## Another Domino Magic Square

Choose 8 dominoes from a set of double-6 dominoes and arrange them into a 4x4 grid, such that each row, column and diagonal has the same sum. Make the common sum as large as possible.

Solutions were received from Joseph DeVincentis, Philippe Fondanaiche, Alan O'Donnell, and Claudio Baiocchi. Joseph's and Claudio's solutions follow.
``` Solution from Joseph DeVincentis:
Use the 4-4, 4-3, and all dominoes with sum 9 or more.
Sum 19 is the largest possible since no 8 dominoes add to 80.

6-6 4-3

4-4 5-6

4 6 4 5
| | | |
5 3 6 5 ```

``` Claudio Baiocchi's solution:
Let us try to minimize the common sum (thus working with smaller numbers);
then replacing each entry X with 6-X we will get the maximum solution.

The smaller pieces in a double-six dominoes are:
- the 0-sum piece (0,0);
- the 1-sum piece (0,1);
- the 2-sum pieces (0,2), (1,1);
- the 3-sum pieces (0,3), (1,2);
- the 4-sum pieces (0,4), (1,3), (2,2);
- the 5-sum pieces (0,5), (1,4), (2,3).

In particular a magic sum 4, requiring 8 pieces with total sum 16, is
impossible; while many choices of 8 pieces give raise to a total sum 20,
thus a priori suitable for a magic sum 5. In fact, almost any 8-tuple of
pieces with total sum 20 gives raise to (many) magic-squares with magic sum
5; the unique exception being that the piece (0,5) must be avoided.

As an example of magic-sum 5 solution and its 6-complement with magic-sum 19
we have the "fully horizontal" solution:

(0,0),(2,3)               (6,6),(4,3)
(1,2),(1,1)               (5,4),(5,5)
(3,0),(2,0)               (3,6),(4,6)
(1,3),(0,1)               (5,3),(6,5) ```

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