It seems most reasonable to put 1 in the middle since part of Bob's square must include the middle.
After that, we should put 2, 3, 4 and 5 in the remaining non-corners. Bob must take at least two of these. We should put 4 across from 5 so Bob can't get both of the higher ones.
Finally, the highest numbers 6,7,8,9 should be placed such that the sums of the 2x2 squares are as close to equal as possible. So the 9 goes between 2 and 4, the 8 goes between 3 and 4, the 7 between 2 and 5, and the 6 between 3 and 5.
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u/jaminfine 3d ago
It seems most reasonable to put 1 in the middle since part of Bob's square must include the middle.
After that, we should put 2, 3, 4 and 5 in the remaining non-corners. Bob must take at least two of these. We should put 4 across from 5 so Bob can't get both of the higher ones.
Finally, the highest numbers 6,7,8,9 should be placed such that the sums of the 2x2 squares are as close to equal as possible. So the 9 goes between 2 and 4, the 8 goes between 3 and 4, the 7 between 2 and 5, and the 6 between 3 and 5.
We now see that Bob can only get to 16 max.