Magic Square

This is very simple if you think in a simple manner.

A magic square is represented in a matrix form. May be all of you know this formula to satisfy magic square rule i.e. for size of matrix n*n sum of each row or column or diagonal should be n[n*n+1]/2.

means

for n=3 , sum(row or column or diag) = 15
for n=4 , sum(row or column or diag) = 34
for n=5 , sum(row or column or diag) = 65​

You can say here n is order of magic square.

Example: for order of magic square 4

16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1


Now Just think it in a programmitically way!!!

Code:

void BuildMagicSquare()
{
	int order=5,loop;
	int MagicMat[5][5];
	int SqrOfOrder = order * order;
	int nCount=0, mid=order/2;
	for(loop=1; loop<=SqrOfOrder ; ++loop) 
	{
		MagicMat[nCount--][mid++] = loop;
		if (loop % order == 0) 
		{ 
			nCount += 2; 
			--mid; 
		}
		else 
		{
			if (mid==order) 
				mid -= order;
			else if (nCount<0) 
				nCount += order;
		}
	}
}

 

I think this will work only for odd Order.

means for n=1,3,5,7, etc

 

I putted wrongly!!! It will work for only Odd Order...
For even order it will not work.

 

Can you delete my post

 

Thanks Shabbir!!!

 

Admin, do you think this should be an Article because it does not work correctly as well as Nothing is explained.

 

Agreed

 

On that time i did'nt think for even order as raised ny Alramesh. I was working on that. but in these days i was busy so to give for even order may be late.

Really i dont know how much explain of any topic should be. I think people will ask If intrested then i will explain.

 

There is nothing personal against you but just my expectation from the Articles from this site is quite high. Look at some of the articles by Shabbir, Pradeep, bashamsc, Karpov2007 and they have code which is well commented and easy to understand even for a layman like me.

 

Sorry to all of you for delayed ... Now this code will work for all order . I have tested on Unix/ solaris

Code:

#include <vector>
#include<iostream.h>
using namespace std;
//There two series will be on even in case of magic square
// One of even order will be for multiple of 4
void BuildDoublyEvenMagicSquare(vector<vector<int> > &mat, int Order);
//Other of even order will be for multiple of 2
void SinglyEvenMagicSquare(vector<vector<int> > &mat, int order);
// For odd order
void BuildOddMagicSquare(vector<vector<int> > &mat, int Order);

// For odd order
void BuildOddMagicSquare(vector<vector<int> > &mat, int Order)
{
  int SqrOfOrder = Order * Order;
  int start=0, mid=Order/2;     // start position
  for (int loop=1; loop<=SqrOfOrder; ++loop)
  {
    mat[start--][mid++] = loop;
    if (loop % Order == 0)
    {
      start += 2;
      --mid;
    }
    else 
    {        
      if (mid==Order) 
        mid -= Order;
      else if (start<0)
        start += Order;
    }    
  }     
}
             
void BuildDoublyEvenMagicSquare(vector<vector<int> > &mat, int Order)
{
  vector<vector<int> > A(Order, vector<int> (Order, 0));
  vector<vector<int> > B(Order, vector<int> (Order, 0));
  int i, j;
  //Building of matrixes I and  J
  int index=1; 
  for (i=0; i<Order; i++)
    for (j=0; j<Order; j++)
    {
      A[i][j]=((i+1)%4)/2;
      B[j][i]=((i+1)%4)/2;
      mat[i][j]=index;
      index++;
    }
  for (i=0; i<Order; i++)
    for (j=0; j<Order; j++)
    {
      if (A[i][j]==B[i][j])
        mat[i][j]=Order*Order+1-mat[i][j];
    }
}

void BuildSinglyEvenMagicSquare(vector<vector<int> > &mat, int order)
{
  int ho=order/2;

  vector<vector<int> > C(ho, vector<int> (ho, 0));

   // For Order is Odd
    if (order%2==1)
      BuildOddMagicSquare(C, order);

   // For Order is Even
   else
   {
    //For Order is Doubly Even Order
    if (order % 4==0)
      BuildDoublyEvenMagicSquare(C, order);
     //For Order is Singly Even Order
    else
      BuildSinglyEvenMagicSquare(C, order);
   }
  int i, j, k;
  for (i=0; i<ho; i++)
    for (j=0; j<ho; j++)
    {
      mat[i][j]=C[i][j];
      mat[i+ho][j]=C[i][j]+3*ho*ho;
      mat[i][j+ho]=C[i][j]+2*ho*ho;
      mat[i+ho][j+ho]=C[i][j]+ho*ho;
    }
  if (order==2)
    return;

  vector<int> A(ho, 0);
  vector<int> B;

  for (i=0; i<ho; i++)
    A[i]=i+1;

  k=(order-2)/4;
  for (i=1; i<=k; i++)
    B.push_back(i);

  for (i=order-k+2; i<=order; i++)
    B.push_back(i);

  int temp;
  for (i=1; i<=ho; i++)
    for (j=1; j<=B.size(); j++)
    {
      temp=mat[i-1][B[j-1]-1];
      mat[i-1][B[j-1]-1]=mat[i+ho-1][B[j-1]-1];
      mat[i+ho-1][B[j-1]-1]=temp;
    }
  i=k;
  j=0;
  temp=mat[i][j]; mat[i][j]=mat[i+ho][j]; mat[i+ho][j]=temp;
  j=i;
  temp=mat[i+ho][j]; mat[i+ho][j]=mat[i][j]; mat[i][j]=temp;
}

int main()
{
  int Order;
  cout<<"Enter the order of square which you wanna: ";
  cin>>Order;
  vector<vector<int> > mat(Order, vector<int> (Order, 0));

  // For order less than 3 is meaningless so printing error
  if (Order<3)
  {
    cout<<" Order Of Square must be greater than 2";
    return -1;
  }

   // For Order is Odd
    if (Order%2==1)
      BuildOddMagicSquare(mat, Order);

   // For Order is Even
   else
   {
    //For Order is Doubly Even Order
    if (Order % 4==0)
      BuildDoublyEvenMagicSquare(mat, Order);
     //For Order is Singly Even Order
    else
      BuildSinglyEvenMagicSquare(mat, Order);
   }


  // Display Results

  for (int i=0; i<Order; i++)
  {
    for (int j=0; j<Order; j++)
    {
      cout<< mat[i][j]<<"  " ;
    }
    cout<<endl;
  }
  return 0;
}

 

can you please try it out again and post it back,

i feel it was a good topic

 

Already explained!!! I dont think so that it should be post back.

 

please post the correct one

 

good one

 

i feel its a nice article i suppose you can post back the one without any problems

 

whatever i post in latter, It has no probs. just check it.