Tiling Problem using Divide and Conquer algorithm
Given a n by n board where n is of form 2k where k >= 1 (Basically n is a power of 2 with minimum value as 2). The board has one missing cell (of size 1 x 1). Fill the board using L shaped tiles. A L shaped tile is a 2 x 2 square with one cell of size 1×1 missing.
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Figure 1: An example input
This problem can be solved using Divide and Conquer. Below is the recursive algorithm.
// n is size of given square, p is location of missing cell Tile(int n, Point p) 1) Base case: n = 2, A 2 x 2 square with one cell missing is nothing but a tile and can be filled with a single tile. 2) Place a L shaped tile at the center such that it does not cover the n/2 * n/2 subsquare that has a missing square. Now all four subsquares of size n/2 x n/2 have a missing cell (a cell that doesn't need to be filled). See figure 2 below. 3) Solve the problem recursively for following four. Let p1, p2, p3 and p4 be positions of the 4 missing cells in 4 squares. a) Tile(n/2, p1) b) Tile(n/2, p2) c) Tile(n/2, p3) d) Tile(n/2, p3)
The below diagrams show working of above algorithm
Figure 2: After placing first tile
Figure 3: Recurring for first subsquare.
Figure 4: Shows first step in all four subsquares.
Input : size = 2 and mark coordinates = (0, 0) Output : -1 1 1 1 Coordinate (0, 0) is marked. So, no tile is there. In the remaining three positions, a tile is placed with its number as 1. Input : size = 4 and mark coordinates = (0, 0) Output : -1 3 2 2 3 3 1 2 4 1 1 5 4 4 5 5
Below is C++ implementation of above idea:
-1 9 8 8 4 4 3 3 9 9 7 8 4 2 2 3 10 7 7 11 5 5 2 6 10 10 11 11 1 5 6 6 14 14 13 1 1 19 18 18 14 12 13 13 19 19 17 18 15 12 12 16 20 17 17 21 15 15 16 16 20 20 21 21
Recurrence relation for above recursive algorithm can be written as below. C is a constant.
T(n) = 4T(n/2) + C
The above recursion can be solved using Master Method and time complexity is O(n2)
How does this work?
The working of Divide and Conquer algorithm can be proved using Mathematical Induction. Let the input square be of size 2k x 2k where k >=1.
Base Case: We know that the problem can be solved for k = 1. We have a 2 x 2 square with one cell missing.
Induction Hypothesis: Let the problem can be solved for k-1.
Now we need to prove to prove that the problem can be solved for k if it can be solved for k-1. For k, we put a L shaped tile in middle and we have four subsqures with dimension 2k-1 x 2k-1 as shown in figure 2 above. So if we can solve 4 subsquares, we can solve the complete square.
This article is contributed by Abhay Rathi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.