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The magic square is obtained by adding the Greek and Latin squares. A peculiarity of the construction method given above for the odd magic squares is that the middle number (n 2 + 1)/2 will always appear at the center cell of the magic square. Since there are (n − 1)! ways to arrange the skew diagonal terms, we can obtain (n − 1)! Greek ...
If the remainder is 3, move 2 to the end of even list and 1,3 to the end of odd list (4, 6, 8, 2 – 5, 7, 9, 1, 3). Append odd list to the even list and place queens in the rows given by these numbers, from left to right (a2, b4, c6, d8, e3, f1, g7, h5). For n = 8 this results in fundamental solution 1 above. A few more examples follow.
A pandiagonal magic square or panmagic square (also diabolic square, diabolical square or diabolical magic square) is a magic square with the additional property that the broken diagonals, i.e. the diagonals that wrap round at the edges of the square, also add up to the magic constant.
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The magic constant or magic sum of a magic square is the sum of numbers in any row, column, or diagonal of the magic square. For example, the magic square shown below has a magic constant of 15. For a normal magic square of order n – that is, a magic square which contains the numbers 1, 2, ..., n 2 – the magic constant is = +.
This question is harder to answer than for Rubik's Cube, because the set of operations on Rubik's Magic does not form a mathematical group. The basic operation (move) consists of transferring a hinge between two tiles T 1 and T 2 , from one pair of edges (E 11 of T 1 and E 21 on T 2 ) to another pair E 12 and E 22 .
With the bent hypotenuse, the first figure actually occupies a combined 32 units, while the second figure occupies 33, including the "missing" square. The amount of bending is approximately 1 / 28 unit (1.245364267°), which is difficult to see on the diagram of the puzzle, and was illustrated as a graphic. Note the grid point where the ...
For instance, the Lo Shu Square – the unique 3 × 3 magic square – is associative, because each pair of opposite points form a line of the square together with the center point, so the sum of the two opposite points equals the sum of a line minus the value of the center point regardless of which two opposite points are chosen. [4]