Discovered an interesting property within Square Hex boards

In a maker-breaker game with n elements and symmetric or equivalent goals for both players, the number of "necessary" elements in a winning set (that is, elements that cannot be removed without making the set non-winning) is at most n/2 if n is even or (n+1)/2 if n is odd.

However, note that for Hex boards greater than size 4x4 (where an 8-necessary-element long chain is possible) and for Y boards greater than size 3, the actual maximum falls increasingly lower than the theoretical limit as size increases. For example, on a 6x6 Hex board, there are at most 11 necessary elements in a winning chain, even though there are 36 total elements, half of which is 18, a difference of 7.

Well, as you may have guessed by now from reading the title and the previous paragraphs, Square Hex boards have the interesting property such that for any size, the maximum number of necessary elements in a winning set is always (n+1)/2 (Square Hex boards always contain an odd number of elements). This happens because unlike on a Hex or Y board, all the corners are "obtuse", and a chain can spiral around the board until it reaches the center, then spiral back the opposite direction, making a chain look like a galaxy.

I don't really know if there is anything mathematically deep about this; I just thought this was pretty neat.

Here is a Y-board with 4 elements and a minimal winning set with 3 elements.

*
/ |
/ |
/ |
*-----*-----*

OK the graphics did not work but one side of the triangle has 3 vertices and the third corner of the triangle is connected to all those 3 vertices.

The formatting is off, so I don't really know what you were trying to display, but how can a Y board have 4 elements? 4 isn't a triangular number. The sizes of Y boards go 1, 3, 6, 10, 15, 21, 28... so on.

I really should quit making conjectures...