This PPT contains a divide and conquer chessboard problem solution and its algorithm.

1. DIVIDE AND CONQUER ALGORITHM

2. CHESS BOARD PROBLEM-
DIVIDE AND CONQUER

3. The Defective Chessboard Problem
GIVEN CONDITIONS:-
1. We have a chessboard of size n x n , where n =2k
2. Exactly on square is defective in the chessboard.
3. The tiles(trominoes) are in L-shape i.e. 3 squares.
OBJECTIVE
Cover all the chessboard with L-shape tiles(trominoes), except the defective square.

4. Is it possible to solve this?
Absolutely, it is possible to cover all non-defective squares.
Let’s see how
• As the size of the chessboard is n x n and n=2k
• Therefore, Total no. of squares =2k x2k =22k
• No. of non-defective squares = 22k-1
• Now, for the value of K,22k-1 is divisible by 3.
• For E.g. K=1 , 22(1)-1 =3 is divisible by 3.
• K=2, 22(2) -1 =15 is divisible by 3.

5. Defective chessboards.
1x1 2 x2 2 x2 2 x2 2 x2
As the Size of these chessboards displayed here, 1x1 and 2x2 we don’t have to place any L-shape tile in the first and
rest can be filled by just using 1 L-shaped tile.

6. For 4x 4 chessboard.

7. 8X8 DEFECTIVE CHESS BOARD
Step-1 One of the cell is
defective
Step- 2 We divide the chess board
into equal sub half's.

8. Creation of defective box
Step- 3 Trick to cover the chess
board with tiles
Step -4 Again creation of defective boxes
as we divide the chess board
DIVISION OF
PROBLEM INTO SUB
PROBLEM

9. Step-5 As we have finally
divided the problem into 2x2
board we will put the tiles.
Step-6 The procedure will
continue until all the sub board
are covered with the tiles.

10. Step-7 The final chess board covered
with all the titles and only left with the
defectives which we created.
Step-7 Here we will cover the
defectives which we have created as in
the last, there should be only one
defective left.
COMBINIG OFALL SUB
PROBLEMS

11. 8x8
4x4 4x4
4x44x4

12. 2x2
2x2
2x2
4x4
2x2

13. 1x1 2x2 4x4 8x8

14. ALGORITHIM
1. For a 2x2 board with defective cell we just need to
add the single tile to it.
2. We will place a L shape tile in the middle such that
it does not cover the sub square in which there is
already defective. All the cell have a defective cell.
3.Repeat this process recursively till we have a 2x2
board.

15. Time Complexity For Defective Chess Board
Problem
T(n) =4T(n/2) +C
By Using Master Theorem,
T(n) =a T(n/b) + Theta(n k log p n ),a>=1,b>1,k>=0 & p is real number.
a=4,b=2,k=0,p=0
a>bk => 4 >20
Case1: If a>bk , then T(n) =Theta(nlog b a).
After putting the values of a and b the Time complexity for this problem
= n2.