Problems & Puzzles: Puzzles Puzzle 267. Talisman Squares "The study of these squares is so new, in fact, that no rules for construction are known, nor are there any mathematical theories...". Joseph S. Madachy, 'Madachy's Mathematical Recreations', 1966. A square of order n, filled with the integers from 1 to n^{2}, has a 'Talisman constant' equal to the minimal difference between each of its elements and the immediate neighbors (diagonal ones included) to each one of it. But not all the squares are 'Talisman Squares'. This name is deserved for those squares that, for a given order n, has a maximal Talisman constant. Example. For n=4
The left one is the well known magic square named "Dürer Magic Square" and has a Talisman constant equal to 1, while the right square has a Talisman constant equal to 3. So the left one square can not be a Talisman Square for this order (4), while we may assert (proof?) that the right one square certainly is a Talisman Square, because 3 is the largest Talisman constant possible for any square of order 4. But, how to construct a Talisman Square for any given order n? Rodolfo Kurchan and Carlos Rivera have been studying this problem the last two months, and they have found a pair of algorithms (one algorithm for even n values, the other one for odd n values) in order to produce (conjecturally) the asked Talisman squares for any given order n. Instead of providing longwinded and boring general instruction rules, we will display the algorithms using a couple of examples and some explanations about them. n = even, K_{TS}(n) = n^{2}/4  1 Example n=6, K_{TS}(6) = 8
As you have noticed, for sure, the filling pattern exhibited in the previous example, divides the numbers 1, 2, 3, …, n^{2} into four sets (S1, S2, S3, S4) of n^{2}/4 consecutive numbers each, as follows: S1 = {1, 2, 3, …, X1} The n^{2}/4 consecutive numbers of each set are allocated in the same general trend:
So, the only important thing you should know in advance is the cells in which the first numbers of each set (1, X, Y & Z) must be allocated, and the answer is: 1 goes in the cell (2, 2), X goes in the cell (1, 2), Y goes in the cell (1, 1) and Z goes in the cell (2, 1). Moreover, if you know to know them in advance, the values of X, Y and Z are, respectively n^{2}/4+1, 2.n^{2}/4+1 and 3.n^{2}/4+1, but this is not necessary at all.
We will call this filling pattern "22A" (22 because it starts in the cell (2, 2) and "A" because the four starting numbers of each set  1, X, Y & Z  describes the profile of an "A" letter) n = odd, K_{TS}(n) = (n.(n1))\4 ^{ (*)} Example: n=7, K_{TS}(7) = 10
As before, here are four sets (S1, S2, S3 & S4) of consecutive numbers. Now the four sets have distinct quantity of integers. Again, the starting number of each set, 1, X, Y & Z are allocated in the four cells in the upperleft corner, but now 1 goes in the cell (2, 1), X goes in the cell (1, 1), Y goes in the cell (2, 2) and Z goes in the cell (1, 2). We will call this pattern "21N" for analogue reasons than before.
The consecutive numbers of the four sets, are allocated in the same general trend than before: 'every two cells, downward and rightward'. But we have a very important difference:
Summarizing: Talisman squares are constructed this way: For n even:
For n odd:
Question: Can you produce a square of any order with a talisman constant greater than the predicted by our algorithms? ___________ (**) As a matter of fact, for the squares of order n even, we have found two more general patterns that produce the same Talisman constant. We have selected this pattern (22A) because it seems appropriate in order to produce Talisman rectangles also. Nevertheless this is a work still in process. Solution: Contribution came from Luke Pebody. On May 18 he wrote " I proved that no talisman square can be produced for even n, with Talisman Constant at least (n^2/4)1, so that you got the correct answer". This his proof:
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