A magic square is a grid of numbers in which the sum of the numbers in each row, column, and diagonal is the same. It is called a "magic" square because the sum is the same no matter which direction you add up the numbers.

To create a magic square in Java, we can use a two-dimensional array to represent the grid of numbers. We can then use a loop to initialize the values in the array and to check if the square is magic.

Here is an example of how we could create a 3x3 magic square in Java:

```
int[][] magicSquare = new int[3][3];
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
magicSquare[i][j] = (i * 3) + (j + 1);
}
}
boolean isMagic = true;
int magicSum = magicSquare[0][0] + magicSquare[1][1] + magicSquare[2][2];
for (int i = 0; i < 3; i++) {
int rowSum = 0;
int colSum = 0;
for (int j = 0; j < 3; j++) {
rowSum += magicSquare[i][j];
colSum += magicSquare[j][i];
}
if (rowSum != magicSum || colSum != magicSum) {
isMagic = false;
break;
}
}
if (isMagic) {
System.out.println("This is a magic square!");
} else {
System.out.println("This is not a magic square.");
}
```

In this code, we first initialize the values in the array using a nested loop. The values are set to the position of each element in the array, starting with 1 in the top left corner and increasing by one for each element to the right and down.

Next, we check if the square is magic by comparing the sum of the numbers in each row, column, and diagonal to the sum of the numbers in the first row. If any of these sums are different, we set the `isMagic`

variable to false and break out of the loop.

Finally, we print a message indicating whether or not the square is magic.

This is just one way to create and check for a magic square in Java. There are many other ways to do it, and you can even create magic squares of different sizes by modifying the size of the array and the loop conditions.

## java

Magic Number vs Happy Number The only difference between magic numbers and happy numbers is that in a magic number we sum up all the digits of the number recursively until we get a signal digit i. This means if you can explore the diagonal next, with only 2 remaining unknowns, by picking 1 value for one diagonal cell say, row 2, col 1 , and computing the other row 1, col 2. If possible place the next number at that position. The values in each row, column, and diagonal must add to 15. This article is contributed by Aarti Rathi.

## for loop

A 2 numbers, usually distinct integers, in a square, such that the n numbers in all rows, all columns, and both diagonals sum to the same constant. The sum of each row and column is 15. The main problem you're having is you are checking if the sum is the same outside the for loop. I believe I have problems in my row and column methods and if anyone could provide me a solution that would be great. I have to write a program that takes in an odd number from the user and creates a magic square. My algorithm continually puts out false according to the isMagic method for when it checks if it's Magic.

## Check given matrix is magic square or not

Then in the main method, the code calls the solve method, which is a private method, like all methods except for the constructor. Similarly, if the calculated position of the column becomes m, then we will wrap it around to 0. All the information you have listed have truly helped me understand what happens behind the scenes! So I commented out most of them. I'm sure you can keep improving this, and at the moment, it doesn't stop after finding the first solution. Similarly, if the calculated column position becomes n, it will wrap around to 0. On the next row, it repeats this with 12 values for the fifth cell, 11 for the next, 10 for the next, and computes the value for the eighth cell. Here is my improved version of my Magic square program from Any help for improvements would be really appreciated.

## Magic square java program

After nine moves, the square is full. After that, you continue with row 1, followed by column 1, then row 2 and column 2, and so on alternating between rows and columns. Let us know in the comments. Space Complexity: The space complexity of the above program is O n 2 , where n is the total number of rows or columns present in the square matrix. In this section, we will discuss what is a magic number and how can we find a magic number through a Java program.