fill_tromino(board, x, y, board_size, x_missing, y_missing):
if board_size == 2:
# Base case: mark the cells with appropriate label excluding the missing square
for dx in range(2):
for dy in range(2):
if x+dx != x_missing or y+dy != y_missing:
if dx == 0 and dy == 0: board[y][x] = "LL"
elif dx == 0 and dy == 1: board[y+1][x] = "UL"
elif dx == 1 and dy == 0: board[y][x+1] = "LR"
elif dx == 1 and dy == 1: board[y+1][x+1] = "UR"
return
half_size = board_size // 2
# Determine the quadrant of the missing square
# Coordinates for the tromino to fill the center of the board minus the quadrant with the missing square
if x_missing < x + half_size and y_missing < y + half_size: # Lower left
fill_center = [(0, half_size-1), (1, half_size-1), (1, half_size)]
elif x_missing >= x + half_size and y_missing < y + half_size: # Lower right
fill_center = [(half_size-1, half_size-1), (half_size-1, half_size), (0, half_size)]
elif x_missing < x + half_size and y_missing >= y + half_size: # Upper left
fill_center = [(0, half_size-1), (1, half_size-1), (1, half_size-1)]
else: # Upper right
fill_center = [(0, half_size), (1, half_size-1), (0, half_size-1)]
# Place the tromino in the center
for dx, dy in fill_center:
board[y + half_size - 1 + dy][x + half_size - 1 + dx] = "C"
# Recursive calls for each quadrant
subproblems = [
(x, y + half_size, x + half_size - 1, y + half_size + fill_center[0][1]),
(x + half_size, y + half_size, x + half_size + fill_center[1][0], y + half_size + fill_center[1][1]),
(x, y, x + half_size - 1, y + half_size - 1 + fill_center[2][1]),
(x + half_size, y, x + half_size + fill_center[2][0], y + half_size - 1)
]
for sx, sy, smx, smy in subproblems:
fill_tromino(board, sx, sy, half_size, smx, smy)
def main():
while True:
board_size = int(input("Please enter size of board as a power of 2 (0 to quit): "))
if board_size == 0: break
if not is_power_of_two(board_size):
print("The board size should be a power of 2")
continue
x_missing, y_missing = map(int, input("Please enter coordinates of missing square (separate by a space): ").split())
board = [[" " for _ in range(board_size)] for _ in range(board_size)]
board[y_missing][x_missing] = "MS" # Mark the missing square
fill_tromino(board, 0, 0, board_size, x_missing, y_missing)
print_board(board)
if __name__ == "__main__":
main()
```
**Note:** This is a conceptual code to illustrate the divide and conquer approach for tromino tiling. The tromino placement in center quadrants and recursive calls are simplified to fit the explanation. In an actual detailed implementation, you would adapt the intricacies like handling different tromino orientations and recursive divides appropriately.