March 2017 - Solution

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March 2017 Challenge

Oscar Volpatti sent us the following solution: Define A, Recursively, as
A(0) = [1];
A(k+1) = [A(k), A(k), A(k); A(k), A(k) + 1, A(k) + 2; A(k), A(k) + 2, A(k) + 1].

Therefore, A(2) is the 9×9 configuration:

1 1 1 1 1 1 1 1 1
1 2 3 1 2 3 1 2 3
1 3 2 1 3 2 1 3 2
1 1 1 2 2 2 3 3 3
1 2 3 2 3 4 3 4 5
1 3 2 2 4 3 3 5 4
1 1 1 3 3 3 2 2 2
1 2 3 3 4 5 2 3 4
1 3 2 3 5 4 2 4 3

Taking any 8×8 minor of A(2) yields an optimal solution which, after any legal sequence of adjustments, will result in no more than 36 watches pointing to a mark.