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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.
Source: Reader Shivani Goyal.

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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