The way to regain the algorithm--board cover

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Subordinate to recursion and division

Board coverage issues

Problem Description:  

In a chessboard of 2^kx2^k squares, if a square is different from other squares, it is a special square and is called a special chessboard. In the board coverage problem, the four ' l ' type dominoes shown cover all squares in a given special checkerboard except for special squares, and any two ' l ' type dominoes cannot overlap. (and know that in any 2^KX2^K board cover, the number of L-type dominoes used is exactly (4^k-1)/3.)

Problem Solving: the problem is solved by a divide-and-conquer strategy,

When k > 0 o'clock, the 2^kx2^k chessboard is divided into 4 2^ (k-1) x2^ (k-1) sub-chessboard, such as:

Special squares must be located in one of the four boys ' chessboard, and there are no special squares in the remaining three chessboard.

In order to cover the remaining three sub-chessboard with the ' L ' type dominoes, a special square is set in each checkerboard,

It can be found that these three squares can form an ' L ' type Domino, such as:

In this way, it is possible to recursively go down until the board is 1x1 in size.

Algorithm code:

<span style= "Font-family:comic Sans ms;font-size:14px;"    >void cheseboard (int lr, int lc, INT sr, int SC, int _size) {if (_size = = 1) return;    int t = tile++, s = _SIZE/2;    Determine if the special lattice is in the upper left corner of the four regions if (SR < LR + S && SC < LC + s) cheseboard (lr,lc,sr,sc,s);        else {board[lr+s-1][lc+s-1]=t;    Cheseboard (lr,lc,lr+s-1,lc+s-1,s);    }//Determine if the special lattice is in the upper right corner of the four regions if (SR < LR + S && SC >= LC + s) cheseboard (lr,lc+s,sr,sc,s);        else {board[lr+s-1][lc+s]=t;    Cheseboard (lr,lc+s,lr+s-1,lc+s,s);    }//Determine if the special lattice is in four areas in the lower left corner area if (SR >= LR + S && SC < LC + s) cheseboard (lr+s,lc,sr,sc,s);        else {board[lr+s][lc+s-1]=t;    Cheseboard (lr+s,lc,lr+s,lc+s-1,s);    }//Determine if the special lattice is in the lower right corner of the four regions if (SR >= LR + S && SC >= LC + s) cheseboard (lr+s,lc+s,sr,sc,s);        else {board[lr+s][lc+s]=t;  Cheseboard (lr+s,lc+s,lr+s,lc+s,s);  }}</span> 

there should be no need to explain a lot, simply say the meaning of each parameter:

LR,LC Horizontal and vertical axis in the upper left corner

Horizontal and vertical axis of SR,SC special lattice

_size size of the chessboard (chessboard to 2^k)

Algorithm Analysis:

Set T (k) is the time required for the algorithm Cheseboard to cover a 2^kx2^k chessboard, which can be calculated from the algorithm's divide-and-conquer strategy:

T (k)= O (1) when k=0

=4*t (k-1) +o (1) when k>0

The recursive equation can be solved by T (k) = O (4^k).

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The way to regain the algorithm--board cover