Given an Grid of integers. The task is to find total numbers of 3 x 3 (contiguous) Magic Square subgrids in the given grid. A Magic square is a 3 x 3 grid filled with all distinct numbers from 1 to 9 such that each row, column, and both diagonals have equal sum.
Examples:
Input: G = { { 4, 3, 8, 4 }, { 9, 5, 1, 9 }, { 2, 7, 6, 2 } }
Output: 1
Explanation: The following subgrid is a 3 x 3 magic square: [ 4 3 8, 9 5 1, 2 7 6 ]Input : G = { { 1, 2, 3, 4, 5 }, { 6, 7, 8, 9, 10 }, { 10, 11, 12, 13, 14 }, { 15, 16, 17, 18, 19 } }
Output : 0
Approach: Let us check every 3 x 3 subgrid individually. For each grid, all numbers must be unique and between (1 and 9) also every rows, columns, and both diagonals must have the equal sum.
Also notice the fact that a subgrid is a Magic Square if its middle element is 5. Because adding the 12 values from the four lines that crosses the center, add up to 60, but they also add up to the entire grid (45), plus 3 times the middle value. This implies the middle value is 5. Hence we can check this condition which help us skip over various subgrids.
You can learn more about Magic_square here or here.
The procedure to check for a subgrid to be a Magic Square is as follows:
The middle element must be 5. The sum of the grid must be 45, and contains all distinct values from 1 to 9. Each horizontal(row) and vertical(column) must add up to 15. Both of the diagonal lines must also sum to 15.
Below is the implementation of above approach:
C++
// CPP program to count magic squares #include <bits/stdc++.h> using namespace std; const int R = 3; const int C = 4; // function to check is subgrid is Magic Square int magic(int a, int b, int c, int d, int e, int f, int g, int h, int i) { set<int> s1 = { a, b, c, d, e, f, g, h, i }, s2 = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }; // Elements of grid must contain all numbers from 1 to // 9, sum of all rows, columns and diagonals must be // same, i.e., 15. if (s1 == s2 && (a + b + c) == 15 && (d + e + f) == 15 && (g + h + i) == 15 && (a + d + g) == 15 && (b + e + h) == 15 && (c + f + i) == 15 && (a + e + i) == 15 && (c + e + g) == 15) return true; return false; } // Function to cound total Magic square subgrids int CountMagicSquare(int Grid[R][C]) { int ans = 0; for (int i = 0; i < R - 2; i++) for (int j = 0; j < C - 2; j++) { // if condition true skip check if (Grid[i + 1][j + 1] != 5) continue; // check for magic square subgrid if (magic(Grid[i][j], Grid[i][j + 1], Grid[i][j + 2], Grid[i + 1][j], Grid[i + 1][j + 1], Grid[i + 1][j + 2], Grid[i + 2][j], Grid[i + 2][j + 1], Grid[i + 2][j + 2])) ans += 1; } // return total magic square return ans; } // Driver program int main() { int G[R][C] = { { 4, 3, 8, 4 }, { 9, 5, 1, 9 }, { 2, 7, 6, 2 } }; // function call to print required answer cout << CountMagicSquare(G); return 0; } // This code is written by Sanjit_Prasad |
Python3
# Python3 program to count magic squares R = 3C = 4 # function to check is subgrid is Magic Square def magic(a, b, c, d, e, f, g, h, i): s1 = set([a, b, c, d, e, f, g, h, i]) s2 = set([1, 2, 3, 4, 5, 6, 7, 8, 9]) # Elements of grid must contain all numbers # from 1 to 9, sum of all rows, columns and # diagonals must be same, i.e., 15. if (s1 == s2 and (a + b + c) == 15 and (d + e + f) == 15 and (g + h + i) == 15 and (a + d + g) == 15 and (b + e + h) == 15 and (c + f + i) == 15 and (a + e + i) == 15 and (c + e + g) == 15): return True return false # Function to cound total Magic square subgrids def CountMagicSquare(Grid): ans = 0 for i in range(0, R - 2): for j in range(0, C - 2): # if condition true skip check if Grid[i + 1][j + 1] != 5: continue # check for magic square subgrid if (magic(Grid[i][j], Grid[i][j + 1], Grid[i][j + 2], Grid[i + 1][j], Grid[i + 1][j + 1], Grid[i + 1][j + 2], Grid[i + 2][j], Grid[i + 2][j + 1], Grid[i + 2][j + 2]) == True): ans += 1 # return total magic square return ans # Driver Code if __name__ == "__main__": G = [[4, 3, 8, 4], [9, 5, 1, 9], [2, 7, 6, 2]] # Function call to print required answer print(CountMagicSquare(G)) # This code is contributed by Rituraj Jain |
1
Time Complexity: O(R * C)
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