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.
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 the first tile
Figure 3: Recurring for the first subsquare.
Figure 4: Shows the first step in all four subsquares.
Examples:
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 the implementation of above idea:
C++
// C++ program to place tiles#include <bits/stdc++.h>using namespace std;int size_of_grid, b, a, cnt = 0;int arr[128][128];// Placing tile at the given coordinatesvoid place(int x1, int y1, int x2, int y2, int x3, int y3){ cnt++; arr[x1][y1] = cnt; arr[x2][y2] = cnt; arr[x3][y3] = cnt;}// Quadrant names// 1 2// 3 4// Function based on divide and conquerint tile(int n, int x, int y){ int r, c; if (n == 2) { cnt++; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { if (arr[x + i][y + j] == 0) { arr[x + i][y + j] = cnt; } } } return 0; } // finding hole location for (int i = x; i < x + n; i++) { for (int j = y; j < y + n; j++) { if (arr[i][j] != 0) r = i, c = j; } } // If missing tile is 1st quadrant if (r < x + n / 2 && c < y + n / 2) place(x + n / 2, y + (n / 2) - 1, x + n / 2, y + n / 2, x + n / 2 - 1, y + n / 2); // If missing Tile is in 3rd quadrant else if (r >= x + n / 2 && c < y + n / 2) place(x + (n / 2) - 1, y + (n / 2), x + (n / 2), y + n / 2, x + (n / 2) - 1, y + (n / 2) - 1); // If missing Tile is in 2nd quadrant else if (r < x + n / 2 && c >= y + n / 2) place(x + n / 2, y + (n / 2) - 1, x + n / 2, y + n / 2, x + n / 2 - 1, y + n / 2 - 1); // If missing Tile is in 4th quadrant else if (r >= x + n / 2 && c >= y + n / 2) place(x + (n / 2) - 1, y + (n / 2), x + (n / 2), y + (n / 2) - 1, x + (n / 2) - 1, y + (n / 2) - 1); // dividing it again in 4 quadrants tile(n / 2, x, y + n / 2); tile(n / 2, x, y); tile(n / 2, x + n / 2, y); tile(n / 2, x + n / 2, y + n / 2); return 0;}// Driver program to test above functionint main(){ // size of box size_of_grid = 8; memset(arr, 0, sizeof(arr)); // Coordinates which will be marked a = 0, b = 0; // Here tile can not be placed arr[a][b] = -1; tile(size_of_grid, 0, 0); // The grid is for (int i = 0; i < size_of_grid; i++) { for (int j = 0; j < size_of_grid; j++) cout << arr[i][j] << " \t"; cout << "\n"; }} |
Java
// Java program to place tilespublic class GFG{ static int size_of_grid, b, a, cnt = 0; static int[][] arr = new int[128][128]; // Placing tile at the given coordinates static void place(int x1, int y1, int x2, int y2, int x3, int y3) { cnt++; arr[x1][y1] = cnt; arr[x2][y2] = cnt; arr[x3][y3] = cnt; } // Quadrant names // 1 2 // 3 4 // Function based on divide and conquer static int tile(int n, int x, int y) { int r = 0, c = 0; if (n == 2) { cnt++; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { if (arr[x + i][y + j] == 0) { arr[x + i][y + j] = cnt; } } } return 0; } // finding hole location for (int i = x; i < x + n; i++) { for (int j = y; j < y + n; j++) { if (arr[i][j] != 0) { r = i; c = j; } } } // If missing tile is 1st quadrant if (r < x + n / 2 && c < y + n / 2) place(x + n / 2, y + (n / 2) - 1, x + n / 2, y + n / 2, x + n / 2 - 1, y + n / 2); // If missing Tile is in 3rd quadrant else if (r >= x + n / 2 && c < y + n / 2) place(x + (n / 2) - 1, y + (n / 2), x + (n / 2), y + n / 2, x + (n / 2) - 1, y + (n / 2) - 1); // If missing Tile is in 2nd quadrant else if (r < x + n / 2 && c >= y + n / 