Could someone kindly point me to references on constructing magic squares of even order? Does a compact formula/algorithm exist?

  • $\begingroup$ I remember, I have seen only of odd order...even ordered exists..? $\endgroup$ – Tapu Oct 27 '11 at 18:37
  • $\begingroup$ @Swapan: Yes: there exist magic squares of any order except $2\times 2$. There are many direct algorithms for odd order magic squares, and if you can do a $k\times k$ and an $m\times m$, then you can do the $km\times km$ by dividing it into an $m\times m$ array of $k\times k$ squares (or a $k\times k$ array of $m\times m$ squares), and solving each of them as the $k\times k$, in the order given by the $m\times m$. This deals recursively with any $n$ not congruent to $2$ modulo $4$. $\endgroup$ – Arturo Magidin Oct 27 '11 at 19:53
  • $\begingroup$ @Arturo, thanks! I read (when I wrote this comment) about the existence from Wikipedia. $\endgroup$ – Tapu Oct 27 '11 at 21:43
  • $\begingroup$ @Swapan: I also missed that you need to deal with the $4\times 4$ and the $8\times 8$ directly. $\endgroup$ – Arturo Magidin Oct 28 '11 at 4:56
  • $\begingroup$ An algorithm for generating even-order magic squares due to Collison is presented here; if you have MATLAB, you can try to find the file magic.m to see the algorithms internally used (discussed briefly here). $\endgroup$ – J. M. is a poor mathematician Oct 28 '11 at 9:36

A very elegant method for constructing magic squares of singly even order $n=4m+2$ with $m\geq1$ is due to J. H. Conway, who calls it the "LUX" method. Create an array consisting of $m+1$ rows of Ls, $1$ row of Us, and $m-1$ rows of Xs, all of length $n/2=2m+1$. Interchange the middle U with the L above it. Now generate the magic square of order $2m+1$ using the Siamese method centered on the array of letters (starting in the center square of the top row), but fill each set of four squares surrounding a letter sequentially according to the order prescribed by the letter. That order is illustrated on the left side of the bellow figure, and the completed square is illustrated to the right. The "shapes" of the letters L, U, and X naturally suggest the filling order, hence the name of the algorithm.

This example should illustrate the "LUX" method:

LUX demo


An method for constructing magic squares of doubly even order $n=4m$ is to draw Xs through each $4×4$ subsquare and fill all squares in sequence. Then replace each entry $a_{(ij)}$ on a crossed-off diagonal by $(n^2+1)-a_{(ij)}$ or, equivalently, reverse the order of the crossed-out entries. Thus in the bellow example for $n=8$, the crossed-out numbers are originally $1, 4, ..., 61, 64,$ so entry $1$ is replaced with $64$, $4$ with $61$, etc.

doubly even example

  • $\begingroup$ Any idea as to what algorithm exist for constructing doubly-even order magic squares? $\endgroup$ – Computist Oct 27 '11 at 23:40
  • 3
    $\begingroup$ @Ross: Of course, this requires you to know how to do a $4\times 4$ and an $8\times 8$ "directly". $\endgroup$ – Arturo Magidin Oct 28 '11 at 4:36
  • 4
    $\begingroup$ Why is the text strangely familiar... $\endgroup$ – J. M. is a poor mathematician Oct 28 '11 at 9:42
  • 1
    $\begingroup$ @pedja: I use to lend my All-seeing Eye to J.M. $\endgroup$ – Ilya Oct 28 '11 at 12:48
  • 6
    $\begingroup$ @pedja, that's not enough! you should always attribute your sources before somebody points it out! $\endgroup$ – t.b. Oct 28 '11 at 12:49

This is 2 years old but here are some new results.

A new/revised class of algorithms for single-even/double-even magic squares can be found on arXiv:



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.