Forming a Magic Square

Problem Statement :

We define a magic square to be an n * n matrix of distinct positive integers from 1 to n^2 where the sum of any row, column, or diagonal of length n is always equal to the same number: the magic constant.

You will be given a 3 * 3 matrix s of integers in the inclusive range [1, 9]. We can convert any digit a to any other digit b in the range [1, 9] at cost of | a - b |. Given s, convert it into a magic square at minimal cost. Print this cost on a new line.

Note: The resulting magic square must contain distinct integers in the inclusive range [1, 9].


$s = [[5, 3, 4], [1, 5, 8], [6, 4, 2]]

The matrix looks like this:

5 3 4
1 5 8
6 4 2
We can convert it to the following magic square:

8 3 4
1 5 9
6 7 2
This took three replacements at a cost of |5 - 8| + |8 - 9| + |4 - 7| = 7.

Function Description

Complete the formingMagicSquare function in the editor below.

formingMagicSquare has the following parameter(s):

int s[3][3]: a 3 * 3 array of integers

int: the minimal total cost of converting the input square to a magic square

Input Format

Each of the 3 lines contains three space-separated integers of row s[i].

s[i][j] belongs to [1, 9]

Solution :


                            Solution in C :

python 3  :

import itertools
s = []
for i in range(3):
    s.extend(list(map(int, input().split(" "))))

min_cost = 1000
best = None
def is_magic(s):
    for i in range(3):
        if sum(s[i*3:i*3+3]) != 15:
            return False
        if sum(s[i::3]) != 15:
            return False
    if s[0] + s[4] + s[8] != 15:
        return False
    if s[2] + s[4] + s[6] != 15:
        return False
    return True

best = None
for p in itertools.permutations(range(1,10)):
    cost = sum([abs(p[i] - s[i]) for i in range(len(s))])
    if cost < min_cost and is_magic(p):
        min_cost = cost
        best = p

Java  :

import java.util.*;
public class crap {
    public static int diff(int[][] s1,int[][] s2){
        int d=0;
        for(int i=0;i<3;i++)
            for(int j=0;j<3;j++)
        return d;
    public static void main(String[] args) {
        Scanner sc=new Scanner(;
        int s[][]=new int[3][3];
        for(int i=0;i<3;i++)
           for(int j=0;j<3;j++)
        int a[][]={{2,7,6},{9,5,1},{4,3,8}};
        int b[][]={{2,9,4},{7,5,3},{6,1,8}};
        int c[][]={{4,3,8},{9,5,1},{2,7,6}};
        int d[][]={{4,9,2},{3,5,7},{8,1,6}};
        int e[][]={{6,1,8},{7,5,3},{2,9,4}};
        int f[][]={{6,7,2},{1,5,9},{8,3,4}};
        int g[][]={{8,1,6},{3,5,7},{4,9,2}};
        int h[][]={{8,3,4},{1,5,9},{6,7,2}};
        ArrayList<int[][]> val=new ArrayList<>();
        int res=Integer.MAX_VALUE;
        for(int[][] x:val){
            res=Math.min(res, diff(x, s));

C++  :

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    int arr[9] =    {4,9,2,3,5,7,8,1,6} ;
    int arr1[9] =   {2,7,6,9,5,1,4,3,8} ;
    int arr2[9] =   {6,1,8,7,5,3,2,9,4};
    int arr3[9] =   {2,9,4,7,5,3,6,1,8};
    int arr4[9] =   {6,7,2,1,5,9,8,3,4};
    int arr5[9] =   {8,1,6,3,5,7,4,9,2};
    int arr6[9] =   {8,3,4,1,5,9,6,7,2};
    int arr7[9] =   {4,3,8,9,5,1,2,7,6};
    int ans[8] = {0};
    for(int i =0;i<9;i++)
        int k;cin>>k;
        ans[0] += abs(k - arr[i]);
        ans[1] += abs(k - arr1[i]);
        ans[2] += abs(k - arr2[i]);
        ans[3] += abs(k - arr3[i]);
        ans[4] += abs(k - arr4[i]);
        ans[5] += abs(k - arr5[i]);
        ans[7] += abs(k - arr7[i]);
        ans[6] += abs(k - arr6[i]);
    int min =ans[0];
    for(int i=0;i<8;i++)
            min = ans[i];
        return 0;

C  :

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <limits.h>

int main() {
    int n = 3;
    int a[3][3] = {4,9,2,3,5,7,8,1,6}; 
    int b[3][3];
    int cost = INT_MAX;
    for(int i=0;i<3;i++) {
        for(int j=0;j<3;j++) {
            scanf("%i", &b[i][j]);
    for(int k=0; k<4; k++) {
        int diff1 = 0, diff2 = 0;
        for(int i=0;i<n;i++) {
            for(int j=0;j<n;j++) {
                diff1 += abs(a[i][j]-b[i][j]);
                diff2 += abs(a[i][n-1-j]-b[i][j]);
            cost = diff1;
            cost = diff2;
        for(int i=0; i<n/2; i++) {
            for(int j=0; j<(n+1)/2; j++) {
                int temp = b[i][j];
                b[i][j] = b[n-1-j][i];
                b[n-1-j][i] = b[n-1-i][n-1-j];
                b[n-1-i][n-1-j] = b[j][n-1-i];
                b[j][n-1-i] = temp;
    printf("%i", cost);
    return 0;

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