2) place(x + n / 2, y + (n / 2) - 1, x + n / 2, y + n / 2, x + n / 2 - 1, y + n / 2 - 1); // If missing Tile is in 4th quadrant else if (r >= x + n / 2 && c >= y + n / 2) place(x + (n / 2) - 1, y + (n / 2), x + (n / 2), y + (n / 2) - 1, x + (n / 2) - 1, y + (n / 2) - 1); // dividing it again in 4 quadrants tile(n / 2, x, y + n / 2); tile(n / 2, x, y); tile(n / 2, x + n / 2, y); tile(n / 2, x + n / 2, y + n / 2); return 0; } // Driver code public static void main(String[] args) { // size of box size_of_grid = 8; // Coordinates which will be marked a = 0; b = 0; // Here tile can not be placed arr[a][b] = -1; tile(size_of_grid, 0, 0); // The grid is for (int i = 0; i < size_of_grid; i++) { for (int j = 0; j < size_of_grid; j++) System.out.print(arr[i][j] + " "); System.out.println();; } }}// This code is contributed by divyeshrabadiya07. |
C#
// C# program to place tilesusing System;class GFG{ static int size_of_grid, b, a, cnt = 0; static int[,] arr = new int[128, 128]; // Placing tile at the given coordinates static void place(int x1, int y1, int x2, int y2, int x3, int y3) { cnt++; arr[x1, y1] = cnt; arr[x2, y2] = cnt; arr[x3, y3] = cnt; } // Quadrant names // 1 2 // 3 4 // Function based on divide and conquer static int tile(int n, int x, int y) { int r = 0, c = 0; if (n == 2) { cnt++; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { if (arr[x + i, y + j] == 0) { arr[x + i, y + j] = cnt; } } } return 0; } // finding hole location for (int i = x; i < x + n; i++) { for (int j = y; j < y + n; j++) { if (arr[i, j] != 0) { r = i; c = j; } } } // If missing tile is 1st quadrant if (r < x + n / 2 && c < y + n / 2) place(x + n / 2, y + (n / 2) - 1, x + n / 2, y + n / 2, x + n / 2 - 1, y + n / 2); // If missing Tile is in 3rd quadrant else if (r >= x + n / 2 && c < y + n / 2) place(x + (n / 2) - 1, y + (n / 2), x + (n / 2), y + n / 2, x + (n / 2) - 1, y + (n / 2) - 1); // If missing Tile is in 2nd quadrant else if (r < x + n / 2 && c >= y + n / 2) place(x + n / 2, y + (n / 2) - 1, x + n / 2, y + n / 2, x + n / 2 - 1, y + n / 2 - 1); // If missing Tile is in 4th quadrant else if (r >= x + n / 2 && c >= y + n / 2) place(x + (n / 2) - 1, y + (n / 2), x + (n / 2), y + (n / 2) - 1, x + (n / 2) - 1, y + (n / 2) - 1); // dividing it again in 4 quadrants tile(n / 2, x, y + n / 2); tile(n / 2, x, y); tile(n / 2, x + n / 2, y); tile(n / 2, x + n / 2, y + n / 2); return 0; } // Driver code static void Main() { // size of box size_of_grid = 8; // Coordinates which will be marked a = 0; b = 0; // Here tile can not be placed arr[a, b] = -1; tile(size_of_grid, 0, 0); // The grid is for (int i = 0; i < size_of_grid; i++) { for (int j = 0; j < size_of_grid; j++) Console.Write(arr[i,j] + " "); Console.WriteLine(); } }}// This code is contributed by divyesh072019. |
Python3
# Python3 program to place tilessize_of_grid = 0b = 0a = 0cnt = 0arr = [[0 for i in range(128)] for j in range(128)]def place(x1, y1, x2, y2, x3, y3): global cnt cnt += 1 arr[x1][y1] = cnt; arr[x2][y2] = cnt; arr[x3][y3] = cnt; def tile(n, x, y): global cnt r = 0 c = 0 if (n == 2): cnt += 1 for i in range(n): for j in range(n): if(arr[x + i][y + j] == 0): arr[x + i][y + j] = cnt return 0; for i in range(x, x + n): for j in range(y, y + n): if (arr[i][j] != 0): r = i c = j if (r < x + n / 2 and c < y + n / 2): place(x + int(n / 2), y + int(n / 2) - 1, x + int(n / 2), y + int(n / 2), x + int(n / 2) - 1, y + int(n / 2)) elif(r >= x + int(n / 2) and c < y + int(n / 2)): place(x + int(n / 2) - 1, y + int(n / 2), x + int(n / 2), y + int(n / 2), x + int(n / 2) - 1, y + int(n / 2) - 1) elif(r < x + int(n / 2) and c >= y + int(n / 2)): place(x + int(n / 2), y + int(n / 2) - 1, x + int(n / 2), y + int(n / 2), x + int(n / 2) - 1, y + int(n / 2) - 1) elif(r >= x + int(n / 2) and c >= y + int(n / 2)): place(x + int(n / 2) - 1, y + int(n / 2), x + int(n / 2), y + int(n / 2) - 1, x + int(n / 2) - 1, y + int(n / 2) - 1) tile(int(n / 2), x, y + int(n / 2)); tile(int(n / 2), x, y); tile(int(n / 2), x + int(n / 2), y); tile(int(n / 2), x + int(n / 2), y + int(n / 2)); return 0size_of_grid = 8a = 0b = 0arr[a][b] = -1tile(size_of_grid, 0, 0)for i in range(size_of_grid): for j in range(size_of_grid): print(arr[i][j], end=" ") print()# This code is contributed by rag2127 |
Javascript
<script>// Javascript program to place tilesvar size_of_grid, b, a, cnt = 0;var arr = Array.from(Array(128), ()=>Array(128).fill(0));// Placing tile at the given coordinatesfunction place(x1, y1, x2, y2, x3, y3){ cnt++; arr[x1][y1] = cnt; arr[x2][y2] = cnt; arr[x3][y3] = cnt;}// Quadrant names// 1 2// 3 4// Function based on divide and conquerfunction tile(n, x, y){ var r, c; if (n == 2) { cnt++; for (var i = 0; i < n; i++) { for (var j = 0; j < n; j++) { if (arr[x + i][y + j] == 0) { arr[x + i][y + j] = cnt; } } } return 0; } // finding hole location for (var i = x; i < x + n; i++) { for (var j = y; j < y + n; j++) { if (arr[i][j] != 0) r = i, c = j; } } // If missing tile is 1st quadrant if (r < x + n / 2 && c < y + n / 2) place(x + n / 2, y + (n / 2) - 1, x + n / 2, y + n / 2, x + n / 2 - 1, y + n / 2); // If missing Tile is in 3rd quadrant else if (r >= x + n / 2 && c < y + n / 2) place(x + (n / 2) - 1, y + (n / 2), x + (n / 2), y + n / 2, x + (n / 2) - 1, y + (n / 2) - 1); // If missing Tile is in 2nd quadrant else if (r < x + n / 2 && c >= y + n / 2) place(x + n / 2, y + (n / 2) - 1, x + n / 2, y + n / 2, x + n / 2 - 1, y + n / 2 - 1); // If missing Tile is in 4th quadrant else if (r >= x + n / 2 && c >= y + n / 2) place(x + (n / 2) - 1, y + (n / 2), x + (n / 2), y + (n / 2) - 1, x + (n / 2) - 1, y + (n / 2) - 1); // dividing it again in 4 quadrants tile(n / 2, x, y + n / 2); tile(n / 2, x, y); tile(n / 2, x + n / 2, y); tile(n / 2, x + n / 2, y + n / 2); return 0;}// Driver program to test above function// size of boxsize_of_grid = 8;// Coordinates which will be markeda = 0, b = 0;// Here tile can not be placedarr[a][b] = -1;tile(size_of_grid, 0, 0);// The grid isfor (var i = 0; i < size_of_grid; i++) { for (var j = 0; j < size_of_grid; j++) document.write(arr[i][j] + " "); document.write("<br>");}// This code is contributed by rutvik_56.</script> |
Output
-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
Time Complexity:
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)
Space Complexity: O(n)
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 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 subsquares 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.
References:
http://www.comp.nus.edu.sg/~sanjay/cs3230/dandc.pdf
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.
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