Magic Squares of Even Order
The Magic Squares of Even Order is a mathematical problem that involves constructing a square matrix of even dimension (order) where the sum of elements in each row, column, and diagonals is the same. In other words, the sum of all the numbers in each row, column, and both diagonals of the square should be equal. This property gives the square a magic quality.
Problem Statement
Given an even integer 'n', the task is to generate a magic square of order 'n'. There are specific methods to construct such magic squares.
Example
Let's take the example of constructing a magic square of order 4 to illustrate the problem.
A standard magic square of order 4 looks like this:
16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1
In this example, each row, column, and diagonal sums up to 34.
Idea to Solve
There are different algorithms to construct magic squares of even order. One common approach is to use the Strachey method for singly even order and the LUX method for doubly even order magic squares.
Pseudocode
magicSquare(n):
if n is even:
if n mod 4 ≠ 0:
Construct a singly even magic square
else:
Construct a doubly even magic square
else:
Print "Magic square of even order is not possible"
Algorithm Explanation

If 'n' is not even, then it's not possible to create a magic square of even order, so the program should handle this case.

If 'n' is even:
 If 'n' is not divisible by 4, construct a singly even magic square.
 If 'n' is divisible by 4, construct a doubly even magic square.

To construct a singly even magic square:
 Divide the grid into four quadrants.
 Fill each quadrant using a counterclockwise pattern starting from the topleft corner.

To construct a doubly even magic square:
 Fill the grid with consecutive numbers in order.
 Reverse the numbers in the topleft, topright, bottomleft, and bottomright corners.
Code Solution

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Time Complexity
The time complexity of the algorithm to construct a magic square of even order is O(n^2), where 'n' is the order of the magic square. The algorithm fills each cell of the grid exactly once, resulting in a quadratic time complexity.
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