I am studying two numbers, related to squares, that can characterize a polygon P:

**MinCoverNumber**= the minimum number of axis-aligned squares required to exactly cover P (the covering squares may overlap, but may not cover points outside P). For example: a square has MinCoverNumber=1. A 4-by-5 rectangle has MinCoverNumber=2, as it can be covered by 2 overlapping 4-by-4 squares. An L-shape has MinCoverNumber=3, if it is fat enough. A triangle has MinCoverNumber=$\infty$, because it cannot be exactly covered by a finite number of axis-aligned squares.**MaxHideNumber**= the maximum number of dots that can be placed inside P, such that no two dots can be covered by a single square. For example, a square has MaxHideNumber=1, a 4-by-5 rectangle has MaxHideNumber=2, etc.

Obivously, MinCoverNumber is an upper bound for MaxHideNumber, for example: if MinCoverNumber=3 (such as a fat L-shape), then for every 4 dots, at least 2 of them are covered by one square, therefore MaxHideNumber$\leq$3. However, I don't know if this is also a lower bound.

If MinCoverNumber=2, then obviously MaxHideNumber=2, but if MinCoverNumber=3 (i.e. a polygon that cannot be covered by 2 squares), I haven't managed to prove that MaxHideNumber=3 (i.e. it is possible to hide 3 dots such that no 2 are coverable by a single square). I also haven't managed to find a counter-example.

So, my question is:

**Is it possible to find an axis-aligned polygon P, such that MaxHideNumber(P) < MinCoverNumber(P)?**

An alternative presentation of the question for MinCoverNumber=3:

**Is it possible to find an axis-aligned polygon P, such that P cannot be covered by 2 squares, but in every set of 3 dots in P, there is a subset of 2 dots that are covered by a single square contained in P?**