Board coverage issues

In a chessboard of 2^k * 2^k squares, if a square is different from other squares, it is called a special square, which is called a special chessboard. It is obvious that special squares appear on the board in the 4^k case. Thus for any k>=0, there are 4^k different special chessboard. The special chessboard shown is one of the 16 special chessboard in k=2.

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In the board coverage problem, the different forms of the L-type dominoes in 4 are used to cover all squares except special squares on a given special chess card, and any 2 L-type dominoes shall not overlap. It is easy to know that in any 2^k * 2^k chessboard, the number of L-type dominoes used is exactly (4^k-1)/3.

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A simple algorithm for solving the checkerboard problem can be designed by using the divide-and-conquer strategy.
When k>0, divide the 2^k * 2^k checkerboard into 4 2^ (k-1) * 2^ (k-1) sub-chessboard as shown.

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Special squares must be located in one of the 4 smaller sub-chessboard, with no special squares in the remaining 3 sub-chessboard. In order to convert the 3 sub-chessboard without special squares into a special chessboard, we can cover the 3 smaller chessboard with an L-shaped domino, as shown above, the squares covered by the L-shaped dominoes on the 3 sub-chessboard become special squares on the chessboard, thus turning the original problem into a 4 smaller checkerboard coverage problem. This segmentation is used recursively until the checkerboard is simplified to a 1x1 checkerboard.

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  Board coverage issues #include<iostream>using namespace std;int board[16][16];int tile=0;  int chessboard (int tr, int tc, int dr, int dc, int  Size) {if (size == 1) return 0;int t=tile++,s=size/2;//  upper left if (dr<tr+s &&  dc<tc+s) chessboard (tr,tc,dr,dc,s); else{board[tr+s-1][tc+s-1]=t; Chessboard (tr,tc,tr+s-1,tc+s-1,s);}   Upper Right if (dr>=tr+s && dc<tc+s) chessboard (tr+s,tc,dr,dc,s); else{board[tr+s][tc+s-1]= T Chessboard (tr+s,tc,tr+s,tc+s-1,s);}   Lower left if (dr<tr+s && dc>=tc+s) chessboard (tr,tc+s,dr,dc,s); else{board[tr+s-1][tc+s]= T Chessboard (tr,tc+s,tr+s-1,tc+s,s);}   Lower right if (dr>=tr+s && dc>=tc+s) chessboard (tr+s,tc+s,dr,dc,s); Else{board[tr+s][tc+s] =t; Chessboard (tr+s,tc+s,tr+s,tc+s,s);}} Int main () {chessboard (0,0,3,2,16); for (int i=0;i<16;i++) {for (int j=0;j<16;j++) cout<< Board[i][j]<< " \ t "; Cout<<endl;cout<<endl;}}

Board coverage issues