Chess board problem(divide and conquer)

Chess board problem(divide and conquer)

• 1.
  DIVIDE AND CONQUERALGORITHM
• 2.
  CHESS BOARD PROBLEM- DIVIDEAND CONQUER
• 3.
  The Defective ChessboardProblem 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 possibleto 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 2x2 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 4chessboard.
• 7.
  8X8 DEFECTIVE CHESSBOARD Step-1 One of the cell is defective Step- 2 We divide the chess board into equal sub half's.
• 8.
  Creation of defectivebox 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 wehave 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 finalchess 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 4x48x8
• 14.
  ALGORITHIM 1. For a2x2 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 ForDefective 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.