C language to achieve the Magic square algorithm (magic square cube singular Rubik's Cube to achieve) _c language

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For example, the third-order demon Phalanx is:

What is the rule of the demon Phalanx?

The Magic square is divided into magical square and even magic square. And even Magic square is divided into 4 multiples (such as 4,8,12 ...). And not multiples of 4 (e.g. 6,10,14 ...). ) Two kinds. The following are described separately.

2 Odd Rubik's Cube algorithm

The rule and algorithm of 2.1 odd Rubik's Cube

Odd Rubik's Cube (Order n = 2 * m + 1,m =1,2,3 ...) ) The following laws:

The number 1 is located in the middle of the first row of the square array;
Number A (1 < a≤n2) Number of rows is 1 less than the number of rows, if the A-1 row number is 1, then A's row number is n;
Number A (1 < A≤N2) is larger than the number of A-1 columns 1, if the A-1 column number is n, then a column number 1;
If A-1 is a multiple of n, the number of rows a (1 < A≤N2) is 1 greater than the number of A-1 rows, and the number of columns is the same as A-1.

The realization of C language of 2.2 odd Rubik's Cube algorithm

Copy Code code as follows:

#include <stdio.h>
author:http://furzoom.com/
n is the cube order
#define N 11

int main ()
{
int a[n][n];
int i;
int col,row;

Col = (N-1)/2;
row = 0;

A[row][col] = 1;

for (i = 2; I <= n*n; i++)
{
if ((i-1)%N = = 0)
{
row++;
}
Else
{
If row = 0, then row = N-1, or row = Row-1
row--;
row = (row+n)%N;

If col = N, then col = 0, or col = col + 1
Col + +;
Col%= N;
}
A[row][col] = i;
}
for (row = 0;row<n;row++)
{
for (col = 0;col < N; col + +)
{
printf ("%6d", A[row][col]);
}
printf ("\ n");
}
return 0;
}

Algorithm 2: Order n = 4 * m (M =1,2,3 ...) ) The laws of the Rubik's Cube are as follows:

By numbers from small to large, that is, 1,2,3......n2 order to the magic square from left to right, from top to bottom to fill;
Dividing the magic square into several 4x4 matrices, and removing the elements on the diagonal of the Sub square;
The removed elements are then filled in the order from large to small to the vacant place in the NxN square.

C Language Implementation

Copy Code code as follows:

#include <stdio.h>
author:http://furzoom.com/
n is the cube order
#define N 12

int main ()
{
int a[n][n];//storage Rubik's Cube
int temparray[n*n/2];//storage of removed elements
int i;//loop variable
int col, row;//col column, row row

Class
i = 1;
for (row = 0;row < N; row++)
{
for (col = 0;col < N; col + +)
{
A[row][col] = i;
i++;
}
}
Remove the diagonal elements from the matrix and discharge them in the order of small to large
i = 0;
for (row = 0;row < N; row++)
{
for (col = 0;col < N; col + +)
{
if ((col% 4 = row% 4) | | (3 = = (col% 4 + row% 4))
{
Temparray[i] = A[row][col];
i++;
}
}
}
To populate the NxN square with the removed elements in order from large to small
i = N*N/2-1;
for (row = 0;row < N; row++)
{
for (col = 0;col < N; col + +)
{
if ((col% 4 = row% 4) | | (3 = = (col% 4 + row% 4))
{
A[row][col] = Temparray[i];
i--;
}
}
}
Output Phalanx
for (row = 0;row < N; row++)
{
for (col = 0;col < N; col + +)
{
printf ("%5d", A[row][col]);
}
printf ("\ n");
}
return 0;
}

3.2 Order n = 4 * m + 2 (M =1,2,3 ...) ) Rubik's Cube (single magic Cube)

Algorithm

Set k = 2 * m + 1; The Rubik's Cube is a more complicated one in the Rubik's Cube.

The Rubik's Cube is divided into a, B, C, D four K-order matrices, the following figure of the four square matrices are singular phalanx, using the method mentioned above in turn A, D, B, c filled as a magic cube.
Exchange A, C Rubik's Cube elements, the middle line of the Rubik's Cube, Exchange from the middle column to the right of the M column corresponding elements; for other rows, swap the corresponding elements from left to right m columns.
Exchange B, D cube elements, Exchange

Copy Code code as follows:

#include <stdio.h>
author:http://furzoom.com/
n is the cube order
#define N 10

int main ()
{
int A[n][n] = {{0}};//storage Rubik's Cube
int i,k,temp;
int col,row;//col column, row row

Class
K = N/2;
Col = (k-1)/2;
row = 0;
A[row][col] = 1;
Create Magic Cube A
for (i = 2; I <= k*k; i++)
{
if ((i-1)%k = = 0)//The previous number is a multiple of 3
{
row++;
}
Else
{
If row = 0, then row = N-1, or row = Row-1
row--;
row = (row+k)%k;

If col = N, then col = 0, or col = col + 1
Col + +;
Col%= K;
}
A[row][col] = i;
}

Generate B, C, D Rubik's Cube according to a
for (row = 0;row < K; row++)
{
for (col = 0;col < K; col + +)
{
A[ROW+K][COL+K] = A[row][col] + k*k;
A[ROW][COL+K] = A[row][col] + 2*k*k;
A[row+k][col] = A[row][col] + 3*k*k;
}
}

Swap A and C
for (row = 0;row < k;row++)
{
if (row = = K/2)//middle row, swap from middle column to right m column, N = 2* (2m+1)
{
for (col = k/2; col < k-1; col++)
{
temp = A[row][col];
A[row][col] = A[row + K][col];
A[row + K][col] = temp;
}
}
else//other rows, switching from left to right m column, N = 2* (2m+1)
{
for (col = 0;col < k/2;col++)
{
temp = A[row][col];
A[row][col] = A[row + K][col];
A[row + K][col] = temp;
}
}
}

Swap B and D
for (row = 0; row < k;row++)//Swap middle column left m-1 column, N = 2* (2m+1)
{
for (i = 0;i < (k-1)/2-1;i++)
{
temp = a[row][k+ K/2-i];
a[row][k+ K/2-I] = A[row + K][K+K/2-i];
A[row + K][K+K/2-i] = temp;
}
}

Output Magic Square
for (row = 0;row < N; row++)
{
for (col = 0;col < N; col + +)
{
printf ("%5d", A[row][col]);
}
printf ("\ n");
}

return 0;
}

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